Week 5

Question 1

How do you synthesise Zintl phases?

Answer 1

Synthesis: The reduction of a less electropositive metal, E (E = Ge, Sn, Pb, In, Tl) by a highly electropositive metal M (M= Na, K) leads to Zintl Phases, MnEx. For example: 4Na + 9Sn → Na4Sn9.
Low T method with condensed NH3(l)
High T method using solid-state techniques

Question 2

How can you crystallise the naked metal cluster?

Answer 2

by sequestering the alkali metal cation with a crown ether or cryptand:
(MnEx) + cryptand n[M(cryptand)]+ + [Ex]n-
[Sn9]4-

Question 3

Why is N - - - N stable but P - - - P is not?

Answer 3

N: 1σ + 2π > 3σ
P: 3σ > 1σ + 2π
Βecause:
- 2nd row elements are small, so good p overlap, s and p bonds close in strength
- 3rd row and later – larger covalent radius, good s overlap, but poorer p overlap, s&raquo_space; p for bond strength

Question 4

CO2 molecular. SiO2 polymeric.
Why is this?

Answer 4

Due to difference in C=O Vs Si=O π bond strengths relative to the respective σ bond strength
C: σ + π bonds more stable than O all s bonds
Si: all σ more stable than σ + π bonds

Question 5

what are the common oxidation states of group 14?
C Si Ge Sn Pb

Answer 5

C +4 -4
Si +4 +2(highly reactive)
Ge +4 +2(highly reactive)
Sn +2 +4 both stable
Pb +2(most common) +4(good oxidising agent)

Question 6

Define Comproportionation

Answer 6

Comproportionation: two reactants containing the same element but with different oxidation states, react to form a product with an intermediate oxidation state.

Question 7

Pb(s) + PbO2(s) + 2 H2SO4(aq) → 2 PbSO4(s) + 2 H2O(l)
Pb4+ + 2e → Pb2+ E0 = 1.46V
Pb2++2e→Pb E0 =-0.13V
Is the reaction spontaneous?

Answer 7

Pb(s) + PbO2(s) + 2 H2SO4(aq) → 2 PbSO4(s) + 2 H2O(l)
Pb4+ + 2e → Pb2+ E0 = 1.46V
Pb2++2e→Pb E0 =-0.13V
Eocell = 1.46 – (-0.13) = 1.59 V
SPONTANEOUS

Question 8

on heating 2GeBr2 →

Answer 8

on heating 2GeBr2 → GeBr4 + Ge

Question 9

Why is ΔHf SiCl4&raquo_space; CCl4

Answer 9

due to a larger ionic contribution to Si-Cl bond, so significant covalent and ionic contribution to bonding in Si-Cl

Question 10

Give the stability of the +2 halides
PbCl2 SnCl2 GeCl2 SiCl2 CCl2

Answer 10

PbCl2 > SnCl2 > GeCl2&raquo_space; SiCl2&raquo_space; CCl2

Question 11

PbF4 is much more stable than PbCl4 – Why?

Answer 11

• bond dissociation energy of F2 is low, strong Pb-F bonds due to large ionic contribution to Pb-F bond
• F- is very hard to oxidise
• PbF4 → PbF2 +F2 hasaverylarge+ve ΔG at STP

Question 12

GeCl4 + 4RMgBr →

Answer 12

GeCl4 + 4RMgBr → GeR4 + 4MgBrCl

Question 13

What is the valency of Group 15

Answer 13

3 and 5