p-Block Elements

Phosphorous acid is written ${\text{H}}_3{\text{PO}}_3$ and clearly contains three hydrogen atoms. How many of them can a base remove?

Two. The acid is dibasic, not tribasic.

The reason is visible only in the structure. One of the three hydrogens is bonded directly to phosphorus, not through an oxygen. A $\text{P}-\text{H}$ bond is not appreciably polar and its hydrogen is not acidic at all, so it never leaves.

$\text{basicity}=\text{number of }\text{P}-\text{OH}\text{groups}$

So hypophosphorous acid ${\text{H}}_3{\text{PO}}_2$, with two $\text{P}-\text{H}$ bonds, is only monobasic, and both it and phosphorous acid are strong reducing agents precisely because of those bonds.

Draw the structure and the answer follows. That is the single most useful habit in the whole of $p$-block chemistry, and it settles basicity, oxidising power, hydrolysis products and shape alike.

1. Group 13: the boron anomaly and electron-deficient bonding

Boron is a non-metal in a group of metals: small, high in ionisation enthalpy, and with only four orbitals available. Its compounds are covalent and electron deficient, which is the source of everything distinctive about them.

Diborane is the standard case. ${\text{B}}_2{\text{H}}_6$ has only $12$ valence electrons but appears to need $16$ for eight bonds.

The resolution is that the two bridging hydrogens each form a three-centre two-electron bond spanning $\text{B}-\text{H}-\text{B}$. Four terminal bonds use eight electrons and the two bridges use the remaining four. The bridge bonds are longer and weaker than the terminal ones, and the bridging angle is close to $83^{\circ }$.

Two further boron facts recur. Boric acid is a Lewis acid, not a proton donor — it accepts a hydroxide from water rather than releasing a proton, and is therefore monobasic despite having three $\text{OH}$ groups. And ${\text{BF}}_3$ is the weakest boron Lewis acid because back bonding from fluorine partly fills its vacant orbital.

Illustration 1

Explain why boric acid turns litmus red and behaves as a monobasic acid despite containing three hydroxyl groups.

It does not ionise by releasing a proton from any of its $\text{OH}$ groups.

Instead it accepts a hydroxide ion from water: $\text{B}{\left(\text{OH}\right)}_3+{\text{H}}_2\text{O}\to {\left[\text{B}{\left(\text{OH}\right)}_4\right]}^{-}+{\text{H}}^{+}$

One proton is released per molecule of water, not per hydroxyl group, so the acid is monobasic.

It is a Lewis acid acting through water. This is why boric acid cannot be titrated directly and is analysed after adding glycerol or mannitol, which form a stronger chelated complex.

Illustration 2

Explain the borax bead test and why it identifies transition metals specifically.

Heating borax drives off water and then decomposes it to sodium metaborate and boric anhydride:

${\text{Na}}_2{\text{B}}_4{\text{O}}_7\to 2{\text{NaBO}}_2+{\text{B}}_2{\text{O}}_3$

The boric anhydride combines with a metal oxide to give a coloured metaborate, for instance a blue $\text{Co}{\left({\text{BO}}_2\right)}_2$ or a green $\text{Cr}{\left({\text{BO}}_2\right)}_3$.

The colour arises from $d$-$d$ transitions, which is why only transition metals give the test. Main-group metals produce colourless beads and cannot be identified this way.

2. Group 14: catenation, and why carbon behaves differently

The dioxide comparison is the classic question. Carbon forms strong double bonds to oxygen, so the molecule is complete and only weak dispersion forces hold molecules together. Silicon cannot form effective $p\pi$-$p\pi$ bonds with oxygen because the orbitals are too diffuse to overlap sideways, so each silicon instead makes four single bonds to four oxygens, and the result is a three-dimensional network melting above $1700^{\circ }$C.

Illustration 3

Explain why ${\text{SiCl}}_4$ is readily hydrolysed by water while ${\text{CCl}}_4$ is not.

Hydrolysis begins with a water molecule attacking the central atom.

