h Web viewIf it is a face centred cubic unit cell, only FOUR (4) atoms will be present

INDEX

Sl. No. Chapter Page No.1 Sold State 2-92 Solutions 10-203 Electro Chemistry 21-314 Chemical Kinetrs 32-445 Sunface Chemistry 45-576 General Principals and Isolation 58-667 P Block Elements 67-858 D Bolck 86-989 Coordination Compounds 99-10710 Organic 108-15111 Bio Molecules 152-16012 Polymers 161-17013 Chemistry in every day life 17114 Value Based Question 172-187

SOLID STATE

GIST :

1. NUMBER OF ATOMS PRESENT IN A UNIT CELL:

(a) If it is a simple cubic unit cell, only ONE (1) atom will be present.

No of atoms present in this unit cell = 8 x 1/8 = 1

( Each Corner atom contributes 1/8th portion to the unit cell)(b) If it is a body centred cubic unit cell, only TWO (2) atoms will be present.

No of atoms present in this unit cell = (8 x 1/8) + 1 = 2 (Each Corner atom contributes 1/8th portion to the unit cell)

( Body centered atom is 1)(c) If it is a face centred cubic unit cell, only FOUR (4) atoms will be present.

No of atoms present in this unit cell = (8 x 1/8) + (6 x ½) = 1 + 3 = 4(Each Corner atom contributes 1/8th portion to the unit cell)

( Each face centered atom contributes ½ portion to the unit cell) 2. RELATION BETWEEN THE LENGTH OF THE UNIT CELL, RADIUS OF THE ATOM AND NEAREST

NEIGHBOUR DISTANCE:

(a). For a Simple cube : a = 2r = d

(b) For B.C.C. : a =

4 r√3 =

2d√3

( c) For F.C.C : a = 2√2 r = √2d

a = Length of the Unit Cell, r = Radius of the Atom, d = Nearest neighbour distance

HOW TO CALCULATE THE EFFICIENCY OF PACKING:

(a) For a simple cubic

a = 2r

∴ Volume occupied by all the spheres in a unit cell = 1 x

43 πr3 (∵ 1 sphere per unit cell)

Volume of the unit cell = (2r)3 (volume = a3)

Volume occupied by all the sphere in unit cell

Hence, packing efficiency = x 100

Volume of unit cell

=

1×43

πr 3×100

(2 r )3

=

227×6 = 52%

(b) In B.C.C

There are two atoms per unit cell.

Packing efficiency =

2×43

πr3×100

( 4√3

r )3

(∵a= 4

√3a

)

= 2 x

43 x

3√34 x 4 x 4 x 100 = 68 %

(c) In F.C.C .

Number of atoms per unit cell = 4

a =

4√2 r = 2√2 r

Packing efficiency =

4× 43

πr3×100

( 4√2

r )3

= 74 %

HOW TO CALCULATE THE DENSITY OF A CRYSTAL:

d =

zMa3 N A

Let us identify each term in it :

d = Density of the crystal z = Number of atoms in the Unit cell M = Mass of the Unit cell

a = Length of the Edge of the Unit cell NA = Avogadro’s Number.

2 x Distance between the two ions = Length of the edge of the Unit cell(a).

(b). To calculate the distance between the two ions, you divide the length of the edge of the

unit cell with 2.

a2 will give the distance between the two ions.

(i) Stoichiometric defects:

The defects in which stoichiometry of the compound is not disturbed. The following two types of point defects are commonly found in the crystalline ionic substances.

Schottky Defects: 1) The absence of a cation or anion from the position which it is expected to occupy in

the periodic arrangement of ions is known as Schottky defect. 2) The vacant positions give rise to vacancies or voids in the structure of the

substance.

3) The Schottky defects are more common in ionic compounds with high coordination numbers and where the positive and negative ions are of similar size.

e.g., NaCl, CsCl, KCl and KBr. 4) Since the solid with a Schottky defect contains lesser number of ions as compared

to a prefect crystal, the density of the crystal exhibiting Schotky defect will be

less as compared to that of the perfect crystal.

Frenkel Defects:

1) In this type of defect, the ion, instead of being in its expected location, is found in one of the interstices.

2) Frenkel defects are more common in ionic crystals where size of the cation is such

that it can be accommodated in the interstitial sites. 3) This type of defect is favoured by a large difference in size between the positive and negative ions and since the size of cation is smaller than that of the anion,

the former can be accommodated in the interstitial position. Examples are ZnS, AgCl, AgBr and AgI. 4) The density of crystals exhibiting Frenkel defect remains unchanged as the ions are present in the interstitial sites without changing the volume of the substance.ii) Non-Stoichiometric point defects:

The defects in which stoichiometry of the compound is disturbed.There are two types 1) Metal excess defect 2) Metal deficiency defect.i) Metal excess defects due to anionic vacancies : i) Alkali halides like NaCl and KCl show this type of defect. ii) When crystals of NaCl are heated in an atmosphere of sodium vapour, the

sodium atoms are deposited on the surface of the crystals as Na+ ions and which traps the Cl- ions from the lattice points. To make crystal electrically neutral electrons released will diffuse into the crystal and occupy anionic sites.

iii) The anionic sites occupied by unpaired electrons are called F-centres (from the German word Farbenzenter for colour centre).

Ex : a)They impart yellow colour to the crystals of NaCl. b) Excess of lithium makes LiCl crystals pink.

c) Excess of potassium makes KCl crystals violet (or lilac).

ii) Metal excess defects due to presence of cations:

Zinc oxide is white in colour at room temperature. On heating it loses oxygen and turns yellow.

ZnO Zn+2 + ½ O2 +2e-

Now there is excess of zinc in the crystal and its formula becomes Zn 1+xO. The excess Zn2+ ions move to interstitial sites and the electrons to neighbouring interstitial sites.

(a) n- type semiconductor:

It is obtained by doping Si or Ge with a group 15 elements (Electron – rich impurities) like P. Out of 5 valence electrons , only 4 are involved in bond formation and the fifth electron is delocalized and can be easily provided to the conduction band.

( b) p – type semi conductor :

It is obtained by doping Si or Ge with a group 13th elements (Electron – deficit impurities) like Gallium which contains only 3 valence electrons. Due to missing of 4th valence electron, electron hole or electron vacancy is created. The movement of these positively charged hole is responsible for the conduction.

Diagrammatic representation of conduction of an conductor, insulator and a semi conductor through a band theory:

II. Magnetic properties:

Diamagnetic solids contain paired electron () and repel the external magnetic field.Ex: H2O, NaCl, C6H6, etc.,

Paramagnetic solids contain unpaired electrons, and are attracted into the applied external magnetic field. Ex : Fe +3, O2, Cu+2, etc.,

In ferromagnetic solids, the metal ions are grouped together into small regions called domains. When the substance is placed in a magnetic field all the domains get oriented in the direction of the magnetic field (). This interaction leads to an increase in magnetic moment.Ex: Iron, Cobalt and Nickel are ferromagtnetic substances.

In antiferromagnetic solids, when the substance is placed in a magnetic field all the domains get oriented in the antiparallel () direction of the magnetic field, which leads to a decrease in magnetic moment. Ex: MnO, MnO2, Mn2O3.

In ferromagnetic solids when the substance is placed in a magnetic field all the domains get oriented randomly () which leads to the presence of uncompensated spins in the opposite direction resulting in some magnetic moment. Ex. Magnetite (Fe3O4), ferrite [M FeO4] where M = Mg (II), Cu (II), Zn (II).

Higher Order Thinking Skills

A compound formed by elements X and Y crystallizes in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face centres. What is the formula of the compound. (1 Mark)

(Ans) For a cubic structure,No. of X atoms at the corner of the unit cell = 8 x1/8 =1No. of Y atoms at the face centre of the unit cell =6 x1/2=3Formula of the compound = XY3

A compound is made of two atoms X and Y. Atom X is arranged in ccp and atom Y occupies tetrahedral site. Find the formula of compound. (2 Marks )

(Ans) No. of atoms of X = 8

No. of atoms of Y = (8/8) + (6/2) = 4

Ratio of X : Y is 2: 1Formula of compound is X2YFerric oxide Crystallises in a hexagonal close packed array of oxides with

two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide.

(1Mark)

Ans: No. of Ferric ions for each oxide ion = 2/3Chemical formula= Fe2/3 O Fe2O3

Find the type of lattice for cube having edge length of 400 pm, atomic wt. = 60 and density = 6.25 g/cc. (3 Marks)

(Ans) Let the no. of atoms in a unit cell = x

Mass of one unit cell =

=

Volume of unit cell = (edge length) 3 =

= 64 x 10-24 cm3

Density =

Mass =

=

= 4

The unit cell has 4 atoms, It is face centered cubic lattice.Analysis shows that nickel node has formula Ni0.98 O1.00. What fractions of Nickel exist as Ni2+ and Ni3+ ion?

Ans: Let no of Ni2+ ions = ‘p’ and The no. of Ni3+ions = ’98-p’.Magnitude of the total Charge present on the Nickel = p (+2) + (98-p)(+3)Magnitude of the total Charge present on the Oxygen = (2) x 100 = 200Total charge present on Metal and oxygen are equal,

(p x 2) + (98-p) x 3 = 2002p + 294 – 3p = 200

p = 94,No of Ni+2 ions = 94, % of Ni+2 ions = (94 / 98) x 100 = 95.91.

No of Ni+3 ions = 98 – p = 98 -94 = 4, % of Ni+3 ions = 4/98 x 100 =4.08.

Classify each of the following as P type and N-type semiconductors.

i) Ge doped with In ans: P-typeii) B Doped with Si : P-type

Account for the following:i) Fe3O4 is ferromagnetic at room temperature but becomes paramagnetic at 850K.ii) Zinc oxide on heating becomes yellow.iii) Frenkel defect does not change the density of AgCl crystals.

Ans: i) As the temperature increases the mixed oxide in which the metal ions in domains randomly distributed. The metal ions Fe+3 or Fe+2 contains unpaired electrons which are weakly attaracted by the magnet, so they become paramagnetic in nature.

ii) Zinc oxide is white in colour at room temperature. On heating it loses oxygen and turns yellow.

ZnO Zn+2 + ½ O2 +2e-

Now there is excess of zinc in the crystal and its formula becomes Zn 1+xO. The excess Zn2+ ions move to interstitial sites and the electrons to neighbouring interstitial sites.

iii) In Frenkel defect one of the lattice point is missing from the regular position and occupies the interstitial position, no lose of cation or anion occurs in the crystal lattice. So, the density does not change.

SOLUTIONS

GIST:

HOW TO CALCULATE THE CONCENTRATION OF A SOLUTION:

1. MOLARITY CAN BE CALCULATED AS : No. of moles of the solute

M =

Volume of solution in liters

2. MOLALITY CAN BE CALCULATED AS : No. of moles of the solute

m =

Volume of the solvent in Kilo grams.

3. NORMALITY CAN BE CALCULATED AS : No. of gram equivalents of solute

N =

Volume of solution in liters

4. MOLE FRACTION OF EACH COMPONENT : No. of moles of the components

XA (or) XB =

Total Number of moles

HOW TO CALCULATE THE NUMBER OF MOLES:

Weight of the component

Number of moles =

Molecular weight of the component

1. .

XA =

nA

nA+nB XB =

nB

nA+nB

Where XA and XB are the mole fractions of the components of A and B

4. XA + XB = 1 . If you calculate XA, you can calculate XB = 1 - XA

Henry’s law:. The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

The solubility of a gas in a liquid solution is a function of partial pressure of the gas. If we use the mole fraction of a gas in the solution as a measure of its solubility, then it can be said that

“The mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.”

p = KH xHere KH is the Henry’s law constant. X is molefraction of gas in the solution.P partial pressure of the gas.

Henry’s law & its applications:

1) Solubility of a gas increases with decrease of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than in warm waters. 2) To increase the solubility of CO2 in soft drinks and soda water, the bottle is

sealed under high pressure.3. Scuba divers must cope with high concentrations of dissolved gases while

breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends, which are painful and dangerous to life.To avoid bends, as well as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).

4) At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition known as anoxia.

Ideal and Non-Ideal Solutions :

a) Ideal Solutions : The solution which obeys Raoult’s law exactly at all concentration and all temperatures.

The ideal solution also have following characteristics:

i) It should obey the Raoult’s law (Total pressure, P = p0A xA + p0B xB ) ii) Heat exchange on mixing is zero ( Δ H mix = 0)iii) Volume exchange on mixing is zero ( Δ Vmix = 0 )

b) Non ideal solutions:

The solution which do not obey Raoult’s law are called non ideal solutions. For these solutions

i) pA # p0A xA and pB # p0B xB , ii) Δ H mixing # 0, iii) Δ V mixing # 0

Types of non ideal Solutions :

A) Non ideal solution showing positive deviation from Raoult’s law :

→ The Mixture contains two components namely A and B, If the interaction between A-B molecules is weaker than the interactions between A-A or B-B or both, then the solution deviate from the ideal behavior and Each component of solution has a partial vapour pressure greater than expected on the basis of Raoult’s law .

→The total vap. Pressure will be greater than corresponding Vap. Pressure expecterd in case of ideal solution of the same composition

pA > p0A xA and pB > p0B xB

Total Vapour pressure P , P = (pA + pB) > ( p0A xA + p0B xB )

→ Example: Mixture of Ethyal alcohol and Cyclohexane.

→Explanation for positive deviation: →In ethyl alcohol the molecules are held together by hydrogen

bonding. H | C2H5 − O ……. H – O……….H – O………

| | C2H5 C2H5

→When Cyclohexane is added to ethyl alcohol, the Cyclohexane molecules occupy spaces between ethyl alcohol molecules. As a result , some hydrogen bonds in alcohol molecules break. The escaping tendency of alcohol and Cyclohexane molecules from solution increases. So there is increase in vapour pressure. In such case i) Δ H mixing > 0, ii) Δ V mixing > 0

B) Non ideal solution showing negative deviations from Raoult’s law:

→ The Mixture contains two components namely A and B, If the interaction between A-B molecules is stronger than the interactions between A-A or B-B or both, then the solution deviate from the ideal behavior and Each component of the solution has a partial vapour pressure less than from pure liquids. As a result, each component of solution has a partial

vapour pressure less than expected on the basis of Raoult’s law. Hence total vap.pressure becomes less than the corresponding vap. pressure expected in case of ideal solution

pA < p0A xA and pB < p0B xB

Total Vapour pressure P , P = (pA + pB) < ( p0A xA + p0B xB )

→ Example: Mixture of Acetone and Chloroform

→Explanation for negative deviation:

→By Mixing Acetone and chloroform, a new attractive forces are formed due to in intermolecular hydrogen bonding. Hence the attractive forces become stronger and the escaping tendency of each liquid from the solution decreases.

CH3 Cl | | C = O-------------H - C - Cl

| | CH3 Cl

→So, the Vapour pressure of the solution is less than expected for an ideal solution . In such case (i) Δ H mixing < 0

(ii) Δ V mixing < 0

Azeotropes:The binary mixtures having the same composition in liquid and vapour

phase and boil at a constant temperature.

Types of Azeotropes:

i) Minimum boiling azeotropes:.

The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition.

Example: Ethanol-water mixture (obtained by fermentation of sugars) on fractional distillation gives a solution containing approximately 95% by volume of ethanol. Once this composition, known as azeotrope composition, has been achieved, the liquid and vapour have the same composition, and no further separation occurs.

ii) Maximum boiling azeotropes:

The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition.

Example: Nitric acid and water is an example of this class of azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393.5 K.

PROBLEMS ON COLLIGATIVE PROPERTIES:

1. Relative Lowering of Vapour pressure :

Xsolute =

Po

solvent−Psolution

Posolvent (or)

w×M solvent

M solute×W =

Po

solvent−Psolution

Posolvent (Xsolute =

w×M solvent

M solute×W )

Let us identify each term in it :

Posolvent = Vapour pressure of the pure Solvent

Psolution = Vapour pressure of the solution.

XB = Mole fraction of the solute

w = Weight of Solute

W = Weight of Solvent

Msolvent = Molecular mass of solvent

Msolute = Molecular mass of solute

NOTE: -

From the value of mole fraction (XB), you can find out the molecular weight of the Non – Volatile solute(MB).

XB =

nB

nA+nB where nB=

W B

MB

Where WB = Weight of the Non – Volatile solute

MB = Molecular weight of the Non –Volatile solute.

2. Problems on Elevation of boiling point:

Msolute =

Kb×1000×wΔT b×W

Let us identify the each term in it :

Msolute = Molecular weight of the non-volatile solute

Kb = Molal Elevation constant

w = Weight of the non – volatile solute

W = Weight of the solvent

ΔT b = Elevation of boiling point.

Students, it is observed that you are doing a mistake while calculating the ΔT b value. Its value always should be in Kelvin only.

ΔT b=T 2 K−T1 K

Where T1K = Boiling point of the pure Solvent, T2K = Boiling point of the solution.

3. Problem on Depression in Freezing Point:

Msolute =

K f×1000×wΔT f×W


Msolute = Molecular weight of the non-volatile solute

Kf = Molal Elevation constant

w = Weight of the non – volatile solute

W = Weight of the solvent

ΔT f = Depression in Freezing point.

NOTE :-

Students it is observed frequently that you are doing mistake while calculating ΔT f value.

Its value always should be in Kelvin only. Some times, ΔT f value is given in the negative sign. But, you should take only the positive value only.

ΔT f=To

f−To

f

Where Tof = Freezing point of pure solvent, Tf = Freezing point of the solution.

4. Problems on Osmotic Pressure : Π=MRT

(or)

Π= nV

RT (or) Msolute =

wRTΠV


Π = Osmatic pressure in atmospheres only.

R = Gas constant = 0.0821 Ltrs. Atms.

V = Volume of the solution in Litres only

T = Temperature in Kelvin only.

n = No. of moles of the Non – Volatile solute.

M = Concentration of the solution in molarity.

NOTE :-

1. Students, the common mistake committed by most of you is substituting the ‘R’ value. It is observed that the value of ‘R’ is taken as 8.314 instead of 0.0821.

2. As the value of ‘R is in Ltrs Atms, care should be taken that the volume of the solution should be in Litres and also the value of Osmotic Pressure in Atmospheres inly.

3. Some times, Osmatic pressure value will be given in mm mercury. Hence, it is to be converted into atmospheres. We know that 760mm Hg = 1 atmosphere.

4. Isotonic solutions were given to find out the molecular weight of the non – volatile solute, Then equate both the Osimatic pressure values and use this formula:

n 1 = n 2

Where n1 , n2 are the number of moles of the two solutes.

HOW TO CALCULATE THE VAN’T HOFF FACTOR (i) , DEGREE OF ASSOCIATION AND DEGREE OF DISSOCIATION.

Normal Moleculat Mass

i =

Observed Molecular mass

NOTE :-

In case of dissociation the value of i < 1 and in case of association the value of i > 1.

To calculate degree of dissociation and association:

1.For Dissociation:

i =

1+α1 (or) i = 1 + α

Let us Know :

i = Van’t Hoff factor, α = Degree of dissociation

% of Dissociation = Degree of dissociation x 100

2.For Association:

i = 1 -

α2

Let us Know : i = Van’t Hoff factor, α = Degree of association

% of Association = Degree of Association x 100

Very Short Answr Type Questions (VSA) One Mark Questions:

1. What will be the mole fraction of water in C2H5OH solution containing equal number of moles of water and C2H5OH?

2. Gases tend to be less soluble in liquids as the temperature is raised. Why?3. What will happen to the boiling point of the solution formed on mixing two miscible liquids

showing negative deviation from Raoult’s law?4. Which type of deviation is shown by the solution formed by mixing cyclohexane and ethanol?5. What is reverse osmosis? Give one large scale use of it.6. What is the value of van’t Hoff factor (i) if solute molecules undergo dimerisation

SA (I) - TYPE QUESTIONS (2 - MARK QUESTIONS)

1. Show that the relative lowering of vapour pressure of a solvent is a colligative property.2. Define azeotropes with one example of each type.3. Conc. H2SO4 has a density 1.9g/ml and is 99% H2SO4 by weight. Find molarity of

solution.4. State Raoult’s law for solutions of volatile liquid components. Taking a suitable example,

explain the meaning of positive deviation from Raoult’s law.5. Calculate the mole fraction of ethanol in 40% of this solution by mass in water.

SA (II) TYPE QUESTIONS (3 - MARK QUESTIONS)

1. (a) State Henry’s Law. (b) If O2 is bubbled through water at 393 K, how many millimoles of O2 gas would be dissolved in 1L of water? Assume that O2 exerts a pressure of 0.95 bar. (Given KH for O2 = 46.82 bar at 393K).2. Given reason for the following :– (a) Aquatic species are more comfortable in cold waters than in warm waters. (b) To avoid bends scuba divers use air diluted with helium. (c) Cold drinks bottles are sealed under high pressure of CO2.3. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, determine the vapour pressure (mm of Hg) of the solution. [Given : Kb for water = 0.76 kg mol–1] [Ans.: 724 mm of Hg] [Hint: ∆Tb = Kb.m ]4. (a) Define osmotic pressure. (b) Why osmotic pressure is preferred over other colligative properties for the determination of molecular masses of macromolecules? (c) What is the molar concentration of particles in human blood if the osmotic pressure is 7.2 atm at normal body temperature of 37°C? [Ans. : 0.283 M]

5. Describe a method of determining molar mass of a non-volatile solute from vapour pressure lowering. (b) How much urea (mol. mass 60 g mol–1) must be dissolved in 50g of water so that the vapour pressure at the room temperature is reduced by 25%? Also calculate the molality of

the solution obtained. [Ans. : 55.55 g and 18.5 m] 6. The vapour pressure of pure liquids A and B are 450 and 750 mm Hg respectively, at 350K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. [Ans. : XA = 0.4, XB = 0.6, YA = 0.3, YB = 0.7]


1. Under what condition molality and molarity of a solution are identical. Explain with suitable reason.

2. Addition of HgI2 to KI (aq.) shows decrease in vapour pressure. Why? 3. Account for the following :– (a) CaCl2 is used to clear snow from roads in hill stations. (b) Ethylene glycol is used as antifreeze solution in radiators of vehicles in cold countries. (c) The freezing point depression of 0.01 m NaCl is nearly twice that of 0.01 m glucose solution. 4. Give reasons for the following :– (a) RBC swell up and finally burst when placed in 0.1% NaCl solution. (b) When fruits and vegetables that have been dried are placed in water, they slowly swell and return to original form. (c ) A person suffering from high blood pressure is advised to take less amount of table salt.5. Glycerine, ethylene glycol and methanol are sold at the same price per kg. Which would be cheaper for preparing an antifreeze solution for the radiator of an automobile?6. Determine the correct order of the property mentioned against them : (a) 10% glucose (p1), 10% urea (p2), 10% sucrose (p3) [Osmotic pressure] (b) 0.1 m NaCl, 0.1 m urea, 0.1 m MgCl2 [Elevation in b.pt.] (c) 0.1 m CaCl2, 0.1 m sucrose, 0.1 m NaCl [Depression in f.pt.]7. Two liquids X and Y on mixing form an ideal solution. The vapour pressure of the solution containing 2 mol of X and 1 mol of Y is 550 mm Hg. But when 4 mol of X and 1 mole of Y are mixed, the vapour pressure of solution thus formed is 560 mm Hg. What will be the vapour pressure of pure X and pure Y at this temperature? [Ans. : X = 600 mm Hg; Y = 400 mm Hg]8. 2g of C6H5COOH dissolved in 25g of benzene shows depression in freezing point equal to 1.62K. Molar freezing point depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms a dimer in solution? [Ans. : 99.2%]9. Calculate the amount of NaCl which must added to one kg of water so that the freezing point is depressed by 3K. Given Kf = 1.86 K kg mol–1, Atomic mass : Na = 23, Cl = 35.5). [Ans. : 0.81 mol NaCl]

(ELECTRO CHEMISTRY)

1. Redox reaction, oxidation and reduction half reactions: the reaction which involves the simultaneous occurrence of both oxidation and reduction processes is called a redox reaction. The redox reaction can be spilt into two half equations, one corresponding to the oxidation process (called oxidation half reaction) and the other corresponding to the reduction process (called reduction half reaction). For example, the oxidation and reduction half reactions of the redox reaction are as follows..

Zn(s) + Cu2+(aq) → Zn2+ (aq) + Cu(s)

Zn(s) → Zn2+(aq) + 2e- (oxidation half reaction)

Cu2+(aq)+ 2e- → Cu ( reduction half reaction)

2. Standard electrode potential(E0): the potential of an electrode assembly is said toe be the standard electrode potential when i) The temperature of the electrode assembly is 250C ii) The ion solution used in the assembly has a concentration of 1 mol L -1 , and iii) The pressure of the gas, if used in the assemble , is 1 atm,

For a particular type of assembly, the standard oxidation potential ( Eoxi or EM/Mn+)

and the standard reduction potential (Ered or EMn+

/M.) are numerically equal but their signs are opposite. Thus,

EM/Mn+= - EM

n+/M

For example,

E0Zn / Zn+2 = -E0

Zn+2 / Zn and E0Cu/ Cu+2 = -E0

Cu+2 / Cu

According to the IUPAC convention, the term standard potential refers to standard reduction potential. Thus, if no subscript is mentioned, E0 represents the standard reduction potential.

3. Standard hydrogen electrode (SHE or NHE): standard hydrogen electrode is set up by passing pure hydrogen gas at 1 atm. Pressure in a solution of H+ (More correctly H3O+) ions of concentration 1 mol L-1 in contact with a Platonized platinum foil and can be represented as

Pt, H2 (g) (1 atm )/H+ (1 mol L-1)

The half cell reactions are as follows

½ H2 (g) → H+(aq) + e-; (when the electrode works as anode)

H+ (aq) + e- → ½ H2(g); (when the electrode works as cathode)

By convention, = = 0

4. E0 cell = E0 cathode- E0 anode

5. Representation of a galvanic cell

Zn|Zn2+(C1)|| Cu2+(C2)|Cu

Anode Cathode

6. The cell reaction:

Zn → Zn2+ + 2e- (at anode)

Cu2+ + 2e- → Cu ( at cathode)

Zn + Cu2+→ Zn2+ + Cu (over all reaction

4. 7. Dependence of E.M.F on concentration and temperature: the electrode potential of an electrode assembly depends upon i) the nature of metal and its ions, ii) the concentration of ions in the solution and iii) temperature.

Nernst equation for a single electrode: for an electrode involving the processes of reduction, Nernst equation can be written as follows.

Where Ecell= reduction potential of the electrode assembly,

E0 = standard reduction potential of the same electrode assembly,

R= gas constant,

n= number of electrons involved in the electrode reduction,

F= 1 Faraday (96500 coulombs)

[Oxidized state]=conc. Of the substance undergoing reduction,

[Reduced state]=conc. Of the substance obtained on reduction.

For the electrode reaction

Mn+(aq) + ne- → M(s)

The above equation can be written as

If T=298 K(250C) the above equation can be written as

For an oxidation electrode involving the process

- ,

the Nernst equation can be written as follows.

8. Equilibrium constant from Nernst equation

E0cell= Kc

Gibbs free energy and E.M.F. of a cell: the E.M.F. of a cell in the standard state is related to the standard free energy change as∆G0 = - n FE0 cell

From this equation following conclusions can be obtained:

i) When Ecell is positive , the cell reaction is feasibleii) When Ecell is negative; the cell reaction is not feasible.

When Ecell is zero, the cell reaction is in equilibrium

Specific conductivity (k): it is defined as the conductance of a conductor whose length is 1 cm and area of cross section equal to 1 cm2. It is equal to the reciprocal of specific resistance (p) of the solution.

( l/A)

Molar conductivity : molar conductivity of an electrolytic solution at dilution Vm is the conductance of the solution containing 1 mole of the electrolyte dissolved in Vm cm3 of the solution.

=

9. Kohlrausch law: The molar conductivity at infinite dilution of an electrolyte is equal to the sum of molar conductance sod its cations and anions, with each conductance term multiplied by the number of respective ion present in the formula unit of the electrolyte.

