UNIT 1 BASIC CONCEPTS OF CHEMISTRY - · PDF fileUNIT 1 BASIC CONCEPTS OF CHEMISTRY ......

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LEVEL WISE QUESTIONS CLASS XI

UNIT 1 BASIC CONCEPTS OF CHEMISTRY

LEVEL- 1

1. State the number of significant figures in (i) 0.436 and (ii) 802.42

Ans. (i) 0.436 has three significant figures

(ii) 802.42 has five significant figures

2. What do you mean by Mole fraction?

Ans. Mole fraction is the ratio of number of moles of one component to the total number of

moles (solute & solvents) present in the solution. It is expressed as ‘x’.

3. Calculate the molecular mass of H2SO4.

Ans. Molecular Mass of H2SO4 = 2 × 1 + 32 + 4 × 16 = 2 + 32 + 64 = 98u

4. What is the SI unit of mass? How is it defined?

Ans. The S.I. unit of mass is Kilogram.

Mass of a substance is the amount of matter present in it.

5. What is limiting reagent?

Ans. The reactant which gets consumed first or limits the amount of product formed is known

as limiting reagent.

LEVEL- 2

1. What is the value of Avogadro number?

Ans. 6.022 × 1023

2. What is the law called which deals with the ratios of the volumes of the gaseous reactants

and products of the volumes of the gaseous reactants and products?

Ans. Gay Lussac’s law of gaseous volumes.

3. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Ans. Molar mass of Na2CO3 = 2 × 23 + 12 × 3 × 16 = 106 g/mol.

0.50 mol Na2CO3 means 0.50 × 106 = 53 g.

0.50 mol Na2CO3 means 0.50 mol i.e. 53 g of Na2CO3 are present in 1 L of the solution.


4. What is the percentage of carbon, hydrogen and oxygen in ethanol?

Ans. Molecular formula of ethanol is C2H5OH

Molar mass of ethanol is:

= 2 × 12 + 5 × 1 + 16 + 1

= 24 + 5 + 17 = 46 g

Mass percent of carbon = 10046

24 = 52.17%

Mass percent of hydrogen = %1.1310046

6

Mass percent of oxygen = 10046

16 = 34.7%

5. Classify the following as pure substances or mixture.

(a) ethyl alcohol (b) oxygen (c) carbon (d) distilled water(e) blood (f) steel

Ans. Pure substance – ethyl alcohol, oxygen, carbon, distilled water mixture – blood, steel

LEVEL – 3

1. What is the difference between 160 cm and 160.0 cm?

Ans. 160 has three significant figures while 160.0 has four significant figures. Hence160.0

represents greater accuracy.

2. In the combustion of methane in air, what is the limiting reagent & why?

Ans. Methane is the limiting reagent because the other reactant is oxygen of air which is always

present in excess. Thus the amounts of carbon dioxide & water formed will depend upon

the amount of methane burnt.

3. Which aqueous solution has higher concentration, 1 molar or 1 molal solution of the same

solute? Give reason.

Ans. 1 molar aqueous solution has higher concentration than 1 molal solution.

A molar solution contains one mole of solute in one litre of solution while one molal

solution contains one mole of solute in 1000 g of solvent. If density of water is 1, then one

mole of solute is present in 1000 mL of water in 1 molal solution while one mole of solute

is present in less than 1000mL of water in 1 molar solution (1000 mL sol = amount of solute

+ amount of solvent). Thus 1 molar solution is more concentrated.

4. Calculate the molarity of water if its density is 1000 kg/m3.

Ans. Molarity of water means the number of moles of water in 1 litre of water

1L of water = 1000 cm3 = 1000 g ( 1000 kg/m3 = 1g/cm3)

1000 g of water = 18

1000

= 55.56 moles

Molarity = 55.56 M


5. A compound made up of two elements A and B has = 70%, B = 30%. Their relative number

of moles in the compound is 1.25 and 1.88. Calculate

a. Atomic masses of the elements A and B

b. Molecular formula of the compound, if its molecular mass is found to be 160.

Ans. Relative no. of moles of an element = mass Atomic

element the of %

Or atomic mass = 5625.1

70

moles of no. lativeRe

element the of %

Atomic mass of B = 88.1

30 = 16

Calculations of Empirical formula

Element Relative no. of moles Simplest molar ratio Simplest whole no. molar

ratio

A 1.25 1.25/1.25 = 1 2

B 1.88 1.88/1.25 = 1.5 3

Empirical formula = A2B3

Calculation of molecular formula –

Empirical formula mass = 2 × 56 + 3 × 16 = 160

n = 1160

160

mass formula

mass

Empirical

molecular

Molecular formula = A2B3

UNIT 2 STRUCTURE OF ATOM

LEVEL – I

1. The atomic number of an element is 5 and mass number is 11. Find the number of

electrons, protons and neutrons present in an atom of it. How can this element be

represented?

Ans : Number of electrons (Z=5)=5

Number of protons (Z=5)=5

Number of neutrons (A=Z)=11-5=6

Since the atomic number is 5 , the element is boron and is represented as : 5B11

2. The Kinetic energy of an electron is 4.55 X 10-25 J. The mass of electron is 9.1X10-31 kg.

Calculate velocity, momentum and the wavelength of the electron.

Ans : Setp-1 Calculation of the velocity of electron.

Kinetic energy = ½ mv2 = 4.55 X 10-25 J = 4.55 X 10-25 kg m2 s-2


V2 = 2𝑋𝐾𝐸

𝑚 =

2 𝑋 (4.55 𝑋 10−25𝑘𝑔𝑚2𝑠−2

9.1 𝑋 10−31𝑘𝑔 𝑋 (103𝑚𝑠−1) =

= 0.728 X 10-6 m = 7.28 X 10-7 m

Ans: We know that,

Number of waves = 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑜𝑟𝑏𝑖𝑡

𝑊𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡 𝑕 =

2𝜋𝑟

𝜆

But 𝜆 = 𝑕

𝑚𝜐

Number of waves = 2𝜋𝑟 x 𝑚𝜐

𝑕 =

2𝜋

𝑕(𝑚𝜐r)

Thus angular momentum for Bohr’s Third orbit is given as:

𝑚𝜐𝑟 = 𝑛𝑕

2𝜋 =

3𝑕

2𝜋

Substituting the value of 𝑚𝜐𝑟 in expression (i)

Thus, number of waves = 2𝜋

𝑕x

3𝑕

2𝜋 = 3

Thus, the number of waves in Bohr’s third orbit = 3.

3. Which has higher energy, 3d or 4s? Explain?

Ans: 3d because n+l for 3d is 3+2=5, but for 4s it is 4+0=5

4. State and explain Hund’s rule of maximum multiplicity?

Ans : It states that pairing of e in the degenerate orbital is not possible until there is one e in

each orbital

5. Define (i) work function (ii) Threshold frequency.

Ans: (i) Work function is the minimum energy of light radiations required to remove an

electron from the metal surface.

(ii) Frequency related to work function is threshold frequency.

6. Half filled and full filled orbitals have extra stability. Why?

Ans: Due to exchange energy

7. What are isoelectronic series?

Ans : Species having same no of electrons.

LEVEL – 2

1. The uncertainty in the position of a moving bullet of mass 10g is 10-5m.Calculate the

uncertainty in the velocity.

Ans:Uncertainty Principle

x.m∆𝑣 = 𝑕/4𝜋 substitute the values and the answer is 5.27 x10-28ms-


2. Light of wavelength 5000 falls on a metal surface of work function 1.9 eV. Find (a) the

energy of photos (b) kinetic energy of photoelectrons?

Ans: Wavelength of light (𝜆) = 5000 = 5000 x 10-10 m = 5 x 10-7 m.

Work function (hv0) = 1.9 eV = 1.9 x 1.6 x 10-19 j.

Step I: Energy of photons E = hv = 𝑕𝑐

𝜆 =

6.626 𝑋 10−34 𝐽𝑠 𝑋 (3 𝑋 108𝑚𝑠−1)

(5 𝑋 10−7𝑚) = 3.97 x 10-19 J.

Step II: Kinetic energy of photoelectrons

KE (1/2 m𝜐2) = hv – hv0 = (3.97 x 10-19 J) – (1.9 x 1.6 x 10-19 J)

= (3.97 x 10-19J) – (3.04 x 10-19 J) = 9.3 x 10-20 J.

3. Which of the following are iso-electronic species?

Na+,, K+, Mg2+, Ca2+, S2-, Ar.

Ans: Na+ and Mg2+ are iso- electrons species (have 10 electrons) K+, Ca2+, S2- are iso-electronic

species (have 18 electrons).

4. (i) Write the electronic configuration of the following ions? (a) H- (b) Na+ (c) O2- (d) F-.

(ii) What are the atomic numbers of the elements whose outermost electronic

configurations are represented by:

(a) 3s1 (b) 2p3 and (c) 3d6 ?

(iii) Which atoms are indicated by the following configurations?

(a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.

Ans: (i) (a) 1s2 (b) 1s2 2s2 2p6 (c) 1s2 2s22p6. (d) 1s22s22p6.

(ii) (a) Na (Z=11) has outermost electronic Configuration = 3s1

(b) N (Z=7) has outermost electronic Configuration = 2p3

(c) Fe (Z=26) has outermost electronic Configuration = 3d6

(iii) (a) Li (b) P (c) Sc

5. Show that the circumference of Bohr orbit for the hydrogen atom is an integral multiple

of the de Broglie wavelength associated with the electron revolving around the orbit.

