"p-Block Elements" — Practice

Question 1

Determine the basicity of $	ext{H}_3	ext{PO}_2$, $	ext{H}_3	ext{PO}_3$, $	ext{H}_4	ext{P}_2	ext{O}_7$ and $	ext{H}_3	ext{BO}_3$, giving a reason for each.

Accepted Answer

$	ext{H}_3	ext{PO}_2$: two $	ext{P}-	ext{H}$ bonds, one $	ext{P}-	ext{OH}$ — **monobasic**.

$	ext{H}_3	ext{PO}_3$: one $	ext{P}-	ext{H}$, two $	ext{P}-	ext{OH}$ — **dibasic**.

$	ext{H}_4	ext{P}_2	ext{O}_7$: no $	ext{P}-	ext{H}$ bonds and four hydroxyls — **tetrabasic**.

$	ext{H}_3	ext{BO}_3$: **monobasic**, but for a quite different reason. It is a Lewis acid that accepts hydroxide from water rather than releasing a proton from any of its three hydroxyls.

Question 2

Assign the oxidation state of sulphur in $	ext{H}_2	ext{S}_2	ext{O}_7$, $	ext{H}_2	ext{S}_2	ext{O}_8$ and $	ext{H}_2	ext{S}_2	ext{O}_3$ using structures.

Accepted Answer

$	ext{H}_2	ext{S}_2	ext{O}_7$ (pyrosulphuric): two tetrahedra sharing a bridging oxygen, no peroxide. Each sulphur is $+6$.

$	ext{H}_2	ext{S}_2	ext{O}_8$ (peroxodisulphuric): contains an $	ext{O}-	ext{O}$ linkage. Counting those two oxygens as $-1$ gives each sulphur $+6$, not the impossible $+7$ the averaging rule returns.

$	ext{H}_2	ext{S}_2	ext{O}_3$ (thiosulphuric): one oxygen replaced by sulphur. The central sulphur is $+6$ and the terminal one $-2$, averaging to $+2$.

The average is meaningless in the third case, since the two sulphurs are in quite different environments.

Question 3

Explain why diborane has a bridged structure while ethane does not, and describe the bonding in the bridges.

Accepted Answer

Ethane has $14$ valence electrons for seven ordinary two-electron bonds, so no bridging is needed.

Diborane has only $12$ electrons but appears to require eight bonds, which would need $16$.

Four terminal $	ext{B}-	ext{H}$ bonds use eight electrons in the ordinary way. The remaining four electrons form **two three-centre two-electron bonds**, each spreading over $	ext{B}-	ext{H}-	ext{B}$.

The bridge bonds are longer ($133$ pm against $119$ pm) and weaker, and the molecule is described as electron deficient.

Question 4

Explain why carbon dioxide is a gas at room temperature while silicon dioxide melts above $1700^{\circ}$C.

Accepted Answer

Carbon forms strong sideways $p\pi$-$p\pi$ bonds to oxygen, so each carbon satisfies its valence with two double bonds and the molecule $	ext{O}=	ext{C}=	ext{O}$ is complete.

Separate molecules are held only by weak dispersion forces, so the substance is a gas.

Silicon's $3p$ orbitals are too diffuse for effective sideways overlap, so it forms four **single** bonds to four different oxygens instead.

Every oxygen bridges two silicons, giving a continuous covalent network that must be broken bond by bond to melt.

Question 5

A silicate has each tetrahedron sharing two corners. Give the empirical formula of the anion and name two structural possibilities.

Accepted Answer

Two unshared oxygens count fully and two shared ones count half each, giving $2+1=3$ oxygens per silicon.

The anion is $\left(	ext{SiO}_3^{2-}ight)_n$.

Two structures are possible: an infinite **chain**, as in the pyroxenes, or a closed **ring**, as in $	ext{Si}_3	ext{O}_9^{6-}$ and the $	ext{Si}_6	ext{O}_{18}^{12-}$ of beryl.

The empirical formula alone cannot distinguish them; the ring size must be determined by diffraction.

Question 6

Explain why nitrogen exists as a diatomic gas while phosphorus forms $	ext{P}_4$ tetrahedra, and why white phosphorus is so much more reactive than nitrogen.

Accepted Answer

Nitrogen is small enough for effective sideways overlap, so it forms a triple bond of $941$ kJ mol$^{-1}$ and needs nothing else.

Phosphorus is too large for $p\pi$-$p\pi$ overlap, so it satisfies its valence with three single bonds, giving a closed $	ext{P}_4$ tetrahedron.