Thus, for HCl,

For CH3COOH,

For BaCl2

Kohlrausch law can also be stated as follows.The equivalent conductance of an electrolyte at infinite dilution is equal to the sum of the equivalent conductances of its cations and anions. Thus.

e.g. for Nacl,

10. Faraday’s laws of electrolysis:

W ∞ I. t

W = Z. I.t

Where, Z is a constant known as electrochemical equivalent of the substance. It is given by

Z =

Criteria for product formation during electrolysis

Electrolysis of conc. Aqueous solution of NaCl

At anode: 2Cl- → Cl2 + 2e- (Oxidation)

At cathode: 2H2O + 2e_ → H2 + 2OH- (Reduction)

Resulting solution is NaOH

Electrolysis of aqueous copper sulphate solution using platinum electrodes: in this case the electrolysis takes place as follows:

At anode; H2O → ½ O2 + 2H+ +2e-

At cathode; Cu2+ + 2e- → Cu

Electrolysis of aqueous copper sulphate solution using copper electrodes

At anode: Cu → Cu2+ + 2e-

At cathode: Cu2+ + 2e- → Cu

Electrolysis of dilute aqueous H2SO4 /Water

At anode H2O → ½ O2 + 2H+ + 2e-

At cathode 2H+ + 2e- → H2

Commercial Batteries

a) Dry Cell

b) Lead Storage Battery:During usage or discharge

During Recharge:Just write the reactions reverse

VSA one Mark

1. What is molar conductivity? Write two unit also2. What is meant by standard electrode potential3. Can you store copper sulphate solutions in a zinc pot?4. Rusting of iron becomes quicker in saline water?5. Name the cell which was used in Apollo space programme.6. How many faradays are required to oxidise 1 mole of H2O to O2.

SA-1 of Two Marks

1. State and explain Kohlrausch law of independent migration of ions2. An electrochemical cell is set-up between cadmium and nickel

Cd(s) | Cd2+ (1M) || Ni2+) (1M) | Ni(s)

If two half-cells work under standard conditions, calculate the cell potential (emf).

E0 cd2+/Cd = -0.40 V, E0Ni2+/Ni=-0.25V

3.

4.

7. Explain effect on concentration on molar conductivity of weak and strong electrolytes using graph.

SA-2 of Three Marks:

1. What are the fuel cells? Explain the working of H2-O2 fuel cell. And write its advantages.2. Conductivity of 0.00241M acetic acid is 7.896 x 10-5S/cm.Calculate its molar conductivity. If

acetic acid is 390.5Scm mol-1, what is the dissociation constant?

Solution:

3. What is the half –cell potential for Ag+/Ag electrode at 250C in which concentration of silver ions is 0.5M? Eo

Ag+/Ag= +0.80 V

Solution:

Ag+(aq) + e- → Ag(s) n=1 [Ag(s)] = 1

EAg+/Ag=E0Ag+/Ag-

=

=+0.80 V- 0.05916 X log

=+0.80 V- 0.05916 X log 2

=+0.80V-0.05916X0.3010

=+0.80-0.0178=+0.7822V

4. Calculate mass of copper deposited when a current 0.3 ampere is passed in aq solution of copper sulphate for two hour.

Solution:

Equivalent wt. of Cu =

Z for Cu =

W = Z i t

=

= 0.7106 g.

5. Calculate standard free energy change for the following chemical reaction –

,

Solution:

It is clear from data that Ag is anode and Cd is Cathode.

=

= 0.80 – (– 0.40)

= 1.20 V

n = 2

=

=

= 231600 J = 231.6 kJ

6. Calculate the emf of cell Mg / Mg+2(0.10) → Cu+2(1X10-3) / Cu at 298K E0Cu/Cu+2 = + 0.34

V and E0Mg+2/Mg = - 2.37 V

Ecell = E0cell – 0.059/2log (Mg+2)(Cu)/(Cu+2)(Mg)

E0cell = 0.34-(-2.37)= 2.81

7.Ans: Low reduction potential high reducing power.

Potassium >Magnesium >Chrmium>Mercury>Silver

8. Ans: During electrolysis when metal deposit at cathode, one mole of metal deposited at cathode is equal to no. of faradays of charge on the metal ion.i) 3Fii) 2F iii) 5F

9.

Ans: m=Zitm= (58/2*96500)*5*20*60

Higher Order Thinking Skills:

1. What is the use of Platinum foil in the hydrogen electrode?Solution:

2. How is cathodic protection of iron different from its galvanization?3. The E0 values for the two metal electrodes are given below:

a. Cr3+/Cr2+ = - 0.4 Vb. Fe3+/Fe2+ = + 0.8V

Comment on the result of treating a solution of Cr (II) with a solution containing Fe (III) ions.

4. How long a current of 3 amps has to be passed through a solution of silver nitrate to coat a metal surface of 80cm2 with a 0.005 mm thick layer? Density 10.5 gm/cc.

5. Account for the following i) Alkaline medium inhibits the rusting of iron.ii) Iron does not rust even if the zinc coating is broken in galvanized iron pipe.iii) Cu2+ + 2e- → Cu ; E0 = + 0.34 V and Ag+ + 1e- → Ag ; E0 = + 0.80 V

Constract a galvanic cell using The above equations.

iii) For what concentration of Ag+ ions will the emf of the cell be zero at 250C,if the concentration of Cu2+ is 0.01M.

6. Resistance of a conductivity cell filled with 0.1M KCl solution is 100 ohms. If the resistance of the same cell when filled with 0.02M KCl solution is 520 ohms, calculate the conductivity and

molar conductivity of 0.02 M KCl solution. Conductivity of 0.1 M KCl solution is 1.29 S m-1.

7.

CHEMICAL KINETICS

Rate of a chemical reaction: the rate of a chemical reaction may be defined as the change in concentration of any of the reactants or any of the products per unit time.

Average rate of a reaction = For Example a hypotheticl reaction

aA + bB → cC + dD

the average rate of reaction = -

Instantaneous rate of a reaction: the rate of change of concentration of any of the reactants or products at a particular instant of time is called the instantaneous rate of the given reaction at that instant.

Instantaneous rate =

-

Factors which affect the reaction rate:Nature of reactants, Concentration of reactants, Temperature, Presence of a catalyst, & Surface area

Rate law and rate constant: The expression which describes the experimentally observed dependence of the reaction rate on the molar concentrations of the reactants is called the rate law or rate equation.aA + bB → cC → products. The rate law is

Rate = And p+q+r = n then Order of a reaction: the power to which the concentration term of a particular reactant in the

rate law is raised is called the order of reaction with respect to that reactant and the sum of all the powers to which all the concentration terms in the rate law are raise

aA + bB → cC + dD ,

Rate = k the experimentally observed then the reaction is said to be of order p with respect to A and of order q with respect to B. the overall order of the reaction is given by

Overall order of reaction = p + q

Molecularity of a reaction: the number of reacting species which must collide simultaneously in Order to bring about a chemical reaction is called the molecularity of that reaction

Molecularity of complex reactions: the reactions which occur in two or more steps are called complex reactions.

Step1.

Step2.

Obviously, step 1 is the rate determining step and therefore the molecularity of the overall complex reaction is 3.

Pseudo-unimolecular reactions: A higher order reaction behaves first order under certain conditions, this is known as pseudo first order reaction.

Units for rate constant: Kn = (mol/lit)1-n /sec where n is the order of reaction.Zero order reaction:

Half life period of Zero Order reaction

First Order:HaHa

Half life period of First Order reaction is Effect of temperature on reaction rate: the rate of a chemical reaction is significantly affected

by a change in temperature. For most of the chemical reactions, the rate increases with increase in temperature. The rate usually becomes doubled or triple for each 10º rise in temperature.

Temperature coefficient: it is defined as the ratio of the rate constant of a reaction at two different temperatures separated by 10 C. the two temperatures generally taken are 35 C and 25 C. thus,

Temperature coefficient = = 2 – 3 Activation energy: the excess energy that the reactant molecules must acquire in order to cross

the energy barrier and to change into the products is called the activation energy of the reaction.

Activation energy = threshold energy-average energy possessed by reactant molecules.

Arrhenius equation K=A e-Ea/RT

Log K=log A –Ea/2.303RT

Rate constant method: if k1 and k2 are the rate constants measured at temperatures T1 and T2

respectively, then on the basis of Arrhenius equation, we can have

Collision theory of reaction rate:

Effective collisions are those collisions which lead to the formation of products. The number of effective collision is governed by the following two factors.

i) Energy barrier: the collisions are effective only when the molecules possess energy greater than or equal to the threshold energy.

ii) Orientation barrier: the reactant molecules must collide with favorable orientation in order to facilitate the breaking of old bond and formation of new bonds.

VSA one mark

1. What is the order of a reaction whose rate constant has same units as the rate of the Reaction?

Ans: Zero order reaction

2. Why does the rate of reaction not remain constant throughout?Solution:

Because the rate of reaction depends upon concentration of reactants which keeps on

decreasing.

3. Identify the reaction order from each of the following rate constants:i) k= 6.2 X 10-5 L mol-1 s-1

ii) k=4.0X10-4 s-1

4. A reaction is 50% complete in 2hours and 75% complete in 4 hours. What is the order of reaction.

SA-1 Two Marks:

1.

Express the rate of following reactions in terms of concentrations of reactants and

products.

i) PCl5 → PCl3 + Cl2

ii) 2NO2 → 2NO + O2

Solution:

i) rate of reaction = -

ii) Rate of reaction = 2.

Ammonia and Oxygen reacted at high temperature as shown below.

4NH3 +5O2 → 4NO+ 6H2O

The rate of formation of NO is 3.6 X 10-3 mol/lit/sec. calculate rate of disappearance of

ammonia and rate of formation of water.

Solution:

Rate =

i) Rate of disappearance of NH3 = Rate of appearance of NO = 3.6 X 10-3 mol/lit/sec

ii) Rate of formation of H2O =

Therefore = X 3.6 X 10-3 mol/lit/sec

3.

Calculate the overall order of reaction which has the rate expression:

i) Rate = k[A]1/2 [B]3/2

ii) Rate = k [A] 3/2 [B] -1

iii) A + B →C is zero order, write rate equation.Solution:

i) Order of reaction = + =2 Second order.

ii) Order of reaction = +(-1) = Half order.iii) For zero order reaction

A + B → C

Rate = k{A]0 [B]0

Order of the reaction = 0 + 0 = 0 Zero order

4.

The rate law of a reaction of A,B and C has been found to be rate = k [A][B][C]2. How are

the rate of reaction change when

i) concentration of C is doubledii) concentration of A is halvediii) concentration of both B and C are doublediv) concentration of each of A, B, C and D are tripled?

Solution:

Suppose the initial concentrations are :

[A] = a mol L-1,

[B] = b mol L-1 Rate = k abc2

[C]= c mol L-1

i) New[C] = 2c

Rate = k ab(2c)2 = 4kabc2

Rate of reaction is four folded (4 times).

ii) New[A] = a/2

Rate = k[ ]bc = kabc

Rate of reaction is halved.

iii) New[B] = 2b New[C] = 2c

Rate = ka(2b) (2c)2 = 8kabc2

Rate of reaction increased to 8 times.

iv) New [A] = 3a, [B] = 3b, [C] = 3cRate = k (3a)(3b)(3c)2 = 81kabc2

Rate increased to 81 times.

5. The decomposition of N2O5 at 320K according to the following equation follows first

order reaction:

N2O5(g) → 2NO2(g) + O2 (g)

The initial concentration of N2O5 was 1.24X10-2 mol L-1 and that after 60 minutes it was 0.20X10-2 mol L-1. Calculate the rate constant of the reaction at 320K.

For a first order reaction

T=60 min.[R]0 =1.24X10-2 mol L-1,[R]

=0.20X10-2 mol L-1

=

=

6. The rate constant for the first order reaction is 60 sec. how much time will it take to

reduce the concentration to 1/16 th value?

Solution:

Let the initial conc.[A]0 = a Final concentration =

Rate constant =60 s-1

= =

SA-II of Three Marks:

1.

. The decomposition of hydrogen peroxide in the presence of iodide ion has been found to

be first order in H2O2:

2H2O2 → 2H2O + O2

The rate constant has been found to be 1.01 X 10-2 min-, calculate the rate of reaction when i) [H2O2] = 0.4 mol/lit ii) [H2O2] = 0.15 mol/lit

iii) What concentration of [H2O2] would give rate of 1.12 X10-2 mol/lit /min

Solution:

The reaction is of first order in H2O2 so that

Rate = k[H2O2]

i) k = 1.01X10-2min-1, [H2O2] = 0.4 mol/L-1

Rate = (1.01X10-2min-1)X(0.4 mol L-1)

ii) When [H2O2] = 0.15 mol L-1

Rate =(1.01X10-2min-1X(0.15 mol L-1)

=1.5X10-3 mol L-1 min-1

iii) To obtain concentration of H2O2 when rate

=1.12X10-2 mol L-1 min-1

[H2O2]=

= 1.11 mol L-1

2. The following data were obtained for the reaction:

2NO(g) + Br(g) → 2NOBr(g)

Experiment initial conc initial rate

[NO] [Br2] (mol L-1 min -1)

I 0.10 0.10 1.3X10-6

II 0.20 0.10 5.2X10-6

III 0.20 0.30 1.56X10-5

Determine i) the orders with respect to NO and Br2 ii) the rate law and

i) Rate constant.Solution:

The rate law may be written as: Rate=k[NO]p[Br2]q

The initial rate becomes (Rate)0 = k[NO]0p[Br2]q

Comparing experiment I and II, we get

(Rate)1=k(0.1)p(0.1)q=1.3X10-6

(Rate)2=k(0.2)p(0.1)q=5.2X10-6

Dividing Eq. (ii) by Eq. (i)

Or (2)p = 4 or (2)p = 22

P=2.

Thus, order with respect to NO is 2.

Comparing experiments II and III

(Rate)2 k(0.2)p(0.1)q = 5.2X10-6

(Rate)3=k(0.2)p(0.3)q=1.56X10-5

Divding Eq. (iv) by Eq.(iii), we get:

Or (3)q = 3

Or q=1

Thus, order with respect can be calculated by substituting the values of rate, [NO] and [Br2] for any experiment

=1.3X10-3 mol-2L2s-1

3. For a first order reaction, show that time require for the completion of 99.9% of

reaction is 3 times time required for completion of 90% of the reaction.

4.


1.

For the reaction 2A + B + C → A2B + C

The rate law has been found to be Rate = k[A][B]2 with k = 2.0 X 10-6 mol -2 L2 s-1 For this reaction, determine the initial rate of reaction with[A]=0.1 mol L- 1 B]=0.2

mol L-1,[C] =0.8 mol L-1.Determine the rate after 0.04 mol L-1 of A has been reacted.

Solution:

Rate law is :

Rate = k[A][B]2

Where k= 2.0 X 10-6 mol-2 L2 s-1

Initial concentration,

[A] = 0.1 mol L-1,[B]=0.2 mol L-1,[C]=0.8 mol L-1

Rate = (2.0X10-6 mol-2 L2 s-1)(0.1 mol L-2)(0.2 mol L-1)2

=8X10-9 mol L-1 s-1

Conc. Of A after 0.04 mol L-1 of A has been reacted

[A] = 0.1-0.04 = 0.06 mol L-1

Now , we know that when 2 moles of A are consumed then one mole of B gets consumed.

Therefore , moles of B consumed when 0.04 moles of A have been consumed=0.04X1/2 = 0.02

mole

[B] = 0.2 – 0.02 = 0.18 mol L-1

Rate = (2.0X10-6 mol-2 L2 s-2)X(0.06 mol L-1)X(0.18 mol L-1)2

=3.89X10-9 mol L-1 s-1

2. How many times a rate of a reaction increases if its temperature is increased by 100, When its activation energy is zero.

3.

4.

5.

SURFACE CHEMISTRY

I. Adsorption The accumulation of molecular species at the surface rather than in the bulk of a solid or

liquid is known as adsorption. It is surface phenomena.

II. Difference between and adsorption and adsorbs.

Absorption Adsorption

1.It is the phenomenon in which the particles of gas or liquid get uniformly distributed throughout the body of the solid.

2. The concentration is the same throughout the material. Therefore, it is a bulk phenomenon.

3. Absorption occurs at uniform rate.

1.it is the phenomenon of higher concentration of particles of gas or liquid on the surface than in the bulk of the solid.

2. The concentration on the surface of the adsorbent is different from that in the bulk.

3. Adsorption is rapid in the beginning and its rate slowly decreases.

III. Thermodynamics behind adsorption

Adsorption of a gas liquid is a ways exothermic process, H=-Ve. Adsorption lowers the degree of randomness of adsorbate (adsorbed substance gas/liquid).

So S=-Ve.

*For a process to be spontaneous. G=-Ve. As the adsorption proceeds, H becomes less –Ve and slowly H becomes equal to T S,

which leads to zero value of G, at this state equilibrium is attend.

IV. Difference between physisorption and Chemisorptions

Physisorption Chemisorption

1) It arises because of Vander Waals’forces.

2) It is not specific in nature.

3) It is reversible in nature

4) It depends on the nature of gas. More easily liquefiable gases are adsorbed readily

5) Enthalpy of adsorption is low (20-40) kJ mol-1) in this case.

6) Low temperature is favorable for adsorption. It decreases with increases of temperature.

7) No appreciable activation energy is needed.

8) It depends on the surface area. It increases with an increase of surface are

9) It results into multimolecular layers on adsorbent surface under high pressure.

1)It is caused by chemical bond formation

2.it is highly specific in nature

3).It is irreversible

4).It also depends on the nature of gas. Gases which can react with the adsorbent show chemisorptions.

5.Enthalpy of adsorption is high (80-240kJ mol-1

In this case.

6). High temperature is favorable for adsorption. It increases with the increase of temperature.

7) High activation energy is sometimes needed.

8) It also depends on the surface area. It too increases with an increase of surface area.

9. It results into unimolecular layer.

V. Important features of solid catalysis:Activity of a catalyst: the ability of a catalyst to increase the rate of a reaction is called activity of a catalyst. The catalytic activity increases we move from group 5 to group11.Maximum activity is shown by groups 7 -9 elements of modern periodic table

SelectivityThe selectivity of a catalyst is its ability to direct a reaction to yielda particular product. For e

example, starting with H2 and CO, andusing different catalysts, we get different products.

CH CH + H2 CH3 –CH3

CH CH + H2 CH2 = CH2

Zeolites as catalysts:

Zeolites are micro porous aluminosilicates with 3D net work of silicates in which some silica (Si) atoms are replaced by aluminum (Al) atoms.

ZSM-5, Zeolite is used to convert alcohols to gasoline by dehydration of alcohols.

Characteristics of enzyme catalysis:

Highly efficient Highly specific Highly active under optimum temperature, pH Activated by co-enzymes Affected by inhibitors and poisons Mechanism of enzyme catalysis: (Lock and Key Mechanism):

Various steps involved in enzyme catalysis are E + S → ESES → EPEP → E + PE = Enzyme. S = Substrate. P = Product.

Catalyst

Catalyst is a substance which alters the rate of reaction(may increase or decrease the rate of reaction)

If the catalyst, reactants and products all are in the same phase, then that is homogeneous catalyst. Examples are as follows

2SO2(g) + O2(g) 2SO3(g)

If the catalyst, reactants and products all are in different phase, then that is heterogeneous catalyst. Examples are as follows

If the catalyst, reactants and products all are in different phase, then that is homogeneous catalyst. Examples are as follows

2SO2(g) + O2(g) 2SO3(g)

N2(g) + 3H2(g) 2NH3

A substance which enhances the efficiency of a catalyst is known as promoter.Mo (molybdenum) acts as promoter in Haber’s manufacture of Ammonia.

N2(g) + 3H2(g) 2NH3

A substance which lowers the efficiency of a catalyst is known as catalytic poison.Arsenic and its compounds like As2O5 act as poison contact processes for manufacture of sulphuric acid.

COLLOIDS

Colloid is heterogeneous system in which dispersed phase particles with 1nm to 1000 nm size are distributed in dispersion medium.

Colloids are classified on the basis of interaction between dispersion medium and dispersion phase are a) Lyophilic and lyophobic colloids. Lyophilic means solvent loving-can be directly prepared by mixing a suitable dispersion phase with dispersion medium-these are reversible-examples-starch sol etc.

Lyophobic means solvent-hating. Can be prepared by indirect methods. Not so stable and ireversible. Eg. Ferric hydroxide sol, aluminium hydroxide sol, arsenic sulphide sol, sulphur sol and all metal sols.

Peptization

The process of converting precipitate into colloidal solution by shaking it with small

Amount of electrolyte and sufficient amount of dispersion medium is known as peptization. Generally colloidal particles adsorbs common ion from added electrolyte either positive or negative.

Examples:

Precipitate of Fe(OH)3 + FeCl3 gives Fe(OH)3 ferric hydroxide sol.

Precipitate of As2S3 + H2S gives As2S3arsenius sulphide sol.

Precipitate of AgI + AgNO3 gives AgI Silver iodide sol.

Tyndal effect: The scattering of light by colloidal particles is known as Tyndall effect, the scattered light is in the form of a cone so it is called tyndal cone. Zigmondy design the first microscope based on tyndal effect.

Tyndal effect is observed only when the following conditions are satisfied.

i) The diameter of the dispersed particles is not much smaller than the wavelength of the light used.

ii) The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude.

Brownian movement: the zig-zag motion of colloidal particles in dispersion medium is known as Brownian movement. Which was observed by British Botanist Robert Brown when pollein grains are left in water.

The primary reason for Brownian movement is constant collision between dispersion medium particles and dispersion phase particles, the secondary is that in general all colloidal particles have similar charge and they may repel with each other.

Electrophoresis: the movement of colloidal particles in eclectic field is know as electrophoresis. If they move towards anode than it is called anaphoresis. If the they move towards cathode it is called cataphoresis.

Electrophoresis is totally different from electro osmosis.

. COAGULATION.

The process of separation of dispersion phase from dispersion medium is known as

coagulation or precipitation or flocculation. This can be done by any of the following base.

By Electrophoresis By mixing two oppositely charged sols By boiling By persistent Dialysis Addition of electrolyte

HARDY-SCHULZE RULES OF COAGULATION.

i) coagulation is cause by ion which is having opposite charge to that of charge present on colloidal particles. This ion which causes coagulations is known as effective ions.

Example:

Fe(OH)3 sol can be coagulated by adding NaCl, in this contest Cl- is effective ion which causes coagulation of positive Fe(OH)3 sol .

ii) The coagulating power of effective ion is directly proportional to its charge (valency).Example:

Fe(OH)3 sol can be coagulated more effectively by Na2SO4 than NaCl. Because the effective ion of Na2SO4 is SO-2 which has more charge.

The coagulating power of Al3+ , Ba2+,Na+ in coagulation negative sols like As2S3 is in the order of Al3+ > Ba2+>Na+ .

EMULSION.

Emulsions are special type of colloidal solutions in which both dispersion medium and

dispersion phase are in liquid state. There are two types of emulsions

a) Oil dispersed in water (O/W) type.b) Water dispersed in oil (W/O) type.The above two emulsions are distinguished are by using dye test or dilution test.

If an emulsion is diluted by water then it will be oil in water type emulsion otherwise it is water in oil type.

The conductivity of oil in water type emulsion increases by adding any ionic compound which is soluble in water.

If a dye is added to oil in water type emulsion color droplets are noticed because added dye is soluble only in oil but not in water.

Very Short Answer Questions of one mark

Q1. Which of the two: absorption or adsorption is a surface phenomenon?

Solution:

Adsorption.

Q2. Compare the heat of adsorption for physical and chemical adsorption?

Solution:

The heat of adsorption for chemical adsorption is high (of the order of 200-400 J/mol)

while the heat of adsorption for physical adsorption is low(of the order of 20-40/mol).

Q3. Why are lyophilic colloidal sols. More stable than lyophobic colloidal sols?

Solution:

The Lyophilic colloidal sols. Are more stable because3 they are highly hydrated in

Solution.

Q5. What is colloidon ?

Solution:

Colloidon is a sol. Of cellulose nitrate in ethyl alcohol.

Q6. Why is adsorption exothermic ?

Q7.Powdered Catalsyt is more effective over solid Catalyst .Why?

Q8.Why is chemisorptions takes place at higher temperature ?

SA- I of two marks

Q1. How the nature of adsorbate affects the adsorption of gases on solids?

Solution:

Easily liquefiable gases like NH3, CO2, HCl get adsorbed to greater extent than

permanent gases like H2,O2,N2 etc,. the ease of liquefaction of a gas depends upon its

critical temperature. Higher is the critical temperature more easily will be adsorbed on

solid.

Q2. The extent of physical adsorption decreases with rise in temperature why?

Solution :

i)The physisorption shows the features of physical process .As physisorption is generally exothermic, the extent of physisorption decreases with rise in temperature.

ii) The physisorption is due to weak wander wall bond between absorbent and adsorbate which starts breaking with rise in temperature.

Q3. Explain the peptization.

Solution:

It is the process of conversion of precipitate in to a colloidal solution by shaking the

Precipitate with suitable electrolyte. Electrolyte used is also called as peptizing agent.

Example: Shaking c Precipitate with FeCl3 solution gives + sol

Shaking As2S3 Precipitate with H2S solution gives – sol

Q4. Delta is generally formed when river meets the Ocean. Why?

Solution:

River water is a colloidal solution of negatively charged silicates, clay and other earthly

Impurities. Sea water consists of Ca2+ ions and Mg2+ ions. So coagulation takes place at

the place where Sea water and river water meets each other which leads to delta

formation.

Q5. What phenomenon will occur when silica gel and anhydrous calcium chloride are placed separately in two corners of a vessel containing water vapour.

Solution:

Adsorption of water takes place in the case of silica gel. Whereas absorption of water

takes place in the case of anhydrous calcium chloride.

Q6. Differentiate between homogeneous and heterogeneous catalysis. Give one example for each.

Solution:

Homogeneous catalysis:-A catalysis in which the catalysts are present in the same phase as the reactants. For example- oxidation of CO gas with O2 gas in the presence of NO gas as catalyst.

Heterogeneous catalysis:-A catalysis in which a catalyst is present in different phase than that of the reactants. For example- manufacture of ammonia from nitrogen and hydrogen gases using solid iron as catalyst.

Q7. What is emulsion? What are their different types?

An emulsion is the colloidal dispersion in which both the dispersed phase and the dispersion mediums are liquids. They can be of two types:-i) Emulsion of oil in water.ii) Emulsion of water in oil.

8 Explain the following terms giving a suitable example in each case:

(i) Emulsification

(ii) Homogeneous catalysis.

9. Complete the following:

i) The liquid-liquid colloidal dispersions are called…….ii) The enthalpy of chemisorption is…… than the enthalpy of physisorption.iii) The scattering of light by colloidal dispersion is called……iv) The zig-zag motion of colloidal particles is called……

Solution:

i) Emulsions ii) more iii) Tyndall effect iv) Brownian movement.

SA-II of three marks

1. Illustrate with examples:

i) Lyophilic and Lyophobic sols.

ii) Multimolecular and Macromolecular colloids.

iii) Homogeneous and Heterogeneous catalysis.

2. Explain the following observations:

i) A beam of light passing through a colloidal solution has a visible path.

ii) Passing an electric current through a colloidal solution removes colloidal

sulphate.

iii) Ferric hydroxide sol coagulates on addition of a solution of potassiumsulphate .

3. Explain the following observations:

(a) Lyophilic colloid is more stable than lyophobic colloid.(b) Coagulation takes place when sodium chloride solution is added to a colloidal solution of

ferric hydroxide.(c) Sky appears blue in color.

4. What are micelles? How do they differ from ordinary colloidal particles? Give two examples of micelles forming substances.

5. (a) In which of the following does adsorption take place and why?

i) Silica gel placed in the atmosphere saturated with water.

ii) Anhydrous CaCl2 placed in the atmosphere saturated with water.

(b) How does BF3 act as a catalyst in industrial process?

(c) Give an example of shape-selective catalysis.

Higher Order Thinking Skill

Q1. Which of the following more effective in coagulating Fe(OH)3 and why ?

Na2PO4,Na2SO4, NaCl.

Solution:i) Fe(OH)3 sol is +vely charged so it will be coagulated by –ve ionic of added electrolyte. ii) Coagulating power is directly proportional to charge , so PO-4 is more effective.