Silicon has vacant $3d$ orbitals that can accept the lone pair, so a five-coordinate intermediate forms readily and the reaction proceeds.

Carbon has no such orbitals available, and its small size leaves no room for a fifth group, so there is no low-energy path for the attack.

${\text{CCl}}_4$ is thermodynamically unstable with respect to hydrolysis but kinetically inert. The distinction between what is favourable and what is achievable is the point of the question.

Silicates and silicones

Every silicate is built from the same ${\text{SiO}}_4^{4-}$ tetrahedron, and the whole classification depends on how many corners each tetrahedron shares:

The physical properties follow directly: sheet silicates cleave into flakes because the sheets are held together only weakly, and double-chain silicates are fibrous for the same reason along one axis.

Silicones are synthetic polymers of repeating ${\text{R}}_2\text{SiO}$ units, made by hydrolysing alkyl chlorosilanes. The number of chlorines controls the architecture: ${\text{R}}_3\text{SiCl}$ terminates a chain, ${\text{R}}_2{\text{SiCl}}_2$ extends it, and ${\text{RSiCl}}_3$ cross-links it. They are water-repellent and thermally stable because the strong silicon-oxygen backbone is shielded by the organic groups.

Illustration 4

A silicate mineral has the empirical formula ${\text{Si}}_2{\text{O}}_5^{2-}$. Deduce its structure type.

Each silicon carries $2.5$ oxygens on average.

An unshared tetrahedron would give four oxygens per silicon; each shared corner halves one oxygen's contribution.

$4-\frac{n}{2}=2.5$ gives $n=3$ shared corners.

The mineral is a sheet silicate, such as mica or talc.

Sheet silicates cleave into thin flakes, because the covalent network extends in only two dimensions and the layers are held by much weaker forces.

Illustration 5

Explain how the chain length of a silicone polymer is controlled during manufacture.

Hydrolysis of ${\text{R}}_2{\text{SiCl}}_2$ gives a unit with two linkage points, which extends the chain indefinitely.

Adding ${\text{R}}_3\text{SiCl}$, which has only one linkage point, caps a growing chain and stops it.

Raising the proportion of the capping reagent therefore shortens the average chain, giving a thinner oil; lowering it gives longer chains and a more viscous product.

Including ${\text{RSiCl}}_3$, with three linkage points, produces cross-links instead, converting the oil into a rubber or a resin.

3. Group 15: nitrogen's uniqueness

Nitrogen differs from the rest of its group for the three usual second-period reasons — small size, no $d$ orbitals, and high electronegativity — and the consequences are dramatic.

${\text{N}}_2$ is inert because its triple bond carries $941$ kJ mol${}^{-1}$, among the strongest known. Phosphorus, unable to form effective $p\pi$-$p\pi$ bonds, exists instead as ${\text{P}}_4$ tetrahedra with $60^{\circ }$ bond angles, and that strain makes white phosphorus dangerously reactive.

Nitrogen forms no pentahalide while phosphorus forms ${\text{PF}}_5$ and ${\text{PCl}}_5$, for lack of $d$ orbitals. Ammonia hydrogen bonds while phosphine does not, so ammonia boils at $-33^{\circ }$C against phosphine's $-88^{\circ }$C. And ammonia is a far stronger base, because its lone pair is concentrated in a small $s{p}^3$ orbital.

Illustration 6

Arrange ${\text{NH}}_3$, ${\text{PH}}_3$, ${\text{AsH}}_3$ and ${\text{SbH}}_3$ by basicity and by boiling point, and explain why the two orders differ.

Basicity: ${\text{NH}}_3>{\text{PH}}_3>{\text{AsH}}_3>{\text{SbH}}_3$, since the lone pair becomes more diffuse and less available down the group.

Boiling point: ${\text{NH}}_3>{\text{SbH}}_3>{\text{AsH}}_3>{\text{PH}}_3$.

Ammonia is out of place in the second list because of hydrogen bonding. Ignoring it, the remaining three rise with molar mass through dispersion forces, exactly as expected.

