2021 9729 P2 Chemistry Suggested Solutions

(a) (i) (63.8/66.6x 31.97 ) + (2.8/66.6 x 33.97) = 32.33
  (ii) FeS2

(c) (i) First IE is the energy required to remove 1 mole of electrons from 1 mole of gaseous S atoms to form 1 mole of S+ ions.


  (ii) Across the period, nuclear charge increase, shielding effect remains relatively constant, first IE increases due to increase nuclear attraction. Sulfur has a pair of e in 3p orbital which results in inter-electronic repulsion.
  (d) (i)

C has 2 sp hybrid orbitals to form sigma bonds w each S and 2 unhybridised p orbitals to form pi bonds w each S respectively.

2 (a) (i) Down the group, quantum shells increases, ionic radius increases.


    (ii) Cs has larger ionic radius, can accommodate more anions around it.


  (b)   Factors: charge and ionic radius
Charge has greater effect as it is multiplication compared to addition (for radius)
  (c) (i)
    (ii) Energy released on forming ion dipole interactions between Na+/Cl w water is sufficient to compensate for energy required to break ionic bonds in NaCl.

In hexane, no ion-dipole interactions are possible / no favourable interactions.



  (d) (i) NH4Cl(s) -> NH4+ (aq) + Cl(aq)

207K is below freezing point of water. Water would exist as ice, unable to have any solvation.

3 (a)   H-F BE = 562 (stronger, less likely to dissociate, so weaker acid)
H-Cl BE = 431


  (c) (i) F / HF buffering system

HF + OH -> F + H2O
F + H+ -> HF


    (ii) Cl is a neutral salt. It is unable to resist a change in pH when a small amt of strong acid/base is added.


    (iii) nH+ = 50/1000 x 0.1 x 2 = 0.01 mol

nF in 200 cm3 = 100/1000 x 1.78 = 0.178 mol

nF in 75 cm3 = 0.178 / 200 x 75 = 0.06675



  (d) (i) Smaller pKa, stronger acid, larger extent of dissociation. CCl3COOH has larger extent of dissociation than CH3COOH.
    (ii) 0.50 (any value smaller than 0.66). F is more electronegative than Cl, so it will stabilise CF3COO more than CCl3COO. CF3COOH is more acidic than CCl3COOH.
  (e) (i) CaF2 ⇌ Ca2+ + 2F  —– (1)

Ksp = [Ca2+] [F ]= (x)(2x)2 = 4x3

X = 2.14 x 10-4

[F ] = 2x = 4.27 x 10-4


    (ii) H+ + F  ⇌ HF  —– (2)

In acidic soln, [H+] increase, eqm (2) shifts right, [F] decreases, eqm (1) shifts right, hence CaF2 becomes more soluble.

4 (a)   All bonds are saturated. C-F bond is also quite strong so unable to perform nucleophilic subn rxn.


  (b)   Ozone layer depletion due to CFC.


  (c)   High temp. Low pressure.


  (d) (i) Covalent bond breaks where each atom obtains 1 e.


    (ii) A: initiation
B: propagation
C: propagation
D: termination 
    (iii) Free radical addition
    (v) ∆H = 150 + 12(610-350) [pi bond only]  – 2(360) – 11(350) = -1300 kJ mol-1


    (vi) P is the major species. Radical can be stabilised via delocalisation into the benzene ring.
5 (a)   Reducing agent. Reduces Fe3+ -> Fe2+


  (b)   Dative bond from lone pair of O to Fe2+
  (c)   To deprotonate HSCH2COOH


  (d)   HSCH2COOH + NH3 -> NH4+ + HSCH2COO


  (e) (i) HSCH2COOH has stronger id-id interactions due to larger e cloud size, more e to be polarised.


    (ii) HOCH2COOH has stronger pd interactions because O is more electronegative than S in HSCH2COOH.