Q2. Which will be adsorbed more readily on the surface of charcoal and why NH3 or

CO2?

Solution:

NH3 because it is more easily liquefiable. Hence, NH3 has greater intermolecular force

and will be adsorbed more readily.

Q3. What is meant by Kraft temperature (Tk) and critical micelle concentration (CMC)?

Solution:

The optimum temperature from which micelle formation starts is Kraft temperature (Tk).

The optimum concentration from which micelle formation starts is critical micelle

concentration (CMC).

Q4. Give any two reasons for the origin of electrical charge on the colloidal particles.

Solution:

The two reasons are:i) Due to electron capture by sol particles during electro dispersion of metals, due to preferential adsorption of ions from solutionii) Dissociation of colloidal sols.

Q5. How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent of adsorption of a gas on a solid?

General Principles and processes of Isolation of Elements

Metallurgy: The entire scientific and technological process used for isolation of the metal from its ores is known as metallurgy

Minerals: Which are naturally occurring chemical substances in the earth’s crust which includes Metal and impurities.

Ores: The Minerals in which metal is extracted profitably called ores. Ex: Copper glance ( Cu2S), Haematite ( Fe2O3)

Gangue: Undesired materials which present along with the metal in the earth crust known as

gangue. The extraction and isolation of metals from ores involve the following major steps:

A) Concentration of the ore (or) Purification of the OreB) Isolation of the metal from its concentrated oreC) Purification of the metal (Or) Refining of the Metal.

Froth Floatation Process: Principle: The mineral particles become wet by oils while the gangue particles by

water.Method : In this process, a suspension of the powdered ore is made with water. To it, collectors and froth stabilisers are added. A rotating paddle agitates the mixture and draws air in it. As a result, froth is formed which carries the mineral particles. The froth is light and is skimmed off. It is then dried for recovery of the ore particlesCollectors : The chemical substances which enhance non-wettability of the mineral

particles. Ex : pine oils, fatty acids, xanthates Froth stabilizers: The chemical substances stabilise the froth. Ex: cresols, aniline.Depressants : It is possible to separate two sulphide ores by adjusting proportion of oil to

water. For example, in case of an ore containing ZnS and PbS, the depressant used is NaCN. It electively prevents ZnS from coming to the froth but allows PbS to come with the froth.

Leaching of Alumina from Bauxite: The ore consists of two chemical impurities a) Iron oxide (Fe2O3) b) Silica (SiO2). These impurities are removed in the following process. The ore is treated with Sodium hydroxide solution (NaOH), in which Iron oxide (Fe2O3) impurity does not react with it and can be removed by filtration.

Al2O3(s) + 2NaOH + 3H2O 2Na[Al(OH)4](aq)The aluminate in solution is neutralised by passing CO2 gas and hydrated Al2O3 is precipitated. 2Na[Al(OH)4](aq) + CO2(g) Al2O3.xH2O(s) + 2NaHCO3

Al(OH)3 is ignited to get pure Bauxite called Alumina. Al2O3.xH2O(s) ------- Al2O3 (s) + 3 H2O (g)

Leaching of Silver Ore : In the metallurgy of silver and that of gold, the respective metal is leached with a dilute solution of NaCN or KCN in the presence of air (for O2) from which the metal is obtained later by replacement.

Roasting: The process in which the ores are heated in presence of excess of oxygen. This process is mainly used for Sulphide ores.

ZnS + O2 ZnO + SO2

Calcination : It involves heating when the volatile matter escapes leaving behind the metal oxide.This process is mainly used for Carbonate and hydroxide ores.

Fe2O3.xH2O(s) Fe2O3 (s) + xH2O(g) ZnCO3 (s) ZnO(s) + CO2(g) CaCO3.MgCO3(s) CaO(s) + MgO(s ) + 2CO2(g)

Extraction of some metals from their metal oxides:

1) Iron from Iron oxide:a) Removal of chemical impurity:

→Silica (SiO2) is a chemical impurity which is acidic in nature present along with the ore.→ A basic flux like calcium oxide (CaO) is added to remove Silica.→ CaO is obtained from the Lime stone (CaCO3) which decomposes to CaO and CO2 in

the furnace due to intense heat.CaCO3 (s) ------- CaO (s) + CO2 (g).

→The impurity SiO2, combines with CaO to form calcium silicate (CaSiO3) called Slag. CaO (s) + SiO2 (s) -------- CaSiO3.

:

Extraction of Iron→ The pure Haematite ore, Ferric oxide (Fe2O3) is reduced to iron in presence of a

reducing agent Coke (carbon).→ Coke (C) combines with CO2 to form carbon monoxide which is a very good reducing

agent which reduces Fe2O3 in to Iron (Fe).C (Coke) + CO2 (g) ------- 2CO (g) Fe2O3 (s) + 3 CO (g) ------- 2Fe (molten) + 3 CO2 (g)

→ The slag calcium silicate ( CaSiO3), is lighter than molten Iron which floats over the molten iron and also which also prevents the molten Iron to oxidize in presence of oxygen in the furnace.

→ The slag and the molten Iron are collected separately as shown in the figure. 2) Aluminium from Alumina (Aluminium oxide): Electrolytic Reduction : Hall’s Process :→The Pure Alumina having high melting point i.e 2348K., it is highly expensive to carry

the electrolytic reduction process at this temperature.

→To reduce the melting point and by increase the electrical conductivity of Alumina a substance called Cryolite (Na3AlF6) is added.

→The electrolytic reduction process carried as shown asCathode reaction : Al+3 + 3e- --- Al

Anode Reaction : 2O-2 - 4e- ----- O2 (g)→ The oxygen gas liberated at anode reacts with graphite anode and form CO2 gas, so

anode is to be replaced from time to time.C(graphite) + O2 (g) ------- CO2(g) .

→ The pure aluminium (Al) is collected from the cathode from time to time.3) Copper from Copper Oxide :

a) Removal of chemical impurity:→The ore contains FeO is a basic impurity which is removed by adding an acidic flux silica.→Iron oxide ‘slags of’ as iron silicate and copper is produced in the form of copper matte. This contains Cu2S and FeS.

FeO + SiO2 FeSiO3

(Slag)→Copper matte is then charged into silica lined convertor. hot air blast is blown to

convert the remaining FeS2, FeO and Cu2S/Cu2O to the metallic copper. Following reactions take place:

2FeS + 3O2 2FeO + 2SO2

FeO + SiO2 FeSiO3

b) Extraction of Copper:Auto reduction/ Self Reduction: Method 1: Process takes place in a reverbatory furnace:→Hot air is blown to convert the Cu2S to Cu2O. The Cu2O formed converted in to

metallic copper in the presence of remaining Cu2S which acts as a reducing agent.2Cu2S + 3O2 2Cu2O + 2SO2 2Cu2O + Cu2S 6Cu + SO2

→The solidified copper obtained has blistered appearance due to the evolution of SO2

and so it is called blister copper.Method 2: Process takes place with the help of a reducing agent C / CO :

→In the Ellingams diagram the Cu2O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (both the lines of C, CO and C, CO2 are at much lower positions in the graph particularly after 500 – 600K).

→ The sulphide ores are roasted/smelted to give oxides2Cu2S + 3O2 2Cu2O + 2SO2

→The oxide can then be easily reduced to metallic copper using coke:Cu2O + C 2 Cu + CO

4) Extraction of zinc from zinc oxide:→The reduction of zinc oxide is done using coke. The temperature in this case is higher

than that in case of copper. For the purpose of heating, the oxide is made into brickettes with coke and clay.

ZnO + C (coke) Zn + CO at 673K→The metal is distilled off and collected by rapid chilling.

Electrolytic refining:

Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode:

Anode: Cu Cu2+ + 2 e–Cathode: Cu2+ + 2e– Cu

Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.

Zone refining :

Principle: The impurities are more soluble in the melt than in the solid state of the metal.Method: Circular mobile heater is fixed at one end of a rod of the impure metal. The

molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated. This end is cut off. This method is very useful for producing semiconductor and other metals of very high purity, e.g., germanium, silicon, boron, gallium and indium.

Vapour phase refining :In this method, the metal is converted into its volatile compound. It is then decomposed to give pure metal. So, the two requirements are:

(i) The metal should form a volatile compound with an available reagent,(ii) The volatile compound should be easily decomposable, so that the recovery is easy.Examples :.1) Mond Process for Refining Nickel: In this process, nickel is heated in a stream of carbon monoxide forming a volatile

complex, nickel tetracarbonyl. Ni + 4CO Ni(CO)4 at 330 – 350 K.

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal

Ni(CO)4 Ni + 4CO at 450 – 470 K.2) Van Arkel Method for Refining Zirconium or Titanium: This method is very useful for removing all the oxygen and nitrogen present in the form

of impurity in certain metals like Zr and Ti. The crude metal is heated in an evacuated vessel with iodine. The metal iodide being more covalent, volatilizes.

Zr + 2I2 ZrI4 The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800K. The pure metal is thus deposited on the filament.

ZrI4 Zr + 2I2

VSA (One mark)

1. Why are cryolite and fluorospar added to alumina during electrolytic reduction?(1 Mark)(Ans) Cryolite and fluorospar are added to alumina during elctrolytic

reduction to reduce the melting point of alumina and to increase its conductivity.

2. Reduction with C for Cu2O can be done at a lower temperature than ZnO. Why? (1 M)

(Ans) In the ellingham diagram the curve for Cu2O lies higher than ZnO i.e. for the reduction of Cu2O with C the negative value of gibbs energy can be reached at a lower temparature than ZnO.

3. Although thermodynamically feasible in practice magnesium metal is not used for the reduction of alumina. Why?

(Ans) Magnesium can reduce alumina at the temperature above 15000c (The intersection point of the curves for Al2O3 and MgO in the Gibbs Energy vs T plot (ellingham diagram)). But the temperature at which this is feasible is too high to be achieved economically and is also technologically difficult. So this reduction is not done.

4. What is the role of depressant in froth floatation process? (Ans) In froth floatation process the depressant selctively prevents one of

the ores from coming to the froth in a mixture of two ores hence enabling the separation of the other one with the froth.

5. State the role of silica in the matallurgy of copper.(Ans) Silica in the metallurgy of copper helps in the removal of iron oxide as iron silicate (slag).

6. What is the role of graphite rods in the electrometallurgy of aluminium? (Ans) In the electrometallurgy of aluminium graphite rods act as anodes in

the electrolytic cell of reduction and are the site for release of oxygen.

7. Which is the purest form of iron and what are its uses?(Ans) The purest form of iron is wrought iron and is used in making anchors, wires, bolts etc.

8: On what principle froth floatation process is carried for sulphide ores?

Ans: Wetting of Ore particles with the oil and the impurities with water

SA –I (two marks)

1. What are minerals and how are they different from ores? (Ans) Minerals are the naturally occurring chemical substances in the

earth’s crust obtained by mining. It is different from ores, as ores are the minerals that are used for the extraction of metals profitably.

2. Name one ore each for iron and copper and give their chemical compositions. (Ans) The ore of iron is haematite- Fe2O3 and the ore for copper is copper pyrites- CuFeS2.

3. What is the purpose of adding collectors and froth stabilisers during froth floatation? Give an example for each.

(Ans) During froth floatation process collectors like pine oil and fatty acids are added to enhance non wettability of the mineral particles and the froth stabilisers like cresol and aniline stabilise the froth.

4. How can the ores ZnS and PbS be separated from a mixture using froth floatation process?

(Ans) During the froth floatation process a depressent like NaCN is added to the tank. The depressent selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps the separation of PbS with the froth.

5. What are minerals and how are they different from ores? (Ans) Minerals are the naturally occurring chemical substances in the

earth’s crust obtained by mining. It is different from ores, as ores are the minerals that are used for the extraction of metals profitably.

6. Name one ore each for iron and copper and give their chemical compositions. (Ans) The ore of iron is haematite- Fe2O3 and the ore for copper is copper pyrites- CuFeS2.

7. What is the purpose of adding collectors and froth stabilisers during froth floatation? Give an example for each.

(Ans) During froth floatation process collectors like pine oil and fatty acids are added to enhance non wettability of the mineral particles and the froth stabilisers like cresol and aniline stabilise the froth.

8. How can the ores ZnS and PbS be separated from a mixture using froth floatation process?

(Ans) During the froth floatation process a depressent like NaCN is added to the tank. The depressent selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps the separation of PbS with the froth.

9. What is vapour phase refining? What are the necessary requirements for the compound to be purified by vapour phase refining?

SA – II (Three marks)1. Name the chief ore of Iron. Write the method of extraction from that

ore. Mention the function of limestone. 2. What is copper matte? How is copper obtained from it’s ore ? Why is

copper found in blister form ?

3. Name the principal ore of Aluminium. Explain the significance of bleaching in the extraction of Aluminium.

4. Explain the role of each of the following(I) NaCN in the extraction of Silver(ii) SiO2 in the extraction of copper(iii) Coke in the extraction of Zinc

H O T S 1: Free energies of formation (∆fG) of MgO (l) and CO(g) at 1273K and

2273K are given below.

∆fG (MgO(s)) = - 941 kj/mol at 1273K, ∆fG(MgO(s)) = -314 KJ/mol at 2273K,

∆fG(CO(g)) = -439KJ/mol at 1273K, ∆fG(CO(g)) = -628KJ/mol at 2273K. On the basis of above data, predict the temperature at which carbon can be used as a reducing agent for MgO (s).

Ans: Reduction of MgO (s) with carbon can be written as MgO (s) + C (s) -- Mg(s) + CO (g)

∆rG= ∆fG(products) -∆fG(reactants)

∆rG= ∆fG(CO(g)) -∆fG(MgO(s))

At 1273K, ∆rG = (-439) – (-941) =502Kj/mol, i.e ∆rG > 0

At 2273K, ∆rG = (-628)-(-314) = -314Kj/mol, i.e ∆rG < 0

At 2273K, ∆rG < 0 So, the reaction is spontaneous.

2: What chemical principle is involved in choosing a reducing agent for getting the metal from its oxide Ore? Considering the metal oxides, Al2O3 and Fe2O3, and justify the choice of reducing agent in each case.

3.How does NaCN act as a depressant in preventing ZnS from forming the froth?

P BLOCK ELEMENTS

In the case of nitrogen, all oxidation states from +1 to +4 tend to disproportionate in acid solution. For example,

It can also be obtained by the thermal decomposition of ammonium dichromate.

On a small scale ammonia is obtained from ammonium salts which

decompose when treated with caustic soda or lime.

On a large scale, ammonia is manufactured by Haber’s process. Ammonia donate lone pair to the metal ions and form different complex

compounds, by this metals ions can be detected.

NH4 NO3 Heat N2O + 2 H2O 4HNO3 + P4O10 4H3PO3 + 2N2O5 On a large scale it is prepared mainly by Ostwald’s process. This method is

based upon catalytic oxidation of NH3 by atmospheric oxygen.

Concentrated nitric acid is a strong oxidizing agent and attacks most metals

except noble metals such as gold and platinum.

Zinc reacts with dilute nitric acid to give N2O and with concentrated acid to give NO2.

Concentrated nitric acid also oxidises non–metals and their compounds. Iodine is oxidised to iodic acid, carbon to carbon dioxide, sulphur to H2SO4, and phosphorus to phosphoric acid.

Brown Ring Test: The familiar brown ring test for nitrates depends on the

ability of Fe2+ to reduce nitrates to nitric oxide, which reacts with Fe2+ to form a brown coloured complex.

It dissolves in boiling NaOH solution in an inert atmosphere giving PH3.

It readily catches fire in air to give dense white fumes of P4O10.

Phosphine is prepared by the reaction of calcium phosphide with water or

dilute HCl.

In the laboratory, it is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2.

it is absorbed in HI to form phosphonium iodide (PH4I) which on treating with KOH gives off phosphine.

When absorbed in copper sulphate or mercuric chloride solution, the corresponding phosphides are obtained.

It reacts with organic compounds containing –OH group such as CH3COOH, C2H5OH.

phosphorous acid on heating disproportionates to give orthophosphoric acid (or phosphoric acid) and phosphine.

SO2 combines with water to give H2SO3, an acid.

Al2O3 reacts with acids as well as alkalies.

it oxidizes lead sulphide to lead sulphate and iodide ions to iodine.

The key step in the manufacture of H2SO4 is the catalytic oxidation of SO2 with O2 to give SO3 in the presence of V2O5 (catalyst).

Sulphuric acid, because of its low volatility can be used to manufacture more volatile acids from their corresponding salts.

Sulphuric acid removes water from organic compounds; it is evident by its charring action on carbohydrates.

metals and non-metals are oxidized by concentrated sulphuric acid, which is reduced to SO2.

In general, a halogen oxidizes halide ions of higher atomic number.

I– can be oxidised by oxygen in acidic medium; just the reverse of the reaction observed with fluorine.

Preparation of Chlorine:

REASONING QUESTIONS IN P BLOCK ELEMENTS

1. Though

nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give reason.

Ans: Nitrogen valence electronic configuration ns2np3 due to absence empty

d- orbitals, it can not extend its valence to 5

2. PH3 has lower boiling point than NH3. Why?

Ans: Unlike NH3, PH3 molecules are not associated through hydrogen bonding in

liquid state. That is why the boiling point of PH3 is lower than NH3.

3. Why are pentahalides more covalent than trihalides ?

Ans: Higher the positive oxidation state of central atom, more will be its

Polarizing power which, in turn, increases the covalent character of bond

formed between the central atom and the other atom.

4. Why is BiH3 the strongest reducing agent amongst all the hydrides of

Group 15 elements ?

Ans: Among hidres of group 15 from NH3 to BiH3 as size of central atom

Increases M—H bond strength decreases. Hence it act as strong reducing

agent.

5. Write the reaction of thermal decomposition of sodium azide.Ans: 2 NaN3 2 Na + 3 N2(g)

6. Why is N2 less reactive at room temperature?

Ans: Due to its high NN bond dissociation energy

7. Why does NH3 act as a Lewis base?

Ans: Due to presence of one lone pair of electron present on Nitrogen

8. Mention the conditions required to maximize the yield of ammonia.

Ans: Optimum Pressure & Optimum Temperature & Removal of ammonia by

cooling.

9. How does ammonia react with a solution of Cu2+?

Ans: Cu2+ + 4 NH3 [Cu(NH3)4]+2 and this is deep blue in color.

10. Why does NO2 dimerise ?

Ans: Due to presence one odd electron on Nitrogen of NO2

11. What is the covalence of nitrogen in N2O5 ?

Ans: Covalency is 4 because N is bonded with 4 bonds

12. In what way can it be proved that PH3 is basic in nature?

Ans: Phosphine is weakly basic and like ammonia, gives phosphonium

compounds with acids e.g., PH3 + HBr PH4Br

Due to lone pair on phosphorus atom, PH3 is acting as a Lewis base in

the above reaction.

13. Bond angle in PH4 + is higher than that in PH3. Why?

Ans: Both are sp3 hybridized. In PH4+ all the four orbital are bonded whereas

in PH3 there is a lone pair of electrons on P, which is responsible for lone

Pair-bond pair repulsion in PH3 reducing the bond angle to less than 109° 28

14. What happens when white phosphorus is heated with concentrated NaOH?

Solution in an inert atmosphere of CO2 ?

15. Why does PCl3 fume in moisture?

Ans: PCl3 react with moisture and gives HCl vapours.

PCl3 + 3 H2O H3PO3 + 3 HCl

16. Are all the five bonds in PCl5 molecule equivalent? Justify your answer.

Ans: PCl5 has a trigonal bipyramidal structure and the three equatorial

P-Cl bonds are equivalent, while the two axial bonds are different and

longer than equatorial bonds.

17. What happens when PCl5 is heated?

Ans: PCl5→ PCl3 + Cl2

18. Why does PCl3 fumes in the air?

Ans: PCl3 hydrolyses in the presence of moisture giving fumes of HCl.

PCl3 + 3 H2O H3PO3 +3HCl

19. How do you account for the reducing behavior of H3PO2 on the basis

of its structure ?

Ans: In H3PO2, two H atoms are bonded directly to P atom which imparts

Reducing character to the acid.

20. What is the basicity of H3PO4?

Ans: Three P–OH groups are present in the molecule of H3PO4. Therefore, its

basicity is three.

21. Phosphorous in solid state is ionic, why?

Ans: In the solid state it exists as an ionic solid, [PCl4]+[PCl6]– in which the

cation, [PCl4]+ is tetrahedral and the anion, [PCl6]–octahedral.

22. Elements of Group 16 generally show lower value of first ionization

enthalpy compared to the corresponding periods of group 15. Why?

Ans: Due to extra stable half-filled p orbital electronic configurations of Group 15

elements, larger amount of energy is required to remove electrons compared

to Group 16 elements.

23. H2S is less acidic than H2Te. Why?

Ans: Due to the decrease in bond (E–H) dissociation enthalpy down the group,

acidic character increases.

24. Write the order of thermal stability of the hydrides of Group 16 elements.

Ans; H2O>H2S>H2Se>H2Te>H2Po As atomic size increases E—H bond

strength decreases, Hence thermal stability decreases.

25. Why is H2O a liquid and H2S a gas ?

Ans: Because of small size and high electro negativity of oxygen, molecules of

water are highly associated through hydrogen bonding resulting in its

liquid state.

Why does O3 act as a powerful oxidizing agent?

Ans: Due to the ease with which it liberates atoms of nascent oxygen (O3 O2 + O),

it acts as a powerful oxidising agent.

27.How is O3 estimated quantitatively? Ans: When ozone reacts with an excess of potassium iodide solution buffered

with a borate buffer (pH 9.2), iodine is liberated which can be titrated

against a standard solution of sodium thiosulphate. This is a quantitative

method for estimating O3 gas.

28.Which form of sulphur shows paramagnetic behavior ? Ans: In vapour state sulphur partly exists as S2 molecule which has two unpaired

electrons in the antibonding * orbitals like O2 and, hence, exhibits

paramagnetism.

29. What happens when sulphur dioxide is passed through an aqueous

solution of Fe(III) salt?

Ans: When moist, sulphur dioxide behaves as a reducing agent. For

example, it converts iron(III) ions to iron(II) ions.

30. Comment on the nature of two S–O bonds formed in SO2 molecule. Are

the two S–O bonds in this molecule equal ?

Ans: Both the S–O bonds are covalent and have equal strength due to resonating

Structures.

.

31. How is the presence of SO2 detected ?

Ans: It is a suffocating odor gas, when SO2 gas passed into lime water it gives

Milky white ppt Ca(OH)2 + SO2 CaSO3 + H2O

32. Write the conditions to maximize the yield of H2SO4 by Contact process.

Ans:

The reaction is exothermic, reversible and the forward reaction leads

to a decrease in volume. Therefore, low temperature and high pressure

are the favourable conditions for maximum yield. But the temperature

should not be very low otherwise rate of reaction will become slow.

33. Why is Ka2 << Ka1 for H2SO4 in water ?

Ans: H2SO4 is a very strong acid in water largely because of its first ionization

to H3O+ and HSO4–. The ionisation of HSO4– to H3O+ and SO42– is very very

small. That is why Ka1 Ka2 .

34. Halogens have maximum negative electron gain enthalpy in the respective

periods of the periodic table. Why?

Ans: Halogens have the smallest size in their respective periods and therefore

high effective nuclear charge. As a consequence, they readily accept one

electron to acquire noble gas electronic configuration.

35. Although electron gain enthalpy of fluorine is less negative as compared

to chlorine, fluorine is a stronger oxidizing agent than chlorine. Why?

Ans: It is due to (i) low enthalpy of dissociation of F-F bond

(ii) high hydration enthalpy of F–

36. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit + 1,

+ 3, + 5 and + 7 oxidation states also. Explain.

Ans: Fluorine is the most electronegative element and cannot exhibit any

Positive oxidation state. Other halogens have d orbitals and therefore,

can expand their octets and show + 1, + 3, + 5 and + 7 oxidation states also.

37. Considering the parameters such as bond dissociation enthalpy, electron gain

enthalpy and hydration enthalpy, compare the oxidizing power of F2 andCl2.

Ans: Oxidizing power depends on bond dissociation enthalpy, electron gain

enthalpy and hydration enthalpy out of these Fluorine has more hydration

Enthalpy and less bond dissociation energy that can compensate high

negative enthalpy of Chlorine .

38. Give two examples to show the anomalous behavior of fluorine.

Ans: Most of the reactions of fluorine are exothermic (due to the small

and strong bond formed by it with other elements). It forms only one

oxoacid while other halogens form a number of oxoacids. Hydrogen

fluoride is a liquid (b.p. 293 K) due to strong hydrogen bonding. Other

hydrogen halides are gases.

39. Sea is the greatest source of some halogens. Comment.

Ans: Various minerals of halogens from soil are soluble in river water hence sea

gets large amount of halogens in the dissolved form.

40. Write the balanced chemical equation for the reaction of Cl2 with hot and

Concentrated NaOH. Is this reaction a disproportionate reaction? Justify.

Ans: 3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O

Yes, chlorine from zero oxidation state is changed to –1 and +5

oxidation states.

41. When HCl reacts with finely powdered iron, it forms ferrous chloride and

not ferric chloride. Why?

Ans: Its reaction with iron produces H2.

Fe + 2HCl FeCl2 + H2

Liberation of hydrogen prevents the formation of ferric chloride.

42. Deduce the molecular shape of BrF3 on the basis of VSEPR theory.

2 lp and 3 bp as VSPER theory it has trigonal bipyramydal shape.

43. Why is ICl more reactive than I2?

Ans: In general, interhalogen compounds are more reactive than halogens due

to weaker X–X1 bonding than X–X bond. Thus, ICl is more reactive than I2.

44. Why are the elements of Group 18 known as noble gases ?

Ans: The elements present in Group 18 have their valence shell orbital

completely filled and, therefore, react with a few elements only under

certain conditions. Therefore, they are now known as noble gases.

45. Noble gases have very low boiling points. Why?

Ans: Noble gases being monatomic have no interatomic forces except weak

dispersion forces and therefore, they are liquefied at very low

Temperatures. Hence, they have low boiling points.

46. Does the hydrolysis of XeF6 lead to a redox reaction?

Ans: No, the products of hydrolysis are XeOF4 and XeO2F2 where the oxidation

states of all the elements remain the same as it was in the reacting state.

47. Why is helium used in diving apparatus?

Ans: It is used as a diluent for oxygen in modern diving apparatus because of its

Very low solubility in blood.

48. Balance the following equation: XeF6 + H2O XeO2F2 + HF

Ans: XeF6 + 2H2O XeO2F2 + 4HF

49. Why has it been difficult to study the chemistry of radon?

Ans: Radon is radioactive with very short half-life which makes the study of

Chemistry of radon difficult.

50. Halogens have maximum negative electron gain enthalpy in the respective

periods of the periodic table. Why?

Ans: Halogens have maximum negative electron gain enthalpy in the

corresponding periods. This is due to the fact that the atoms of these

elements have only one electron less than stable noble gas configurations.

51. AgCl is soluble in water, where as it is soluble in aqueous ammonia

solution.

Ans: Ammonia react with AgCl forms [Ag(NH3)2]Cl complex which is soluble

in water.

52. Give the reason for bleaching action of Cl2.

Ans: Chlorine water on standing loses its yellow colour due to the

formation of HCl and HOCl. Hypochlorous acid (HOCl) so formed, gives

nascent oxygen which is responsible for oxidising and bleaching

properties of chlorine.

53. Why does the reactivity of nitrogen differ from phosphorus?

Ans: Nitrogen can form multiple bonds and has more bond dissociation energy

due to which it is less reactive. Phosphorous can not form multiple bonds.

54. Why does NH3 form hydrogen bond but PH3 does not?

Ans: Electronegetivity of N is more than P.

55. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Ans: Due to more electronegative nature of Nitrogen it causes more repulsion,

hence bond angle is more.

56. Why does R3P = O exist but R3N = O does not (R = alkyl group)?

Ans: Nitrogen can not form d-p multiple bond due absence of empty

d - orbitals.

57. Explain why NH3 is basic while BiH3 is only feebly basic.

Ans: Due to small size of ‘N’ it has more electron density and act as strong base.