Those bonds are forced to $60^{\circ}$ angles, far from the $109.5^{\circ}$ they would prefer, so the molecule carries considerable **angle strain**.

That strain makes white phosphorus spontaneously flammable in air, while nitrogen's exceptionally strong triple bond makes it almost inert.

Question 7

State the conditions used in the contact process and explain why each is chosen.

Accepted Answer

The oxidation $2	ext{SO}_2+	ext{O}_2ightarrow2	ext{SO}_3$ is exothermic and reduces three gaseous moles to two.

**Temperature about $450^{\circ}$C:** a compromise. Lower would improve the equilibrium yield but be impractically slow; higher would be fast but shift the equilibrium backwards.

**Pressure about $2$ atm:** higher pressure would help the equilibrium slightly, but the yield is already above $95\%$ and higher pressures are not worth the cost.

**Vanadium pentoxide catalyst:** raises the rate without affecting the equilibrium position at all.

Question 8

Explain why ozone is a much stronger oxidising agent than oxygen, and describe its structure.

Accepted Answer

Ozone is bent at about $117^{\circ}$ with two equal bonds of $128$ pm, intermediate between a single and a double bond because of delocalisation over the three atoms.

It readily gives up one oxygen atom to leave the much more stable dioxygen molecule, and that step is strongly exothermic.

Molecular oxygen has no such easy route, since breaking it requires severing a double bond entirely.

Ozone therefore oxidises substances that oxygen leaves untouched, including iodide to iodine and lead sulphide to lead sulphate.

Question 9

Arrange $	ext{HF}$, $	ext{HCl}$, $	ext{HBr}$ and $	ext{HI}$ by acid strength and explain why the order is not that of electronegativity.

Accepted Answer

Acid strength: $	ext{HF}\ll	ext{HCl}<	ext{HBr}<	ext{HI}$.

Electronegativity would predict hydrogen fluoride to be the strongest, since its bond is by far the most polar.

What actually matters is the bond enthalpy: $567$, $431$, $366$ and $299$ kJ mol$^{-1}$ respectively. The weaker the bond, the more readily the proton is released.

Hydrogen fluoride is further weakened as an acid by hydrogen bonding in the liquid, which stabilises the undissociated molecule and even lets it form the stable $	ext{HF}_2^-$ ion.

Question 10

Predict the shapes of $	ext{ICl}_2^-$, $	ext{ICl}_4^-$ and $	ext{ClF}_3$, and explain why fluorine never appears as the central atom.

Accepted Answer

$	ext{ICl}_2^-$: two bonds and three lone pairs, five domains. **Linear**, with lone pairs equatorial.

$	ext{ICl}_4^-$: four bonds and two lone pairs, six domains. **Square planar**.

$	ext{ClF}_3$: three bonds and two lone pairs, five domains. **T-shaped**.

Fluorine has no accessible $d$ orbitals and cannot expand beyond an octet, so it can never accommodate more than one bond. The larger, more polarisable halogen is always central.

Question 11

Write the equations for the complete and partial hydrolysis of $	ext{XeF}_6$, and give the shape and hybridisation of each xenon product.

Accepted Answer

Complete: $	ext{XeF}_6+3	ext{H}_2	ext{O}ightarrow	ext{XeO}_3+6	ext{HF}$

Partial: $	ext{XeF}_6+	ext{H}_2	ext{O}ightarrow	ext{XeOF}_4+2	ext{HF}$

$	ext{XeO}_3$: three bonds and one lone pair, four domains, $sp^{3}$, **trigonal pyramidal**.

$	ext{XeOF}_4$: five bonds and one lone pair, six domains, $sp^{3}d^{2}$, **square pyramidal**.

Xenon remains at $+6$ throughout, which is what identifies the starting fluoride from either product.

Question 12

Explain why hypochlorous acid is the weakest of the chlorine oxyacids yet the most effective bleaching agent.

Accepted Answer

Acidity depends on how well the conjugate base spreads its negative charge. Hypochlorite has only one oxygen, so the charge is poorly delocalised, the anion is unstable and the acid is weak.

Oxidising power depends on how readily the species gives up oxygen. Precisely because hypochlorite is poorly stabilised, it does so easily.

Perchlorate, with four oxygens, is the mirror image: exceptionally stable, so the acid is very strong and the anion a poor oxidant.

The two trends therefore run in opposite directions, which is why bleaching powder is based on hypochlorite rather than any higher oxyacid.

Practice — "p-Block Elements"

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