1993 AL Chemistry Paper I Marking Scheme

1993 AL-CHEM MS – page 1

1993 AL Chemistry paper 1 marking scheme 1.(a)(1) X is / helium nucleus/ alpha particle, charge = +2, mass = 4 amu [1m] Y is / fast moving electron/ beta particle, charge = -1, mass = 0 amu [1m] (2) The path of alpha and beta particles are deflected / bent by a magnetic field but they are deflected in

opposite directions [1m + 1m] OR: [diagram showing the bent motion and correct directions: 2m]

(b)(i)

[2m] (ii)

[2m] (c)(i) Step 3. It is the intermediate step with highest activation energy. Therefore It is the slowest step [2m] (ii) Exothermic Since the potential energy of the products is lower than that of the reactant. i.e. ∆H is negative. [1m] (Exothermic Since ∆H is negative 1m only) (d)(i) 2MnO2(s) + PbO2(s) + 4H+(aq) (or à) 2MnO4

-(aq) + 3Pb2+(aq) + 2H2O(l) [1.5m] (ii) 3H2O2(aq) + 2Cr(OH)4

- + 2OH-(aq) 2CrO42-(aq) + 8H2O(l) [1.5m]

(e)

2.(a)(i) Al(s) [1m] Equations: or [Al(OH)4]- 2Al(s) + 2OH-(aq) + 2H2O(l) à 2AlO2

-(aq) + 3H2(g) [1m] AlO2

-(aq) + H+(aq) + H2O(l) à Al(OH)3(s) [1m] (ii) CO3

2-(aq) [1m] Equation: CO3

2-(aq) + 2H+(aq) à CO2(g) + H2O(l) [1m] (iii) S2O3

2-(aq) [1m] Equation: S2O3

2-(aq) + 2H+(aq) à S(s) + 2SO2(g) + H2O(l) [1m] (iv) AlCl3(s) [1m] Equation:


AlCl3(s) + 3H2O(l) Al(OH)3(s) + 3HCl(g) [1m] (v) Na(s) [1m] (b)(i) The outer electronic oribitals of transition metal ion have only small enregy difference [1/2m] and many

of these ions have unpaired electrons(s) in the outermost orbital. [1/2m] These unpaired electrons can absorbed electromagnetic radiation [1/2m] and undergo transition in

the visible region and hence the ions are coloured. [1/2m] (ii) CrO3 < Cr2O3 < CrO [1m] Increase in oxidation state [1/2m], increases the polarizing power [1/2m] of chromium, thus the

electron density of oxygen decreases [1/2m] and its tendency to react with proton also decreases[1/2m] (iii)

Cr

Cl

Cl

enen

+

Cl- Cr

Cl

en

en

Cl

Cl-

+

(and its mirror image) [1m] trans-dichlorobis-(ethane-1,2-diamine)chromium(III) chloride or cis-dichlorobis-(ethane-1,2-

diamine)chromium chloride [1m] (Name should agree with structure drawn.)

(iv) Fe3+(aq) + SCN-(aq) Fe(SCN)2+(aq) [1m] Fe(SCN)2+(aq) + 6F-(aq) [FeF6]3-(aq) + SCN-(aq) [1m] [FeF6]3-(aq) + 3OH-(aq) Fe(OH)3(s) + 6F-(aq) [1m] 3.(a)(i) a: sp3 b:sp c:sp2 [3 x 1/2m] NOT sp3 NOT sp NOT sp2 (ii) α: 109° β:180° θ:120° [3 x 1/2m]

(iii) CH3CH2CH2COOH or C O

2

H

[1m] (Zero mark for CH3CH2CH=CHCHO) Pentanoic acid or n-pentanoic acid [1m] (b) R-CO2Na + H+ à RCO 2H + Na+ [1m] insoluble OR CnH2n+1CO2

-Na+(aq) + H+(aq) CnH2n+1COOH(s) + Na+(aq) insoluble in H2O OR The carboxylic acid formed (from the water soluble salt) is insoluble in water (c)

(d) Alcohol molecules are held together by H-bond (or intermolecular force). [1/2m] Or In alcohols


H | R-O-H .... :O-R etc.