58. Nitrogen exists as diatomic molecule and phosphorus as P4. Why?

Ans: ‘N’ can form multiple bonds where as P can not hence its tri valency

it utilized in tetrahedron structure.

59. Give the disproportionate reaction of H3PO3.

Ans: phosphorous acid on heating disproportionates to give orthophosphoric acid

(or phosphoric acid) and phosphine.

60. Can PCl5 act as an oxidizing as well as a reducing agent? Justify.

Ans: In PCl5 oxidation state of P is +5, it can not increase its oxidation

state more than it hence it can not act as reducing agent. But it can

change its oxidation state to +3 and act as oxidizing agent.

61. Why is dioxygen a gas but sulphur a solid?

Ans: Due to small size Oxygen can form multiple bonds where as S can’t hence

S bond with S8 molecule and exist in solid.

62. Knowing the electron gain enthalpy values for O O– and O O2– as –141

And 702 kJ mol–1 respectively, how can you account for the formation of a

Large number of oxides having O-2 species and not O–?

Ans: Higher second electron gain enthalpy is compensated by high hydration

energy and more lattice energy.

63. Why are halogens strong oxidizing agents?

Ans: Since halogens are having high electron gain enthalpy they act as strong

oxidizing agents.

64. Explain why fluorine forms only one oxoacid, HOF.

Ans: ‘F’ can form only -1 oxidation state, due to small size and absence of

empty d – orbitals , hence it forms only one oxoacid HOF

65. Explain why in spite of nearly the same electro negativity, oxygen forms

Hydrogen bonding while chlorine does not.

Ans: Due to small size of Oxygen as compared to Sulphur.

66. Why are halogens colored?

Ans: All halogens are coloured. This is due to absorption of radiations in visible

region which results in the excitation of outer electrons to higher energy

level. By absorbing different quanta of radiation, they display different

colours. For example, F2, has yellow, Cl2 , greenish yellow, Br2, red and

I2, violet colour.

67. What inspired N. Bartlett for carrying out reaction between Xe and PtF6?

Ans: In March 1962, Neil Bartlett, then at the University of British Columbia, observed

the reaction of a noble gas. First, he prepared a red compound which is formulated as

O2+PtF6–. He, then realised that the first ionization enthalpy of molecular oxygen (1175 kJmol–1) was almost identical with that of xenon (1170 kJ mol–1). He made efforts to prepare same type of compound with Xe and was successful in preparing another red colour

compound Xe+PtF6 – by mixing PtF6 and xenon. After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised.

68. With what neutral molecule is ClO– isoelectronic? Is that molecule a Lewis

base?

Ans: ClF is isoelectronic with ClO- and it has lone pairs hence act as Lewis

base.

69.. Arrange the following in the order of property indicated for each set:

(i) F2, Cl2, Br2, I2 - increasing bond dissociation enthalpy.

(ii) HF, HCl, HBr, HI - increasing acid strength.

(iii) NH3, PH3, AsH3, SbH3, BiH3 – increasing base strength.

Ans: (i) Cl2> Br2>F2>I2 As atomic size increases bond dissociation energy

Decreases and since F is extremely small size and cause repulsion

Between lone pairs of two F atoms of F2 molecule.

(ii) HF<HBr<HCl<HI as atomic size increases from F to I bond strength

decreases and acidity increases.

(iii) BiH3<SbH3<AsH3<PH3<NH3 as size increases from N to Bi electron

density decreases and basesity decreases.

70. Give the formula and describe the structure of a noble gas species which

is isostructural with:

(i) ICl4- (ii) IBr2- (iii) BrO3-

Ans: (i) XeCl4 ( 2lp + 4 bp =6, sp3d2occupied with 2 lone pairs over all

geometry is square planar)

(ii) XeBr2 (2bp+4lp=6, sp3d2occupied with 2 lone pairs over all

geometry is linear)

(iii) XeO3 (3bp+ 1lp=4, sp3 occupied by one lone pair, pyramidal shape.

71. Why do noble gases have comparatively large atomic sizes?

Ans: Noble gases exist as mono atomic gases, there is no overlapping of atomic

orbitals like in other compounds. Here atomic radius is considered as

vandrwaal’s radii. Hence atomic radii are comparatively large.

72.Why does the reactivity of nitrogen differ from phosphorus? Ans: Due to small size, multiple bond formation, absence of empty d- orbitals.

73. The increase in the atomic size from As to Bi is small as compare from

N to P, justify your answer.

Ans: There is a considerable increase in covalent radius

from N to P. However, from As to Bi only a small increase in covalent radius is observed. This is due to the presence of completely filled d and/or f orbitals in heavier members.

74. Why N differs in its properties as compare with rest of elements of goup 15.

Ans: Nitrogen differs from the rest of the members of this group due to

its smaller size, high electronegativity, high ionisation enthalpy and

non-availability of d orbitals.

75. Except N other elements of group 15 don’t form p- p multiple bond.

Explain?

Ans: Heavier elements of this group do not form p- p bonds as their atomic orbitals

are so large and diffuse that they cannot have effective overlapping.

76. Why N–N bond is weaker than the single P–P bond?

Ans: N–N bond is weaker than the single P–P bond because of high inter electronic

repulsion of the non-bonding electrons, owing to the small bond length.

77.Red P is chemically less reactive compare with White P. Give reason.Ans: Due to its polymeric structure it will have more bonds to break in

chemical reactin.

78. oxygen exists as diatomic molecule (O2) whereas sulphur exists as

polyatomic molecule (S8). Give reason.

Ans: oxygen exists as diatomic molecule (O2) whereas sulphur exists as

polyatomic molecule (S8) because due to large size of S it cannot form

multiple bonds.

79. Why oxygen shows anomalous behavior as compare with its congeners ?

Ans: It is due to its small size, High Electro negativity, More Ionization

Enthalpy and absence of empty d- orbitals.

80. The stability of + 6 oxidation state decreases down the group and stability of

+ 4 oxidation state increase in group 16, Explain.

Ans: It is due to inert pair effect.

81. Oxidising ability of halogens is F2>Cl2>Br2>I2 give reason.

Ans: The decreasing oxidising ability of the halogens in aqueous solution

Down the group is evident from their standard electrode potentials

which are dependent on the parameters indicated below:

82. Ionic character of metal halides decreases from MF to MI. Give reason.

Ans: The ionic character of the halides decreases in the order

MF >MCl > MBr > MI where M is a monovalent metal.

As electronegetivity/Electron gain enthalpy decreases Ionic character

increases.

83. Ionic character of Metal chlorides decreases as oxidation state increases,

Explain.

Ans: As oxidation state increases Ionization energy increases hence ionic

Character decreases.

the halides in higher oxidation state will be more covalent than the one in

lower oxidation state .For e.g., SnCl4, PbCl4, SbCl5 and UF6 are more

covalent than SnCl2, PbCl2, SbCl3 and UF4 respectively.

D & f block elements

The transition elements may be defined as the elements whose atoms or simple ions contain at least one partially filled d –orbitals.

Properties of these elements that are transitional between s & p block elements, there fore they are called as transition elements.

General electronic configuration of these elements as ns1-2 (n-1)d1-10 . Zn, Cd and Hg have full d 10 configuration in their ground state as well as in their common

oxidation states and hence, are not regarded as transition metals. Half and completely filled sets of orbitals are relatively more stable. Lanthanides and actinides are called inner transition elements. Their general electronic

configuration is ns2(n-1)d0-1 (n-2)f1-14

Transition metals are having strong metallic bonding due to presence of large no. of valence electrons and greater nuclear charge. Hence these are hard, possess high densities and high enthalpy of atomization.

Transition elements are having high melting and boiling points due to presence of large no. of half filled d – orbitals which causes strong inter particle attractions.

Ionization energy increases from left to right due to increase in the nuclear charge, as the increased nuclear charge is partly cancelled by screening effect there is no much increase.

The reduction potential values of transition elements varies irregularly because it depend up on Enthalpy sublimation, Ionisation energy and hydration energy. All these values are irregularly changing hence their reduction potential values are also varies irregularly.

Atomic and ionic radii decreases to middle and become constant in middle and at the end it increases. The variation is not much because increased nuclear charge is partly cancelled by screening effect.

Heat required to break the metal lattice to get free atoms is called enthalpy of atomization. Transition elements exhibit high enthalpies of atomization. This is because of large number of unpaired electrons in their atoms which results in stronger interatomic attraction and hence stronger bonding between atoms.

The atomic radii of the second and third transition series are almost the same. This is due to lanthanoid contraction (imperfect shielding of one electron by another in the 4f orbital). Thus, Zr and Hf have very similar and chemical properties.

It is difficult to obtain oxidation state greater than two for Cu, Ni and Zn because of the high values of third ionization enthalpies of these elements.

Third ionization enthalpy of Mn is high because the third electron has to be removed from the stable half filled 3d orbital (3d 5).

Third ionization enthalpy for change from Fe 2+ to Fe 3+ is very small because loss of third electron gives the stable configuration of 3d 5.

The first ionization enthalpies of 5d elements are higher as compared to those of 3d and 4d elements. This is because the weak shielding of nucleus by 4f electrons in 5d elements results in greater effective nuclear charge acting on the outer valence electrons.

Transition elements possess variable oxidation states due to involvement of both ns electrons and (n-1)d electrons in bonding. Lower o.s is zero and highest is +8, +2 is the most common o.s.

Transition metals forms complex compounds due to small size, high charge density and presence of empty d- orbitals.

Transition elements forms colored complexes due to d – d transition.

The transition metals ions generally contain one or more unpaired electrons in them and hence their complexes are generally paramagnetic. If a metal ion does not have at least one electron it behave as diamagnetic.

Transition metal ions and their compounds known to act as catalysts. The catalytic activity of compounds is due to variable oxidation states, and providing surface for adsorption.

Transition metals are almost similar size, there fore these elements can mutually substitute their positions in their crystal lattices and forms alloys.

Transition elements are capable of entrapping smaller atoms of other elements such as H, C & N in the interstitial position and the trapped atoms get bonded with transition elements. And these compounds are known as interstitial compounds. Example Steel.

The large positive E 0 value for Mn 3+/Mn 2+shows that Mn 2+ is much more stable than Mn 3+ (Mn 3+ can be reduced to Mn 2+ ). Negative value of E 0 for Cr 3+/Cr2+shows that Cr 2+ is less stable than Cr 3+ (Cr 2+ can be oxidized to Cr 3+ ).

Cu2+ (aq) is more stable than Cu1+ (aq). This is due to the much more negative standard enthalpy of hydration value of Cu2+ (aq) than Cu1+ (aq), which more than compensates for the second ionization enthalpy of copper.

a) Lanthanoid contraction: The regular decrease in the atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. As we move along the lanthanoid series, the nuclear charge increases by one unit at each successive element. The new electron is added into the same subshell (4f). As a result, the attraction on the electrons by the nucleus increases and this tends to decrease the size. Further, as the new electron is added into the f subshell, there is imperfect shielding of one electron by another in this subshell due to the shapes of these f orbitals. The imperfect shielding is unable to counterbalance the effect of the increased nuclear charge. Lanthanoid contraction causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. As a result, these elements occur together in nature which makes the separation of the elements in the pure state difficult.

b) The ionization enthalpies of early actinoids are lower when compared to the ionization enthalpies of the early lanthanoids. When 5f orbitals are beginning to be occupied, they will penetrate less into the inner core of electrons. The 5f electrons will therefore, be more effectively shielded from the nuclear than the 4f electrons of the corresponding lanthanoids. Because the outer electrons are less firmly held, they are available for bonding in the actinoids.

Preparation of potassium dichromate from chromite involves the following steps The chromite ore is finely ground and heated strongly with molten alkali in

the presence of air.

2FeCr2O4 + 8NaOH + 7 ½ O2 4Na2CrO4 + Fe2O3 + 4H2O

The solution of sodium chromate is filtered and acidified with dilute sulphuric

acid so that sodiumdichromate is obtained.

2NaCrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O

A calculated quatity of potassium chloride is added to a hot concentrated

solution of sodium dichromate. Potassium dichromate is less soluble therefore

it crystallizes out first.

Na2Cr2O7+ 2KCl K2Cr2O7 + 2NaCl

In alkaline solution dichromate which is orange in colour is converted to chromate which is yellow.The acidification reverses the reaction.

2CrO4 2- + 2H+ 2HCrO4- Cr2O7 2- + H2O (reversible)

a)It oxidizes potassium iodide to iodine

Cr2O7 2- + 14H+ + 6I- 2Cr3+ + 7H2O + 3I2

b)It oxidizes iron (ii) solution to iron(III) solution

Cr2O7 2- + 14H+ + 6Fe2+ 2Cr3+ + 7H2O + 6Fe3+

It oxidizes H2S to S

Cr2O7 2- + 8H+ + 3H2S 2Cr3+ + 7H2O + 3S

Pyrolusite ore is fused with alkali in the presence of air when potassium manganate is formed.2MnO2 + 4KOH + O2 2K2MnO4 + 2H2O

Potassium manganate is oxidized by using either CO2, ozone or chlorine to potassium permanganate.

2K2MnO4 + Cl2 2KMnO4 + 2KCl

Potassium permanganate is crystallized from the solution

a) It oxidizes iron(II)salts to iron(III) salts.MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+

b)It oxidizes SO2 to sulphuric acid

2MnO4- + 5SO2 + 2H2O 5SO4 2- + 2Mn2+ + 4H+

b) It oxidizes oxalic acid to CO2 and H2O2MnO4

- + 16H+ + 5C2O42- 2Mn2+ + 8H2O + 10CO2

Elements which have partly filled f-orbitals.There are two series of inner transition elements.In the first series of inner transition elements the 4f orbitals are incomplete and electrons are progressively filled in these orbitals as atomic number increases.These elements are called lanthanoids.In the second series of transition elements the electrons are progressively filled in 5f- orbitals as atomic no. increases.These elements are called actinoids.

As we move from La to Lu there is a gradual decrease in size .The steady decrease in size from La to Lu is called lanthanide contraction.This is due to poor shielding effect of 4f orbitals.The important consequences are

1. Similarities in the properties of Y and heavier lanthanides .2. Similar atomic radii of second and third transition series.3. Separation becomes difficult4. Causes small differences in the properties like basicity, solubility of salts, formation of

complexes, etc. Colour of the salts and ions in solution Most of the lanthanide trivalent ions are coloured in

solid as well as in the solution phase. The ions containing x and (14 – x) electrons show the same colour. The colour of the salts or ions is due to the f – f transition of electrons.

.

Extraction of Iron: Concentration by hydraulic washing, Calcination and Reduction in Blast furnace.

Extraction of Copper: Concentration by froth floatation, Roasting, Bessemerisation and Electrolytic refining.

Extraction of Al: Concentration by Bayer’s method, Electrolytic reduction. Extraction of Sn: Concentration of tin stone by hydraulic washing, Reduction of coke,

Refining by liquation; Extraction of Lead: Concentration of PbS by froth floatation, Roasting, Reduction by coke,

and Refining by liquation. Extraction Silver: Concentration of Ag2S by Leaching method , Refining by electrolytic

refining method. 4Ag2S + 8 NaCN + O2 + 2 H2O →4Na [Ag (CN)2] + 4 NaOH + 2S

2Na [Ag (CN)2] + Zn →Na2 [Zn (CN)4] + 2Ag

Extraction of Hg: Concentration of Cinnabar by Froth floatation method, Roasting and reduction by heating, Refining of Hg by distillation method.

Extraction of Zn: Concentration of ZnS by froth floatation, Roasting, Reduction by coke, and refining by distillation.

. (In brief each step can be described for extraction)

Q. 1. Give two use of lanthanide compounds.

Ans. (i) Misch metal is pyrophoric and used in gas lightness, tracer bullets and shell.

(ii) Oxides of neodymium and praseodymium are used for making colour

glasses.

Q. 2. why the melting points of transition elements are high ?

Ans. The melting points of transition elements are high due to the presence of strong

Intermetallic bonds and covalent bonds.

Q. 3. Why Zn, Cd and Hg are not regarded as transition elements ?

Ans. Because they have the completely filled d-subshell with outer electronic

configuration (n –1)d10 ns2.

Q. 4. Why is K2Cr2O7 generally preferred over Na2Cr2O7 in volumetric analysis

although both are oxidising agents ?

Ans. Because Na2Cr2O7 is hygroscopic, hence it is difficult to prepare its standard

solution for volumetric analysis, but because of non hygroscopic nature of

K2Cr2O7 its standard solution can be prepared.

Q. 5. Write any two uses of pyrophoric alloys.

Ans. Pyrophoric alloys contain rare earth metals and are used in the preparation

of ignition and shells and flints for lightness.

Q. 6. Why do Zr and Hf exhibit similar properties ?

Ans. Because of the lanthanide contraction Hf has similar size to Zr, therefore both

Zr and Hf exhibit similar properties.

Q. 7. In the transition servies, with an increase in atomic number the atomic radius

does not change very much, why is it so ?

Ans. In the transition series, the effect of increasing nuclear change is partly

cancelled by the increased screening effect of the d- electrons of penultimate

shell. Because of this reason, the atomic radius does not change very much in the

transition series.

Q. 8. Why is the third ionization energy of mangnese unexpectely high ?

Ans. Third electron of the manganese is removed from 3d-orbitals which have half

filled configuration, thus have extra stability. Due to this reason high enegy is

required to remove third electron from Mn.

Q. 9. All scandium salts are white. Why ?

Ans. Because they have no electron in d-orbital, thus no d - d transition is possible.

Due to this reason all salts of Sc3+ are white.

Q.10. The first ionisations energies of the 5d-transition elements are higher then

those of 3d and 4 d transition elements.

Lanthanoid contraction

Q. 11. Why do the d-block elements exhibit a longer number of oxidation states than

f-block elements ?

Ans. Because the energy of ns-electron and (n – 1) d-electrons are nearly same,

therefore, ns electrons as well as (n – 1) d-electrons can take part in bond

formation in transition elements. In fblock elements last electron goes to the

f-orbitals of second order outer most shell, thus the difference between the

energy of ns-electron and (n – 2) f-electrons increases. Due to this reasons

all the (n – 2) f-electrons cannot take point in bond formation.

Q. 12. Explain why the first ionisation energies of the elements of the first transition

service do not vary much with increasing atomic numbers.

Ans. With the increasing atomic number, d-electrons add one by one in (n – 1) shell

or penultimate shell. The screening effect of these d- electrons shield the outer

s-electrons from inward nuclear pull. The effect of the increase in nuclear charge

with the increase in atomic number is opposed by the shielding effect of

d-electrons . thus due to these counter effect there is a vary little variations in the

values of ionization energies of first transition series.

Q. 13. Explain why transition metals are paramagenetic ?

Ans. The ions of transition metals generally contain one or more unpaired electrons

hence the compounds of transition metal are paramagnetic i.e., they are weakly

attracted by magnetic field. The paramagnetic character is directly related to the

value of magnetic moment, m which intern depends upon the number of

unpaired electrons (n) i.e., B.M. n + (n + 2)

Q. 14. Give plausible reason for the fact that transition metals have high enthalpy of

atomization.

Ans. Transition elements are metallic in nature and form strong metallic bonds.

Moreever, they have incomplete d-orbitals, hencey they can form covalent

bonds also. Due to these two reasons transition metals have strong force of

attraction. Therefore, they have high enthalpy of atomization.

Q. 15. What is the effect of pH on the color of the solution of potassium dichromate ?

Ans. A lower pH, the colour of the solution is orange due to the solution is orange due

To the presence of dichromate ions (Cr2O7 2–). But in alkaline PH, the colour of

the solution changes to yellow due to the conversion of dichromate ions to

chromate ions.

2CrO4 2- + 2H+ ↔ 2HCrO4- ↔ Cr2O7 2- + H2O (reversible)

Q. 16. Why the compounds of transitions elements are coloured ?

Ans. The colour of compounds of transition elements depend upon the unpaired

electrons present in d-orbitals fo transition element. If d-orbitals are completely

vacant or completely filled the compounds will be colourless, but if any

unpaired electron is present in d-orbitals, the compound will be coloured due to

d d transition. The unpaired electron is excited from one energy level

to another energy level with in the same d-sub-shell. For this purpose, the energy

is absorbed from vissible region of radiation. The complementary part of the

absorbed light i.e., reflected light will decide the colour of the compound.

Q. 17. Why the transition elements act as catalyst ? Give two examples.

Ans. (a) Transition metal show variable oxidation states, therefore, they can from

intermediate products of difficult reactant molecules.

(b) Transition elements are capable to form interstitial compounds due to which

they can absorb and activates the reacting molecules.

Example :

(a) V2O5 is used for the oxidation of SO2 to SO3 in contact procsss of H2SO4.

(b) Ni is used as a catalyst in the hydrogenaton of alkanes and alkynes

Q. 18. Why transition elements form (a) interstitial compounds and (b) Alloys ?

Ans. (a) Interstitial compounds : Transition elements form large number of intestitial

compounds. In these compounds small size atoms like hydrogen, carbon,

nitrogen , nitrogen, boron etc. occupy the empty space of metal lattice. The

small entraped atom in the interstics form the bonds with metals due to which

malleability and ductility of the metals decrease, whereas tensile

strengthincreases.

(b) Alloys : Tranisition element forms alloys with each other because they

have alomost similar sizes. Due to similar sizes atoms of one metal in the

crystal lattice can easily take up the position of the atom of transition

elements. Alloys are more resistant to corrosion than the

constituent elements, and usually harder with higher melting point.

Q.19. Why are Ni2+ compounds thermodynamically more stable than Pt2+ compounds,

whilst Pt4+ compounds are rerlatively more stable than Ni4+ compounds ?

Ans. Thermodynamic stability of the compounds can be judged on the basis of the

magnitude of ionization energies. The sum of the two first ionization energies of

Ni2+ is lower than the sum of first two ionization energies of Pt2+, therefore, Ni2+

compounds are more statble than Pt2+. On the other hand, sum of first four

ionization energies of Pt4+ lower than the sum of the first four ionization energies

of Ni4+, therefore, Pt4+ compounds are more stable than Ni4+.

Q.20. Write the chemical equations for oxidizing action of dichromate of following

substances.

a) Iodide in to Iodine b) Ferrous salt into ferric c) oxalate into carbon di oxide

Q.21. Name a transition metal which does not exhibit variation in oxidation state in its

compounds.

Ans. Zinc in its compounds shows an oxidation state of + 2 only.

Q.22. Why is + 2 oxidation state of manganese (Z = 25) more stable than its + 3

oxidation state, while the same is not true for iron (Z = 26)?

Ans. The electronic configurations of Mn2+, Mn3+, Fe2+ and Fe3+ are,

Mn2+ : 1s2 2s2 2p6 3s2 3p6 3d 5 4s0 Mn3+ : 1s2 2s2 2p6 3s2 3p6 3d 4

Fe2+ : 1s2 2s2 2p6 3s2 3p6 3d 6 4s0 Fe3+ : 1s2 2s2 2p6 3s2 3p6 3d 5

Due to the extra stability of the half-filled orbitals d 5 configuration is more stable than d 4 or d 6. Mn2+ and Fe3+ both have d 5 configuration. Therefore, Mn2+ is more stable than Mn3+ and Fe3+ is more stable than Fe2+.

Q.23. For the first row of the transition metals the E° value are:

Metal Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu

E0 [M /M] 2– 1.18 – 0.91 – 1.18 – 0.44 – 0.28 – 0.25 + 0.34 volts (V)

Give suitable explanation for the irregular trend in these values.

Ans. The irregular trend in the

(M /M) E values for the first row transition metals is due to the irregular

variation in the ionisation and sublimation energies across the series.

Q.24. Of the ions Co2+, Sc3+ and Cr3+ which ones will give coloured aqueous

solutions and how will each of them respond to a magnetic field and why?

(Atomic numbers: Co = 27, Sc = 21, Cr = 24)

Ans. The electronic configurations of the given ions are,

Co2+ : 1s2 2s2 2p6 3s2 3p6 3d7 No. of unpaired electrons = 3

Sc3+ : 1s2 2s2 2p6 3s2 3p6 3d0 No. of unpaired electrons = 0

Cr3+ : 1s2 2s2 2p6 3s2 3p6 3d3 No. of unpaired electrons = 3

Co2+ and Cr3+ ions will give coloured aqueous solutions. Co2+ and Cr3+ are

paramagnetic, and Sc3+ is diamagnetic. Therefore, Co2+ and Cr3+ ions will get

attracted to the magnetic field, whereas Sc3+ ion will be repelled by the

magnetic field.

Q.25. Why is it that anhydrous copper(II) chloride is a covalent while anhydrous

copper(II) fluoride is ionic in nature?

Ans. This is because in halides of transition metals, the ionic character decreases with

increase in atomic mass of the halogen. Fluorine being the most electronegative,

forms ionic salt with copper. Cupric chloride consists of infinite chains

consisting of bridging Cl atoms and has a covalent character.

Q.26. (a) Assign reason for each of the following:

(i) Ce3+ can be easily oxidised to Ce4+.

(ii) E° for Mn3+/Mn2+ couple is more positive than for Fe3+/Fe2+.

(iii) Transition metals exhibit higher enthalpies of atomization.

(iv) The transition elements form interstitial compounds.

(b) Mention two uses of potassium permanganate in the laboratory.

(Atomic number: Mn = 25, Fe = 26, Ce = 58)

Ans. (i) Ce3+ has only one electron in its 4f orbitals. Due to extra stability of

completely empty orbitals belonging to an energy level as compared to

having only one electron in it, Ce3+ tends to lose its only electron from

4f orbital and get oxidised to Ce4+.

(ii) Mn3+ has a d 4 configurtion, so it has greater tendency to accept one

electron to acquire d 5 configuration. On the other hand, Fe3+ has a d5

configuration which is more stable than the d 6 configuration of Fe2+.

As a result, reduction of Fe3+ to Fe2+ is not favoured. Since,

E° values reflect the reduction tendency, therefore, E° value for

Mn3+/Mn2+ couple is more positive than Fe3+/Fe2+.

(iii) Transition metals exhibit high enthalpies of atomisation. This is because

the atoms in these elements are held together by strong metallic bonds.

The metallic bond is formed as a result of interaction of electrons in the

outermost shells. In general, greater the number of valence electrons

stronger is the metallic bond.

(iv) The transition metals form a number of interstitial compounds in which

small atoms of light elements such as, H, C and N occupy the voids in

their lattices. The products obtained in this way are hard and rigid. For

example, steel and cast iron become hard due to the formation of an

interstitial compound with carbon.

(b) The uses of potassium permanganate in the laboratory are

(i) As an oxidising agent,

(ii) In volumetric estimation of reducing agents such as Fe2+ salts,

oxalic acid etc.

27. Defferentiate the properties of Lanthanoids and actinoids?

Lanthanoids Actinoids

1. The el.confgn. is [Xe]4f 0-14 5d 0-1 6s2.2. Shows oxidn state +3 only except in few

cases where it is +2 or +4

3The ionic radii of M3+ ions in lanthanides show a regular change in the size of ions with increase in atomic no.

3. These posses lesser tendency for complexation

Confign. is [Rn] 5f 0-14 6d 0-1 7s2

2.Actinides show higher oxidation state such as +4,+5,+6,+7 also in addition to +3

3.There is a gradual decrease in the size of atoms or M3+ ions across the

series.

4.These possess greater tendency for complexation

28. What happen when K2Cr2O7 is heated and write its chemical equation.

29. Describe with chemical reaction for the preparation of potassium

permanganate from pyrolusite ore.

30. Chromate and dirchromate are interconvertable by variation of PH. Write

chemical equation for this.

31. Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their

+3 state?

32. Calculate the magnetic moment of a divalent ion in aqueous solution

if its atomic number is 25.

Ans: With atomic number 25, the divalent ion in aqueous solution will have

d5 configuration (five unpaired electrons). The magnetic moment, μ is

√5.92BM

33. Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration.

Ans: Cr2+ is reducing as its configuration changes from d4 to d3, the latter

having a half-filled t2g level (see Unit 9) . On the other hand, the change

from Mn2+ to Mn3+ results in the half-filled (d5) configuration which has

extra stability.

34. Why transition metals show horizontal similarity in their properties?

Ans: Due to similar valence electronic configuration i.e ns electrons.

