**CHE203A Solid State and Inorganic Chemistry** (TSS)

*Practice Problems 1 (Answers) *

*Complete by Tuesday 12 noon Week 8 *

- 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]3*d*^{10}4*s*^{2}4*p*^{1} and Al [Ne]3*s*^{2}3*p*^{1}. Even though valence shell of Ga is further from nucleus than Al and lower electronegativity would be expected , 10 electrons in 3*d* orbitals of Ga shield nuclear charge (*Z*) poorly so effective nuclear charge (*Z**) is greater for Ga than Al.

- Use electronegativities to predict the nature of bonding in SnI
_{4.} Do the physical properties of SnI_{4} confirm your prediction?

F_{P} (Sn(IV)) = 2.0 and F_{P} (I) = 2.7 therefore ‘F_{P} = 0.7 hence bonding in SnI_{4 } is predicted to be polar COVALENT *i.e.* bond polarity but net diploe is

zero (tetrahedral). Hence expect SnI_{4 }to be a low melting point (m.p. 144°C) solid and soluble in organic solvents.

- Why do Lewis acidities decrease in the following order: BBr
_{3} > BCl_{3} > BF_{3} even though elec onega i i are F > C l > Br?

X (F, Cl, Br, filled n*p *o bi al) B (emp 2*p*) S back-bonding is most effective if sizes and energies of over-lapping orbitals are similar hence most effective in BF_{3} (F has filled 2*p* orbitals) hence since B 2*p* orbital is now quite electron-rich, it is the least able to act as a Lewis acid (lone pair acceptor). In BBr_{3} S back-bonding Br (4*p*) B (2*p*) is least effective; note also increased bond distance and so less effective sideways

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

- State whether the following neutral boranes are
*nido* or *arachno*: B_{7}H_{11}; B_{13}H_{19}; B_{16}H_{20}; B18H22.

B7H11 *nido *(BnHn+4)

B13H19 *arachno *(BnHn+6)

B16H20 *nido *(BnHn+4) B_{18}H_{22} *arachno *(B_{n}H_{n+6})

- Write down the
*styx* code for pentaborane(11), B_{5}H_{11}, using the following topological diagram.

3203

**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 B_{6}H_{10}, [B_{3}H_{8}] , C_{2}B_{10}H_{12 }and [C_{2}B_{3}{Fe(CO)_{3}}H_{5}]. State the formulae of the parent *closo* borane in each case and *if appropriate* the corresponding neutral borane formula for each of the heteroboranes.

# B6H10

**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: [B**_{7}H_{7}]^{2- }

**[B3H8]– **

**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: [B**_{5}H_{5}]^{2- }

# C2B10H12

**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: [B**_{12}H_{12}]^{2- }

**[C**_{2}B_{3}{Fe(CO)_{3}}H_{5}]

**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 **

# [B6H6]2-

2. B_{5}H_{11} 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 B_{5}H_{11}.

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

**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: [B_{7}H_{7}]^{2- }

3. **Electron-precise Rules**

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

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

Show if B_{3}N_{3}H_{6} and P_{4} obey all of the above rules. Explain how these rules can be used to distinguish the difference between a *ring* and a *cluster*.

| | |

(a) | Yes | Yes |

(b) | B/N Yes | Yes |

(c) | No (ring) | Yes (cluster) |

(d) | Yes B/N-H bond | Yes P lpe |

(e) | 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 *

- Write down the formulae of the nitrides of Li, Mg, Al, Na and Ca and explain why only the first three are ionic. [
*r*_{ionic} / Å: N^{3 };1.32; Li^{+} 0.73; Mg^{2+} 0.71; Al^{3+} 0.53; Na^{+} 1.13; Ca^{2+} 1.14]

Li_{3}N; Mg_{3}N_{2}; AlN; Na_{3}N; Ca_{3}N_{2}. Li^{+}, Mg^{2+} and Al^{3+} 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 N^{3} from N_{2}; Na^{+} and Ca^{2+} are too big and the lattice energies are not large enough.

- Draw mechanisms for the reaction of borazine with HBr and Br
_{2}.

- 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.

- Why does NO
_{2} readily dimerise to produce N_{2}O_{4} but NO only reluctantly dimerises (in solid state) to produce N_{2}O_{2}. (Hint: think about the number and types of bonds in reactants and products).

- Explain the order in O N O bond angles: NO
_{2}^{+} (180°) > NO_{2} (134°) > NO_{2} (115°).

NO_{2}^{+ }is isoelectronic with CO_{2} therefore no unpaired electron(s) at central N atom therefore linear; NO_{2} has one more electron than NO_{2}^{+} 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; NO_{2} is isoelectronic with O_{3} and has one more electron than NO_{2} *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.

- Tl in TlF
_{3} is in the +3 oxidation state but Tl in TlI_{3} 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 6*s*^{2} electrons) but because Tl F bonds are strong, Tl can overcome the promotional energy required to promote an electron from 6*s* to 6*p* through the two strong extra Tl F bonds formed to give trivalent TlF_{3} (*c.f.* BF_{3}). 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^{+}I_{3 }.

- Why is P
_{4}N_{4}Cl_{8} puckered while P_{4}N_{4}F_{8} is planar? Note for (PNR_{2})_{n} Q(P N) decreases along the series R = F > Cl > Br > Me.

P_{4}N_{4}F_{8} is planar to maximise Sbonding from 2*p*_{z }(N) to 3*d* (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 3*d* (P) with 2*p* (N).

P_{4}N_{4}Cl_{8} is puckered because Sbonding is not important and therefore steric factors dominate structure of ring.

**STUDENT SIGNATURE: DATE:**

1. Explain the following periodic trends:

(a) melting point of BiF_{3} (650 C) >> melting point of PF_{3} (–150 C); [2 *marks*]

(b) SnF_{2} exists, but SiF_{2} does not. [2 *marks*]

2. A *closo* borane *anion* has the following structure

(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 [C_{2}B_{3}H_{7}]. 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 S_{8}(g) to 4 S_{2}(g) is +401 kJ and the average S–S bond energy in S_{8} is 211 kJ mol^{–1}

(a) Calculate the energy of the S_{2} bond. [3 *marks*]

(b) Explain why the conversion of S_{8} to S_{2} is endothermic. [2 *marks*]

4. (a) Use VSEPR theory to draw the shape of the [XeF_{5}]^{+} cation. [2 *marks*]

(b) Construct the appearance of the low temperature ^{129}Xe NMR spectrum for [XeF_{5}]^{+}.

[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 (F_{3}B←NH_{3}).

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

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

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