Ans: According to Bohr’s theory,

m𝜐r = 𝑛𝑕

2𝜋 (n = 1, 2, 3 ……. so on)

or 2𝜋r = 𝑛𝑕

m𝜐 or m𝜐 =

𝑛𝑕

2𝜋r ….(i)

According to de Broglie equation,

𝜆 = 𝑕

m𝜐 or m𝜐 =

𝑕

𝜆 ….(ii)

Comparing (i) and (ii),

𝑛𝑕

2𝜋=

𝑕

𝜆 or 2𝜋r = n𝜆

Thus, the circumference (2𝜋r) of Bohr orbit for hydrogen atom is an integral

multiple of the de Broglie wavelength.


6. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will

experience more effective nuclear charge from the nucleus?

Ans: Configuration of the two elements are:

Al (Z=13) : [Ne]103s23p1 ; Si (Z=14) : [Ne]103s23p2

The unpaired electrons in silicon (Si) will experience more effective nuclear

charge because the atomic number of the element Si is more than that of Al.

7. The electronic energy is negative. How will you account for that?

Ans: Electron just out of atom has zero energy. It enters in the atom due to attraction of

atom so it loses energy.

8. How many unpaired electrons are present in Pd (Z=46)?

Ans: Zero

9. Write electronic configuration of Ca(z=24) and Cu(z=22).

Ans: Ca= 1s2 2s2 2p6 3s2 3p6 4s2

Cu=1s2 2s2 2p6 3s2 3p6 3d10 4s1

LEVEL – 3

1. Show that Heisenberg’s uncertainty principle is of negligible significance for an object of

mass 10-6 kg (Given h/4𝜋 = 0.528 x 10-34 kg m2 s-1)?

Ans: According to uncertainty principle,

∆𝑥. 𝑚∆𝜐 = 𝑕

4𝜋 or ∆𝑥∆𝜐 =

𝑕

4𝜋𝑚 ;

𝑕

4𝜋 = 0.528 x 10-34 kg m2 s-1;m = 106 kg

∆𝑥 ∆𝜐 = (0.528 x 10-34 kg m2 s-1) x 1

(10−6𝑘𝑔) = 0.528 x 10-28 m2 s-1 .

Since this value comes out to be very small (negligible), this means that uncertainty

principle is also of negligible significance.

2. How many electrons in an atom may have the following quantum numbers?

(a)n=4 , ms=-1/2 (b) n=3, l=0

Ans: (a) 16 (b) 2

3. Write down all four quantum numbers for (i) 19th electron of 24 Cr (ii) 21st electron for

21Sc.

Ans: (i) 19th electron of 24 Cr.

Cr(Z=24): 1s2 2s2 2p6 3s2 3p6 4s1 3d5

The 19th electron is present in the 4s orbital. For the 4s electron

n = 4, l = 0, m = 0, s = +1

2 or -

1

2

(ii) 21stelectron for21 Sc.

Sc(Z=21) 1s2 2s2 2p6 3s2 3p6 4s2 3d-1

The 21st electron is present in the 3d orbital. For the 3d electron

n = 3, l = 2, m = -2, -1, 0, +1, +2, s+1

2 or -

1

2


4. Draw the shape of all d orbitals.

Ans:

5. Consider an element Z that has two naturally occurring isotopes with the following percent

abundances: the isotope with a mass number of 19.0 is 55.0% abundant; the isotope with a

mass number of 21.0 is 45.0% abundant. Calculate the average ATOMIC MASS for element Z.

Ans: Mass x Abundance

19.0 x 0.55 = 10.45 AMU (Note: 55.0% = 0.55)

21.0 x 0.45 = 9.45 AMU (Note: 45.0% = 0.45)

19. 90 AMU is the Atomic Mass for element Z

6. Complete the following table.

Symbol Atomic # Mass # Protons Neutrons Electrons Net Chg. Atom/Ion

32 40 32

78 120 +2

56 82 54

Iron - 55

I—1 74 54


UNIT – 3 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

LEVEL 1

1. What would be the IUPAC name and symbol for the element of atomic no.120?

Ans: Unbinilium (Ubn)

2. State Mendeleev’s periodic law.

Ans: Physical and chemical properties of elements are the periodic function of their atomic

masses.

3. How many groups are there in the p-block of periodic table?

Ans: 6

4. Explain the term electron gain enthalpy.

Ans: The enthalpy change taking place by attracting an electron towards its outermost shell of an

isolated gaseous atom.

5. What are lanthanides and actinides?

Ans:

(a) Lanthanides: Elements in which last electron enter in 4f sub-shell of antepenultimate shell.

(b) Actinides: Elements in which last electron enter in 5f sub-shell of antepenultimate shell.

6.What is the general electronic configuration of d-block elements?

Ans: 𝑛 − 1 𝑑1−10𝑛𝑠0−2

7. Arrange the following in increasing order of ionization enthalpy:-

B, C, N, O.

Ans: B < 𝐶 < 𝑂 < 𝑁

8. Arrange the elements in increasing order of their size:- F- ,Li + ,Na+,Cl-.

Ans: 𝐿𝑖+ < 𝑁𝑎+ < 𝐹− < 𝐶𝑙−

9. Which has the highest value of electron gain enthalpy fluorine or chlorine?

Ans: Chlorine

10. Name the first element having an incomplete 3d-level.

Ans: Na


LEVEL-2

1. What is diagonal relationship?

Ans: Due to similar charge/size ratio element of second period are diagonally related with

the third period element like Li is diagonally related to Mg.

2. How are Li and Mg related to each other in the periodic table? Write the name of another

pair having such a relationship?

Ans: Diagonal relationship. Be and Al

3. Write down the main characteristic of p block elements?

Ans: In these elements the last electron entering p sub-shell of last shell.

4. Noble gases have zero electron affinity. Why?

Ans: due to completely filled duplet or Octet.

5. Account for the low values of electron affinities of nitrogen and phosphorus?

Ans: Half filled p subshell in N and P.

6. What are f block elements? By which alternative name are they known and why?

Ans: Elements in which last electron enter in f sub-shell of antepenultimate shell. They are

called inner transition elements.

7. What are isoelectronic ions? Account for the decrease in size of following species oxide

ion >Fluoride ion>sodium ion>magnesium ion

Ans: The ions having same number of electrons are called isoelectronic ions. the above order

is due to increasing order of effective nuclear charge.

8. Why ionization of d block elements does remain constant?

Ans: Due to Screening/shielding effect.

LEVEL -3

1. What is modern periodic law? How does it in the removal of some of the anomalies of the

Mendeleev’s periodic table?

Ans: Modern Periodic law states that “the physical and chemical properties of the elements are periodic functions of their atomic numbers”. When atomic number is taken as fundamental property then abnormalities like position of isotope, anomalous pair itself removed.


2. Explain

a. Dobereiner triads

b. Newlands law of octaves.

Ans: Dobereiner triads: He make a group of three elements in such a way that the average of

atomic masses of 1st and 3rd elements is equal to the atomic mass of the middle one such as Li,

Na and K.

Newlands law of Octaves: When elements were arranged in such a way that the physical and

chemical properties of each eighth element were resembling.

3. Classify the elements into four blocks on the basis of electronic configuration?

Ans: On the basis of electronic configuration

a) s block elements

b) p block elements

c) d block elements

d) f block elements

4. Describe the characteristics of s-block elements?

Ans: Characteristics of s-block elements:

i) All the elements are soft metals.

ii) They have low melting and boiling points.

iii) They are highly reactive in nature.

iv) Most of them impart characteristic colours to the flame.

v) They form ionic compounds by losing their valence electrons and, thus, show

oxidation states of +1 and +2 in their compounds.

vi) The elements are good reducing agents.

vii) All of them are good conductors of heat and electricity.

5. With reference to periodic table indicate,

a. Element that is in group 13 and third period both.

b. First transition element of the fourth period.

Ans: (a) Al (b) Sc

6. What do you mean by atomic radii? Discuss its variation in periodic table?

Ans: Atomic radius may be defined as the distance from the centre of the nucleus to the

outermost shell containing the electrons.


The exact determination of the atomic radius of an element is not as simple as expected

because:

Along a period the atomic radii of the elements generally decreases from left to the right.

It is due to increased nuclear charge from left to the right.

The atomic radii of the elements in every group of the p[periodic table increases as we

move downwards.

7. What do you mean by electro negativity? Discuss its variation in periodic table?

Ans: The electronegativity of an element is defined as the relative electron attractive tendency

of its atom for a shared pair of electrons in a chemical bond.

It depends upon the nature of the atom with which a particular atom is linked. Different

scales of comparison have been given for this purpose.

It increases along the period and decreases down the group.

8. The electronic configuration of A is 1s22s22p63s2

a. To which period does it belong?

b. Write down the formula of the nitrate of A.

c. state whether a solution of nitrate is conductor of electricity or not

Ans: (a) Third period

(b) A NO3 2

(c) Conductor of electricity

9. The ionic radii of alkaline earth metals are smaller than those of alkali metals. Why?

Ans: Effective nuclear charge increases so the radius of alkaline earth metal decreases.

UNIT -4 CHEMICAL BONDING AND MOLECULAR STRUCTURE

LEVEL-1

1. Arrange the following in order of increasing bond angles around central atom SF6,CF4,BF3

Ans: SF6,CF4,BF3

2. A single bond between two atoms is a sigma bond. Comment?

Ans: Bonding taking place along the nuclear axis.