*******

Coordination Compounds

Double salts dissociate completely into simple ions when dissolved in water. However, complex ions such as [Fe (CN) 6]4- of K4 [Fe (CN) 6], do not dissociate into Fe2+ and (CN) 1- ions. Mohr’s salt and Potash alum are double salts

Main Postulates of Werner’s Theory : i) Metals show two types of valences in coordination compounds- Primary & Secondary. ii) The Primary valences are normally ionizable and are satisfied by negative ions. iii) The Secondary valences are non ionizable and are satisfied by neutral molecules or negative ions (Ligands). iv) The ions/groups bound by the secondary linkages to the metal have characteristics spatial arrangements (coordination polyhedra) corresponding to different coordination numbers

A coordination entity comprises of metal atom or ion bounded with fixed no. of molecules or ions. Ex. K4 [Fe (CN) 6]

Ions or molecules bound to the central metal atom or ion is known as ligand. When a ligand is bound to a metal ion through a single donor atom, as with Cl–, H 2O or NH3,

the ligand is said to be unidentate. When a ligand can bind through two donor atoms as in H2NCH2CH2NH2 (ethane-1,2-diamine) C2O4

2– (oxalate), the ligand is said to be didentate and when several donor atoms are present in a single ligand as in N(CH2CH2NH2)3, the ligand is said to be polydentate. Ethylenediaminetetraacetate ion (EDTA4–) is an important hexadentate ligand. It can bind through two nitrogen and four oxygen atoms to a central metal ion.

The number of coordinating or ligating groups present in a ligand is called the denticity of that ligand. When a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. Ligand which can ligate through two

different atoms is called ambidentate ligand (e.g. NO2- and SCN

- ions).

The coordination number of a metal ion in a complex can be defined as the number of ligand donor atoms to which the metal is directly bonded. The coordination number of the complex ions, [PtCl6] 2-, [Ni(NH3) 4] 2+,

[Co (en) 3] 3+ and [Fe (C2O4)3] 3- are 6, 4, 6 and 6 respectively.

It is important to note that the coordination number of the central atom/ion is determined only by the number of sigma bonds formed by the ligand with the central atom/ion. Pi bonds, if formed between the ligand and the central atom/ion, are not counted for this purpose.

Complexes in which a metal is bound to only one kind of donor groups are known as homoleptic complexes.e.g. [Co (NH3)6] 3+. Complexes in which a metal is bound to more than one kind of donor groups are known as heteroleptic complexes. e.g. [Co (NH3) 4 Cl2] +.

Chelate effect : When a polydentate ligand forms a coordination complex having a ring structure, it is called chelate complex. A chelate complex is always more stable than the

complex formed by monodentate ligands. The stabilization of coordination compound due to chelation is called chelate effect.

The coordination number (CN) of a metal ion in a complex can be defined as the number of ligand donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl6]2– and [Ni(NH3)4]2+, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C2O4)3]3– and [Co(en)3]3+, the coordination number of both, Fe and Co, is 6 because C2O4 2– and en (ethane-1,2-diamine) are didentate ligands.

The central atom/ion and the ligands attached to it are enclosed in square bracket and is collectively termed as the coordination sphere.

The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and

tetrahedral. For example, [Co(NH3)6]3+ is octahedral, [Ni(CO)4] is

tetrahedral and [PtCl4]2– is square planar.

IUPAC NAMES OF SOME COORDINATION COMPOUNDS

1. [Co (NH3)4(H2O) 2] Cl3 = Tetraamminediaquacobalt(III) chloride2. [Cr(en)3] Cl3 = Tris(ethane-1,2-diamine)chromium(III) chloride

3. [Pt(NH3)BrCl(NO2)]-= Amminebromidochloridonitrito-N-platinate(II)

4. [PtCl2(en)2](NO3)2 = Dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate5. Fe4[Fe(CN)6]3 = Iron(III) hexacyanoferrate(II)6. [Co(NH3)5Cl]Cl2 = Pentaamminechloridocobalt(III) chloride7. K3[Fe(C2O4)3] = Potassium trioxalatoferrate(III)8. K2[PdCl4] = Potassium tetrachloridopalladate(II)9. [Pt(NH3)2Cl(NH2CH3)]Cl = Diamminechlorido(methylamine)platinum(II) chloride10. [Co(NH3)6]2(SO4)3 = Hexaamminecobalt(III) sulphate11. [Co(NH3)5(ONO)]2+ = Pentaamminenitrito-O-cobalt(III) ion12. [Co(NH3)4Cl2]3[Cr(CN)6] = Tetraamminedichloridocobalt(III)hexacyanochromate(III)13. Na2[Fe(CN)5NO] = Sodium pentacyanonitrosoniumferrate(II)14. K3[Co(CN)5NO] = Potassium pentacyanonitrosylcobaltate(II)15. Na2[CrF4O] = Sodium tetrafluoridooxochromate(IV)16. [Cr(H2O)4Cl2]NO3 = Tetraaquadichloridochromium(III) nitrate17. (NH4)3[Cr(SCN)6] = Ammonium hexathiocyanato-S-chromate(III)18. Na2[Cr(CH3COO)4(en)] = Sodium ethylenediaminetetraacetatochromate(II)19. [Co(NH3)5(CO3)]Cl = Pentaamminecarbonatocobalt(III) chloride20. [Pt(py)4][PtCl4] = Tetrapyridineplatinum(II)tetrachloridoplatinate(II)21 LiAlH4 Lithium tetra hydrido Aluminate (III)

VBT:

Some limitations of VBT: i) It does not explain the colour exhibited by coordination compounds. ii) It does not distinguish between weak and strong ligands. iii) It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds. iv) It does not give quantitative interpretation of magnetic data

CRYSTAL FIELD THEORY: The CFT considers the metal—ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions and as dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion are degenerate. The degeneracy of the d orbitals is lifted when ligands approach the metal atom/ion. This is due to ligand-electron metal-electron repulsions. It results in splitting of the d orbitals into lower energy and higher energy orbitals. The pattern of splitting depends upon the nature of the crystal field.

If Äo < P, the fourth electron enters one of the eg orbitals giving the configuration t2g3eg

1 . Ligands for which Äo < P are known as weak field ligands and form high spin complexes. (ii) If Äo > P, it becomes more energetically favourable for the fourth electron to occupy a t2g orbital with configuration t2g 4eg

0. Ligands which produce this effect are known as strong field ligands and form low spin complexes.

The arrangement of ligands in order of their increasing CFSE values is known as spectrochemical series. The ligands with small CFSE values are called weak field ligands, whereas those with large value of CFSE are called strong field ligands

Synergic bonding interactions in a carbonyl complex : The bond between metal and CO in metal carbonyls is strong due to the following reasons: i) The M—C sigma bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant d orbital of the metal. ii) The M—C pi bond is formed by the donation of a pair of electrons from a filled d orbital of metal into the vacant antibonding orbital of CO. Thus, the Metal to Ligand bonding creates a synergic effect which strengthens the bond between CO and the metal

Stability of a complex & factors affecting stability of a complex ion :. i) Greater the charge and smaller the size of the central metal ion, more is its charge density and greater will be the stability of the complex. ii) For different metal ions with same charge, stability of the complexes increases as the size decreases. iii) More the basicity of the ligand more is the stability of the complex. iii) For anionic ligands the complexes are more stable if ligand has greater charge and smaller size. iv) Chelating ligands form more stable complexes.

ISOMERISM IN COORDINATION COMPOUNDS:

Ionization isomerism : e.g. [Co(NH3)5Cl]2+SO42- and [Co(NH3)5SO4]+Cl-

Coordination isomerism : e.g. [Co(NH3)6][Cr(C2O4)3] and [Cr(NH3)6][Co(C2H4)3] .Linkage isomerism: e.g.[Co(NH3)5ONO]Cl2 and [Co(NH3)5NO2]Cl2

Nitritopentammine Nitropentammine

Cobalt(III)chloride Cobalt(III)chloride

Solvate isomerism (hydrate isomerism) arises when water is involved as a solvent. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice

Geometrical isomerism or cis-trans isomerism:

Optical isomerism :

VSA of One Mark

1. (each carry one mark)

2.

3. Define Spectrochemical series?

4. What is CFSE?

5. Draw the CFT for octahedral complexes.

6. Differentiate strong and weak field in coordination compounds?

7. Draw the structure of multi dentate ligand which is used for lead poisoning.

SA-1 of Two Marks

1. [NiCl4]2- is more likely to be tetrahedral while [Ni (CN) 4]2- is more likely to be square planar. Explain

Ans:

2.

3.

4. Predict the Isomerism present in complexe like

SA-2 of Three Marks

1.

2. What is meant by CFSE on the basis of crystal field theory write the electronic

configuration of d4 in terms of t2g & eg in an octahedral field when i) 0> P

ii) ) 0<P

3. Name three factors that favours the formation of complexes by metal ion?

4. Describe the hybridization in following complexes.

(i) [Fe(CN)6]4– (ii) [FeF6]3–

HOTS

1.

2

3.

4. Predict the Isomerism present in complexe like

SL.NO Name of the test Identification of compounds

01 Neutral ferric chloride Only phenols(violet or purple color)02 Sodium bicarbonate Only carboxylic acid(brisk effervescence)03 Iodo form test All 2-ketones,2-alohols, ethanal, ethanol

(yellow color crystals)04 Lucas test To distinguish10 20 & 30 alcohols05 Hinsberg test 10 20 &30 amines06 Carbylamines test Only primary amines07 Sodium metal test Alcohols(effervescence)08 2,4 DNP test Both aldehydes & ketones (yellow color

crystals)09 Tollen’s test Only aldehydes(silver mirror formation)10 Fehiling’s test Only aldehydes(red ppt)11 Schiff’s test Aldehydes & Ketones

(Organic) DESTINGUISH TESTS

DISTINGUISH BETWEEN

How can we distinguish organic compounds?

Test for Alcohols:

1. Ester test : Alcohol form ester with carboxylic acid. Ester can be identified by sweet smelling. R-CH + R’-COOH↔R-COOR’ + H2O

2. Do not turn Blue litmus in to red but gives hydrogen gas with sodium metal.

Different types of Alcohol:

Lucas Test: Alcohol treated with Lucas reagent. (HCl + ZnCl2)

If turbidity appears immediately --- Tertiary alcohol

If turbidity appears after 5 min --- Secondary alcohol

If turbidity not appears --- Primary alcohol

Iodoform test: etyl alcohol and all 2-hydroxy alcohol gives this test.

Reactions:

Test for Phenol:

Phenol when reacts with neutral ferric chloride, it produces violet colour.

Test for aldehydes and Ketones:

a) Test given by both aldehydes and ketones.i). 2,4-dinitrophenyl hydrazine test: carbonyl compound react with 2,4-dinitrophenylhydrazine to form yellow or orange crystals.

ii). Sodium bisulphate test: Aldehydes and ketones both form crystalline addition product with NaHSO3.

iii). All aldehyde and ketones having CH3-C═O structure gives iodoform test positive.

b) Test given by aldehydes only.i). Tollen Test: On warming with ammonical silver nitrate solution(Tollen reagent) in alkaline medium, aldehyde gives silver mirror on the side of test tube.

RCHO + 2[Ag(NH3)2]+ + 3OH- → RCOO- + 2Ag↓ + 2H2O + 4NH3.

ii). Fehling solution Test: Fehling solution is alkaline solution of copper sulphate containing sodium potassium tartarate. Aldehydes gives reddish brown precipitate of Cu2O with Fehling solution. RCHO + 2Cu+2 + 5OH-→ RCOO- + Cu2O + 3H2O

Test for Carboxylic acid:

1. Litmus test( phenol also giving this test)2. Ester test 3. Sodium bicarbonate test RCOOH + NaHCO 3 → RCOONa +CO2 + H2O

Test for Amines:

i). Litmus test

i) Hinsberg Test : Amine when shaken with benzene sulphonyl chloride-

1. A clear solution in NaOH solution which on acidification gives an insoluble material indicates primary amine.

C6H5-SO2Cl + H-NH-R –HCl > C6H5SO2-NH-R NaOH> [C6H5-SO2-NR]- Na+

Benzene sulphonyl chloride N-alkylbenzene sulphonamide(insoluble)

2. A precipitate (insoluble) which is insoluble in NaOH solution indicates secondary amine.

3. Tertiary amine does not react with benzene sulphonyl chloride. It forms soluble substance (addition product).

Specific test for Primary amine:

1. Azo-dye test : for aromatic primary amine only.

2. Carbylamine test : for aliphatic and aromatic primary amine both.RNH2 + CHCl3 + 3KOH → RNC + 3KCl +3H2O

C6H5-SO2Cl + NR3 +HCl > [R3NH]+Cl-

C6H5-SO2Cl + H-NR-R –HCl > C6H5SO2-NR-R

Mechanism in Organic Chemistry

1. SN2 Mechanism:

2. SN1 mechanism ;

3. Hydration of Alkenes:

4. Dehydration of alcohols:

5. Dehydration of alcohols to give Ethers:

6. Reaction of ether with HI

7. Nucleophilic addition mechanism of aldehydes & ketones.

8. Esterification mechanism;

*******

Organic Chemistry

Name Reactions

1. Aldol Condensation : condensation between two molecule of an aldehyde or a ketone having atleast one α-hydrogen atom to form a β-hydroxyaldehyde or a β-hydroxyketone is known as aldol condensation.

Aldol condensation takes place in presence of dil base.

2. Cannizzaro Reaction : The disproportionation (self-redox) of aldehydes lacking α-hydrogen atom (as C6H5CHO, HCHO, R3C.CHO etc.) in presence of strong base to form salt of an acid & a primary alcohol is known as Cannizzaro reaction.

3. Carbylamine test : When a primary amine is heated with alcoholic caustic potash and chloroform, an offensive smelling compound called carbylamine ( alkyl or arylisocyanide) is formed.

4. Clemmension Reduction : The reduction of >C=O group to methyl group (>CH2) with amalgamated zinc and conc. HCl is known as Clemmension reduction.

5. Coupling Reaction : The reaction in which a diazonium salt condenses with an aromatic compound having an electron rich group eg, aniline, phenol or their derivatives to form an azo compound (Ar-N=N-Ar) is termed as coupling reaction.

6. Etard Reaction : Chromyl chloride ( CrO2Cl2) oxidizes methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde. It is called Etard reaction.

7. Esterification Reaction : Reaction of an alcohol with a carboxylic acid in the presence of a small quantity of conc.H2SO4 to form an ester is called esterification.

Esterification process is generally reversible.

8. Finkelstein Reaction : Alkyl iodides can be prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone.

9. Friedel-Craft Reaction : Introduction of an alkyl (-R) or an acyl (RCO-) group in to the benzene ring of an aromatic compound in the presence of of a lewis acid catalyst (eg.anhydrous aluminium chloride or Zinc chloride) is called as Friedel-Craft reaction).

Introduction of an acyl group (RCO-) is called acylation.

10. Gabriel pthalimide synthesis : This method is used to prepare primary amine. The various steps involved are:i) pthalimide is treated with alcoholic solution of KOH to form

potassium pthalimide. ii) The potassium salt is treated with an alkylhalide. iii) The product N-alkyl phthalimide is hydrolysed with dilute HCl to

form a primary amine.

11. Gattermann Reaction : Gattermann reaction is used for obtaining chlorobenzene or bromobenzene from benzenediazonium chloride by treating it with Cu/HCl or Cu/HBr respectively.

12. Gattermann-Koch Reaction : When benzene or its derivative is treated with carbon monoxide and HCl in the presence of anhydrous aluminium chloride or CuCl, it gives benzaldehyde or substituted benzaldehyde.

13. Iodoform test : The compound containing methyl group bonded to carbonyl group (CH3-CO-) or (CH3-CH.OH-) reacts with aquous NaOH and iodine solution gives yellow ppt of Iodoform.

14. Hell-Volhard-Zelinsky Reaction : When aliphatic carboxylic acid containing α-hydrogen are reacted with chlorine or bromine in presence of small amount of red phosphorous, the corresponding α-haloacids are obtained.

15. Hoffmann-Bromamide Reaction : When an amide is heated with bromine and an alkali, a primary amine containing one carbon less than the amide is obtained. This reaction is called Hoffmann-Bromamide reaction. This reaction is very useful for converting a higher homologue to next lower one.

16. Kolbe’s-Electrolysis process : Preperation of higher a;kanes by the electrolysis of sodium or potassium salt of lower fatty acids is called Kolbe’s electrolysis reaction.

17. Riemer-Tiemann Reaction : The reaction of pheonal with chloroform or carbon-tetrachloride in the presence of aqueos alkali at 340k followed by hydrolysisof the resulting product gives salicyldehyde and salicylic acid respectively.

18. Rosenmund Reduction : Reduction of acid chloride (RCOCl)to the corresponding aldehyde with hydrogen using Pd/BaSO4 as catalyst is known as rosenmund reaction.Here Pd/BaSO4 used as negative catalyst and prevent further reduction to alcohol.

19. Sandmayer Reaction : The convesion of benzene diazonium salt into halogen of cyano derivative of the parent aromatic hydrocarbon by treating it with a mixture containing the corrsponding salt and the acid is called sandmeyer reaction.

20. Stephen Reaction : Nitriles can be reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde. This reaction is called Stephen reaction.

21. Swart’s reaction : The synthesis of alkyl fluorides is accomplished by heating an alkyl chloride/ bromide in the presence of a metallic fluoride such as AgF,Hg2F2 etc.

22. Williamson’s synthesis : In williamson synthesis,when an alkoxide or a phenoxide is made to react with an alkyl halide,an ether is obtained.In this method,haloarenes can not be used for the preparation of alkyl-aryl ethers because of the low reactivity of aryl halides.

23. Wolf-Kishner reaction : A carbonyl compound on heating hydrazine and pottassium hydroxide(KOH)in a high boiling polar solvent such as ethylene glycol,gets reduced to give a hydrocarbon.

24. Wurtz-Fittig Reaction : This reaction is used for obtaining higher alkane from the halogen derivatives by using sodium.

HALO ALKANES AND HALO ARENES

In halo alkanes as size of halogen increases C—X bond strength decreases hence reactivity increases.

Preparation of Halo alkanes from alcohols using Thionyl chloride is preferred because the other two products are escapable gases.

Preparation of Halo alkanes from alcohols using hydrogen halides, the order of reactivity of alcohols with a given halo acid is 3°>2°>1°.

The above method is not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than a single bond.

Markovnikov’s rule: When un symmetrical alkene under goes addition with un symmetrical reagent, the major product of alkyl halide one in which, the negative part of addendum will be added to the carbon having less no. of hydrogen atoms.

Anti Markovnikov’s rule: (Peroxide Effect) (Kharausch Effect) When un symmetrical alkene under goes addition with un symmetrical reagent, the major product of alkyl halide one in which, the negative part of addendum will be added to the carbon having more no. of hydrogen atoms.

Alkyl halides are having more boiling point as compare with hydro carbons of equal molecular mass it is due to dipole- dipole attractions present in it.

For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.

The boiling points of isomeric halo alkanes decrease with increase in branching due to decrease in spherical surface area which in turn decreases vander waal’s force.

Boiling points of isomeric di halo benzenes are very nearly the same. However, the para-isomers are high melting as compared to their ortho and meta-isomers. It is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho- and meta-isomers.

Groups like cyanides and nitrites possess two nucleophilic centres and are called ambident nucleophiles.

Halo alkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. It is because KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through carbon atom and not through nitrogen atom since C—C bond is more stable than C—N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

SN1 reaction mechanism is followed by those alkyl halides whose carbo cation is stable like tertiary alkyl halides, Allyl halides & Benzyl halides. It forms carbo cation intermediate, which is triangular planar & SP2 hybridized intermediate. If alkyl halide is optically active after substitution it forms racemic mixture.

The order of undergoing SN1 reaction is Benzyl Halides, Ally halides>30>20>10

SN2 reaction mechanism is followed by primary alkyl halide (whose carbanion is stable). It forms carbanion intermediate, which is sp3d hybridized and pentavalent. During SN2 mechanism optical activity is reversed this is known as optical inversion.

The order of undergoing SN2 reaction is Benzyl Halides, Ally halides<30<20<10

Molecules which are able to rotate plane polarized light are called optically active compounds. Non super imposable mirror images are called Enantiomers.

Substances which rotate plane polarized light are called dextro and one which rotates towards left is known as leavo rotatory compound.

In a molecule if Carbon atom is surrounded by four different groups called asymmetric carbon or chiral carbon which is must for optical active compound.

1:1 ratio of dextro and leavo mixture is known as racemic mixture. The process of conversion of enantiomer into a racemic mixture is known as racemisation. Secondary alkyl halides undergoes both SN1 & SN2 mechanism. If there is possibility of formation of more than one alkene due to the availability of more than

one α-hydrogen atoms, usually one alkene is formed as the major product. These form part of a pattern first observed by Russian chemist, Saytzeff who in 1875 formulated a rule which can be summarized as

“in dehydrohalogenation reactions, the preferred product is that alkene which has

the greater number of alkyl groups attached to the doubly bonded carbon atoms.”

.

Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the Resonance effect (partial double bond ), halogen bonded to sp2 carbon which is more electronegative and less bond length, instability of phenyl cation & repulsion between electron rich arenas, nucleophile.

Nucleophilic substitution on aromatic rings becomes easy in the presence of electron with drawing groups like nitro which stabilize the carbanaion. This effect is not shown when nitro group is present at meta position.

Chloroform is slowly oxidized by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in closed dark colored bottles completely filled so that air is kept out.

ALCOHOLS, PHENOLS & ETHERS

In phenols, the –OH group is attached to sp2hybridised carbon of an aromatic ring. The carbon– oxygen bond length (136 pm) in phenol is slightly less than that in methanol. This is due to (i) partial double bond character

Hydroboration-oxidation of alkenes gives addition of water to alkene as per anti Markovnikov’s rule.

The boiling points of alcohols and phenols are higher in comparison to other classes of compounds, namely hydrocarbons, ethers, haloalkanes and haloarenes of comparable molecular masses. The high boiling points of alcohols are mainly due to the presence of intermolecular hydrogen bonding.

Solubility of alcohols and phenols in water is due to their ability to form hydrogen bonds with water molecules. The solubility decreases with increase in size of alkyl/aryl (hydrophobic) groups.

Alcohols are weak Bronstead acids i.e. they can donate proton to a stronger base.

The acidic character of alcohols is due to the polar nature of O–H bond. An electron-releasing group (–CH3, –C2H5) increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid strength. Hence acidity of alcohols is 1 0>20>30

alcohols. Alcohols act as Bronsted bases as well. It is due to the presence of unshared electron pairs on

oxygen, which makes them proton acceptors. Alcohols and phenols react with Na metal gives hydrogen gas it supports the acidic nature. But

Phenol even reacts with NaOH hence phenol is strong base than alcohol. It is due to the resonance stability of Phenoxide ion.

the relative ease of dehydration of alcohols follows the following order: Tertiary > Secondary > Primary it is due to stability of carbocation.

The commercial alcohol is made unfit for drinking by mixing in it some copper sulphate (to give it a color) and pyridine (a foul smelling liquid). It is known as denaturation of alcohol.

The dehydration of secondary and tertiary alcohols to give corresponding ethers is unsuccessful as elimination competes over substitution and as a consequence, alkenes are easily formed.

The order of reactivity of hydrogen halides is as follows: HI > HBr > HCl. Reaction of anisole with HI gives Phenol & methyl iodide but not other products because there is

a partial double bond character between Benzene ring and Oxygen. Tertiary butyl methyl ether react with HI gives Methanol and tertiary butyl Iodide it is due to

stability of tertiary carbocation. 10,20 &30 alcohols are distinguished by Lucas Test. Lucas reagent is HCl + ZnCl2 tertiary

amines react with Lucas reagent and cloudiness forms immediately, secondary amines react with Lucas reagent and cloudiness forms after 5 mts, primary amines react with Lucas reagent and cloudiness forms only by heating.

ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses. It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions. Also, their boiling points are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding.

The lower members of aldehydes and ketones such as methanal, ethanal and propanone are miscible with water in all proportions, because they form hydrogen bond with water. However, the solubility of aldehydes and ketones decreases rapidly on increasing the length of alkyl chain which dominates non polar nature.

The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance and hence it is less reactive than propanal.

Sodium hydrogensulphite adds to aldehydes and ketoses to form the addition products by equilibrium reaction. The position of the equilibrium lies largely to the right hand side for most aldehydes and to the left for most ketones due to steric reasons. The hydrogensulphite addition compound is water soluble and can be converted back to the original carbonyl compound by treating it with dilute

mineral acid or alkali. Therefore, these are useful for separation and purification of aldehydes.

Addition of ammonia derivatives to the aldehydes & ketones takes place under controlled PH because reaction takes place by acid catalyzed mechanism, if acidity is more it react with ammonia derivatives (bases).

Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding.

Characteristic reactions of aldehydes and ketones are not given by carboxylic acids in spite of having C=O it it due to resonance stabilization of carboxylic group.

Aliphatic carboxylic acids having more than 15 carbon atoms are known as fatty acids. Quantitatively acidity of carboxylic acids are compared by using Ka acid dissociation constant or

PKa values. Higher Ka value or less PK

a represents strong acids. Carboxylic acids are strong acids than Phenol it is due to high stability of carboxylate ion as

compare to Phenoxide ion. Negative charge is distributed on two equally electronegative atoms where as in phenol it is distributed on two un equal atoms.

Electron withdrawing groups increase the acidity of carboxylic acids by stabilizing the conjugate base through delocalization of the negative charge by inductive and/or resonance effects. Conversely, electron donating groups decrease the acidity by destabilizing the conjugate base.

AMINES

Ammonolysis of alkyl halides gives amines, but it has the disadvantage of yielding a mixture of primary, secondary and tertiary amines and also a quaternary ammonium salt. However, primary amine is obtained as a major product by taking large excess of ammonia. The order of reactivity of halides with amines is RI > RBr >RCl.

Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water.

Equal molecular masses of amines & alcohols, alcohols are more polar than amines and form stronger intermolecular hydrogen bonds than amines.

This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows:

Primary > Secondary > Tertiary

Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base. Basic character of amines can be better understood in terms of their Kb and pKb values.

Larger the value of Kb or smaller the value of pKb, stronger is the base. Basisity of amines in gaseous state increased by +ve inductive effect, but in aqueous state apart

from +I effect it also depends on hydration effect & steric effect. The order of basicity of amines in the gaseous phase follows the expected order: tertiary

amine > secondary amine > primary amine > NH3. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in

aqueous solution is as follows: (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3

(CH3)2NH > CH3NH2 > (CH3)3N > NH3

pKb value of aniline is quite high than ammonia . It is because in aniline or other aryl amines, the -NH2 group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation.

In case of substituted aniline, it is observed that electron releasing groups like –OCH 3, –CH3

increase basic strength whereas electron withdrawing groups like –NO2, –SO3, –COOH, –X decrease it.

Reaction of nitrous acid with different amines gives different products.Primary aliphatic amines gives alcohols, primary aromatic amines gives diazonium chloride, secondary aliphatic and aromatic amines gives N-Nitroso amines, tertiary aliphatic amines gives quaternary ammonium nitrite where as tertiary aromatic amines gives p-nitroso aniline.

10,20 & 30 amines are distinguished by Hinsberg’s test i.e. primary amine react with Hinsberg’s reagent (Benzene Sulphonyl Chloride) and resulting compound is soluble in KOH, secondary amine react with B.S.C and resulting compound is insoluble in KOH, tertiary amines do not react with B.S.C.

Aniline is more reactive for electrophilic substitution due to involvement of lone pair on nitrogen makes benzene ring activated towards electrophilic substitution. This effect is minimized by adding acyl group on nitrogen. The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below: Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of –NHCOCH3 group is less than that of amino group.

Nitration aniline gives ortho, para & meta nitro aniline mixture of compounds. Though –NH 2 is a o-p directing it forms substantial amount of meta product also it is due to in strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.

Sulphonation aniline forms sulphonilic acid which forms zwitter ion. The conversion of primary aromatic amines into diazonium salts by reacting with nitrous acid at

0-50C is known as diazotization . Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para

position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling reaction. Similarly the reaction of diazonium salt with aniline yields p-amino azo benzene. This is an example of electrophilic substitution reaction.

