Week 2

Question 1

Describe the electronegativity for metals and molecular compounds

Answer 1

Metals - low electronegativity (high orbital energy) relatively large, diffuse non-directional orbitals: 3D packing
Molecules - High electronegativity (low orbital energy) relatively small directional orbitals: low co-ordination numbers

Question 2

What are the general characteristics of ionic compounds?

Answer 2

• 3D (extended) structures
• High melting points and boiling points
• Can conduct electricity when molten
  but not when a solid

Question 3

What are the general characterisitcs of polar covalent molecules

Answer 3

• Structures with discrete molecules
• Relatively low melting points and
  boiling points
• Unable to conduct electricity when solid
  or molten

Question 4

Describe the 3 lattice structures

Answer 4

1. Body centred cubic
   - Co-ordination number: 8
   Radius ratio: <0.732
2. Face centred cube
   - Co-ordination number: 6, octahedral holes
   Radius ratio: 0.732 - 0.414
3. Face centred cubic
   - Co-ordination number: 4 tetrahedral holes
   Radius ratio: 0.414 - 0.225

Question 5

What is the ratio of
spheres : octahedral : tetrahedral holes ?

Answer 5

1 : 1 : 2

Question 6

What are the assumptions in calculating lattice energy?
What is a common occurrence that makes the lattice energy less accurate?

Answer 6

• Bonding is purely ionic
• Cations and anions are undistorted spheres
  Less accurate: Small, highly charged cations are strongly polarizing (high charge density e.g. Be2+, Th4+). These cations will strongly attract electron density from polarizable anions (large, low charge) that leads to some degree of covalency.

Question 7

Give the outcomes and reason for the reaction below:
Na2O +H2O ->
MgO +H2O ->

Answer 7

Highly ionic oxides are bronzed bases so can deprotonate water
Na2O +H2O -> 2NaOH
MgO +H2O -> Mg(OH)2

Question 8

Give the outcomes and reason for the reaction below:
Molecular covalent oxides on dissolution in water are bronsted bases
SO2 + H2O ->
H2SO3 + HSO3 ->

Answer 8

SO2 + H2O -> H2SO3
H2SO3 + HSO3- -> SO32-

Question 9

How could you synthesise Hydrogen?

Answer 9

Group 1/ metal + water
2M + 2H2O -> 2M(OH) + H2

Metal+acid :
2M+2HX->2MX+H2

Industrially, water-shift reaction
CH4 + H2O -> CO + 3H2

Question 10

What is a hydridic substance?

Answer 10

Hydridic substances react either as hydride ion (H–) donors or contain a d- hydrogen
There are 2 types: ionic and molecular

Question 11

What are the general trends of ionic hydrides?

Answer 11

• solid with high m.p
• Highly reactive
• strong bronsted bases

Question 12

8LiH + Al2Cl6 ->
4NaH + BCl3 ->

Answer 12

8LiH + Al2Cl6 -> 2LiAlH4 + 6LiCl
4NaH + BCl3 -> NaBH4 + 3NaCl

Question 13

Why is BH4 less reactive than AlH4?

Answer 13

B-H bond stronger than AlH
χ(H) > χ(B) > χ(Al)
so Al-H bond is more polarised

Question 14

What are the conditions to convert a cyclic ketone to a secondary alcohol?

Answer 14

1. LiAlH4 , dry THF 0*c
2. NaΒH4 , EtOH / H2O 25*c

Question 15

What is the trend of thermodynamic stability of covalent hydrides down a group?

Answer 15

Thermodynamic stability of hydrides decreases down a group

Question 16

Why does acidity of covalent hydrides increase down a group?

Answer 16

Weaker E-H bonds due to the poorer overlap between 1s orbital of H and the increasingly large np orbital (covalent bond strength decreases down the group)

Question 17

Both NaH and CaH2 adopt ionic structures. Describe the structures in terms of close-packed spheres and filling of holes.
r(Ca2+) = 114 pm,
r(Na+) = 116 pm,
r(H-) = 160 pm.
Which of the two hydrides will be more hydridic (H-)?

Answer 17

CaH2 : r+/r-= 114 pm/160 pm = 0.71 …. C.N = 6 NaH will be similar R+/r- ratio (0.725), so also C.N = 6.
0.414 — 0.225 4-coord
Lattice of H- (larger ion) with Na+ in the octahedral holes.
For NaH the atomic ratio is 1:1, the same as the ratio of spheres:oct. holes, so all holes are occupied.
For CaH2 the Ca2+ cations are in 1⁄2 Oct holes => Ca1⁄2H => CaH2
Ca2+ is more polarising than Na+, so CaH2 is more covalent therefore NaH is the better H- source.

Question 18

For each of the pairs below which is more covalent?
BF3 LiF
CsF LiF
NaI NaF
AlN NaF

Answer 18

BF3 > LiF
B3+ very polarising cation (higher charge) compared with Li+

CsF < LiF
Li+ (smaller) more polarising than Cs+

NaI > NaF
Same cation, so consider polarizability of the anion: I- (larger) more easily polarised than F-

AlN > NaF
Al3+ (high charge) is more polarising than Na+ & N3- (high “excess” charge) will be more easily polarised than F-

Question 19

Suggest identities for A, B in the reaction sequence below of balanced reactions, and propose a reagent C.
Identify the ionic packing for B, given that the radii of the cation and anion are 126 and 179 pm, respectively.
Mg + H2 → A
2 A + NaBO2 → B + 2 MgO
B + 4 C → Na[B(OEt)4] + 4H2

Answer 19

A must be the hydride of Mg, so MgH2.
B can be obtained from the fact that the equations are said to be balanced, so: (2MgH2 + NaBO2) – 2MgO = NaBH4
C: One of the products is H2, we know NaBH4 is a source of H-, so we require H+ to form H2, and need EtO- for the other product, so ethanol would work:
NaBH4 + 4EtOH → Na[B(OEt)4] + 4H2
The radius ratio for the ions in B is 126pm/179pm = 0.704, so expect 6 coordination, ie a lattice of (BH4)- anions with Na+ cations in the octahedral holes (i.e. like NaCl).