3. Name an inorganic molecule showing sp3 and sp2 hybridization.

Ans: NH3 , BCl3

4. The two 0-0 bond distances in ozone molecule are equal justify.

Ans: Due to resonance


5. Define bond enthalpy and bond order?

Ans: Bond Enthalpy: the amount of energy required to break one mole bonds of a particular type

between the atom in the gaseous state of a substance.

Bond Order: According to Lewis concept, in a covalent bond, the bond order may be given

as the number of bonds between two atoms in a molecule.

6. What is the shape of NH4+?

Ans: Tetrahedral

7. Which has larger dipole moment NH3 orNF3?

Ans: NH3

8. How many sigma and pi bond are present in C2H4 molecule?

Ans: 5 sigma and one pi bond

LEVEL-2 1. Explain the formation of chemical bond?

Ans: Due to the transfer or sharing of electrons.

2. Write octet rule and its significance and limitation.

Ans: The atoms of different elements take part in chemical combination in order to complete

their octet or duplet in some cases such as H, Li, Be etc. This is known as octet rule.

Limitations of Octet rule:

i) Odd number of electrons.

ii) Incomplete octet.

iii) Expanded octet

3. What is dipole moment? What is its significance?

Ans: Dipole moment is the product of Charge and distance between both the nucleus.

It significance: (i) Polarity between the charge

(ii) Ionic nature.

4. What is hybridization of atomic orbital?

Ans: The phenomenon of mixing of orbitals of the same atom with slight difference in energies

so as to redistribute their energies and give new orbitals of equivalent energy and shape.

The new orbitals which get formed are known as hybrid orbitals.

5. Discuss the shape of sp and sp2 hybrid orbital.

Ans: sp – linear and sp2 – trigonal planar


6. What are two different types of hydrogen bonding?

Ans: Inter and Intramolecular hydrogen bonding

LEVEL-3

1. Why the bond angle H-N-H is 1070 and H-O-H is 1050, instead of 109028’ characteristic of

tetrahedral structure?

Ans: lone pair lone pair repulsion in H2O has more magnitude than that of lone pair bond pair

repulsion in NH3.

2. What type of bonding would you expect between

a. A metal and a non metal?

b. A metal and another metal?

c. A non metal and a non metal?

Ans: a. Ionic bonding

b. metallic bonding

c. covalent bonding

3. Which one of the sigma or pie bond is stronger and why?

Ans: Sigma bond is more stronger than pi bond because of more extent of overlapping.

4. Differentiate between a polar and a non polar covalent bond?

Ans: Polar covalent bond is formed between two non-metals of different electro negativities.

Non-polar covalent bond is formed between two non-metals of same or nearly same electro

negativities.

5. Explain the VSEPR theory.

Ans:

i. The shape of the molecule depends upon the number of electron pairs (bonded or non-

bonded) around the central atom.

ii. The electron pairs around the central atom tend to repel one another since the electron

clouds are negatively charged.

iii. The electron pairs in space tend to occupy such positions that they are at maximum

distance apart and the repulsive interactions are minimum.

iv. A multiple bond is treated as if it is a single bond and the electron pairs which constitute

the bond may be regarded as single pairs.

v. Where two or more resonance structures can represent a molecule, the VSEPRmodel is

applicable to any such structure.


vi. Gillespie and Nyholmn stated that the electron pairs existing as lone pairs cause greater

repulsive interactions as compared to bonded electron pairs. In the light of this, the

repulsive interactions follow the order:

Lone pair – Lone pair > Lone pair – Bond pair > Bond pair – Bond pair

UNIT 5 STATES OF MATTER

LEVEL -1

1. At what condition surface tension vanishes?

Ans: At critical temperature.

2. Why He is used in balloons in place of hydrogen?

Ans: Due to the incombustible nature of He.

3. Below what temp. A gas does not obey ideal gas laws?

Ans: Below Boyle’s Temperature.

4. At what temp. Vol. of a gas is supposed to be zero?

Ans: At 0 K.

5. What is the molar volume of a gas at 00C& 1 bar pressure?

Ans: 22400mL

6. Name the temperature above which a gas cannot be liquefied byany amount of pressure?

Ans: Critical temperature

7. What is the effect of increase of temperature on surface tension and viscosity in a liquid?

Ans: Both decreases with increase in temperature .

8. Why vegetables are cooked with difficulty at hill station?

Ans:At hill station the atmospheric pressure is less and so boiling point decreases.

9. What is the value of Z for an Ideal gas?

Ans:For ideal gas,Z =1.


LEVEL -2

1. What type of graph will you get when P V is plotted against P at constant temperature?

Ans: A straight line parallel to pressure axis.

2. What would have happened to the gas if the molecular collision were not elastic?

Ans: On every collision there is loss of energy, so molecule s would haveslowed downand

settled down in vessel and pressure reduce to Zero.

3. At a particular temperature, why vapour pressure of acetone is less than ether?

Ans: Molecular force of attraction in acetone is stronger than those present in ether.

4. Why liquids diffuse slowly as compared to gases?

Ans: In liquids the molecules are less free than in gas i.e intermolecular forces in liquid are

greater than in gas.

5. In terms of Charles law, explain why – 273 oC is the lowest temperature?

Ans: At -273 oC , the volume of gas become equal to ZERO i.e gas ceases to exist .

6. Name two phenomena that can be explained on the basis of surface tension.

Ans: surface tension can explain 1) capillary action (!!) spherical shape of small drops of liquid.

7. What correction is applied to obtain pressure of dry gas with total pressure and aqueous

tension?

Ans: P(dry gas) = P(Total) - Aqueous tension

8. Write vander Waals equation for n mole of gas.

Ans: P+ an2/v2)(V-nb)= nRT.

LEVEL – 3

1. (a) Why aerated water bottle kept under water during summer?

(b) Why do real gases deviate from ideal behaviour?

(c) Why is moist air lighter than dry air?

Ans: (a) To reduce the temperature so as to reduce pressure, otherwise bottle may burst.

(b) Real gas deviate from ideal behavior due to force of attraction and volume of

molecules of gases are negligible.

(c) Moist air has water vapours which lower vapour density, so it I lighter.

2. An oxygen cylinder has 10 L oxygen at 200 atm. If patient takes 0.5 mL of oxygen at 1 atm in

one breathes at 370C. How many breaths are possible?

Ans: 𝑃1𝑉1 = 𝑃2𝑉2


200 × 10 = 1 × 𝑉2

𝑉2 = 2000𝐿

Number of breath = 𝑇𝑜𝑡𝑎𝑙𝑉𝑜𝑙𝑢𝑚𝑒

𝑉𝑜𝑙𝑢𝑚𝑒𝑓𝑜𝑟 1 𝑏𝑟𝑒𝑎𝑡 𝑕=

2000𝐿

0.5×10−3𝐿= 4 × 106

3. Account for the following:

a) The size of weather balloon becomes longer and larger as it ascends into higher altitude.

b) Critical temperature of CO2& CH4 gases are 31.1 0◦ & 81.9 0◦C respectively which of these

has strong intermolecular forces & why?

Ans:. a) At higher altitude atmospheric pressure decreases but inside the balloon gas exerts

pressure & size becomes larger.

b)CO2 has stronger van der Waals forces of attraction than CH4 because of greater polarity

& high molar mass.

4. Why a does sharpened edge become smooth on heating up to melting point?

Ans: On heating the glass it melts & takes up rounded shape at edges which has minimum

surface area due to surface tension.

5. Why a liquid drop is spherical in shape?

Ans: A liquid tries to keep minimum no. of molecules in its surface & sphere has minimum

surface area.

UNIT 6 THERMODYNAMICS

LEVEL-1

1. Can the absolute value of internal energy can be determined? Why or why not?

Ans: No, because it is the sum of different types of energies some of which cannot be

determined.

2. Separate out the following into extensive and intensive:

Volume, Temperature, Pressure, Boiling point, Free energy

Ans: Volume and free energy are extensive, other are intensive.

3. Under what condition ∆H becomes equal to ∆U?

Ans: ∆H=∆U during a process which is carried out in a closed vessel (∆V=0) or number of moles

of gaseous products = number of moles of gaseous reactants or the reaction does not involve

any gaseous reactant or products.

4. Water can be lifted into the water tank at the top of the house with the help of a pump. Then

why is not considered to be spontaneous?


Ans: A spontaneous process should continue taking place by itself after initiation. But this is not

so in the given case because water will go up so long as the pump is working.

5. For the process to occur under adiabatic conditions, the correct conditions is:

(i) ∆T=0 (ii) ∆p=0 (iii) q=0 (iv) w=0

Ans: (iii)

6. A reaction A + B→ C+ D +q is found to have a positive entropy change. The reaction will be

(i) possible at high temperature (ii) possible only at low temperature

(iii) not possible at any temperature (iv) possible at any temperature

Ans: Here ∆H = - ve and ∆S= +ve. ∆G=∆H-T∆S. For the reaction to be spotneous, ∆G should

be –ve which will be so at any temperature i.e., option is (iv)

7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system.

What is the change in the internal energy of the process?

Ans: Internal energy of the system increases by 307 J

8. Why endothermic reactions are favoured at high temperature?

Ans: If temperature is high, then T∆S will be much greater than ∆H in magnitude so that ∆G is

highly negative.

9. A real crystal has more entropy than an ideal crystal.

Ans: A real crystal has some disorder due to presence of defects whereas ideal crystal has no

disorder.

LEVEL-2

1. State the first law thermodynamics and derive a mathematical expression for it.

Ans: Definition and expression for the same i.e., ∆U= q+w where ∆U is change in internal energy,

q is the heat given to the system and w is the work done on the system.

2. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110,-393,-81 and 9.7 KJ/mol

respectively. Find the value of ∆rH for the reaction:

N2O4(g) +3 CO(g) → N2O(g)+ 3CO2(g)

Ans: - 777.7 KJ

3. For an isolated system, ∆U=0, what will be ∆S?

Ans: ∆S>0


4. The equilibrium constant for a reaction is 10. What will be the value of ∆G0 ? R=8.314J K-1 mol-

1 T=300K

Ans: ∆G0 = -2.303RTlogK = -5744.1J

5. what is bond energy? Why is it called enthalpy of atomization?

Ans: Amount of energy required to dissociate one mole of bonds present between the atoms in

gaseous molecules. As the molecules dissociate completely into atoms in the gaseous state,

therefore bond energy of diatomic molecule is called enthalpy of atomization.

6. Why standard entropy of an elementary substance is not zero whereas standard enthalpy of

formation is taken as zero?

Ans: Entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in

the formation of one mole of the substance from its elements. An element formed from itself

means no heat change, i.e∆fH0 =0

7. Enthalpy of combustion of carbon to CO2 is -393.5 KJmol-1 . Calculate the heat released upon

the formation of 35.2 g of CO2 from carbon and dioxygen gas.

Ans: Heat released when 44gCO2 is formed = 393.5 KJ

Hence heat released when 35.2 g CO2 is formed= 393.5× 35.2/44 = 314.8 KJ

LEVEL-3

1. Show that in the isothermal expansion of an ideal gas, ∆U= 0 and ∆H=0

Ans: For 1 mole of an gas, Cv= dU/dT or dU= CvdT

For a finite change, ∆U= CvdT

As here ∆T=0 , hence ∆U=0

∆H=∆U +∆(PV)= ∆U+∆(RT)= ∆U+ R∆T

But as ∆U=0 and ∆T=0( for isothermal process), hence ∆H=0

2. Justify the statement:

An exothermic reaction is always thermodynamically spontaneous.

Ans: As the reaction of the type accompanied by decrease of randomness, the heat released is

absorbed by the surroundings so that the entropy of the surroundings increases to such an

extent that ∆Stotal is positive.

3 Justify the statement:

The entropy of a substance increases on going from liquid to vapour state at any temperature.

Ans: The molecules in the vapour state have greater freedom of movement and hence greater

randomness than those in the liquid state. Hence, entropy increases in going from liquid to

vapour state.


4. Justify the statements:

(a)Reactions with ∆G0<0 always have an equilibrium constant greater than 1.

(b)Many thermodynamically feasible reactions do not occur under ordinary conditions.

Ans: (a) -∆G0=RTlnK. Thus, if ∆G0 is less than zero,i.e., it is –ve and hence K will be greater than 1.

(b) Under ordinary conditions, the average energy of the reactants may be less than the

threshold energy. They require some activation energy to initiate the reaction.

5. Give reasons:

(a) Neither q nor w is a state function but q+w is state function.

(b) The dissolution of ammonium chloride in water is endothermic still it dissolves in water.

Ans:

(a) q+w = ∆U. As ∆U is a state function, hence q+wis a state function.

(b) On dissolution, entropy increases i.e., ∆S is +ve. If T∆S>∆H, then ∆G will be –ve. Hence , the

process is spontaneous.

6. Calculate the Standard enthalpy of formation of CH3OH (l) from the following data:

(i) CH3OH (l) +3/2 O2 (g) →CO2(g) +2H2O(l) ; ∆rH0= -726 KJmol-

(ii) C (s) + O2 (g)→ CO2(g) ∆cH0= -393 KJmol-

(iii) H2(g) +1/2 O2(g)→ H2O(l) ; ∆fH0= -286 KJmol-

Ans: Aim: C (s) + 2H2 (g)→ CH3 OH(l) ∆fH0= ?

Eqn. (ii) + ×2 eqn. (iii) – eqn. (i) gives the required Eqn. with ∆H = - 393 +2(-286)-(-726) = -239

KJmol-1

7. Write the expression for the work done by 1 mole of the gas in each of the following cases:

(i) For irreversible expansion of the gas from V1 toV2

(ii) For reversible isothermal expansion of the of the gas from V1 toV2

Ans:

(i) Irreversible expansion takes place when external pressure is constant

Wirr = - Pext(V2 – V1)= - Pext∆V

(ii) Reversible expansion takes place when internal pressure is infinitesimally greater than

external pressure at every stage

Wrev= -nRTlnV2/V1

UNIT-7 EQUILIBRIUM

LEVEL – I 1. Mention the factors that affect equilibrium constant.

Ans: Temperature, pressure, catalyst and molar concentration of reactants and products.

2. what is ionic products of water?

Ans: Kw = [H+] [OH-]


3. Write conjugates acids of H2O & NH3.

Ans: H3O+& NH4+.

4. Define Arrhenius acids.

Ans: Arrhenius acids are the substances that ionize in water to form H+.

5. Define the term degree of ionization.

Ans: Extent up to which an acid/base/salt ionize to form ions.

6. What are Buffer solutions?

Ans: The solutions which resist change in pH on dilution or with the addition of small amounts of

acid or alkali are called Buffer Solutions.

7. Write Kc for the gaseous reaction- N2 + 3H2⇌2NH3

Ans: . Kc=[NH3]2/[N2] [H2]3

8. Out of H2O & H3O+ which is stronger acid?

Ans: H3O+.

9. What is common ion effect?

Ans: Shift in equilibrium on adding a substance that provides more of an ionicspecies already

present in the dissociation equilibrium.

10. Write relationship between Kp and Kc for the gaseous reaction :

N2 + O2 ⇌2NO

Ans: Kp = Kc as Δn is zero for the above said reaction.

LEVEL – 2 1. What is effect of catalyst on equilibrium constant “Kc”?

Ans: A catalyst does not affect equilibrium constant because it speeds upboth forward and

backward reactions to the same extent.

2. State Le Chatelier’s principle.

Ans: It states that a change in any of the factors that determine the equilibriumconditions of a

system will cause the system to change in such a manner so as to reduce or to counteract the

effect of the change.

3. What is meant by conjugate acid –base pairs? Explain.


4. Classify the following bases as strong and weak bases: NaHCO3, NaOH,KOH, Ca(OH)2, Mg(OH)2.

Ans: Strong bases: NaOH, KOH ; Weak bases: NaHCO3,Ca(OH)2, Mg(OH)2.

5. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3M.What is its pH ?

Ans: pH = – log[3.8 × 10–3]

= – {log[3.8] + log[10–3]}

= – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42

Therefore, the pH of the soft drink is 2.42and it is acidic.

6. The species: H2O, HCO3–, HSO4

– and NH3can act both as Bronsted acids and bases.For each

case give the corresponding conjugate acid and conjugate base.

Ans:

-

7. Explain Lewis acids and bases with suitable examples.

Ans: Lewis acids are lone pair (of e-) accepters while Lewis bases are lone pair donators.AlCl3 is a

Lewis acid while NH3 is a Lewis base.

8. What is difference between alkali and bases? Give examples.

Ans: An alkali is a water soluble base. All the alkalis are bases but all the bases are not alkali. Ex-

NaOH is an alkali/base.Ca(OH)2 is a base but not an alkali.

9. Explain homogeneous and heterogeneous equilibrium giving examples.

Ans: If all the reactants and products present in an equilibrium mixture are insame

phase→homogeneous equilibrium.

If all the reactants and products present in an equilibrium mixture are indifferent phase→

heterogeneous equilibrium.

𝑁2 𝑔 + 3𝐻2 𝑔 ⇌ 2𝑁𝐻3(𝑔)homogeneous equilibrium

𝐶𝑎𝐶𝑂3 𝑠 ⇌ 𝐶𝑎𝑂 𝑠 + 𝐶𝑂2(𝑔)heterogeneous equilibrium

LEVEL – 3 1. The pH of some common substances is given bellow. Classify the substances as acidic/basic

Name of fluid pH

Lime water 10

Milk of magnesia 10

Human saliva 6.4


Lemon juice 2.2

Sea water 7.8

Vinegar 3

milk 6.8

Ans: acidic-Human saliva, Lemon juice, milk, vinegar

Basic- Lime water, sea water, milk of magnesia.

2. Explain general characteristics of acids and bases.

Ans: Most of the acids taste sour. Acids are known to turn blue litmus paper into red and liberate

dihydrogen on reacting with some metals.

Bases are known to turn red litmus paper blue, taste bitter and feel soapy.

3. Water is amphoteric in nature. Explain.

Ans: Water can react with acid as well as base

H2O + HCl → H3O+ +Cl- water is basic

H2O + NH3 → OH- + NH4+ water is acidic

4. Describe the effect of :

a) addition of H2

b) addition of CH3OH

c) removal of CO

d) removal of CH3OH

on the equilibrium of the reaction:

2H2(g) + CO (g )⇌CH3OH (g)

Ans:

a) addition of H2 equilibrium will shift on RHS

b) addition of CH3OH equilibrium will shift on LHS

c) removal of CO equilibrium will shift on LHS

d) removal of CH3OH equilibrium will shift on RHS

5. Classify the following species into Lewis acids and Lewis bases and show how these act as

such:

(a) HO– (b)F – (c) H+ (d) BCl3

Ans: (a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:OH– ).

(b) Fluoride ion acts as a Lewis base as it can donate any one of its four electronlone

pairs.

(c) A proton is a Lewis acid as it can accept a lone pair of electrons from baseslike

hydroxyl ion and fluoride ion.