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ORGANIC CONVERSIONS (text book)

1. 1-Bromopropane to 2-bromopropane1-Bromopropane (alc KOH) → Propene (HBr) →2-bromopropane

2. Toluene to benzyl alcoholToluene (Cl2/Fecl3) → phCH2Cl (aq KOH) → benzyl alcohol

3. Benzene to 4-bromonitrobenzeneBenzene (Br2/FeBr3) → Bromo Benzene (HNO3/H2SO4) → 4-bromonitrobenzene

4. Ethanol to propanenitrileEthanol (SOCl2) → CH3CH2Cl (KCN) →Propane Nitrile

5. Aniline to chlorobenzene6. Aneline ( HNO2, 0-50c) → Diazonium Chloride (CuCl)→ Choro Benzene7. 2-Chlorobutane to 3, 4-dimethylhexane

Wurtz reaction Na metal dry ether

8. 2-Methyl-1-propene to 2-chloro-2-methylpropaneHCl addition

9. Ethyl chloride to propanoic acid10. CH3CH2Cl (KCN) →Propane Nitrile (H2O/H+)→ propanoic acid11. But-1-ene to n-butyliodide12. 2-Chloropropane to 1-propanol

2-Chloropropane (alc KOH) → Propene (B2H6/H2O2/OH-) → propan-1-ol

13. Isopropyl alcohol to iodoformNaOH/I2

14. Chlorobenzene to p-nitrophenolChlorobenzene (623k/NaOH) followed by hydrolysis →Phenol (dil HNO3) → p-nitrophenol

15. 2-Bromopropane to 1-bromopropane16. 2-Bromopropane (alc KOH) → Propene (HBr/peroxide) →1-bromopropane17. tert-Butyl bromide to isobutyl bromide

alc KOH

18. Aniline to phenylisocyanideCHCl3/alc KOH

19. Phenol into o-nitro PhenolDil HNO3

20. Phenol into Tri nitro phenolConc. HNO3

21. Propene → Propan-2-ol.(H2O/H+)

22. Benzyl chloride → Benzyl alcohol.Aq KOH

23. Ethyl magnesium chloride → Propan-1-ol.HCHO, hydrolysis

24. Methyl magnesium bromide → 2-Methylpropan-2-ol.Acetone, hydrolysis

25. (Ethanol to 3-Hydroxybutanal PCC, NaOH Aldol Condensation

26. Ethanoic acid into methanamineEthanoic acid (NH3, heat)→Ethanamide (NaOH, Br2) → Methanamine

Some Important Conversions

1. Propene to propan-1-olPropene (B2H6/H2O2/OH-) → propan-1-ol

2. 2- Bromo Propane into 1 – Bromo Propane 2- Bromo Propane → Propene →1 – Bromo Propane (1. alcoholic KOH, 2. HBr, Peroxide)

3. Propan-2-ol into Propan -1- ol Propan-2-ol (Conc sulphuric Acid, 440K →Propene (B2H6, H2O2/OH-) → Propan -1- ol

4. Formaldehyde into Propan -1- ol5. Formaldehyde (Ethyl Magnesium Bromide followed by hydrolysis) → Propan -1- ol6. Acetaldehyde into Propan -2-ol7. Acetaldehyde (Methyl Magnesium Bromide followed by hydrolysis)→ Propan -2-ol8. Acetone into t-Butyl alcohol

Acetone (Methyl Magnesium Bromide followed by hydrolysis) → t-Butyl alcohol

9. Aneline into Phenol10. Aneline (HNO2 0-50c) → Diazonium Chloride (H2O/H+) → Phenol11. Cumene into Phenol

Cumene (arial oxidation O2) → Cumene peroxide (H+) → Phenol

12. Phenol into salicylic AcidPhenol (CHCl3/NaOH followed by hydrolysis) → salicylic Acid

13. Phenol into 2 – Acetoxy Benzoic Acid ( Aspirine)Phenol (CO2/NaOH followed by hydrolysis) → salicylic Acid (Acetyl Chloride) → salicylic Acid

14. Phenol into Benzaldehyde Phenol (CHCl3/NaOH followed by hydrolysis) →Salcilaldehyde (Zn dust) → Benzaldehyde

15. Ethanol into diethyl etherEthanol (Conc. Sulplhuric Acid , 413 K) → diethyl ether

16. Benzene into Benzaldehyde17. Benzene (CH3Cl, FeCl3) → Toluen (Cl2/hv) → Benzyl Chloride (aq KOH)→ Benzyl alcohol (PCC) →

Benzaldehyde18. Acetic acid to acetaldehyde19. Acetic acid (SOCl2) → Acetyl Chloride (H2/Pd/BaSO4) → acetaldehyde20. Acetic acid into acetone Acetic acid (SOCl2) → Acetyl Chloride (dimethyl Cd) → Acetone

21. Acetaldehyde into crotanaldehyde (But-2enal)Aldol Condensation

22. Acetaldehyde into Butan-1-ol23. Acetaldehyde (NaOH / heating )→ Crotanaldehyde (H2/Ni) → Butan-1-ol24. Acetaldehyde into But-2enoic acid Acetaldehyde (NaOH / heating )→ Crotanaldehyde (CrO3) → But-2enoic acid

25. Benzene into Benzoic acidBenzene (CH3Cl/FeCl3) → Toluene ( KMnO4/OH-) → Benzoic Acid

26. Benzoic acid into Aneline Benzoic acid (NH3, heat) → Benzamide ( NaOH,Br2) → Aneline

27. Chloro Benzene into Benzoic acid Chloro Benzene (CH3Cl, Na) → Toluene ( KMnO4/OH-) → Benzoic Acid

28. 4-methyl Acetophenone into terphthalic acid 4-methyl Acetophenone ( KMnO4/OH-) → terphthalic acid

29. Benzene into AnelineBenzene ( HNO3/H2SO4) → Nitro Benzene (Sn/HCl) →Aneline

30. Ethyl chloride to propanamine Ethyl chloride (KCN) → Propane Nitrile (H2/Ni) → propanamine

31. Ethyl amine into Ethyl Carbyl amineCHCl3/alc KOH

32. Aneline into 4-bromo AnelineAneline (Acetic Anhydride) → Acetanilide (Br2/water) → 4-Bromo Acetanilide (H2O/H+) →4-bromo Aneline

33. Aneline into 4-Nitro AnelienAneline (Acetic Anhydride) → Acetanilide (HNO3) → 4-Nitro Acetanilide (H2O/H+) →

4-Nitro Aneline

34. Aneline into Sulphanilic acid35. Aneline (H2SO4) → Sulphanilic acid36. Aneline into Acetanilide Aneline (Acetic Anhydride) → Acetanilide

37. Phenol into Phenyl AcetatePhenol (Acetic Anhydride) → Phenyl Acetate

38. Aneline into Floro BenzeneAneline ( HNO2, 0-50c) → Diazonium Chloride (HBF4)→ Floro Benzene

39. Aneline into Phenol Aneline ( HNO2, 0-50c) → Diazonium Chloride (H2O/H+) → Phenol

40. Aneline into Benzoic AcidAneline ( HNO2, 0-50c) → Diazonium Chloride (CuCN)→ Benzo Nitrile (H2O/H+) → Benzoic Acid

41. Aneline into BenzeneAneline ( HNO2, 0-50c) → Diazonium Chloride (H2O/H3PO2)

42. Acetic Acid to MethanolAcetic Acid (NH3, heat)→Ethanamide (NaOH, Br2) →Methanamine (HNO2) →Methanol

43. Benzoic Acid into Aneline Benzoic Acid (NH3, heat)→ Benzamide (NaOH, Br2) →Aneline

44. Benzaldehyde into alpha hydroxyl phenyl Acetic acidBenzaldehyde (HCN) → PhCH(OH)CN (hydrolyisis) → alpha hydroxyl phenyl Acetic acid

45. Benzaldehyde into 3-Phenyl Propan-1-ol46. Benzaldehyde (Acetaldehyde/NaOH) → phCH=CHCHO ((H2/Ni) → 3-Phenyl Propan-1-ol47. Propanone into Propene

Propanone (NaBH4) →Propan-2-ol (85% H3PO4 /440k) → Propene

48. Ethyl Chloride into Propanoic acidEthyl Chloride (KCN) → Propane Nitrile (H2O/H+)→ Propanoic acid

49. Aniline to 2,4,6-tribromo AnelineAniline (Bromine water ) → 2,4,6-tribromo Aneline

SOME WORD PROBLEMS

1.Write structural formulae of the organic compounds ‘A’,’B’,’C’and ’D’ in the following sequence of reactions: ‘A’ +CH3MgBr — H2O→ CH3CH2CH(OH)CH3 —conc H2SO4→ ‘B’ — Br2 → ‘C’ — alc.KOH→ ‘D’

2. Etherical solution of an organic compound ‘A’ when heated with magnesium gave ‘B’. ‘B’ on treatment with wthanal followed by acid hydrolysis gave 2-propanol. Identify the compound ‘A’. what is ‘B’ known as?

3. A organic compound ‘A’ (molecular formula C3H6O) is resistant to oxidation but forms a compound ‘B’ (C3H8O) on reduction. ‘B’ reacts with HBr to form a bromide ‘C’ which on treatment with alcoholic KOH forms an alkene ‘D’ (C3H6). Deduce the structures of A,B,C and D.

4. A compound ‘A’ which has characteristic odour, on treatment with NaOH forms two compounds ‘B’ and ‘C’. compound ‘B’ has the molecular formula C7H8O which on oxidation gives back compound ‘A’. Compound ‘C’ is the sodium salt of an acid , ‘C’ when heated with soda lime yields an aromatic hydrocarbon ‘D’. Deduce the structures of A,B,C and D.

5. An organic compound ‘A’ has the molecular formula C5H10O. it does not reduce Fehling solution but forms a bisulphate compound. It also gives positive iodoform test. What are possible structures of ‘A’?

6. A compound ‘A’ with molecular formula C5H10 O gave a positive 2,4-DNP test but a negative Tollen,s test. It was oxidized to carboxylic acid ‘B’ with molecular formula C3H6O2. Sodium salt of ‘B’ gave a hydrocarbon ‘C’ on Kolbe’s electrolytic reduction. Identify A,B and C and write the chemical equations for the reaction.

7. When phenol heated with Zn it forms ‘A’. ‘A’ converts in to ‘B’ when reacts with conc.HNO 3 in presence of sulphuric acid. ‘B’ converts in to ‘C’ when it reacts with Br2. ‘C’ converts into ‘E’ by hydrogenation followed by reaction with nitrous acid/HCl. Finally it converts into ‘F’ by acidic hydrolysis. Identify A,B,C,D,E and F.

8. An aliphatic compound ‘A’ with a molecular formula of C3H6O reacts with phenyl hydrazine to give compound ‘B’. Reaction of ‘A’ with I2 in alkaline medium on warming, gives a yellow ppt ‘C’. Identify compound A,B and C.

9. Two moles of organic compound ‘A’ on treatment with a strong base gives two compound B and C. Compound ‘B’ on hydrogenation with Cu gives ‘A’ while acidification of C yields carboxylic acid ‘D’, having molecular formula of CH2O2. identify the compounds A,B,C and D.

10. An organic compound ‘A’ having molecular formula C8H16O was hydrolysed with dilute sulphuric acid to give a carboxylic acid ‘B’ and an alcohol ‘C’. oxidation of ‘C’ with chromic acid produces ‘B’. Write IUPAC names for possible structures of ‘A’ and equations for the reactions involved.

11. A compound ‘A’ of molecular formula C3H7O2N on reaction with Fe and conc.HCl gives a compound ‘B’ of molecular formula C3H9N. compound ‘B’ on treatment with NaNO2 and HCl gives another compound ‘C’ of molecular formula C3H8O. Compound ‘C’ gives effervescence with Na. on oxidation with CrO3, the compound ‘C’ gives a saturated aldehyde containing three carbon atoms. Deduce the structures of A,B and C and write the equations for the reactions involved.

12. A compound ‘X’ having molecular formula C3H7NO, reacts with Br2 in presence of NaOH gives compound ‘X’. the compound ‘X’ reacts with HNO2 to form ethanol and N2 gas. Identify the compound X and Y and write the reactions involved.

13. An aliphatic compound ‘A’ with molecular formula C2H3Cl on treatment with AgCN gives two isomeric compounds of unequal amounts with the molecular formula C3H3N. The minor of these two products on complete reduction with H2 in the presence of Ni gives a compound ‘B’ with a molecular formula C3H9N. identify the compounds A,B and write the reactions involved.

14. Iodomethane reacts with KCN to form a major product ‘A’. Compound ‘A’ on reduction in presence of LiAlH4 forms a higher amine ‘B’. Compound B on treatment with CuCl2 forms a blue colour complex ’C’. Identify the compounds A,B and C.

15. An organic compound ‘A’ having molecular formula C2H3N on reduction gave another compound ‘B’. The Compound ‘B’ on treatment with HNO2 gave ethyl alcohol. ‘B’ on warming with CHCl3 and alcoholic potash gave an offensive smelling substance ‘C’. Identify A,B and C.

16. Two isomeric compounds A and B having molecular formula C4H11N, both loose N2 on treatment with HNO2 and gives compound C and D respectively. C is resistant to oxidation but immediately responds to Lucas reagent, whereas ‘D’ responds to Lucas reagent after 5 minutes and gives a positive iodoform test. Identify A and B.

Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b).

Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When

(a) is reacted with sodium metal it gives compound (d), C8H18 which is different

from the compound formed when n-butyl bromide is reacted with sodium.

Give the structural formula of (a) and write the equations for all the reactions.

An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative,

reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation,

it gives 1,2-benzenedicarboxylic acid. Identify the compound.

An organic compound (A) (molecular formula C8H16O2) was hydrolysed with

dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation

of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene.

Write equations for the reactions involved.

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen.

The molecular mass of the compound is 86. It does not reduce Tollens’ reagent

but forms an addition compound with sodium hydrogensulphite and give positive

iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write

the possible structure of the compound.

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating

forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’

of molecular formula C6H7N. Write the structures and IUPAC names of compounds

A, B and C.

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Organic Reagents

S.No. Reagent Function

1 PCl3, PBr3, PI3 Alcohols into Alkyl halides

2 SoCl2, PCl5 Alcohols into Alkyl chlorides & Carboxylic acids into Acid Chlorides

3 HCl/ZnCl2,HBr,HI Alcohols into alkyl halides

4 Cl2/Fe or FeCl3 Cl group substituting on Benzene

5 NaNO2 /HCl 0-50C Diazotisation

6 CuCl,CuBr,CuCN,KI,H2O,

H3PO2

Diazonium Cholride into Chlor Benzene, Bromo Benzene, Benzo nitrile, Iodo Benzene , Phenol,Benzene respectively

7 HBF4 or NaBF4 Diazonium Chloride into Floro Benzene

8 AgF or Hg2F2 or SbF3 or CoF2

Alkyl halides into alkyl florides

9 Na / dry ether Alkyl halides into alkanes

10 NaOH 623/443/368K Chloro benzene to phenol

11 Br2 /FeBr3 Bromination of Benzene

12 Cl2 /FeCl3 Chlorination of Benzene

13 CH3Cl /AlCl3 alkylation of benzene and its derivatives

14 CH-CO-Cl /AlCl3 Acylation on benzene

15 H2SO4 /HNO3 Nitration of benzene

16 (CHCO)2O /AlCl3 O – Acylation of Phenol

17 H2SO4 Sulphonation on Benzene

18 H2O/ H2SO4 alkenes into alcohols

Aq KOH Alkyl halide into alcohol

19 BH3 /H2O2 /OH- Alkenes into alcohols (Anti Markownikoff product)

20 NaBH4/ LiAlH4(LAH) Aldehydes, ketones, acids into alcohols, Nito & Cynides , Isocyanides into amines

21 H2/ Ni or H2/Pd reduction of aldehydes, ketones and cynides

22 RMgX/H3O+ Aldehydes, ketones into alcohls

23 O2/H+ Cumene to phenol

24 Na Alcohol or phenol into Sodium alkoxide/Phenoxide

25 (CHCO)2O/ CH-CO-Cl O acylation on phenol or N acylation on Aneline or amine

26 Conc.H2SO4/443K Conversion of primary alcohols into Alkenes

27 Conc.H2SO4/410K Conversion of alcohols into Ethers

28 85% H3PO4 / 440K Secondary alcohol into alkene

29 20% H3PO4/358K Tertiary alcohol into alkene

Alcoholic KOH Alkyl halide into alkene

30 CrO3/KMnO4 or K2Cr2O7 in acidic medium

oxidation of alcohols into acids

31 Cu /573k dehydrogenation of alcohols gives 10 alcohols into aldehydes and 20 alcohols

into ketones & 30 alcohos into alkenes

32 Dil. HNO3 Mono nitration of Phenol

33 Conc.HNO3 tri nitration of phenol

34 Br2 /H2O tri bromination of phenol

35 Br2 /Cs2 mono bromination of phenol

36 NaOH /CO2 Phenol to salicilic acid

37 CHCl3 /NaOH Phenol to salcilaldehyde

38 Zn dust Phenol to Benzene

39 Na2Cr2O7 /H2SO4 or air Phenol to Benzo quinone

40 Zn/Cr2O3 200to 300 atm 573 – 673K

CO & H into methanol

41 Invertase Sucrose into Glucose or Fructose

42 Zymase Glucose or Fructose into ethanol

43 HI Ether into alcohol & alkyl halide

44 PCC alcohol to aldehyde

45 Pd /BaSO4,H2 acid chloride into aldehyde

46 SnCl2/HCl/H3O+ Cyanides into aldehydes

47 AlH(i-Bu)2/H2O Cyanides into aldehydes

48 DIBAL-H/H2O Esters into aldehydes

49 CrO2Cl2/H2O Toluene to aldehyde

50 CrO3 /(CH3CO)2O Toluene into Benzaldehyde

51 Cl2/hv Chlorination on alkyl group of Benzene or alkane

52 CO, HCl anhydrous AlCl3 Benzene to Benzaldehyde

53 (CH3)2Cd acid chloride into ketones

54 RMgX/H3O+ Cyanides into ketones

55 HCN Carbonyl compound into cyanohydrin

56 NaHSO3 addition to aldehyde and ketone

57 H2NOH carbonyl compound into oxime

58 H2N-NH2 carbonyl compound into hydrazone

59 H2N-NH-Ph carbonyl compound into Phenyl hydrazone

60 2,4DNP carbonyl compound into 2,4 dinitro phynyl hydrazone

61 H2N-NH-CO-CH3 carbonyl compound into semi carbazide

62 ROH/HCl Aldehydes & ketones into hemiacetal and acetal

63 HO-CH2-CH2-OH/HCl Aldehyde or ketone into ethelene glycol ketone

64 Zn-Hg/HCl carbonyl compound into alkane

65 H2N-NH2 /KOH carbonyl compound into alkane

66 KMnO4/OH-/ K2Cr2O7 /H2SO4 or HNO3

Ketones into mixture of carboxylic acids on prolonged oxidation

67 (Ag(NH3)2)NO3+NaOH Tollen’s test

68 Cu(OH)2 Fehiling’s test

69 NaOH+I2 Iodoform

70 NaOH or Ba(OH)2 aldal condensation

71 Conc KOH or NaOH Cannizaro’s reaction

72 KMnO4 /KOH Toluene/alkyl Benzene into Benzoic Acid

73 H2O/H+ Cyanides into carboxylic acids, amides into carboxylic acids, esters into carboxylic acis and alcohols, acid chlorides or anhydrides into carboxylic acids

74 NaOH Saponificaiton of ester, acid into salt of acid

75 Na2CO3 or NaHCO3 Carboxylic acid test

76 P4O10 or P2O5 Dehydration of acids into anhydride, amides into nitriles

77 ROH/conc H2SO4 Caroxylic acids into esters

78 PCl3, SoCl2, PCl5 Carboxylic acids into acid chlorides

79 NH3 heating Carboxylic acids into amides

80 NaOH/CaO Decarboxylation (acids into alkanes)

81 LiAlH4 Carboxylic acids into alcohols, amides into amines

82 Cl2 /red.P4 HVZ reaction

83 Sn /HCl or Fe /HCl, H2/Pd Reduction of nitro compounds into amines

84 NH3 Alkyl halides into amines

85 H2/ Ni or H2/Pd LiAlH4 Amides into cyanindes

86 KOH/R-X Phthalamide into amine

87 NaOH /Br2 Hoffman bromamide, amide into amine with one ‘C’ less

88 KOH,CHCl3 Amines into Carbyl amines

89 NaNO2/HCl 10 aliphatic amines into alcohols

90 NaNO2/HCl 0 – 50C Aniline into diazonium chloride

91 C6H5SO2Cl Distinguishing 10,20 & 30 amines

92 Br2/H2O Aneline into tri bromo aniline

93 Br2/ CH-CO-Cl /(CHCO)2O Aniline into Bromo Aniline

94 HNO3/ CH-CO-Cl /(CHCO)2O

Nitro aniline

95 H2SO4 Sulphonation on aniline

96 CuCl,CuBr,CuCN,KI,H2O,

H3PO2 or CH3-CH2-OH

Diazonium Cholride into Chlor Benzene, Bromo Benzene, Benzo nitrile, Iodo Benzene , Phenol,Benzene respectively

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REASONING QUESTIONS IN ORGANIC CHEMISTRY1. Arrange each set of compounds in order of increasing boiling poings.

(i) Bromo Ethane , Bromoform, Chloro methane, Dibromo methane

(ii) 1-Chloro Propane, Isopropyl chloride, 1-chloro butane

Ans: Chloro Methane(CH3Cl) < Bromo Ethane(CH3-CH2-Br) < Dibromo

Mehane(CH2Br2) < Bromoform (CHBr3)

CH3CHCl-CH3 < CH3-CH2-CH2Cl < CH3-CH2-CH2-CH2-Cl

As atomic mass increases boiling point increases. And branching compounds has less spherical contact as compare with straight chain compounds. Hence bp less

for branching chain alkyl halides.

2. Which of the following undergo SN2 faster?

(i) Cl CH2Cl

(i) I Cl

Ans: CH2Cl is more faster under go SN2 because it is Primary alkyl halide

I is more faster to under go SN2 mechanism because I is better

leaving group due to its large size.

3. Predict the order of reactivity of the following compounds in SN1 & SN2 reactions.

(i) The four isomeric bromo butanes

(ii) C6H5CH2Br, C6H5CH2 (C6H5 )Br, C6H5CH2 (CH3)Br, C6H5CH2 (CH3 )(C6H5 )Br

Ans: CH3-CH2-CH2-CH2-Br< CH3- CHCH3- CH2- Br < CH3-CH2-CHBr –CH3<

C(CH3)3Br order for SN1 mechanism

CH3-CH2-CH2-CH2-Br> CH3- CHCH3- CH2- Br > CH3-CH2-CHBr –CH3>

C(CH3)3Br order for SN2 mechanism

C6H5CH2Br< C6H5CH (CH3)Br< C6H5CH (C6H5 )Br< C6H5C (CH3 )(C6H5 )Br

Order for SN1 mechanism

C6H5CH2Br> C6H5CH (CH3)Br> C6H5CH (C6H5 )Br> C6H5C (CH3 )(C6H5 )Br

Order for SN2 mechanism

4. Although Chlorine is an electron with drawing group yet it is orho para directing in

electrophilic aromatic substitution reactions. Why?

Ans: Cl is electron with drawing by inductive effect, and it electron relesing by resonance

and creates negative charge at ortho and para positions. And here resonance effect

is more as compare with inductive effect.

5. Which alkyl halide from the following pairs would you expect to react more rapidly

by an SN2 mechanism? Explain your answer?

1. CH3-CH2-CH2-CH2-Br & CH3-CH2-CHBr –CH3

2. CH3-CH2-CHBr –CH3 & C(CH3)3Br3. CH3- CHCH3- CH2- CH2- Br & CH3- CH2- CHCH3- CH2- Br

Ans: 1. CH3-CH2-CH2-CH2-Br primary alkyl halide is more reactive than secondary

alkyl halide

2. CH3-CH2-CHBr –CH3 Secondary alkyl halide is more reactive than Tertiary

alkyl halide

3. CH3- CHCH3- CH2- CH2- Br As methyl group is far from Br it will less stabilize

the carbo cations and facilitate the SN2 mechanism.

6. Which of the following compound go faster SN1 reaction:

1. &

Cl

2. & Cl

Ans: 1. first compound ismore stable because tertiary is more reactive than secondary

alkyl halide for SN1 mechanism.

2. . first compound is more stable because Secondary is more reactive than

primary for SN1 mechanism.

7. Which one of the following has highest boiling point;

CH2Cl2 CHCl3 CCl4

Ans: net Dipole moment of CH2Cl2 & CCl4 is zero due to cancellation dipole moment of

bonds due to its symmetric structure. Where as net dipole moment of CHCl3 is not

zero. Because its structure is not symmetric.

8. Halo alkanes react with KCN to form alkyl cyanides as main product while AgCN

forms isocyanides as the chief product. Explain.

Ans: KCN is predominantly ionic and provides cyanide ions in solution. Although both

carbon and nitrogen atoms are in a position to donate electron pairs. The attack

takes place mainly through carbon atom not through nitrogen atom since C—C is

more stable than C—N bond. However, AgCN is mainly covalent innatrue and

nitrogen is free to donate electron pair forming isocyanide as the main product.

9. Vinyl chloride is less reactive towards nucleophilic substitution. Explain.

Ans: Due to resonance between C—Cl bond in vinyl chloride show partial double bond

character.

CH2=CH—::Cl:

10. Chloroform stored in dark colored bottles by completely filling it it. Explain.

Ans: Chloroform react with air in presence of sun light it forms poisonous phosogene.

2 CHCl3 + O2 2 COCl2+ 2 HCl

11. Out of C6H5CH2Cl & C6 H5CHCl C6H5 Which is more easily hydrolyzed by KOH?

Ans; The second comound is more resonance stabilized hence easily forms carbo cation

and react faster by SN1 mechanism.

12. The treatment of alkyl halides with aqueous KOH leads to the formation of alcohols

but in the presence of alcoholic KOH alkenes are major products. Explain

Ans: Substitution of alkyl halide gives alcohols and elimination gives alkenes and there is

a competition between substitution and elimination. Heavy nucleophile in Alcoholic KOH prefers elimination, where as in aq KOH nucleophile is light OH- it prefers substitution.

13. Arrange the following sets of compounds in order of their increasing

boiling points:

(a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.

(b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.

Ans: (a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.

(b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.

14. Arrange the following compounds in increasing order of their acid strength:

Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol,

phenol, 4-methylphenol.

Ans: Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3,5-dinitrophenol,

2,4, 6-trinitrophenol.

15. The relative dehydration of alcohols is Tertiary>Secondar>Primary give reason.

Ans: Dehydration carryout by carbocation mechanism, and stability of carbocation is teriary>Secondary>Primary

16. Alcohols are comparatively more soluble in water than hydrocarbons of

comparable molecular masses. Explain this fact.

Ans: Due to hydrogen bond with water molecules

17. Preparation of ethers by dehydration of alcohols is not suitable for the using

of secondary and tertiary alcohols give reason.

Ans: dehydration of secondary and tertiary alcohols to give corresponding ethers is

successful as elimination competes over substitution and as a consequence,

alkenes are easily formed.

18. Which of the following is appropriate for the preparation of anisole?

a) Bromo Benzene & Sodium Methoxide b) Phenoxide & CH3Br

Ans: Set b is appropriate because in set a nucleaophilic substitution on aromating

ring is difficult due to partial double bond character.

19. Phenol is stronger acid than alcohol. Explain.

(Or)

Phenol is weak proton acceptor than alcohol. Explain.

Ans. Acidity of the substance depends on stability of anion after loosing H+ ion. Since phenoxide ion undergoes resonance stabilization compare to

alkoxide, phenol is more acidic than alcohol.

20. Arrange the following compounds in increasing order of their acidic strength.

Propan-1-ol, 4-Methyl phenol, Phenol, 3,5 dinitrophenol, 2,4,6

Trinitrophenol.

Ans. Propan-1-ol< 4-Methoxy phenol< Phenol< 3,5 dinitrophenol<

2, 4, 6 Trinitrophenol

Phenol is more than alcohol due to resonance stability of phenoxide ion.

