Solid State and Inorganic Chemistry Questions

CHE203A Solid State and Inorganic Chemistry (TSS)


Practice Problems 1 (Answers)

Complete by Tuesday 12 noon Week 8


  1. Explain the anomaly at gallium (Ga) which has a higher electronegativity than aluminium (Al) (Hint: Write out the electronic configurations for Ga and Al).


Ga [Ar]3d104s24p1 and Al [Ne]3s23p1. Even though valence shell of Ga is further from nucleus than Al and lower electronegativity would be expected , 10 electrons in 3d orbitals of Ga shield nuclear charge (Z) poorly so effective nuclear charge (Z*) is greater for Ga than Al.


  1. Use electronegativities to predict the nature of bonding in SnI4. Do the physical properties of SnI4 confirm your prediction?

word image 849

FP (Sn(IV)) = 2.0 and FP (I) = 2.7 therefore ‘FP = 0.7 hence bonding in SnI4 is predicted to be polar COVALENT i.e. bond polarity but net diploe is

zero (tetrahedral). Hence expect SnI4 to be a low melting point (m.p. 144°C) solid and soluble in organic solvents.


  1. Why do Lewis acidities decrease in the following order: BBr3 > BCl3 > BF3 even though elec onega i i are F > C l > Br?


word image 850 X (F, Cl, Br, filled np o bi al) B (emp 2p) S back-bonding is most effective if sizes and energies of over-lapping orbitals are similar hence most effective in BF3 (F has filled 2p orbitals) hence since B 2p orbital is now quite electron-rich, it is the least able to act as a Lewis acid (lone pair acceptor). In BBr3 S back-bonding Br (4p) B (2p) is least effective; note also increased bond distance and so less effective sideways

overlap and B 2p orbital is quite electron-poor and so better able to act as a Lewis acid.


  1. State whether the following neutral boranes are nido or arachno: B7H11; B13H19; B16H20; B18H22.


B7H11 nido (BnHn+4)

B13H19 arachno (BnHn+6)

B16H20 nido (BnHn+4) B18H22 arachno (BnHn+6)


  1. Write down the styx code for pentaborane(11), B5H11, using the following topological diagram.


word image 3200


CHE203A Solid State and Inorganic Chemistry (TSS)


Practice Problems 2

Complete by Tuesday 12 noon Week 9


1. U e Wade e c a f he f g B6H10, [B3H8] , C2B10H12 and [C2B3{Fe(CO)3}H5]. State the formulae of the parent closo borane in each case and if appropriate the corresponding neutral borane formula for each of the heteroboranes.




Total no. of electrons = (3e x 6) + (1e x 10) = 28 e


Substract 2c-2e B-H bonds: 6 x 2e = 12 e


Therefore: (28 – 12) e = 16 e


SEP = 16 / 2 = 8


(n+2) therefore nido


Equivalent closo structure: [B7H7]2-




Total no. of electrons = (3e x 3) + (1e x 8) + 1 e = 18 e


Substract 2c-2e B-H bonds: 3 x 2e = 6 e


Therefore: (18 – 6) e = 12 e


SEP = 12 / 2 = 6


(n+3) therefore arachno


Equivalent closo structure: [B5H5]2-




Total no. of electrons = (4e x 2) + (3e x 10) + (1e x 12) = 50 e


Substract 2c-2e C/B-H bonds: 12 x 2e = 24 e


Therefore: (50 – 24) e = 26 e


SEP = 26 / 2 = 13


(n+1) therefore closo


Equivalent closo structure: [B12H12]2-





Total no. of electrons = (4e x 2) + (3e x 3) + (8e x 1) + (2e x 3) (1e x 5) = 36 e


Substract 2c-2e C/B-H bonds: (5 x 2e) + (2e x 3) + (6e, t2g of Fe) = 22 e


Therefore: (36 – 22) e = 14 e


SEP = 14 / 2 = 7


(n+1) therefore closo





2. B5H11 has a has a styx code (3203). Determine the number of SEP, classify this borane and state the formula of the parent closo borane. Construct a topological diagram for B5H11.


SEP = (s + t + y +x) = (3 + 2 + 0 +3) = 8



word image 3201

Total no. of electrons = (3e x 5) + (1e x 11) = 26 e


Substract 2c-2e B-H bonds: 5 x 2e = 10 e


Therefore: (26 – 10) e = 16 e


SEP = 16 / 2 = 8


(n+3) therefore arachno (as predicted by BnHn+6 formula)


Equivalent closo structure: [B7H7]2-





3. Electron-precise Rules

The following rules apply to electron-precise rings and clusters.


  1. Simple localised bonding treatments are adequate.
  2. Each atom in the cluster obeys the octet rule.
  3. Each vertex of the cluster is three coordinate (a connectivity of three).
  4. Each cluster atom has two electrons in a lone pair or a bond to an exocyclic (external) group.
  5. An n-atom cluster will have 5n electrons of which 3n are involved in cluster bonding.