Due to steric interaction [1/2m], 3° alcohol forms less strong or weaker intermolecular hydrogen bonds, so its boiling point is lower. [1/2m]

Further, the molecules are less closely packed, and therefore results in lower density [1/2m] (It must be stated that 2-methylpropan-2-ol is a 3° alcohol and that butan-1-ol is not in the same way.) (e)(i) ONLY TWO structural isomers are possible for W [1/2m]

[1.5m] (ii)

CH3 - CHCH2CO2H | CH3

Br2

P

Br |CH3 - CHCHCO2H | CH3

excess

NH3

HCl*Valine

[1/2m] [1/2m] [1/2m][1/2m]

Correct structure (iii)(1) Under acidic conditions: +NH3-CH2COOH or HOOCCH2NH2

.HCl [1m] (Must show an ammonium salt in some way) (2) Under basic conditions: H2NCH2CO2

- or H2NCH2COONa etc. [1m] (Must show a carboxylic acid salt in some way, zero mark for +NH3-CH2-CO2

- in both cases.) (f)(i) Acidity depends on the equilibrium

HB(aq) H+(aq) + B-(aq) [1/2m] Both compounds have PHENOL functional group which can be involved in above equilibrium. [1/2m]

In case of phenols, equilibrium proceeds to right due to resonance stabilization of phenoxide ion: [1/2m]

i.e.

[1/2m] (ii) Y is the stronger acid [1/2m] This is because the electron withdrawing >C=O group further stabilize the phenoxide ion

or show by [1.5m]

4.(a)


[Preparation of chlorine: 2m Drying + Reaction of Cl/S: 1m + 1m Collection (condenser or ice water): 1m Disposal of excess chlorine (fume cupboard or inverted funnel in NaOH): 1m]

(b) If heated too strongly, the products may decompose back to sulphur and chlorine. [1m] (c) Distillation / Fractional distillation [1m] (d) One of the hydrolysis product is S which is oxidised to SO4

2- by nitric acid [1m] Yellow precipitate (Sulphur) is formed. (e) (1) Transfer a known volume of the test solution to a conical flask using a pipette. [1m] (2) Add to the conical flask excess CaCO3(s) to neutralise HNO3 present. [1m] (3) Add approximately 1 cm3 K2CrO4(aq) as indicator. (4) Titrate the chloride solution using standard silver nitrate [1m] (5) Red coloration occurs at the end-point. [1m] (6) Repeat the titration to obtain 2-3 consistent sets of data [1m] (f) No. of moles of Ag+ required to react with all the Cl- presnet in the hydrolysed product = 0.1 x 31.25 x 10-3 x 10 = 3.125 x 10-2 No. of moles of Cl present = 3.125 x 10-2 [1m] No. of moles of S used = 0.8 / 32.06 = 0.025 [1m] No. of moles of S : No. of moles of Cl = 2.5 x 10-2 : 3.125 x 10-2 [1m] = 1 : 1.25 Empirical formula of Binary Compound is SCl1.25 or S4Cl5 [1m] (If SCl: 0m ; the product is a mixture of S2Cl2 and SCl2) Essay type questions 1. 1.Chemical knowledge Core marks(max 6) -Emission spectrum of hydrogen consists of series of discrete lines which converge in different parts of the electromagnetic radiation(1) -e- in an atom can exist in certain ‘allowed’ orbits i.e., energy of e- is quantised.(1) - Upon absorption of energy, an e- moves from a lower energy orbit to a higher energy orbit. When falls

from a higher energy orbit to a lower energy orbit, it emit energy in form of a photon, the energy of which corresponds to a line in the emission spectrum.(1)

- The energy of photon emitted equals to the energy difference between two orbitals..(1) - The convergence limit corresponds to the ionization of hydrogen - Relationship between emission spectrum and electronic structure.(1) Optional marks(max 4) - Formation of the H emission spectrum: the discharge tube(1) - Names and position of the different series of spectrum lines.(1)

