Week 3

Question 1

What are the general properties of s-block metals?

Answer 1

Soft, reactive metals
Smooth trends in most properties
Low electronegativity, so valence orbital energies high
Bonding is predominantly ionic

Question 2

4 Li + O2 ->
2 Na + O2 ->
K/Rb/Cs + O2 ->

Answer 2

4 Li + O2 -> 2 Li2O Oxide
2 Na + O2 -> Na2O2 Peroxide
K/Rb/Cs + O2 -> MO2 Superoxide

Question 3

When is salt dissolution favoured and spontaneous?

Answer 3

For dissolution to be spontaneous
ΔGdissolution must be negative:
ΔGdissolution = ΔH - TΔS

Favoured:
ΔGsolvation > ΔGLatt

Question 4

What is ligand field stabilisation?

Answer 4

ligand field stabilization refers to the stabilization of metal ions in coordination complexes due to the specific arrangement of ligands around the metal ion, which affects the energy levels of the metal ion’s d orbitals

Question 5

Describe the process of dissolution of a salt in water

Answer 5

On dissolution “lone pairs” of solvent molecules act as a Lewisbase(egH2O,NH3)& coordinate to metal ion (Lewis acid).
If M is charge dense (i.e. small, and/or high charge) it can polarize the M-OH2 bond), this weakens the O-H bonds in water (so that dissociation of H+ occurs to a greater extent)

Question 6

Why is lithium different in dissolution and E values?

Answer 6

Li has the smallest primary hydration sphere of all the metals, but has the largest secondary hydration sphere, and total hydration number, as it has the highest charge density.
Lithium is out of line and has the most reducing Eo value because the greater solvation energy overcomes the higher ionisation energy (compared to other group 1 metals

Question 7

What is the pKa trend down groups 1 and 2

Answer 7

Group 1 pKa decreases down the group
Increasing covalency to water down the group - becomes more acidic

Group 2 pKa increases down the group - becomes less acidic

Question 8

What is the formula for pKa?
What does a low pKa mean?

Answer 8

pKa = -log( [H3O+][A-] / [HA]
low pKa: stronger acid

Question 9

What are the 2 equations for:
ΔG = -RΤlnβ beta is reaction quotient
ΔG =

Answer 9

ΔG = -RΤlnβ
beta is reaction quotient and highly +ve
ΔG = ΔH - TΔS

Question 10

Why is a multi dentate ligand more stable?

Answer 10

If a multidentate ligand is used then a more stable complex is formed due to the macrocyclic effect

Question 11

Will group 2 salts generally be more or less soluble than Group 1?

Answer 11

Lattice energy is proportional to z+z–/r. Group 2 metal ions are smaller and higher charge than group 1. So G2 salts have larger lattice enthalpies than G1, therefore less soluble.

Question 12

Both MCO3 or MSO4 decompose on heating. Will they form oxides, superoxides or peroxides?

Answer 12

G2 ions are smaller than G1, so expect smallest anion ie. oxide. MCO3 ->MO+CO2 MSO4 ->MO+SO3

Question 13

Will the 1st I.E. for Ra be higher or lower than Ba? (kJ mol-1)
Ca 589
Sr 550
Ba 503
Ra = ?

Answer 13

Ra = 509
Expect 7s to be effected by relativity.

Question 14

Describe the properties of Be

Answer 14

Extremely small,
common CN = 4
highly polarizing, significant covalent contribution to bonding.
“salts” are very acidic in water and covalent

Question 15

Be(OH)2 + 2NaOH ->
Be(OH)2 + H2SO4 ->

Answer 15

Be(OH)2 + 2NaOH -> Na2[Be(OH)4]
(reacts with acid & base)
Be(OH)2 + H2SO4 -> BeSO4 + 2H2O

Question 16

What is the most common oxidation state of group 12?

Answer 16

+2

Question 17

Why is Hg an outlier in ionisation energy?
(much higher than expected)

Answer 17

The relativistic contraction of the 6s orbital results in a decreased shielding effect from the inner electron shells, leading to a contraction of the atomic size and higher ionization energy than expected.

Question 18

2ZnS + 3O2 ->
ZnO + C ->
HgS+O2 ->

Answer 18

2ZnS + 3O2 -> 2ZnO + 2SO2
ZnO + C -> Zn + CO
HgS+O2 ->Hg+SO2

Question 19

3 PhMgBr + PBr3 →
3PhLi + PBr3 →

Answer 19

3 PhMgBr + PBr3 → PPh3 + 3 MgBr2 (in Et2O) 3PhLi + PBr3 → PPh3 +3LiBr (inEt2O)

Question 20

Use the following data :
Tl3+ + 2e → Tl+ Eo = 1.25 V,
0.5 I2 + e- → I- Eo = 0.54 V,
to determine if the following reaction spontaneously occur?
TlI3 → TlI + I2

Answer 20

In the chemical reaction written above Tl3+ is reduced to Tl+ (OILRIG) and I- is
oxidised to I2.
So the cell potential (Ecell) = Ered – Eox = 1.25 V – 0.54 V = +0.71 V.
As Ecell is positive so DG (DG = -nFE) is negative, therefore spontaneous.
Thinking about the reaction backwards, the energy released by forming two additional (weak) Tl-I bonds in making TlI3 does not compensate for cleaving of I-I and ionization of Tl(I) to Tl(III). An example of the inert pair effect.

Question 21

For which halogen will the reaction above favour the left hand side, ie TlX3?

Answer 21

In the reaction I- is oxidised to I2 by Tl3+, so need a halide that can’t be easily oxidised (i.e. loose electrons), most obvious other halide would be F-, so TlF3, ie match a hard acid (metal ion with a high charge) with a hard base (small anion)

Question 22

For the following sequence of reactions identify A, B, C and D when (i) M = Li and (ii) M = Cs.
A <-H2O– M –O2-> C –heat-> D
_NH3_> B

Answer 22

Li : small and polarising so high degree of covalent character to bonding
Cs: large, mainly ionic bonding.
TlF3
Li
A LiOH (slow reaction)
B [Li(NH3)4]+ + e-(NH3)6
C Li2O (oxide)
D Li2O

Cs
A CsOH (explosively fast)
B[Cs(NH3)6]+ + e-(NH3)6
C CsO2 (superoxide)
C Cs2O

Question 23

How would you prepare the following compounds? Write a balanced equation in each case.
i) PhLi (starting with Li)
ii) Ph2PH (starting with PCl3)
iii) Li2CO3 (starting with Li)
iv) ND3 (from D2O) [D = deuterium = 2H]

Answer 23

i) PhLi (starting with Li)
PhBr + 2Li → PhLi + LiBr
ii) Ph2PH (starting with PCl3)
PCl3 + 2PhLi → PPh2Cl
PPh2Cl + LiAlH4 → PPh2H
iii) Li2CO3 (starting with Li)
Li + O2 → Li2O
Li2O + CO2 → Li2CO3
iv) ND3 (from D2O)
6Li + N2 → 2Li3N
Li3N + 3D2O → ND3 + 3LiOD

Question 24

Identify the compounds A, B, C and D in the scheme below:
2ButBr + 2Mg → 2A + BeCl2 → B → heat → D + 2Me2C=CH2 +C
(m.p. 710 °C)

Answer 24

A Bu tMgBr.
B BeCl2
C MgBrCl
D BeH2