(d) BCl3 acts as a Lewis acid as it can accept a lone pair of electrons from species like

ammonia or amine molecules.


6. For the equilibrium,2NOCl(g) ⇌2NO(g) + Cl2(g)the value of the equilibrium constant, Kc is 3.75

× 10–6 at 1069 K. Calculate the Kpfor the reaction at this temperature?

Ans: We know that,Kp= Kc(RT)Δn

For the above reaction,Δn= (2+1) – 2 = 1

Kp= 3.75 ×10–6 (0.0831 × 1069)

Kp= 0.033.

7. Hydrolysis of sucrose gives, Sucrose + H2O →Glucose + Fructose

Equilibrium constant Kc for the reaction is 2 ×1013 at 300K. Calculate ΔG0at300K.

Ans: ΔG0= – RT lnKc

ΔG0= – 8.314J mol–1K–1J x 300K × ln(2×1013)

ΔG0= – 7.64 ×104 J mol–1

8.Explain the following :

(i) Common ion effect (ii) solubility products (iii) pH

Ans: (i) Suppression of ionization of weak electrolyte by adding a strong electrolyte having an ion

common.

(ii) Product of the molar concentrations of the ions in a saturated solution,each concentration

term raised to the power equal to the no. of ions produced.

(iii) Negative logarithm of hydrogen ion concentration.

FIVE MARKS QUESTIONS

1. At 473 K, equilibrium constant Kc for decomposition of phosphoruspentachloride,PCl5 is 8.3

×10-3. If decomposition is depicted as,

PCl5 (g) ⇌PCl3 (g) + Cl2 (g) ΔrH0= 124.0 kJ mol–1

a) Write an expression for Kc for the reaction.

b) What is the value of Kc for the reverse reaction at the same temperature?

c) What would be the effect on Kc if (i) more PCl5is added (ii) pressure isincreased(iii) the

temperature is increased ?

Ans: (a) 𝐾𝑐 =[𝑃𝐶𝑙3][ 𝐶𝑙2]

𝑃𝐶𝑙5

(b)120.48

(c) (i) equilibrium will shift on RHS

(ii) equilibrium will shift on LHS

(iii) equilibrium will shift on RHS

2. Dihydrogen gas is obtained from natural gas by partial oxidation with steam As per following

endothermic reaction:CH4 (g) + H2O (g) ⇌CO (g) + 3H2 (g)

(a) Write as expression for Kp for the above reaction.

(b) How will the values of Kp and composition of equilibrium mixture be Affected by(i) increasing

the pressure(ii) increasing the temperature(iii) using a catalyst?


Ans:

(a) Kp=p(CO).p(H2)3/ p(CH4).p(H2O)

(b)

(i) Value of Kp will not change, equilibrium will shift in backward direction.

(ii) Value of Kp will increase and reaction will proceed in forward direction.

(iii)No effect.

3. What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following

species:HNO2, CN–, HClO4, F –, OH–, CO32–, andS2–

Ans: The acid-base pair that differs only by one proton is called a conjugate acid-base pair

UNIT-8 REDOX REACTIONS

LEVEL – 1 1. Define oxidation and reduction in terms of oxidation number.

Ans: Increase in Oxidation Number is Oxidation and decrease in Oxidation Number is called

reduction.

2. What is meant by disproportionation? Give one example.

Ans: In a disproportionation reaction an element simultaneously oxidized and reduced.

P4 + 3OH- +3H2O→ PH3 +3H2PO2-

3. What is O.N. of sulphur in H2SO4?

Ans: +6

4. Identify the central atom in the following and predict their O.S. HNO3

Ans: central atom:- N; O.S. +5

5. Out of Zn and Cu which is more reactive?

Ans: Zn.

6. What is galvanization?

Ans: Coating of a less reactive metal with a more reactive metal e.g.- coatingof iron surface with

Zn to prevent rusting of iron.


7. How is standard cell potential calculated using standard electrode potential?

Ans: E0cell = E0

cathode − E0anode

8. What is O.S. of oxygen in H2O2?

Ans: - -1.

9. The formation of sodium chloride from gaseous sodium and gaseous chlorideis a redox

process justify.

Ans: Na atom get oxidize and Cl is reduced.

LEVEL – 2 1. Write the balanced redox reaction .

(I) MnO4–(aq) + Fe2+(aq) → Mn2+(aq)+ Fe3+(aq) [acidic medium]

(II) Cr2O72– + Fe2+ →Cr3+ + Fe3+ [Acidic medium]

Ans:

(i) MnO4–(aq) +5Fe2+(aq) + 8H+(aq) → Mn2+(aq)+ 5Fe3+(aq) + 4H2O(l)

(ii) Cr2O72– +6Fe2+ + 14H+→ 2Cr3+ + 6Fe3+ + 7H2O

2. Identify the strongest & weakest reducing agent from the following metals:

Zn, Cu, Na, Ag, Sn

Ans: Strongest reducing agent: Na, weakest reducing agent: Ag.

3. Determine the oxidation no. of all the atoms in the following oxidants:KMnO4,

K2Cr2O7 and KClO4

Ans :

In KMnO4 K = +1, Mn = +7, O = -2

In K2Cr2O7K = +1, Cr = +6, O = -2

In KClO4K = +1, Cl = =+7, O = -2

4. Determine the oxidation no. of all the atoms in the following species:Na2O2and OF2.

Ans: In Na2O2Na = +1, O = -1

InOF2, F = -1, O = +2

5. Is it possible to store:

(i) H2SO4 in Al container?

(ii) CuSO4 solution in Zn vessel?

Ans : (i) yes. (ii) No.

6. Calculate the standard e.m.f. of the cell formed by the combination of Zn/Zn2+⎤⎤Cu2+/Cu.

Ans: E0cell = E0

cathode − E0anode

=0.34 – (-0.76) = 1.10V.


7. Identify the oxidizing and reducing agents in the following equations:

(i) MnO4–(aq) +5Fe2+(aq) + 8H+(aq) → Mn2+(aq)+ 5Fe3+(aq) + 4H2O(l)

(ii) Cr2O72– +6Fe2+ + 14H+→ 2Cr3+ + 6Fe3+ + 7H2O

Ans :

(i) O.A. = MnO4– ; R.A.= Fe2+

(ii)O.A.=Cr2O72–; R.A.= Fe2+

8. Predict all the possible oxidation states of Cl in its compounds.

Ans: 0, -1, +1, +3, +5, +7

9. Formulate possible compounds of ‘Cl ‘in its O.S.is: 0, -1, +1, +3, +5, +7

Ans: Cl2, HCl, HOCl, HOClO, HOClO2, HOClO3 respectively.

10. List three measures used to prevent rusting of iron.

Ans:

(i) galvanization(coating iron by a more reactive metal)

(ii) greasing/oiling

(iii) painting.

LEVEL – 3 1. Write short notes on :

(a) Electrochemical series (b) redox reactions (c) oxidizing agents

Ans :

(a) Electrochemical series :- arrangement of metals(non-metals also)in increasing order of their

reducing power or vice versa.

(b) Reactions in which both Oxidation and reduction take place simultaneously are REDOX

reactions.

(c) Oxidizing agents : chemical specie which can oxidize the other one or can reduce itself.

2. Calculate O. S. of sulphur in the following oxoacids of S :

H2SO4, H2SO3 , H2S2O8and H2S2O7

Ans : +6, +4, +6 and +6 respectively.

3. Explain role of salt bridge in Daniel cell.

Ans :

(a) it completes the electric circuit in the cell.

(b) It maintains the electric neutrality in the cell.

4. Account for the followings :

(i) sulphur exhibits variable oxidation states.

(ii) Fluorine exhibits only -1 O.S.

(iii) Oxygen can’t extend its valency from 2.


Ans.

(i) Due to the presence of vacant d orbitals in sulphur.

(ii) It is most electronegative element

(iii) Small size/unavailability of vacant d orbitals in O

5. Complete and balance the following equations:

(i) H+ + Cr2O72-+ Br-→ 2Cr3+ + Br2+ -----

(ii) H2O2 + Cl-→ OH- + Cl2

(iii) Zn + Cu2+→ ?

Ans : (i) 14H+ + Cr2O72-+6 Br-→ 2Cr3+ + 3Br2 + 7H2O

(ii) H2O2 + 2Cl-→ 2OH- + Cl2

(iii) Zn + Cu2+ → Zn2+ + Cu

6. Identify the oxidizing and reducing agents in the following equations:

(i) Fe + H2SO4→FeSO4 + H2

(ii)H2 + Cl2 →2HCl

(iii) MnO2 + 4HCl→MnCl2 + 2H2O + Cl2

Ans : (i) O.A. =H2SO4 ; R.A.= Fe

(ii) O.A. = Cl2; R.A.=H2

(iii)O.A. = MnO2; R.A. =HCl

7 . Arrange the following in increasing order of their reducing power:

Cu, Ag, Au, Zn, Fe, Al, Na, Mg, Pt(SHE), Hg, Ca, K

Ans : Au, Hg, Ag, Cu, Pt(SHE), Fe, Zn, Al, Mg, Na, Ca, K

LEVEL – 4

1.What is SHE? What is its use?Draw its diagram.

Ans :Standard Hydrogen Electrode (SHE) has been selected to have zero standard potential at all

temperatures. It consists of a platinum foilcoated with platinum black (finely divided platinum)

dipping partially into an aqueous solution in which the activity (approximate concentration 1M)

of hydrogen ion is unity and hydrogen gas is bubbled through the solution at 1 bar pressure. The

potential of the other half cell is measured by constructing a cell in which reference electrode is

standard hydrogen electrode. The potential of the other half cell is equal to the potential of the

cell.