Show if B3N3H6 and P4 obey all of the above rules. Explain how these rules can be used to distinguish the difference between a ring and a cluster.


word image 3202

word image 3203







B/N Yes



No (ring)


Yes (cluster)


Yes B/N-H bond


Yes P lpe


5 x 6 B/N toms = 30 e


B (3 e x 3) + N (5 e x 3) + H (1 e x 6) = 30e


3n = 3 x 6 B/N atoms = 18 e i.e. 6 (2c-2e) B-N sigma bonds 3 (2c-2e) , da e N B

(2e x 3) = 18 e



5 x 4 P atoms = 20 e


5e x 4 P atoms = 20 e


3n = 3 x 4 P atoms in cluster bonding =

12 e i.e. 6 2c-2e P-P bonds = 12 e


CHE203A Solid State and Inorganic Chemistry (TSS)


Practice Problems 3 (ANSWERS)

Complete by Tuesday 12 noon Week 10


  1. Write down the formulae of the nitrides of Li, Mg, Al, Na and Ca and explain why only the first three are ionic. [rionic / Å: N3 ;1.32; Li+ 0.73; Mg2+ 0.71; Al3+ 0.53; Na+ 1.13; Ca2+ 1.14]


Li3N; Mg3N2; AlN; Na3N; Ca3N2. Li+, Mg2+ and Al3+ are small and so the large lattice energies in the formation of these ionic nitrides can overcome the large inputs of energy required to form N3 from N2; Na+ and Ca2+ are too big and the lattice energies are not large enough.


  1. Draw mechanisms for the reaction of borazine with HBr and Br2.


word image 851



word image 852


  1. Write down two redox half-equations and an overall redox equation for the preparation of hydroxylamine from the reaction of nitrite ions with bisulfite ions under basic conditions.


word image 3204


  1. Why does NO2 readily dimerise to produce N2O4 but NO only reluctantly dimerises (in solid state) to produce N2O2. (Hint: think about the number and types of bonds in reactants and products).


word image 853


  1. Explain the order in O N O bond angles: NO2+ (180°) > NO2 (134°) > NO2 (115°).


NO2+ is isoelectronic with CO2 therefore no unpaired electron(s) at central N atom therefore linear; NO2 has one more electron than NO2+ therefore 1 electron at central atom and since single electron bond pair repulsion is greater than bond pair bond pair repulsion the molecule is bent; NO2 is isoelectronic with O3 and has one more electron than NO2 i.e. a lone pair and since lone pair bond pair is greater than single electron bond pair and bond pair bond pair repulsion the molecule is bent further.


  1. Tl in TlF3 is in the +3 oxidation state but Tl in TlI3 is in the +1 oxidation state. Explain why.


Thalli i i G 13 a d bjec he i e ai effec (i hi ca e la 6s2 electrons) but because Tl F bonds are strong, Tl can overcome the promotional energy required to promote an electron from 6s to 6p through the two strong extra Tl F bonds formed to give trivalent TlF3 (c.f. BF3). However, Tl I bonds are weak and the promotional energy required is not compensated by the formation of two extra weak Tl I bonds. Note TlI = Tl+I3 .


  1. Why is P4N4Cl8 puckered while P4N4F8 is planar? Note for (PNR2)n Q(P N) decreases along the series R = F > Cl > Br > Me.


P4N4F8 is planar to maximise Sbonding from 2pz (N) to 3d (P) interaction which is strong because F is more electronegative than Cl and so makes P atom more electron-deficient and lowers their energy to enable better energy match of 3d (P) with 2p (N).


P4N4Cl8 is puckered because Sbonding is not important and therefore steric factors dominate structure of ring.



1. Explain the following periodic trends:

(a) melting point of BiF3 (650 C) >> melting point of PF3 (–150 C); [2 marks]

(b) SnF2 exists, but SiF2 does not. [2 marks]

2. A closo borane anion has the following structure

word image

(a) Write down the formula of the closo borane anion. [1 mark]

(b) Write down the number of skeletal electron pairs (SEP) in the anion. [1 mark]

(c) Determine the formula of the neutral boron hydride that can be formed by removing

two of the vertices of the above borane anion. [2 marks]

(d) Use Wade’s rules to classify [C2B3H7]. Is it related to neutral boron hydride in your answer

to (c) above? [3 marks]


3. Given that the enthalpy change for the conversion of S8(g) to 4 S2(g) is +401 kJ and the average S–S bond energy in S8 is 211 kJ mol–1

(a) Calculate the energy of the S2 bond. [3 marks]

(b) Explain why the conversion of S8 to S2 is endothermic. [2 marks]

4. (a) Use VSEPR theory to draw the shape of the [XeF5]+ cation. [2 marks]

(b) Construct the appearance of the low temperature 129Xe NMR spectrum for [XeF5]+.

[2 marks]

(c) How does the spectrum change as the sample is warmed up? [1 mark]

5. Construct labelled line diagrams to predict the appearance of the following NMR spectra for ammonia boron trifluoride (F3B←NH3).

(a) 19F NMR spectrum (11B, I = 3/2, 80.1%, but assume as a first approximation 11B is 100% abundant). [2 marks]

Modify your predicted spectrum to now show it would differ given the actual abundance of 11B, I = 3/2, being only 80.1%. [1 mark]

(b) 1H-{11B} [1 mark]

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