Electron releasing groups like (O-CH3) methoxy decreases acidity.

Electron withdrawing groups like (-NO2) nitro groups increases acidity.

More number of electron withdrawing groups more acidity.

21. O- nitro phenol & p-nitro phenol can be separated by distillation? Explain.

Or

Boiling point of p-nitro phenol is more than the o- nitro phenol. Explain?

Ans. O-nitro phenol is a stream volatile due to intermolecular Hydrogen bond

where as p-nitro phenol is having inter molecular Hydrogen bond.

Inter molecular hydrogen bond decreases Boiling point

Inter molecular hydrogen bond increases Boiling point

22. t- Butyl methyl ether reacts with HI forms methanol & t- butyl iodide explain?

Ans. Since tertiary carbocation is more stable and reaction is following

carbocation mechanism it gives t- Butyl iodide.

23. The commercial ethanol is mixed with copper sulphate & pyridine. Explain.

Ans. Commercial ethanol is mixed with CuSo4 & pyridine to make it unfit for

drinking. It is known as denaturation of alcohol.

24. Anisole react with HI gives phenol & methyl iodide but not iodo benzene and

methanol. Explain.

Ans. Nucleophilic substitution on aromatic ring is difficult due to partial double

Bond character between oxygen and carbon of benzene ring.

25. While separating a mixture of ortho & para nitro phenols by steam distillation,

name the isomer which is Steam volatile. Give reasons?

Ans:O-nitro phenol is steam volatile due to inter molecular hydrogen bond.

26. Explain why is propanal higher boiling than butane?

Ans:Due to intermolecular hydrogen bond present in alcohols,they possess high

Boiling point.

27. Arrange the following compounds in increasing order of their boiling points?

CH3-CHO , CH3-CH2-OH , CH3-O-CH3 , CH3-CH2-CH

Ans:Their molecular masses are almost all similar but

CH3-CH2-OH > CH3-CHO > CH3-O-CH3 > CH3-CH2-CH3

Inter molecular more polar than presence of no H-bond &

H-bond ethers polarization

28. Preparation of Alkyl halides from alcohols by using SOCl2 is preferable.Explain?

Ans.When SOCl2 is used,side products are gases and forms pure alkyl halides.

R-OH+SOCl2 R-Cl + SO2+ HCl

29. Would you expect benzaldehyde to be more reactive or less reactive in

nucleophilic addition reactions than propanal? Explain your answer.

Ans: The carbon atom of the carbonyl group of benzaldehyde is less

electrophilic than carbon atom of the carbonyl group present in

propanal. The polarity of the carbonyl group is reduced in benzaldehyde

due to resonance as shown below and hence it is less reactive than propanal.

30. Carboxylic acids are having higher boiling points than aldehydes, ketones and even alcohols of comparable molecular masses. Explain.

Ans: Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses. This is due to more extensive

association of carboxylic acid molecules through intermolecular hydrogen bonding. The hydrogen bonds are not broken completely even in the vapour phase. In fact,

most carboxylic acids exist as dimer in the vapour phase or in the aprotic solvents.

31. Carboxylic acids are stronger acids than pheonol thought both possess

resonance stabilization of respective anions.

Ans: The higher acidity of carboxylic acids as compared to phenols can be understood similarly. The conjugate base of carboxylic acid, a carboxylate ion, is stabilised by two equivalent resonance structures in which the negative charge is at the more electronegative oxygen atom. The conjugate base of phenol, a phenoxide

ion, has non-equivalent resonance structures in which the negative charge

is at the less electronegative carbon atom. Thus, the carboxylate ion is more stabilised than phenoxide ion, so carboxylic acids are more acidic than phenols.

32. Arrange the following in increasing order of their relativity towards nucleophilic addition reactions.

H- CHO, CH3 -CHO, CH3 -CO- CH3

Ans: CH3 -CO-CH3 <CH3-CHO< H-CHO

Electron releasing effect

Steric effect

33. Carboxylic acid are having more boiling point than alcohols of same molecular

mass. explain

Ans. In carboxylic acids strength of H bond is stronger than alcohols and forms

dimmer in carboxylic acids which increases boiling point.

34. Carboxylic acid can not give characteristic reactions of aldehydes and ketones

though both are having carbonyl group CO?

Ans. Carbonyl group in carboxylic acid is resonance stabilized as carboxyl ate

ion . Hence it can not give nucleophilic addition reactions given by

carbonyl group.

35. Reactivity order of carboxylic acid derivatives are

Acid chlorides>Anhydrides>Esters>Amides.Explain the reason?

Ans.I) Basisity of leaving group

II) Resonance effect36. Acid chlorides give pungent smell in air.Explain?

Ans. Acid chlorides undergo hydrolysis with atmospheric moisture and gives HCl fumes,which are pungent.

37. Give plausible explanation for each of the following:

(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-

Trimethylcyclohexanone does not.

(ii) There are two –NH2 groups in semicarbazide. However, only one is involved

in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in

the presence of an acid catalyst, the water or the ester should be removed

as soon as it is formed.

Ans: Due to steric effect and electron releasing effect.

The other two NH2 groups involve resonance with C=O

To prevent back ward reaction

38. Aryl amines cannot be prepared by Gabriel Phthalamide synthesis. Explain.

Ans: Aromatic primary amines cannot be prepared by this method because aryl

halides do not undergo nucleophilic substitution with the anion formed by

phthalimide.

39. The basic strength of amines in aqueous solutions is different from gaseous phase give reason.

It is due to inductive effect, hydrogen bond and hydration effect

40.

41. Generally aniline direct bromination gives tri substituted product to get mono stustitued product what is to be done.

Ans: This can be done by protecting the -NH2 group by acetylation with acetic

anhydride, then carrying out the desired substitution followed

by hydrolysis of the substituted amide to the substituted amine.

42. Account for the following:

(i) pKb of aniline is more than that of methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated

ferric oxide.

(iv) Although amino group is o– and p– directing in aromatic electrophilic

substitution reactions, aniline on nitration gives a substantial amount of

m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic

amines.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Ans: More Pkb less basic, less basisity of aniline is due to involvement of lone pair in

resonance.

In Aneline aromatic nature makes insoluble in water.

CH3NH2 + H2O → CH3NH3+ OH-

These hydroxide ions can make Ferric hydroxie.

in the strongly acidic medium, aniline is protonated to form the

anilinium ion which is meta directing.

Aryl amines can not under go nucleophilic substitution.

43. Give plausible explanation for each of the following:

(i) Why are amines less acidic than alcohols of comparable molecular masses?

(ii) Why do primary amines have higher boiling point than tertiary amines?

(iii) Why are aliphatic amines stronger bases than aromatic amines?

Ans: Nitrogen is less Electronegetive than Oxygen which makes less polarization of

N—H bond as compare with O—H bond.

There is no H bond in Tertiary amines.

In aromatic amines lone pair involve in resonance.

44. Why does silver chloride dissolve in methyl amine solution?

Ans. Methyl amine forms a complex compound with silver chloride and that is

soluble in water

2CH3NH2 + AgCl [ Ag (CH3NH2)2] Cl

45. Why does amides less basic than amines?

Ans. Lone pair present in amides is involved in resonance hence it is less basic than amines.46. Methyl amine in water reacts with ferric chloride to precipitate ferric hydroxide.Why?

Ans.Methyl amine is a proton acceptor from water and forms hydroxyl ion which reacts with ferric ions forms ferric hydroxide.

CH3-NH2 + H2O CH3-NH3+ + OH-

FeCl3 + 3OH- Fe(OH)3 + 3Cl-

47. Tertiary amines are having low boiling point compare with primary and secondary amines.Explain?

Ans.Due to absence of Hydrogen bond in tertiary amines,they possess low boiling point as compare with primary and secondary.

48. In solutions,basic strength of amines is secondary>tertiary>primary.Explain?

Ans. Basic strength of amines depends on ammonium cation in solutions .

Ammonium stability not only depends on electron releasing effect but also H-bonding,stearic factor. Hence the above order of basic strength is correct .

49. Nitration of benzene gives substantial amount of meta product,though –NH2 is

o-pdirecting group?

Ans.Nitration takes place in the presence of H+ ions which protanates on Amine and forms which behaves as electron withdrawing group,hence it forms some amount of meta product.

*******

BIO MOLECULES

Carbohydrates : Polyhydroxy aldehydes and ketones are known as carbohydrates.

the carbohydrates may be defined asoptically active polyhydroxy aldehydes or ketones or the compoundswhich produce such units on hydrolysis. Monosaccharides: Glucose, Fructose, Ribose, GalactoseOligo Saccharides: Sucrose, Maltose, LactosePoly Saccharides: Starch, Cellulose, GlycogenGlucose Preparation:

Properties of Glucose:

It suggests presence of carbonyl group.

It indicates presence of carbonyl group and primary alcohol.

It indicates Presence of five hydroxyl groups

It indicates presence of carbonyl group and primary alcoholOpen chain structure of Glucose

Reasons Which Could Not Explain Open Chain Structure –(Ring Structure)

Ring Structure of Glucose:

Ring Structure of Fructose:

Di Saccharides:

Poly Saccharides:Starch: monomer is alpha D glucose and linkage is C1-C4 & C1-C6 linkages.It has two components like amylase which is water soluble and linear chain polymer of alpha D glucose. And other is Amylopectin which is water insoluble and it is branched chain polymer of alpha D glucose.Cellulose: Its monomer is Beta D glucose with C1-C4 linkage.Glycogen is similar structure as starch and it is also known as animal starch.

PROTEINS:Proteins are vital molecules which are made of amino acids. There are 20 amino acids with which all proteins are made. These amino acids are 10 essential amino acids which are important but not synthesized in human body. They must taken through diet. Ex. Valine, LeucineNon essential amino acids synthesized in human body. Ex Glyciene, Alanaine.

Zwitter ion Structure of Amino Acids

In acid medium Amino Acid has Cationic structureIn Basic medium it has anionic structure. At certain PH it has neutral structure this is known as Isoelectric point.Di Peptide:Two amino acids joined by amide linkage between COOH and NH2 this is also known as peptide linkage.

Structure of Protein:Its structure is studied under four categories

1. Primary Structure: It is sequential order of various amino acids present in protein chain.

2. Secondary Structure: It explains type of hydrogen bond between carbonyl and amino group. It has two forms.

a) Alpha helix due to intra molecular H bondb) Beta Pleated sheet structure: Inter molecular H Bond

3. Tertiary Structure: It explains the folding up of protein chain and gives shape to it. Like Globular protein (water soluble) and Fibrous protein(water in soluble). And inter particle forces are di sulphide, vander waals, hydrogen bond electrostatic forces.

4. Quarternary Structure: Some of the proteins are composed of two or more polypeptide chains referred to as sub-units. The spatial arrangement of these subunits with respect to each other is known as quaternary structure.

Denaturation of Proteins: Proteins are sensitive to heat and PH, when it is subjected these it looses its biological activity. And coverts water soluble globular protein into water insoluble fibrous protein. During denaturation primary structure is not affected but higher order structure is affected.Enzymes: enzymes are known as biological catalysts. These can increase the rate of biochemical reactions by many times. Ex. Maltase, zymage liphase.

Vitamines: organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism.

Vitamin A,D,E & K are water insoluble but fat soluble. And B (Except B12 ) & C are water soluble. They must be taken regularly.

Nuceic Acids:

1. 2 deoxy Ribo Nucleic Acid DNA2. Ribo Nucleic Acid RNA

DNA unit is Nucleotide and made up of 2-de oxy ribose, Nitrogen base (A,G,C,T) and Phosphate group.

RNA unit is has Ribose sugar and N base like A, G,C U and Phosphate group.

One can understand linkage of nucleic acid by following phosphor di ester linkage from the diagram

Note: Structures are for understanding linkage but not important.

DNA Structure: 1. Primary Structure: Sequential order of nucleotides 2. Secondary structure: Double helix structure of two strands of Nucleic Acid

chains. Two chains are not idential but complimentary to each other. Due to Base pair rule i.e A=G, C≡T

Functions of DNA:1. Replication: DNA molecule is capable of self duplication during cell

division and identical DNA strands are transferred to daughter cells.

2. Protein Synthesis: it is in two steps Transcription and Translation

VSA of One

Mark:

1.Write two main functions of carbohydrates in plants.

2. What is meant by inversion of sugar ?

3. Define the term native state as applied to proteins.

4. What type of bonding helps in stabilising the -helix structure of proteins?

5. What is the effect of denaturation on the structure of proteins?

6. The two strands in DNA are not identical but are complementary. Explain.

SA-1 of Two Marks:1. Answer the following questions briefly

(i) What are any two good sources of vitamin A

(ii) What are nucleotides?

2. What are essential and non essential aminoacids?Give two example of each

3. Differentiate between (i) Primary strcture and secondary strcture of

S.No.

DNA RNA

1 Sugar moiety is 2-Deoxy -D(-) -ribose

Sugar moiety is D(-) ribose

2 Cotains thymine and cytosine as pyrimidine bases and guanine and adenine as purine bases

Contains cytosine and uracil as pyrimidine bases and guanine and adenine as purine bases

3 Has double stranded helix structure

Has single stranded helix structure

4 Chiefly occurs in nucleus of cell

Mainly occurs in the cytoplasm of the cell

5 Very large molecules - molecular mass may very from 6 million to 16 million

Smaller than DNA – molecular mass varies from 20000 to 40000

6 Has unique property of replication

Does not replicate

7 Controls transmission of hereditary effects

Controls synthesis of protein

proteins4. What do you understand by the term glycosidic linkage?

5. What is glycogen? How is it different from starch?

6. Why Sucrose is Non Reducing sugar?

7. What are the hydrolysis products of (i) sucrose and (ii) lactose?SA-2 of Three Marks:1. Answer the following questions briefly.

(i) What are reducing sugars? (ii) What is meant by denaturation of protein? (iii) How is oxygen replenished in our atmopsphere?

2. Define the following terms (i) nucleotides

(b) List out main functions of carbohydrates in organisms.

3. Write the chemical reactions of Glucose with (i) NH2OH (ii) (CH3CO)2O.Also draw simple Fischer projection of D – Glucose and L - Glucose. (b) Name the food sources and deficiency diseases caused due to lack of any two of the vitamins A,C,E and K.4. Classify the following into monosaccharides and disaccharides. Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.5. Define the following as related to proteins

(i) Peptide linkage (ii) Primary structure (iii) Denaturation.

Polymerisation

POLYMERS

Polymers: A polymer is a compound of high molecular mass formed by the combination of a large no. of small molecules.

The small molecules which form therepeating unitin a polymer are called asmonomers.

The process by which the monomers are converted into polymers is called as polymerization.e.g.

n CH2 = CH2 – (CH2 – CH2 – )n

ethene Polyethene

(monomer) (Polymer)

Homopolymers and Co-polymers: Depending upon the nature of repeating structural unit, the polymers are divided into two categories:

1. Homopolymers: Polymers whose repeating structural unit is derived from only one type of monomer unit are called as homopolymers.

e.g. in case of polyethene

n CH2 = CH2 – (CH2 – CH2 – )n

ethene Polyethene

2. Co-polymers: Polymers whose repeating structural unit is derived from two or more type of monomer units are called as copolymers. e.g. in case of Nylon 66,

n H2N – ( CH2 )6 – NH2 + n HOOC –(CH2)4 –COOH

HexamethylenediamineAdipic acid – n H2O

(monomer) (monomer)

[– HN – (CH2)6– NH – OC –(CH2)4 – CO –]n

III Classification based upon Synthesis: On the basis of synthesis, polymers can be classified as :

1. Addition polymers: In these polymers, the monomer units simply add on one another, without losing molecule such as water, alcohol etc. This generally occurs among molecules, which contain double bond. e.g.Polyethene, polystyrene, polyvinyl chloride etc.

473 K1500 atm

340 K, 6-7 atm(C2H5)3Al + TiCl4

2. Condensation polymers: In these polymers, the monomer units combine together with simultaneous loss of simple molecule such as water, alcohol etc.e.g. Nylon,Terylene, Bakelite etc.

IV Classification based on Molecular Forces:

1. Elastomers: “Polymers in which the intermolecular forces of attraction are weakest are called as elastomers.”

. Examples of elastomers are Vulcanized rubber, Buna-S etc.

2. Fibers: “Polymers in which the intermolecular forces of attraction are strongest are called as fibres.”

Examples of fibers are nylons, terylene, decron, polyacrylonitrile etc.

3. Thermo plastics: “Polymers in which the intermolecular forces of attraction are in between those of elastomers and fibers are called as thermoplastics”. Due to less intermolecular forces of attraction, they can be easily moulded on heating. Examples are polyethene, polystyrene, PVC , Teflon etc.

4. Thermo setting polymers: These are the polymers which undergo permanent change on heating and become hard and infusible. On heating, they form cross-links and become very hard. e.g. bakelite, melamine-formaldehyde resin etc.

Some Commercially important Polymers

I Addition polymers :

1. Polyethene or polyethylene:

(a) Low density polyethene (LDP):

n CH2 = CH2 (– CH2 –CH2 –)n

ethane poly ethene

Uses : (i) As a packing material in the form of thin plastic films , bags etc.

(ii) In insulating wires & cables.

(iii) In the manufacture of toys, bottles etc.

(b) High density polyethene(HDP) : It is prepared by the polymerization of ethene in the presence of Triethylaluminium& Titanium tetrachloride ( known as Zeigler - Natta catalyst).

n CH2 = CH2 (– CH2 –CH2 –)n

Uses : in the manufacture of buckets, tubs, bottles, toys, pipes etc.

4. Polyacrlonitrile (PAN) or Orlon:

n CH2 = CH (–CH2 – CH–)n

I I

CN CN

Vinyl cyanide (acrylonitrile) Polyacrylonitrile

Uses: for making blankets, sweater, carpets etc.

6. Poly vinyl Chloride (PVC):

n CH2 = CH (–CH2 – CH–)n

I I

Cl Cl

Vinyl chloride Polyvinylchloride (PVC)

Uses: (i) In making rain coats, hand bags, plastic dolls, shoe-soles, vinyl flooring etc.

(ii) As an insulating materials for wires & cables.

(iii) In making gramophone records & hose pipes. etc.

heatpress.

7. Polytetraflouro ethylene (PTFE or Teflon) :

n CF2= CF2 (–CF2 –CF2–)n

tetraflouroethylenepolytetraflouroethylene(teflon)

Uses: (i) As a martial resistant to heat & chemical attack.

(ii) For coating of non- stick cook wares.

(iii) For making gaskets, pump, packing, valves, seals, non-lubricated bearings etc.

II. Condensation Polymers: These are of followingthree types:

(A) Polyesters (B) polyamides (C) Formaldehyde resins

O

II

(A) Polyesters: They involves ester ( – C – O – ) linkage. e.g.

1. Terylene (or Decron) :It is a polymer of ethylene glycol and terephthalic acid.

n HO – CH2 – CH2–OH + n HOOC COOH –nH2O

ethylene glycol terephthalic acid 425K

–[O – CH2 – CH2– O – OC CO –]n

Terylene (or Decron)

HOOC COOH

OC CO–H2O

Uses: (i) For making cloth by mixing with cotton.

(ii) For making magnetic recording tapes, tyrecord , seat belts etc.

2. Glyptal: It is a polymer of ethylene glycol and phthalic acid

n HO–CH2–CH2–OH + n – O–CH2–CH2–O–

Uses: (i) In the manufacture of paints and lacquers.

(ii) For making building material such as asbestos, cement etc.

O

II

(B) Polyamides: These involve amide (–C– NH –) linkage. e.g.

1. Nylon – 6,6 : It is a polymer of hexamethylenediamine and adipic acid.


HexamethylenediamineAdipic acid – n H2O

[– HN – (CH2)6– NH – OC –(CH2)4 – CO –]n

Uses: (i) in making bristles for brushes.

(ii) In textiles e.g. in making soaks and sweaters by mixing with cotton.

2. Nylon 6, 10 : It is a polymer of hexamethylenediamine and sebacic acid .


HexamethylenediamineSebacic acid – n H2O

[– HN – (CH2)6– NH – OC –(CH2)8 – CO –]n

Glyptal

O

HNN

O NOH

3. Nylon- 6 ( orPerlon) : Its monomer is called as caprolactum, which is obtained from cyclohexane, Caprolactum is hydrolysed to get ε-amino caproic acid which is then polymerized to obtain Nylon-6.

H⃗ 2SO 4N⃗H 2 OH–H 2OO⃗xidation

cyclohexanecyclohexanonecyclohexanoneoximecaprolactum

[– HN – (CH2)5–CO –]n

Δ⃗, PolymerisationH 2 O

H2N – (CH2)5 – COOH

Uses: In making tyres, fabric, ropes etc.

2. Melamine–formaldehyde Resin: Melamine and formaldehyde undergo copolymerization to form Melamine –formaldehyde resin as shown:

Uses: It is used in making unbreakable crokerye.g cup, plate, tray etc.

Natural Rubber: It is prepared from Latex, which is obtained from Rubber trees. It is a polymer of 2- methyl 1, 3 butadiene (isoprene)

ε -AminocaproicacidNylon-6

H 2 O, Δ

SS

SS

S

S

S

Vulcanized Rubber: As natural rubber becomes soft in summer and hard in winter, therefore it is improved by the process Vulcanization. It involves heating of natural rubber with 3-5% suplhur. Sulphur forms cross- links between polymer chain and thus make rubber harder & stronger.

Synthetic Rubbers:

1. Neoprene:

Uses: (i) In making bells , hoses , shoe-heals, stoppers etc.

(ii) In making containers for storing petrol & oils.

2. Buna–S : ( or SBR) : It is prepared by copolymerization of 1,3 butadiene and styrene , in the presence of Sodium.

3. Buna-N: Itis a copolymer of 1,3- butadiene and acrylonitrile (vinyl cyanide),

S⃗ulphurHeat

uses: (i) For storage of petrol, lubricating oils and organic solvents.

(ii) In making oil seals, hoses and tank livings

Some Commercially Important Polymers:

BIODEGRADABLE POLYMERS

VSA of One Mark

1. What is the Monomer present in Un Breakable Plastics.2. Name the monomer present in Non stick appliances.3. Give one application of Bekalite.4. Name the monomer present in Nylon-6?5. What does designation 6,6 mean in the name Nylon 6,6?

SA-1 of Two Marks

1. Arrange the following polymers in an increasing order of their intermolecular

forces.

(i) Nylon- 6,6, Buna – S , Polythene (ii) Nylon- 6 , Neoprene, Polyvinyl chloride. Ans: Buna - S,Polythene, Nylon- 6,6

Neoprene, Polyvinyl chloride ,Nylon- 6 2. What are the biodegradable polymers? Give one example of a biodegradable aliphatic polyester

.

Write the structures of monomers used and one use of each of 3. the following polymers. (a) Teflon (b) Buna – N 4. Differentiate between addition and condensation polymers with suitable example

of each.

5.

SA-2 of Three Marks:

1. Explain following process with example Chain growth polymerization. Step growth polymerization.2. What is Vulcuonization? How does it changes the properties of natural rubber?

3

4. Write the information asked for in the following polymers:

i) Bakelite – materials used for preparation

ii) PVC – Monomer Unit.

5. Describe Free radical mechanism in polythene.

CHEMISTRY IN EVERY DAY LIFE

Drug: Chemicals of low molecular mass which interact with macromolecular targets and produce a biological response. Ex. Paracetamol.

Medicine: Drugs which cure diseases i.e. which cause therapeutic biological response . Ex. Equanil.

Antacid: Chemical substances which remove the excess acid and raise the pH to appropriate level in stomach . Ex. Mg(OH)2.

Antihistamines/ antiallergic drugs: are the drugs used to control the allergy effects produced by histamines. Ex. Chloropheniramine.

Tranquilizers/psychotherapeutic drugs: Chemical substances used for treatment of stress, mild and severe mental diseases. These relieve anxiety, irritability or excitement by inducing a sense of well being. Ex. Equanil.

Analgesics: Chemical substances used to relieve pain without causing impairment of consciousness, mental confusion, incoordination or paralysis or some other disturbances of nervous system. Ex. Aspirin.

Antidepressant drugs: Chemical substances used for mood changing or to counteract the effect of depression. Ex. Iproniazid.

Pathogen: Any organism that causes disease. Ex. Bacteria, viruses

Allosteric site: The site of an enzyme other than active site to which a drug binds.

Receptors: Proteins that are crucial to body’s communication process which are mainly embedded in the cell membrane with their active site projecting outside.

Antagonists: Drugs that inhibit natural function of receptors by binding to their active sites.

Antipyretic: A chemical substance used to lower body temperature when there is fever . Ex. Paracetamol.

QUESTION BANK

1. What causes depression? What drug do you suggest for counteracting depression? How does it decrease depression?

2. What is a hypnotic? Give an example.Ans: Sleep producing chemical substances. Ex. Luminal

3. What are barbiturates? Give an example.Ans: Tranquilzers derived from barbituric acid. Ex. Luminal

4. Describe two types of analgesics with an example for each. Ans: Non-narcotic/non–addictive analgesics: Analgesics which relieve pain without causing unconsciousness or without inducing sleep. Ex. Aspirin.

Narcotic drugs: Analgesics which relieve severe pain by causing unconsciousness or inducing sleep. Ex. Heroin.

5. Name an important antiseptic added to toilet soap.Bithional.

6. Give a point of difference between a bactericidal and bacteriostatic with an example each.

Bactericidal: antibiotics which kill organisms in the body Ex. Penicillin

Bacteriostatic: antibiotics which inhibit the growth of microorganisms in the body. Ex.Tetracycline.

7. Antibiotic therapy is nothing but “setting one thief against another”. Explain. Chemical substances produced by micro organisms (bacteria, fungi and mould) that inhibit the growth or even destroy micro organisms are called antibiotics.

Ex. Penicillin

Antibiotics themselves are product of microbial growth and hence it is nothing but setting one thief against another.

8. Describe broad spectrum antibiotics and narrow spectrum antibiotics with an example each.

Broad spectrum antibiotics: Antibiotics which are effective against several different microorganisms and thus capable of curing several infections .

Ex. Tetracycline.

Narrow spectrum antibiotics: Antibiotics which are effective only against limited number of types of harmful microorganisms. Ex. Penicillin.

9. Name two antioxidants used to preserve wine and beer.SO2 and sulphites.

10.Name the first artificial sweetener. For which disease is it a life saver? Saccharin is the first artificial sweetener. It is a life saver for diabetes.

11.Some chemicals which do not have any nutritive value of food are added to food stuff. What are those? Are they harmful to us? If so how?

Those are artificial sweeteners. Yes they are harmful to us. They do not have any nutritive value, but our digestive system (stomach) works on it and hence if taken in large quantity it weakens digestive system.

12.Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor . Why?

Most of the drugs taken in higher doses than recommended may cause harmfull effect and act as poison. Therefore a doctor should always be consulted before taking medicine.

13.With refernce to which classification has the statement “ rantidine is an antacid” has been give. ?

This classification refers to the classification according to pharmacological effect of the drug because any drug which will be used to counteract the effct of excess of acid in stomach will be called antacid.

14.Mention one use each of following drugs(i) Rantidine (ii) Paracetamol (iii) Tincture of iodine

Ans: Rantidine : used as an antihistamine

Paracetamol : used as an antipyretic.

Tincture of iodine : used as an antiseptic

15.What are Detergents? And name the types of detergents with example. And how detergents are advatnge over soaps?

Sodium or Potassium salts of sulphonic acids obtained from petroleum products or ammonium salts with long hydrocarbon chain used for cleaning are called detergents Ex. Sodiumalkylbenzene sulphonate.

Detergents are advantageous to soaps . They work with hard and soft water.

Three types of detergents are (i) anionic detergents (ii) cationic detergents (iii) non-ionic detergents.

16.Whenever a doctor prescribes a medicine of aspirin content directs patient not to take in empty stomach. Why?

Aspirin contains a carboxylic acid group and hence when it interacts as a drug it causes acidity and higher acidity caused by it some times may lead to ulcers also.

17.What is an antioxidant? How does it preserve food? Give an example. The food additives used to prevent or retard oxidation of food items

and thereby help in food preservation.