2. We spend crore of Rupees and even thousands of lives every year due tocorrosion. How can

it be prevented? Explain.

Ans : (i) By Galvanization: Coating of a less reactive metal with a more reactive metal e.g.-

coating of iron surface with Zn to prevent rusting of iron.

(ii) By greasing /oiling (to keep away the object from the contact of air&moisture.)

(iii)By painting (to keep away the object from the contact of air &moisture.)


3. Balance the equation MnO4– + I-→ Mn2+ +I2 + + H2O by ion electron method in acidic medium.

Ans :

Step-I Balancing of reduction half reaction by adding protons and electrons on LHS and more

water molecules on RHS:

8H++ MnO4– +5e-→ Mn2+ + 4H2O

Step-II Balancing of oxidation half reaction by adding electrons on RHS:

2I-→ I2 +2e-

Step-III To multiply the OHR by 5; RHR by2 and to add OH & RH reactions toget overall redox

reaction(cancellation of electrons of RH & OH reactions):

[8H+(aq)+ MnO4–(aq) +5e-→ Mn2+(aq) + 4H2O(l)] x 2

[ 2I-→ I2 +2e-] x 5

_________________________________________________________

MnO4–(aq) +5Fe2+(aq) + 8H+(aq) → Mn2+(aq)+ 5Fe3+(aq) + 4H2O(l)

UNIT – 9 HYDROGEN

LEVEL – 1

1.What happens when peroxydisulphuric acid is hydrolysed with water?

Ans: Sulphuric acid and hydrogen peroxide.

2. Name the gases produced when water reacts with:

(a) Aluminium nitride

(b) Aluminium carbide

Ans: (a)ammonia gas is formed

(b) methane is formed

3.What is meant by autoprotolysis of water?

Ans: The self oxidation reduction of water.

4. Why can’t sea animals live in distilled water?

Ans: Due to the absence of free oxygen in the distilled water.

5. Which cations cause hardness in water?

Ans: Calcium and magnesium cations.

6. What is meant by Hydrogenation?

Ans: The addition of hydrogen is called hydrogenation.

7. What is ‘syn’ gas ?

Ans: Syn is a mixture of CO and H2.


LEVEL – 2 1. How is hydrogen prepared :

(i) From water by using a reducing agent?

(ii) In laboratory in the pure form?

(iii) From hydrocarbon?

Ans: (i) It is usually prepared by the reaction of granulated zinc with dilute hydrochloric acid. 2𝑁𝑎 + 2𝐻2𝑂 ⟶ 2𝑁𝑎𝑂𝐻 + 𝐻2

(ii) By the electrolysis of water.

(iii) 𝐶𝐻4 + 𝐻2𝑂 𝑔 ⟶ 𝐶𝑂 𝑔 + 3𝐻2(𝑔)

2. How does H2O2 behave as a bleaching agent?

Ans: On decomposition it produces nascent oxygen which decolourises the colour.

3. Discuss the structure of common form of ice.

Ans: Due to the formation of hydrogen bonding it produce cage like structure.

4. How is heavy water prepared from normal water?

Ans: Due to continuous electrolysis of water.

5. Write the chemical reaction to show the amphoteric nature of water.

Acidic character: 𝑁𝐻3 + 𝐻2𝑂 ⟶ 𝑁𝐻4+ + 𝑂𝐻−

Basis character: 𝐻𝐶𝑙 + 𝐻2𝑂 ⟶ 𝐻3𝑂+ + 𝐶𝑙−

6. What are the advantages of using hydrogen as fuel?

Ans: (i) Produce no pollution.

(ii) High calorific value

(iii) Produces drinking water.

7. Name the isotopes of hydrogen. What is the importance of heavier isotopes of hydrogen?

Ans: Deuterium and tritium. The importance of heavier isotopes of hydrogen is in nuclear

reactor.

LEVEL – 3

1. When hydrogen is passed over a black solid compound A , the products are brown metal B

and a colourless liquid C .The metal B displaces Silver from Silver nitrate and reacts with

dilute hydrochloric acid to give hydrogen.

(i) Name A, B, and C.

(ii) Give the chemical equation involved.

Ans: (i) A is CuO, B is Cu and C is H2O

(ii) 𝐶𝑢𝑂 + 𝐻2 ⟶ 𝐶𝑢 + 𝐻2𝑂 , 𝐶𝑢 + 2𝐴𝑔𝑁𝑂3 ⟶ 𝐶𝑢 𝑁𝑂3 2 + 2𝐴𝑔


2. What do you understand by the terms?

(i) Hydrogenation

(ii) Syn gas

(iii) Water gas-shift reaction

Ans: (i) The addition of hydrogen is called hydrogenation.

(ii) The mixture of CO and H2 is known as syn gas.

(iii) 𝐶𝑂 + 𝐻2𝑂 ⟶ 𝐶𝑂2 + 𝐻2

UNIT – 10 THE s-BLOCK ELEMENT

LEVEL – 1

1. Name the alkali metal which form superoxide when heated in excess of air.

Ans: Potassium

2. LiCl is more soluble in organic solvent, why?

Ans: It is due to covalent in nature.

3. Explain why Sodium is less reactive than Potassium?

Ans: Due to less metallic character of sodium

4. Why are Potassium, and Cesium , rather lithium can be used in photoelectric cell ?

Ans: Due to lesser ionisation enthalphy.

5. Lithium is the only alkali metal to form nitride.

Ans: Due to diagonal relationship with Mg.

6. Name the radioactive element of group 1 and 2.

Ans: Fr and Ra

7. Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction?

Ans: They are better reducing agent than C and CO.

LEVEL – 2 1. Explain the following :

(i) The mobility of the alkali metal ion in aqueous solution are :

Li+ ˂ Na+ ˂ K+ ˂ Rb+ ˂ Cs+

(ii) Lithium is the only alkali metal to form nitride directly.

Ans: (i) The magnitude of hydration enthalphy decreases down the group so the size of the

hydrated ion decreases down the group.

(ii) Due to diagonal relationship with Mg.


2. ‘The chemistry of beryllium is not essentially ionic’. Justify the statement by making a

reference of beryllium.

Ans: It is due to high polarising power of Be cation its compound are covalent in nature.

3. When alkali metal dissolves in liquid ammonia, the solution gives deep blue colour which is

conducting in nature .Why?

Ans: Due to ammoniated cation and ammoniated electron.

4. Describe the three industrial uses of caustic soda.

Ans: Industrial uses of caustic soda:

(i) In soap industries

(ii) In the laboratory

(iii) In cotton industries.

(iv)

5. Why it is necessary to add gypsum in the final stages of preparation of cement?

Ans: To increase the time of setting of the cement.

6. Beryllium and Magnesium do not give colour to flame whereas other alkaline earth metals

do so why?

Ans: Due to high ionisation enthalpy of Be and Mg.

7. What happens when

(i) Sodium metal is dropped in water?

(ii) Sodium metal is heated in free supply of air?

(iii) Sodium peroxide is dissolved in water?

Ans: (i) NaOH is formed.

(ii) Na2O2 is formed

(iii) NaOH and H2O2is formed.

8. What happen when

(i) Calcium nitrate is heated?

(ii) Chlorine reacts with slaked lime?

(iii) Magnesium is burnt in air?

Ans: (i) CaO and O2 and NO2

(ii) bleaching powder is formed.

(iii) MgO is formed.

LEVEL – 3 1. Write balanced equation for the reaction between :

(i) KO2 and water

(ii) BeO + H2SO4

(iii) Ca(OH)2 + Cl2


Ans: (i)2𝐾𝑂2 + 2𝐻2𝑂 ⟶ 2𝐾𝑂𝐻 + 𝐻2𝑂2 + 𝑂2

(ii) 𝐵𝑒𝑂 + 𝐻2𝑆𝑂4 ⟶ 𝐵𝑒𝑆𝑂4 + 𝐻2

(iii) 𝐶𝑎 𝑂𝐻 2 + 𝐶𝑙2 ⟶ 𝐶𝑎𝑂𝐶𝑙2 + 𝐻2𝑂

2. Give a brief account of the following :

(i) Blue colour solutions of alkali metals in liquid ammonia.

(ii) Tendency to form divalent compounds of alkaline earth metals than monovalent.

(iii) Tendency to give flame colours by alkali metals.

Ans: (i) Due to the presence of ammoniated electron.

(ii) In divalent oxidation state the octet is complete.

(iii) Due to less ionisation enthalpy of alkali metals.

UNIT – 11 THE p- BLOCK ELEMENTS

LEVEL – 1

1. Why is boron used in nuclear reactions? Ans: Because Boron can absorb neutrons .

2. By giving a balanced equation show how B(OH)₃ behaves as an acid in water. Ans: B(OH)₃ + H0H → B(OH )⁻₄ + H₃ O⁺ 3. Name the element of group 14 which exhibits maximum tendency for catenation. Ans: Carbon 4. What is the basic building unit of all SILICATES? Ans: SiO⁴₄⁻ 5. What happens when NaBH₄ reacts with iodine ? Ans: 2 NaBH₄ + I₂ → B₂H₆ + 2 NaI + H₂

6. What is producer gas? Ans: Producer gas is a mixture of CO and N₂ in the ratio 2:1 7. What happens when BORIC acid is heated? Ans: H₃BO₃ → HBO₂ → H₂B₄O₇ 8 .Ga has high ionization enthalpy than Al. Explain. Ans:Due to poor shielding effect of d-electrons in Ga effective nuclear charge increases as compared to Al thus IE is higher than Al. 9.What are Fullerene ? How they were prepared Ans: Fullerene are the allotropes of carbon Its structure is like a soccer ball. They were prepared by heating graphite in electric arc in presence of inert gases He or Ar.