NaOH


e.g. Lyman series: UV region Balmer series: Visible region -Mathematical relationship concerning the wavelengths of the spectral lines(1) (out of syllabus) -Meaning of ground state and excited state(1) -Definition of principal quantum number(1) -E=hv or E= hc/λ to calculate energy of a photon(1) -Zeeman effect (1) 2. 1. Chemical knowledge 1. Electron configuration(I mark) From Ti-Cu, they form ions an incomplete d-electron subshell e.g. Cu2+ (Ar) 2. Atomic properties(2 marks) The variation of the atomic radius (I.E., electronegativity) from one member to another is small Nuclear increases from Ti to Cu, but electron is added to the inner d-subshell. This gives better shield of outer electrons form the increasing nuclear charge and consequently the atomic radius decreases but less rapidly than in Na to Ar 3. Physical properties(2 marks) Strong metallic bonds through close-packed structure and small atomic radius, therefore high melting points: boiling points: densities anf heat of fusion and vaporization than main group metals High tensile strength and good mechanic properties Thus used as constructional material. 4. Chemical properties( 1 mark) Most of these metals react slowly with dil. Acids owing to protection by a thin impervious and unreactive oxide layer. Ions of these metals have a high charge/radius ratio. Thus there is highly polarsing which results in (compared with s-block elements): (1) Oxides and hydroxides in oxidation state +2, +3 are less basic and less soluble (2) Salts less ionic and less thermally stable (3) Salts and aquesous ions more hydrated and more readily hydrolysed forming acid salts (4)ions more easily reduced 5. Other properties of the transition metals/transition metal ions(4 marks) (1) Variable oxidation state -Exhibit more than 1 oxidation state since energy difference is small for 3d and 4s levels -Common oxidation states: +2, .+3 -Up to Mn, max oxidation number = electrons outside Ar core -Highest oxidation state occurs in compound of O and F. -Ti, V, Cr, Mn, never form simple ions in highest oxidation state since ions of extremely high charge density will result, these compounds exhibit highest oxidation state either covalently bonded TiO2, V2O5 or complex ions VO3

-,MnO4-

-Stability of ion related to 1/2 filled shell: Mn2+ and Fe3+ more stable. (2) Complex ion formation –High charge density on central metal ion helps formation of complex where co-ordination no. is often 6. Hydrated complexes dissociate Fe(H2O)6

3+ = (Fe(H2O)3OH)2+ + H+ Hence aq. Solution are acidic. For oxidation state. +3, oxo-anions formed, thus CrO4

2-, MnO4-

- If C.N = 6- complex is octahedral arrangement - C.N=4- complex is usually tetrahedral arrangement /square planar - Isomerism- structural, geometric, optical, examples

(3) Colored compounds - Energy level splitting in d-blocks is small and absorption falls in the visible region of the spectrum

therefore salts are colored. (4) Catalytic properties -Explained in terms of the ability to exist in variable oxidation state or co-ordination compound formation 1. Mark for each property


3. 1. Chemical knowledge Introduction: A good will begin with a definition of what a polymer is Basically: a very large molecule obtained from monomers. Types pf polymers This essays requires that some kind of classification be given. There are many ways of doing this. This part also be accompnmied by examples of each types. Addition polymers General reaction:

G **

G G G Gn

characterized by repeating sub-units G=H, polyethylene G=Cl, PVC G=CN, orlon

G=CO2Me Perspex, others Teflon(from CF2 CF2 ), Natural Rubber (from= )

CO Polymers\ Similar to above but with 2 different monomers. Condensation polymers Formed from intermolecular reactions in which sub-units joined through elimination of small molecules e.g. H2O, ROH Examples: Polyamides Though some examples given in the original marking scheme are out of the syllabus. Thus the following will just include the name of it.(they are in syllabus) 1. Equations of Nylon 2. equations of formation of terylene Mechanisms of their formation Addition polymers: Mechanisms are generally Chain reaction involving initiation, propagation, termination. Addition may occur via the following 1. Radical polymerization 2. Cationic polymerization (Out AL) 3. Anionic polymerization (Out AL) Condensation Polymer mechanisms: nucleophilic addition followed Elimination

Sources: CH2 CH2 - cracking of petroleum

CH2CHCl

- addition of HCl to HC CH from petroleum source

Aromatic compound-

OH OH

Nylon

terephthalic

All from petroleum sources This could be elaborated with reactions Suggested allocations of marks


Definition/Introduction max0.5 Well explained types -up to 1.5 for each max3.5 Mechanisms max3.5 Sources max1 Examples(with 2 formula, name 2 for 1/2)

1993 AL Chemistry Paper I Marking Scheme

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