These act as sacrificial materials i.e. these are more reactive towards than are the materials that they protect. They also reduce rate of involvement of free radicals in the aging process. In some cases they are least reactive towards oxygen and hence protect the material with which they are put. Ex. Dintrogen in chips packets. Ex. BHT(Butylated Hydroxy Toluene)

18.While antacids and antiallergic drugs interfere with the function of histamines, why do these not interfere with the function of each other?

Antiallergic and antacid drugs do not work on the same receptor at a time i.e. they work on different receptors.

19.Low level of noradrenaline is the cause of depression. What type of drugs is needed to cure this problem? Name two drugs.

Antideppressant drugs are needed to cure this problem.

E.g. (i) Iproniazid (ii) Phenelzine

20.Name a substance which can be used as an antiseptic as well as disinfectant.Phenol solution.(0.2% Phenol solution is used as an antiseptic and 1% Phenol solution is used as a disinfectant)

21.Give examples for the following.(i) Cationic detergents (ii) anionic detergents (iii) non- ionic detergents

Ans: Cationic detergents Cetyltrimethylammonium bromide.

anionic detergents Sodiumlaurylsulphate

non- ionic detergents Polyethyleneglycol stearate

VALUE BASED QUESTIONS

1. Manu went with his father to a Shopkeeper who showed them two types of batteries, one with lead plates and the other with cadmium plates. The battery with cadmium plates was more expensive than thelead battery. Manu’s father wanted to purchase lead battery as it was cheaper.

a. As a student of chemistry, why would you suggest to Manu’s father to buy the expensive cadmium plate battery. Give two reasons. (2)

b. What are the values associated with the above decision? (1)

Ans: a. Cadmium plate battery though expensive is not as strong a pollutant as lead. Lead salts being insoluble in water if ingested into our system cannot be excreted out. Whereas Cd and Ni salts are water soluble, therefore get excreted and hence do not get biomagnified inthe body. b. Keeping the environment safe from pollution due to lead.2. Kalavati wanted to give her baby a medicine for fever. She added boiled and cooled water as per the instruction, to the contents of the bottle, upto the mark. She shook the bottle. Then gave a spoonful of the medicine to the baby. As a student of chemistry answer the following questions: a. Why did she shake up the contents? What is the process called?(2) b. What is the value associated with selling medicine in this form?(1)

Ans:

a. She shook the contents of the bottle to bring the contents into the form of a sol. Absorption of medicine are easier in the colloidal form. This process of agitating a precipitate into colloidal sol is called as peptization.b. When the medicine is sold in anhydrous form it has a higher shelf life and thus can be stored for a longer time. This is a way of being thrifty by not wasting the available resources.

3. Kartick went to a sugar producing factory. He noticed an alcohol producing unit associated with it. Generally alcohol is prepared industrially in places where sugar is extracted from sugarcane. a. As a student of chemistry can you tell why? (2) b. What is the value derived from this fact? (1)

Ans:a. The molasses that are bi-products of sugar industry can be used in making alcohol.b. Recycling of industrial waste keeps the environment clean.

4. Ashraf is 50 years old and has diabetes. He uses saccharine as sweetening agent in tea and coffee and sugar free in sweets. Lakshmi too is diabetic. She controls her sugar level in diet by using less sugar and by exercising. a. Who is able to handle diabetes more efficiently and why? (1)b. What value do you derive from this? (1)c. What are the harmful effects of artificial sweeteners?(1)

Ans:

a. Lakshmi is able to handle diabetes better, because exercises activate the pancreases to produce insulin. Exercise keeps one fit and fine.b. It is necessary to lead a disciplined life.c. Researches have shown that the artificial sweeteners have harmful effect on the body because they are not excreted easily.5. An innovative washer woman while washing a copper miner’s clothes found that sand and similar dirt particle fell to the bottom, while the ore particles stuck to the soapsuds and came to the top. The washer woman discussed thismatter with a client who was a chemist.a. What is the reason for this observation? (2)b. What value do you get from this episode? (1)

Ans:

a. The miner’s clothes had particles of CuS / Cu2S on it. This adhered to the froth and came up.b. Keen observation can lead to great discoveries.

6. Manoj went to a paper industry. The manager, paper industry H2O2 insisted that H2O2 be used for bleaching instead of Chlorine in bleaching. Manoj had learnt that Cl2 is also a bleaching agenta. Then why is H2O2 used instead of Cl2 (2)a. What value do you derive from this? (1)Ans:a) H2O2 after bleaching, the product formed is water. While whenusing chlorine the byproduct is HCl.b) Seeing the products of the reaction we should select thereagents, so that there is minimum pollution.7. Rahul’s father recovered from a massive heart attack. As follow up doctorgave him Aspirin.a. Why? After some time he started to develop some stomach ache. Then his wife gave him some butter milk (1)b. Why? (1)c. What value do you get from this? (1)

Ans:

a) Aspirin is a blood thinner. Hence for a patient suffering from heart attack is given Aspirin. Aspirin is acetylated salicylic acid. In thepresence of acid in the stomach it undergoes hydrolysis to form two acids salicylic acid and acetic acid. This increases the acidity in thestomach. (1)b) To neutralize this acidity butter milk is also given.c) Being dedicated and faithful.

8. Manu and his father went to a shop to purchase a battery for their inverter. Shopkeeper showed them two types of batteries, one with lead plates and the other with cadmium plates. The battery with cadmium plates was more expensive than the lead battery. They decided to purchase lead battery as it was cheaper. After reading the above passage, answer the following questions: a) As a student of chemistry, why would you suggest to Manu and his father to buy the expensive cadmium plate battery. Give two reasons. b) Which two values will you be promoting through your suggestions?

9. 10. Almelu did not like the costlier brand of dish washer because she was not vsatisfied by using less quantity of the dishwasher, so she bought the cheaperbrand of dish washer and used large amounts of it. a. As a student of chemistry what would you advise Almelu to use? (1)b. Why? Explain. (1)c. What value did you impart to Almelu? (1)

Ans:a. I would advise Almelu to use smaller quantities of the costlier detergent.b. Almelo was pouring detergents into the drain. These detergents are not biodegradable. Branched detergents are highly nondegradablebecause the microbes cannot attack it. Straight chain detergents are being prepared these days to reduce the pollutionproblem. The costlier detergent contains straight chainhydrocarbon.c. The value imparted was to use environmentally friendlysubstances.

SOLID STATEQ.(1) In a sample of salt, Mohit detected some MgCl2. Based on that he predicted cation vacancy imperfection but Rohit said metal excess defect. Whom would you support and why? Write at least two values each Mohit and Rohit do possess. Q.(2) Rajesh visited Bheraghat (Jabalpur, M.P.) last summer vacation. There he observed a definite cleavage pattern on marble rocks. He started of thinking as to which class of solids does marbles belong to? Now name the class and write down the general properties of this class of solids? In your opinion what value is reflected in this case? Q.(3) Roja and Ramya were studying about the properties of solids. The solid which was under investigation showing properties like insulator in solid as well as in molten state, very hard, very high melting point. Roja thought it was crystalline solid but Ramya said it would be covalent solid. Whom do you favour and why? What value is indicated by this? Q.(4) For practical purposes Ravish wanted to make a permanent magnets from one magnet he has. He has some metals stripes of Cu, Zn, Co and Nickel. Which metal do you think he may choose and why? Write the value associated with. Q.(5) Shailesh being student of Chemistry and Physics well known about components of digital camera, computers, Television sets etc. An important component common to all of these gadgets is a material which is a combination of group 13 and group 15 elements with group 14 elements. Name this component and also the class of solid. Shailesh wrote all the answers very correctly in the examination. What moral values does shailesh possess write any two?

UNIT 2SOLUTION

Q.(1) A container is divided by a semipermeable membrane. In one compartment 1 molal glucose solution is taken and in another compartment 1 molal NaCl solution is taken. Shruti expect the level of 1 molal NaCl solution will increase but Prema says level of 1 molal glucose solution will increase. Whom do you support and why? Which value is associated with it?

Q.(2) Persons need replacement of body fluid or nutrients which can not be fed orally are administrated solutions by intravenous infusion. To prevent crenation or hemolysis of red blood cells the I.V. solution must be isotonic with intracellular fluids of the cells. Why is it so? What values (any two) are found to be defeated if a Person suggested to feed above item orally?

Q.(3) The formula of common salt is NaCl and its calculated molar mass is 58.5. When the molar mass of common salt is determined by measuring the osmotic

pressure of its 0.5 molal aq. solution it comes to be different than 58.5. Bhargavi thought it because of calculation mistake but Bharat says it there is no calculation mistake and he can correct the answer. How can Bharat correct the answer? What moral value do you find? . (4) John and Praisy were on their way to Laddakh in their winter break holidays by their own car. On the way they stayed for refreshment, mean while John was filling water in the radiator of the car. On seeing Praisy suggested to mix up some ethyl alcohol she had with her. John surprisingly asked why? Answer the query of John and also write the value associated with it.Q.(5) Akshay along with his father were going to Srinagar by their own car. Their journey was interrupted by road block due to heavy snow fall. Get got down and observing the situation, mean while his father said it is melting property of water responsible for road block and it’s melting property of water which will help to clear the raod. Hearing this Akshay looked his father with question mark in his mind. Explain the statement of his father and write one value associated with it.

UNIT 3ELECTROCHEMISTRY

Q.(1) Manu and his father went to a shop to purchase a battery for their inverter. Shopkeeper showed them two types of batteries, one with lead plates and another with cadmium plates. The battery with cadmium plates was more expensive than lead battery. Manu’s father wanted to purchase lead battery as it was cheaper. After reading the above passage, answer the following questions: a) As a student of chemistry, why would you suggest to manu’s father to

buy the expensive cadmium plate battery. Give two reasons b) What are the values associated with the above decision?

Q.(2) In newspaper Yogesh read that iron pipes of sewerage system of the town will be protected by cathodic protection method (sacrificial method). His friend Juned said in this method something will replace iron to protect it from rusting. Explain the above method for Yogesh and also write the value associated with. Q.(3) Sanjay went to buy dry cell for wall clock. His father instructed him to check date of manufacturing before buying the cells. Sanjay was wondering if buying new cells with seal packed what to do with manufacturing date. If Sanjay is write? Give your opinion and the value associated with. Q.(4) Uday was performing an experiment to show deposition of sodium metal. He need a compound of sodium as an electrolyte. He selected cheapest and easily available common salt for that. He was suggested by his teacher that not to use solid NaCl or its aq. solution for that. What then you suggest for him? Explain your answer and write the value you get from this.

Q.(5) Ramesh and Suresh went to buy lead storage battery for their vehicle. They went to a shop. Ramesh was interested to read the details like internal volume, no of plates of Pb and PbO2 etc. What do these information’s have to do with efficiency of the battery? Explain your answer and write one value associated with.

UNIT 4CHEMICAL KINETICS

Q.(1) Akash and his friends were going on picnic at Shillong peak. Along with other requirement he packed pressure cooker also for cooking food. Anish objected, saying at high altitude it is of no use as atmospheric pressure would be less and cooking of food will take much time. He was insisting to have readymade food items. Akash said pressure cooker will work effectively there also. How far you are agree with Akash? Write the value associated with each of Akash and Anish. Q.(2) Anusha’s mother was trying hard to burn a big lump of coal faster for cooking food. Anusha wanted to help her mother in this regard. But she was not getting any idea. After studying chemical kinetics chapter she got the idea. What idea came in her mind? Explain and also write the value associated. Q.(3) The half life for radioactive decay of 14C is 5730 years. An archaeological artifact contained wood had only 80% of the 14C found in a living tree. Estimate age of the sample. Also write two more importance of radio-isotopes in our every day life. Mention the value associated with the use of isotopes in our daily life. Q.(4) In the lesson chemical kinetics Robin studied about need of activation energy and energy barrier for every chemical reaction. What is the significance of energy barriers in chemical reactions in our life? How does this help to stop destruction of life on the earth surface? What value is conveyed by this situation? Q.(5) Anuj and Sukhbeer visited a car manufacturing company. They saw different sections and found that different parts of cars are made in very large number but the rate of manufacturing of car is very less than their number because each part is not made in equal number. Does this concept help us to understand the concept of order of a reaction? Explain your answer reflecting a suitable value. UNIT 5 SURFACE CHEMISTRY Q.(1) Wasim observed that the barber applied alum on the face of his customer. He also observed her mother dissolving alum to purify muddy water. How do these two observation relate to the concept of colloidal solution and its properties. Write the value associated with. Q.(2) “Shake well before use” is a very common precaution written on syrup take as medicine. Why is it so? If Raghvendra didn’t followed the instruction ever after reading the precaution, what value do you observe defeated in the this case. Q.(3) People who prepare for fire fighting use mask preferably in place of covering their face completely. Why do they so? Write the value associated with.

Q.(4) There was crowded place where fire broke out abruptly, a person Mr. David who was standing there advised to other person to put Gas Mask in, so that poisonous gases could be adsorbed. Does it adsorb gases like N2 and O2 equally? Explain your answer citing a suitable value possessed by Mr. David. Q.(5) Milk and Cold cream are the examples of colloidal system. Which type of colloid are they? Can we dilute them by adding water? Mrs.Nainy being student of chemistry brought the samples of Milk and Cold cream and tried to dilute them and performed experiment before writing answer. Why did she do

so is there any value do you find here. UNIT 6

GENERAL PRINCIPAL & PROCESSES OF ISOLATION OF ELEMENTS

Q.(1) Rohit found that Steel is the most useful variety of iron. A High Quality steel can be made by adding desired proportion of metals such as Cr, Ni, Mn, W, V etc. One of the varieties of steel is used in making of electromagnets and cables for oceans. Another variety of steel is used for making pipes for carrying acids. While using steel for the purpose of steel pipes to carry Acid and prepare Electromagnets which type steel should Rohit get and doing so what are the main points Rohit must think over and what values do you find justifies the selection in this case. Q.(2) The prime ore of some important metals like Cu, Fe, Zn, Al etc. are CuFeS2, Fe2O3, ZnS and Al2O3 respectively. Can these metal be extracted form any other ore? If yes explain your answer and write value associated with it. Q.(3) Iron exists in three commercial varieties. They vary in their carbon contents. One variety is used for casting articles like pipes, gutter and raiway sleepers, second variety is used for making magnets, nails, bolts, chain etc. and the third variety is most useful and its quality can be improved by adding metals like Ni, W, Mn, Cr etc. based on above uses, write the important properties of these varieties as per utilization of iron for different purpose. What value (any two) are essential while selecting the best fit iron in above case? Q.(4) Gangue is an undesirable part of ore. Based on the nature of the ore different concentration methods are used. Silica is a common type of impurity. If you go through metallurgy of different metals you will find that silica is always not undesirable but useful also. Explain the above fact with suitable example and also mention the value behind it if Rohit finds that even a few undesirable material could be proven the most useful. What value does this thinking of Rohit possess? Q.(5) In the process of froth floatation collectors (pine oil) and depressants (NaCN) are used for different purposes. Explain how the use of depressants technically and economically beneficial for us. Write any one value related with it.

UNIT 7

‘p’ BOLCK ELEMENTSQ.1 Sumit in the market,while shopping, heard two people talking about purchasing good amount of Nitric Acid from market to take to the local dense forest, and got an alert & immediately followed and observed all of the activities, very carefully till they were in the market. In your opinion why did Sumit get an alert , why did he suspect and follow them? Write any two values Sumit possesses, the what probable doubt had Sumit in his mind . Q.2 Rakesh and Kamal were in discussing about the acidic strength of halogen acid. Rakesh said that HCl is stronger acid so it should be used for etching of glass in making of thermometers, burettes etc, HF should be used. Whom do you favour and why? Give the value which emerges from your answer. Q.3 Ankita and Sweta visited an art exhibition. They appreciated the paintings displayed there. Sweta observe black spots on white paint in a nice painting. She tried to remove them using cloth but she could not, then she recalled what she learned in ‘p’ block elemets. She thought that pale blue colour gas found in upper atmosphere can remove these stains. You are expected to appreciate Sweta and explain the process to remove the black spots from white painting. What value is reflected from this? Q4. In their final practical exam Ajay and Ankit both required ammonia solution they asked the lab attendant. He gave them sealed bottle of ammonia solution. Ankit was about to open the seal instantly but Ajay stopped him to do so. Who is right? What ajay wanted to do? Explain it reflecting the value we may get from the answer. Q5. While working in chemistry laboratory, Accidently, Kavita’s upper arm came in contact of conc. H2SO4, she cried in pain. Sonali rushed with NaOH solution to pour on her hand but She preferred to keep the affected area under running water. Whom you will favour and why? What value you learn from this?

UNIT 8‘d’ AND ‘f’ BLOCK ELEMENTS

Q.1 An orange colour compound of chromium is very useful to detect presence of alcohol in air breathed out. The colour of the compound changes in the presence of traces of alcohol. Write the colour which develops. And also explain the process with suitable chemical equation. Q.2 A mixture of K2Cr2O7 crystals and conc. H2SO4 was added in four test tube containing KI, H2S, CH3CHO and NaBr respectively. It was followed by strong heating. Identify in which tubes, the following changes will be observed. Support your answer with suitable chemical equations. Write associated values. Q.3 Gas ‘A and gas ‘B’ both turns K2Cr2O7, in acidic medium, to green. Gas ‘A’ also turns lead acetate paper black.as ‘A’ is passed into gas ‘B’ When in an aq. solution, yellowish white turbidity appears. Identify gas ‘A’ and gas ‘B’ and explain the reaction. Write associated values. Q.4 In our every day life we use a good number of magnetic and electronic

devices. They are made of different chemical compounds. In your opinion which type of elements/compounds are used to make such devices and why? Q.5 Lanthanoids were known as ‘rare earth elements’. Now a day this name is irrelevant. These metals are seldom used in pure state. Give reason for above fact. Cerium is used in glare reducing spectacles. What property of cerium is reflected from this? Also write the value associated with it.

UNIT 9COORDINATION COMPOUNDS

Q1. There are some complex compounds which are known as chelate complexes. There extra stability is quite useful in medical science to cure cases of metal poisoning. Explain and identify the value being expected. Q2.As per the Medical Report, Mohit found to be affected with the dearth of Hemoglobin in his Blood, Mohit being the student of Chemistry was known to the fact that there are many complex compounds found in nature with specific metal atom/ion. Such a few complexes are Chlorophyll, Blood pigment, Vitamin B12 etc. Will this knowledge help, any way, to recover from the dearth of Hemoglobin to Mohit, what sort of value you find associated with, in this particular instance? Q3.Anurag has selected a science project to estimate the hardness of water samples collected from different places. He went to water testing laboratory of his city. There he came to know bout some specific terms like ‘titration’ and ‘EDTA’. Explain these terms. How far his project may be useful for the local habitants? Q4. When Raghav came to that his sister is suffering form deficiency of iron he innocently brought some iron powder and asks his sister to eat? Is he right? Justify your answer. In what ways iron ion is absorbed in our body? Q5. In four different beakers solutions of [ Ni(H2O)6]2+, [ Fe(H2O)6]2+,[ Ni(CN)4]2+ and [ Ni(CN)6]-4 are kept and white light from same source is allowed to fall on them. Would they appear similar in colour? Do [Ni(H2O)6]2+and [ Ni(CN)4]2+ only appear same? Explain your answer. Write the value associated with it.

UNIT 10HALOALKANES AND HALOARENES

Q1.Antiseptic are antimicrobial applied on living tissues. Tincture of iodine is a common example. Rewati had read that Iodoform also possesses an appreciable antiseptic property. How will you explain this? Identify the value associated with it. Q2.Some halogen containing compounds are useful in daily life. Some compounds of this class responsible for exposure of fauna and flora to more and more UV light which causes destruction to a great extent. Name the class of the compound. In your opinion, what steps to be taken to minimize harmful effects of these compounds? Q.4 Chlorine gas was used earlier to bleach paper. Tetrafluoroethane was used as

dry cleaning solvent. Now a day their used has been minimized. How would you justify this? What value is plays important role in this case? Q5. A sweet smelling organic compound ‘A’ is slowly oxidized by air in the presence of light to a highly poisonous gas. On warming with silver powder, it forms a gaseous substance ‘B’ which is also formed by the action of calcium carbide on water. Gas ‘B’ is also used to make metallic objects reusable. Identify ‘A’ and ‘B’. Write the values associated to use the reagent as minimum possible to conduct this chemical reaction. UNIT 11 ALCOHOLS , PHENOLS AND ETHERS Q1. Consumption of alcoholic beverage is harmful for our health. Still people consume them. Alcohol supplied for industries, hospitals and for laboratories are also consumed by addict persons. How can this practice be stopped or minimized? Suggest some ways. Q2.We often read in news paper that many people die or hospitalized due to consumption of poisonous alcoholic beverage. What actually is this? How can the effect of poisoning be neutralized /cured? Write the value associated with it Q3. Phenol is very important compound. Name its natural source along with the process followed to prepare it commercially. It has disinfectant property. Can this be used as antiseptic? Explain your answer and indicate the value associated with it. Q4. Power alcohol and Natalite have a common use. Write their chemical composition. Does this information help you any way? Explain and write the value associated with it. Q5. An organic compound with molecular formula C6H6O is widely used in preparation of plastics and related materials. It is also used to prepare antiseptic/disinfectants. It is basically extracted from coal tar. Its vapors are corrosive to the eyes, the skin, and the respiratory tract. Write its structural formula and reaction with benzene diazonium chloride. What is the importance of this reaction? Write any one value associated with it?

UNIT 12ALDEHYDES KETONES AND CARBOXYLIC ACIDS

Q1. An organic aromatic carboxylic acid is used as food preservative. How can this be differentiated form another aromatic compound used as disinfectant at home? How can this test help you in your everyday life? Raj kept this disinfectant opened home despite knowing the fact that he has his younger brother of age 5years in home, what values does Raj lack here? Q2. An aromatic compound gives brisk effervescence on reacting with baking soda. It is also used as an analgesic, has molecular formula C9H8O4. It is also supposed to give positive iodoform test. Identify the compound and write its structural formula. Sunny being chemistry student used this Analgesic when suffered with pain. Do you justify the act of Sunny what good and against the moral values you find assess with the act on part of Sunny in this case? Q3. Two chemical compounds as colourless liquids available in our home. One is a component of nail polish remover and another is a component of vinegar.

Write their structural formula and also suggest how will you identify them in case of no labeling. Does this help to imply practical use of your learning in school? Q4. Ajay a students of class XII was carrying out Nucleophilic addition reaction of ammonia derivatives with Acetdehyde under the direction of his teacher. What moral values are essential for a students to obey his teachr. His teacher asked to maintain pH near about 3.5 to get the desired products. What would happen if pH range increases or decreases than that of 3.5? Q5. Ranjeet despite frequent teaching in the class room by the teacher forgot that an aromatic compound with molecular formula C7H6O is very commonly known as oil of bitter almonds. It is used to manufacture malachite green dye. It does not give iodoform test but give positive silver nitrate test. But Rajesh well understood the same instantly during the first attempt in the class Rajesh even learnt that how does it react with NaHSO3? Now identify the compound and write the chemical reaction involved in this context and also write one each moral value associated with:

(i) Ranjeet (ii) Rajesh

UNIT 13ORGANIC COMPOUNDS CONTAINING NITROGEN

Q1. A solution contains 1 g mol each of p-toluenediazonium chloride and p- nitrophinyldiazonium chloride. To this 1 g mol of alkaline solution of phenol is added. Predict the major product. Explain your answer. While performing this chemical reaction Rohit utilized resources as minimum as possible. Why did he do so, justify your answer in tow lines, writing down any one basic value associated. Q2. Dyes are an important class of organic compounds. A class of organic compounds containing nitrogen are used as a starting material for the manufacture of azo dyes. It reacts with nitrous acid to form diazonium salt, which can undergo coupling reaction to form azo compound. Low molecular weight compounds are toxic, and some are easily absorbed through the skin. Many higher molecular weight compounds are, biologically, highly active. Identify the the compounds and write the value associated with it. Q3. Rahgu and Raghav took organic compound synthesis as their chemistry project. They prepare benzenediazonium chloride. Raghu wanted to store it and synthesis dye from it next day but Raghav said we can store its and need to use immediately after preparation. Whom do you think is right and why? Write the value associated with it. Q4. Creatinine is a break-down product of creatine phosphate in muscle, and is usually produced at a fairly constant rate by the body. It belong to amines. Its presence can be tested in our blood and urine. Does this test help us to be aware of our health? Explain.

Write the value associated with it. Q5. A primary amine RNH2 can be reacted with CH3-X to get secondary amine R- NHCH3 but the only disadvantage is that 3 ⁰ amine and quarternary ammonium salts are also obtained as side products. Can you suggest a method where RNH2 forms only 2 ⁰ amine? Does knowing special method of preparation of a compound someone has any of 02 values think and write.

UNIT 14BIOMOLECULES

Q1. Starch is one of the most important food material for all of us. But Rajesh was advised by his Physical not to avoid it, as he was suffered with Diabetes Mellitus, but one day while his trip to forest he strayed from the rout and keep on going to find the right path but in vain, finally he completely tired off and had appetite, on searching food he found nothing to eat except a few sweet potatoes he carried in a small bag with him by mistake from his home. Now answer whether to eat these sweet potatoes, justify your answer supporting the any 2 values associated with. Q2. Shudha when taught by the teacher in the class room that vitamin is B12

contains a metal ion, now after the class when student will go to home they will find the tell the name of the Metal ion tomorrow, but Sudha immediate after during the interval time went to the Computer lab and requested the teacher to surf the net and traced the name of the metal ion, now write down the name of the metal ion state at least 2 moral virtues possessed by Sudha Q3. DNA and RNA are the two important nucleic acids present in our body. They are very important in our life. Study of DNA is very useful in investigation of criminal cases and finding. In what ways study of DNA is useful to us? Explain. Write the value associated with this. Q4. In our diet we include good amount of plant products rich in cellulose but such diet is not nourishing. Explain the value you have learned about biomolecules. Q5. On medical checkup Snajit got to know that his vision is getting poor slowly. As soon his younger brother came to know the fact he rushed to the market to bring some mangoes, carrot & papaya, Why, what values do you find his younger brother possesses. UNIT 15 POLYMERS Q1. Synthetic polymers can not be degraded in he nature for a long time. How can biodegradable synthetic polymers be made? Why do we need Synthetic polymers at all? What value you assign to replace Synthetic polymers by biodegradable Synthetic polymers. What values are associated when a person selects biodegradable polymers. Q2. Low density polythene and high density polythene, both are polymers of ethene but there is marked differences in their properties. Explain keeping certain value in mind. Q3. Ramesh went along with his father to buy new tyre for their vehicle. He found

that his father is more interested to know about the chemical composition of the rubber used, presence of cross linkages in the used rubber. How far you agree with his father’s questions? Does it have any value? Q4. The electrical switches used in our home are made of bakelite? Why? What are values there in selecting this polymer in this case? Q5. What is the primary feature necessary for a molecule to make it useful in a condensation and addition polymerization? Polymerization is very useful synthesis. Write its advantages and disadvantages of it. What value is associated with it?

UNIT 16CHEMISTRY IN EVERYDAY LIFE

Q1. Synthetic detergents have advantage over usual soaps as far as cleansing power is concerned. But the use synthetic detergents over a long period should be avoided. What value you conveyed by this Question? How can we minimize the concerned problems? Q2. Ravi brought an artificial sweetener to prepare tea for his diabetic father, on seeing Aspertame in the level of the packet his father advised not to use that artificial sweetener. What reason do you find behind his advice? What value is indicated by this? Q3. Narcotic drugs are banned by the govt. but it is grown and produced under supervision of govt. Why is it so? Do you find any value behind this? Write about. Q4. Sugar is the main source of energy as it produces energy on metabolic decomposition. But this days low calorific drinks are more popular, why? Which value is associated with it? Q5. Aspirin is an analgesic but it is also advised to heart patients. Does it work as analgesic for them? Support your answer with suitable reason and also indicate value associated with it.

h Web viewIf it is a face centred cubic unit cell, only FOUR (4) atoms will be present

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Transcript of h Web viewIf it is a face centred cubic unit cell, only FOUR (4) atoms will be present