LEVEL – 2

1. Give biological importance of Ca and Mg. Ans: Calcium and magnesium ions are essential for the transmission of impulses along nerve fibres. Magnesium is present in chlorophyll in green plants, calcium ions also regulates the beating of the heart.

2. Name the chief factors responsible for anomalous behaviour of lithium. Ans: (i) Small size atom and ion, (ii) High ionization enthalpy (iii) Absence of d-orbital in its Valence shell.

3. Describe two similarities and two dissimilarities between B &Al. Ans: Similarities:

(i) Both have same number of valence electrons. (ii) Both have similar electronic configuration.

Dissimilarities: (i) B is a non - metal whereas Al is a metal. (ii) B forms acidic oxides whereas Al forms amphoteric oxides.

4. Give reason for the following observations.

(a) The tendency for catenation decreases down the group. (b) The decreasing stability of +3 oxidations state with increasing atomic number in group 13. (c) PbO₂ IS a stronger oxidizing agent than SnO₂. (d) Molten aluminium bromide is a poor conductor of electricity. Ans:

(a) It is due to decrease in bond dissociation energy which is due to increase in atomic size (b) It is due to Inert Pair Effect. (c) PbO₂ is stronger oxidizing agent than SnO₂ because Pb²⁺ is more stable than Pb⁴⁺ whereasSn⁴⁺ is more stable than Sn²⁺. (d) Molten AlBr₃ is poor conductor of electricity because it is covalent compound.

LEVEL – 3

1. Silanes are few in number where as alkanesare large in number .Explain. Ans: Carbon has the maximum tendency for catenation due to stronger C-C bonds as aresult it forms a large number of alkanes Silicon on other hand due to weaker Si-Si bonds has much lesser tendency for catenation and hence forms only a few silanes.

2. What are SILICONES?

Ans: SILICONES are the synthetic organo-silicon polymers having general formulae ( R₂ SiO )n in which R = alkyl ( methyl, ethyl or phenyl )


3. What are Zeolites? Ans: Zeolites are 3D silicates in which the Si atoms are replaced by Al3⁺ions and negative charge is balanced by cations such as Na⁺, K⁺,Ca²⁺ ( NaAlSi₂O₆H₂O)

4. When aqueous solution of borax is acidified with hydrochloric acid,a white crystalline solid is formed which is soapy in touch .Is this solid acidic or basic in nature? Explain.

Ans: When an aqueous solution of borax is acidified with HCl boric acid is formed Na₂B₄O₇ + 2 HCl + 5 H₂O → 2NaCl + 4 H₃BO₃ (BORIC ACID) Boric acid is not a proticacid it does not give H⁺ ion accepts a pair of electrons and acts as a Lewis acid 5 .Out of CCl₄ and SiCl₄ which one reacts with water? Ans: Due to presence of d-orbitals in Si, SiCl₄ reacts with water but due to the absence of d-orbitals in C , CCl₄ does not react with water.

6. Which compound is known as inorganic benzene? Ans: Borazine ( or borazole ) B₂H₆.2NH₃ when heated it gives B₃N₃H₆( Borazine) because of its structure is very similar to C₆H₆ ( Benzene) and hence BORAZINE (OR BORAZOLE ) is known as inorganic benzene.

UNIT – 12 ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES

LEVEL – 1

1. Suggest a method to purify a liquid which decomposes at its boiling point?

Ans: The process Distillation Under reduced pressure is used to purify aliquid which decomposes

at its boiling point.

2. How will you separate a mixture of O-nitrophenol and P-nitrophenol?

Ans: O-nitrophenol is steam volatile therefore it can be separated by Steam distillation.

3. Lassaigne's test is not shown by diazonium salt .Why?

Ans: On heating diazoniumsalt , loses Nitrogen and could not fuse with the Sodium metal

therefore diazonium salt does not show Positive Lassaigne's test for Nitrogen .

4. Why alcohols are weaker acids than Water?

Ans: The alkyl group in alcohol is has +I effect due to which electron density increases on

oxygen atom which makes the release of H⁺ ion more difficult from alcohol R→ O – H

5. Why is nitric acid is added to Sodium Extract before adding Silvernitrate for testing halogens?

Ans: Nitric acid is added to decompose NaCN& Na₂ S .

NaCN + HNO₃ → NaNO₃ + HCN

Na₂ S + HNO₃ → 2NaNO₃ + H₂ S


6. Write the hybridized state of C atoms in the following

𝐶𝐻2 = 𝐶𝐻 − 𝐶 ≡ 𝑁

Ans: SP² , SP² ,SP

7. Draw the structures of the following compounds

(a) Hex – 3 – enoic acid

(b) 2-Chloro-2-methylbutan-1-ol

Ans: (a) O

CH₃-CH₂-CH=CH-CH₂-C-OH

Cl

(b) CH₃-CH₂-C-CH₂-OH

CH₃

8. Which bond is more polar in the following pairs of molecules :

(i) H₃C-NH₂, H₃C-OH

(ii) H₃C-OH , H₃C-SH

Ans:

(i) C-O is more polar because O is more electronegative than N

(ii) C-O is more polar because O is more electronegative than S

9. In which C-C bond of CH₃CH₂CH₂Br , the inductive effect is expected to be the least.

Ans: Magnitude of inductive effect diminishes as the intervening bonds increases. Hence, the

effect is least in the bond between C-3 and C-2.

10. Why (CH₃)₃C⁺ carbocation is more easily formed than ( CH₃)₂C⁺H carbocation ?

Ans: Tertiary butyl carbocation is more stable than isopropyl carbocation.

11. Diphenylketone ( C₆H₅-CO-C₆H₅ ) shows tautomerism. Explain.

Ans: It contains two alpha H atoms.

12. What are Electrophiles and Nucleophiles?

Ans: Electrophiles: The species which have one atom the octet of which is incomplete they

attract nucleophiles are called Electrophiles

e.g. BF₃ ,Cl⁺ etc

Nucleophiles: The species which carry negative charge or a loan pair of electrons on an

atom for donation are called Nucleophiles

e.g. HS⁻ , (CH₃)₃N: etc


13. Write resonance structure of CH₂ = CH-CHO.

Ans: Proper structure I,II,III

14. Explain Hyperconjugation.

Ans: Hyperconjugation is a general stabilising interaction. It involves delocalisation of 𝜎electrons

of C—H bond of an alkyl groupdirectly attached to an atom of unsaturatedsystem or to an atom

with an unsharedp orbital. The s electrons of C—H bond of thealkyl group enter into partial

conjugation withthe attached unsaturated system or with the unshared p orbital.

Hyperconjugation is a permanent effect.

15. Why does C-H bond length decrease in the order C₂H₆ > C₂H₄ >&C₂H₂ ?

Ans: Because the hybridisation on the C-atoms of these are SP³ , SP² and SP respectively .The p-

orbitals are bigger in size and p-character decreases in the order SP³ > SP² > SP

LEVEL – 2

1. Write the IUPAC name of the following compound

CH₃ CH₃

CH₃- C –CH = CH - CH₂- CH -CH₃

CH₃

Ans: The IUPAC name is 2,2,6-trimethylhept-3-ene.

2. Write IUPAC names for the following compounds.

(i) ClCH₂CH₂OH

(ii) CH₂=CH-CHO

(iii) CH ≡C-COOH

Ans:

(i) 2-Chloroethan1-ol

(ii) Prop-2-en-1-al

(iii) Prop-2-yn- 1-oic acid

3. Write the IUPAC name of

CH₃ CN

Ans: 3-Methylpent-3-ene-1-nitrile


4. Write the IUPAC name of

CH₂CHO

Br

Ans: 2-(2-Bromophenyl)ethanol.

4. How Sodium fusion extract is prepared ?

Ans: A small piece of dry Sodium metal is heated with an organic compound in a fusion tube for

2-3 minutes and the red hot tube is plunged in to distilled water contained in achina dish The

contained of the china dish is boiled ,cooled and filtered . The filtrate is known as Sodium fusion

extract.

5. Explain the principle of Paper Chromatography.

Ans: Paper Chromatography is based on the difference in the rates at which the components of

a mixture are adsorbed. The material on which different components are adsorbed is called

Stationary phase which is generally made up of Alumina, Silica gel or Activated Charcoal.The

mixture to be separated is dissolved in asuitable medium and it is called Moving phase. The

moving phase is run on the stationary phase, the different compounds are adsorbed on

stationary phase at different rates.

LEVEL – 3

1. Write a short note on RESONANCE EFFECT.

Ans: The polarity produced in the molecule by the interaction of two pi bonds or between a pi

bond and loan pair of electron present on an adjacent atom is known as RESONANCE EFFECT

There are two types of resonance effect:

(1) Positive resonance effect (Transfer of electron is away from an atom or substituent group

attached to the conjugated system ) +R effect

e.g. –OH , -OR , -NH₂

(2) Negative resonance (Transfer of electron is towards the atom or substituent group attached

to the conjugated system ) –R effect

e.g. –COOH , -CHO , -CN

2. Differentiate between the principles of estimation of nitrogen in an organic compound by

(i) DUMAS method

(ii) KJELDAHL'S method


Ans: (i) Dumas method : A known mass of organic compound is heated with excess of CuO in an

atmosphere of CO₂, when nitrogen of the organic compound is converted into N₂ gas The volume

of N₂ gas thus obtained is converted into STP and the percentage of nitrogen is determined by

the formula.

P₁ V₁ X273 Volume of Nitrogen at STP = ------------------------------

760 X T₁

28X vol of N₂ at STP X100

% of N = --------------------------------------------

22400 X mass of organic compound

(ii) Kjeldahl's: A known mass of organic compound is heated with conc H₂ SO₄ in presence of

K₂ SO₄ and little CuSO₄ or Hg in a long necked flask called Kjeldahl'sflask . When nitrogen present

in the compound is quantitatively converted into ( NH₄)₂ SO₄ . ( NH₄)₂ SO₄ thus obtained is boiled

with excess of NaOH solution to liberate NH₃ gas which is absorbed in aknown excess of a

standard acid such as H₂ SO₄ or HCl

1.4 X Molarity of the acid X Basicity of the acid X Vol of the acid used

% N = ------------------------------------------------------------------------------------------------------

Mass of organic compound taken

UNIT 13 HYDROCARBONS

LEVEL – 1

1. Why is cyclproprane more reactive than propane?

Ans: In Cyclopropane,bond angle is 60o which is much less compared to the normal tetrahedral

bond angle of 109.5o for sp3 hybridized carbon. Therefore, the molecule is very much strained

and hence reactive.

2. Why is Wurtz reaction not preferred for preparation of alkanes containing odd no of carbon

atoms?

Ans: In this type of preparation we get mixture of hydrocarbons and they cannot be separated

out due to about same boiling points.

3. In the presence of peroxide addition of HBr to prepare propene takes place according to anti

Markovnikov’s rule but peroxide effect if not seen in the case of HCl and HI. Explain.


Ans: Homolysis of HCl does not takes place and in the case of HI homolysis takes place but the

iodide free radical combine with the similar free radical to form I2 molecule .Thus peroxide effect

is observed only in case of HBr.

4.Arrange the different conformations of ethane in decreasing order of stability?

Ans: Staggered> Skew >eclipsed.

5.What is the basic differences between conformational isomers and configurational (like

geometrical isomers) isomers?

Ans: One conformational isomer is changes into other without any bond rearrangement i.e.

due to single bond rotation while geometrical isomers are converted in to one form to other by

bond rearrangement i.e. bond breaking and bond making.

6.Which of the following has the highest boiling point?

(i) 2-Methylpentane (ii) 2,3-Dimethylbutane

Ans: 2- Methylpentane has highest boiling point because it has the least branched chain

structure as compared to 2,3-Dimethylbutane . Therefore, it has largest surface area and hence

has the highest boiling point.

7. Name two tests to test the presence of double bond in a compound.

Ans: . (i) Decolourises brown colour of bromine water

(ii) Decolourising pink colour of Baeyer’s reagent.

8. Name the process which may be used to locate the position of triple bond.

Ans: .Ozonolysis

9.Why benzene is extraordinarily stable though it contains three double bonds?

Ans: Due to resonance in benzene the π –electron cloud gets delocalized resulting stability of

molecule.

LEVEL -2

1. Why alkynes do not show geometrical isomerism?

Ans: Alkynes have linear shape and therefore, do not show geometrical isomerism.

2.How many isomers are possible for monosubstituted and disubstituted benzene?

Ans: There is one mono substituted benzeneand three di substituted benzene i.e. ortho, para

and metal.

3. Arrange benzene,n-hexane and ethyene in decreasing order of acidic behaviour.

Ans: Ethyne> benzene> n-hexane


4. An alkene’A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and

IUPAC name of ‘A’ .

Ans: CH3-CH=C(C2H5)2 , 3-Ethylpent-2-ene.

5. Out of benzene, m-dinitro benzene and toluene which will undergonitration most easily and

why?

Ans: CH3- is a ERG while –NO2 is a EWG .Therefore electron density is more in toluene than in

benzene and the electron density in m-dinitrobenzene will be less than in benzene therefore

toluene will undergo easily in nitration reaction.

6. What effect does branching of an alkane chain has on its boiling point?

Ans: On increasing branching surface area decreases and approaches that of a sphere .Since

sphere has minimum surface area, therefore, van der Walls forces of attraction are minimum

and hence boiling point decreases with branching.

7.How will you prepare acetaldehyde from acetylene?

Ans: C2H2 + H2O (Hg+2/H2SO4) → CH=CHOH → CH3CHO

8. Why alkanes do not dissolved in water but dissolve in benzene?

Ans: Because alkanes are non-polar and water is a polar solvent.

9.What is the function of CaO in soda lime?

Ans: It helps in the fusion of reaction mixture.

10.Can pyridine be regarded as an aromatic compound?

Ans: . Yes because it has planar structure and has (4n+2 )π electrons i.e 6 π electrons.

LEVEL-3

1.What are the necessary conditions for any system to be an aromatic?

Ans: (i) The molecule should contain cyclic cloud of delocalized electrons above and below the

plane of the molecule.

(ii) For the delocalization of pi electrons the ring must be planer to allow cyclic overlap of p-

orbitals.

(iii)It should contain (4n+2) π electrons where n=0,1,2,3........ This is known as Huckel rule.

2. Name the alkane which cannot be prepared by Wurtz Reaction?

Ans: Methane.


3.What is the cause of geometrical isomerism in alkanes?

Ans: Alkanes have π –bond and the restricted rotation around the π-bond gives rise the

geometrical isomerism.

4.Why alkanes are called paraffins?

Ans: They have little affinity for chemical reactions so they are called as paraffins.

5.What is Markownikoff’s rule?

Ans: During the addition across unsymmetrical multiple bond ,the negative part of the

addendum joins with that double bonded carbon which has less no of hydrogen.

6. Find out terminal and non-terminal alkynes from the following:

Ethyne, Pent -2-yne ,but-1-yne , propyne.

Ans: Ethyne,but-1-yne , propyne because these molecules have triple bond at the end of the

carbon chain.

7.What is the use of BHC (benzene hexachloride)?

Ans: It is used as an insecticide under the name of Gammexane or lindane.

8.What is the product of cyclic polymerization of ethyne?

Ans: . Benzene.

9. Name the two extreme type of conformation of ethane.

Ans: Staggered and eclipsed.

10. Suggest name of any other Lewis acid instead of anhydrous aluminium chloride which can be

used for ethylation of benzene?

Ans: .FeCl3

UNIT 14 ENVIRONMENTAL CHEMISTRY

LEVEL -1

1.Name the region of atmosphere which contains Ozone layer?

Ans: Stratosphere.

2. Name the chemical responsible for acid rain?

Ans: Oxide of Sulphur , Oxide of Nitrogen.

3.What is CFC stand for ?

Ans: Chlorofluorocarbon or Freons.


4. Name the gases responsible for green house effect?

Ans: CO2 ,CFC, Oxide of Nitrogen, Water vapours.

5. What is PAN.

Ans: Peroxyacyl nitrate, a toxic substance.

6. Name the sink of carbon dioxide in atmosphere.

Ans: Ocean and green plants.

LEVEL-2

1. Define environmental chemistry.

Ans: Environmental chemistry is defined as the branch of science which deals with chemical

process occurring in the environment It involves the study of origin , Transport, reaction, effect

of chemical species in the environment.

2. What do you mean by Biochemical Oxygen Demand (BOD)?

Ans: It is a measure of the dissolved oxygen that would be needed by micro-organisms and

various pollutants.

3. What is eutrophication?

Ans: Eutrophication is the process of nutrient enrichment of water bodies and subsequent loss

of biodiversity.

4. What is abbreviations FGD and CFC?

Ans: FGD means Fuel Gas Desulpherisation and CFC means Chloroflurocarbon.

5. How do pollutant and contaminant differ?

Ans: Pollutant is a substance already present in atmosphere which spoils the environment by its

concentration due to human activities.

Contaminant is a substance that does not occur in the nature but is introduced in significant

amount into the atmosphere by human action or accidentally.

6. How can domestic waste be used as manure?

Ans: Domestic waste comprises of two types of materials, biodegradable such as leaves, rotten

foods, etc, and non-biodegradable such as plastic, glass, metal scrap, etc , The non-

biodegradable waste is sent to industry for recycling. The biodegradable waste should be

deposited in the landfills. With the passage of time, it is converted into compost manure.


LEVEL 3

1.Explain giving reasons “ The presence of CO produce the amount of haemoglobin available in

the blood for carrying oxygen to the body cells “ ?

Ans: CO combine with haemoglobin of the red blood corpuscles ( RBC ) about 200 times more

easily than Oxygen to form carboxyhaemoglobin reversibly as follows ;

Hb + CO = HbCO

Thus it is not able to combine with oxygen to form oxy haemoglobin and transport of oxygen to

different body cells cannot takes place.

2. Give three examples in which green chemistry has been applied?

Ans: (a) In dry cleaning use of liquefied CO2 in place of tetrachloroethene ( Cl2C=CCl2 )

(b) In bleaching of paper using H2O2 in place of chlorine.

(c) In the manufacture of chemicals like ethanol using environment friendly chemicals

and conditions.

3. What is Pneumoconiosis? How does it occur?

Ans: Pneumoconiosis is a disease of lungs such as lung cancer, bronchital asthma, chronic

bronchitis, etc. it is caused by small sized particulates which enter into lung through nose and

provide a large surface area for adsorption of carcinogenic compounds such as polynuclear

hydrocarbons, asbestos, etc.

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