Illustration 7

Explain why ${\text{H}}_3{\text{PO}}_2$ is a stronger reducing agent than ${\text{H}}_3{\text{PO}}_4$.

Hypophosphorous acid contains two $\text{P}-\text{H}$ bonds, phosphoric acid none.

A hydrogen bonded to phosphorus is easily given up along with its electrons, so the acid is readily oxidised to a higher phosphorus oxidation state.

Phosphorus is already at $+5$ in phosphoric acid, its maximum, so no further oxidation is possible.

Reducing power tracks the $\text{P}-\text{H}$ count exactly, which is the same structural feature that fixes the basicity.

4. Group 16: ozone, and the sulphur oxyacids

Ozone is bent at $117^{\circ }$ with two equal bonds of $128$ pm, intermediate between single and double, and is a powerful oxidant because it readily gives up one oxygen atom.

Sulphuric acid is made by the contact process, oxidising sulphur dioxide over vanadium pentoxide at about $450^{\circ }$C and $2$ atm. The moderate temperature is a compromise: the reaction is exothermic, so a higher temperature would be faster but would lower the yield.

Sulphur's oxyacids illustrate two structural rules. Peroxo acids contain an $\text{O}-\text{O}$ linkage, as in Caro's acid ${\text{H}}_2{\text{SO}}_5$ and Marshall's acid ${\text{H}}_2{\text{S}}_2{\text{O}}_8$. Thio acids replace an oxygen by sulphur, as in ${\text{H}}_2{\text{S}}_2{\text{O}}_3$. Counting the $\text{S}-\text{OH}$ groups gives the basicity in every case.

Illustration 8

Give the oxidation state of sulphur in ${\text{H}}_2{\text{S}}_2{\text{O}}_8$ by structure rather than by the average rule.

The naive average gives $\frac{2\times 1+8\times \left(-2\right)}{-2}$ arithmetic yielding $+7$, which is impossible for sulphur.

Drawing the structure reveals a peroxide linkage, where each oxygen is $-1$ rather than $-2$.

With two peroxide oxygens at $-1$ and six normal ones at $-2$, each sulphur comes out at $+6$, its ordinary maximum.

The average rule fails whenever a peroxide or a direct element-element bond is present. Drawing the structure is the only reliable route.

5. Group 17: the fluorine anomaly and the oxyacids

Hydrofluoric acid is weak despite fluorine being the most electronegative halogen, because the $\text{H}-\text{F}$ bond enthalpy of $567$ kJ mol${}^{-1}$ far exceeds the others, and because hydrogen bonding in the liquid stabilises the undissociated molecule.

Among the oxyacids of chlorine, acid strength rises with oxidation state while oxidising power falls:

$\text{HOCl}<{\text{HClO}}_2<{\text{HClO}}_3<{\text{HClO}}_4\text{in acidity}$ $\text{HOCl}>{\text{HClO}}_2>{\text{HClO}}_3>{\text{HClO}}_4\text{in oxidising power}$

More oxygens withdraw more electron density, stabilising the anion and so raising acidity; but the same delocalisation stabilises the molecule against giving up oxygen, lowering its oxidising power.

Illustration 9

Predict the shapes of ${\text{ClF}}_3$, ${\text{BrF}}_5$ and ${\text{IF}}_7$, and explain why no fluorine analogue exists for any of them.

${\text{ClF}}_3$: five domains with two lone pairs, T-shaped.

${\text{BrF}}_5$: six domains with one lone pair, square pyramidal.

${\text{IF}}_7$: seven domains with none, pentagonal bipyramidal.

Fluorine cannot be the central atom in any interhalogen, because it has no accessible $d$ orbitals and cannot expand its octet.

Only the larger halogen is ever central, which is why the formulas are always written with the heavier element first.

6. Group 18: noble gas compounds

Xenon reacts with fluorine because it is large enough for its ionisation enthalpy to be comparable with oxygen's. The three fluorides follow VSEPR exactly.

${\text{XeF}}_2$ is linear with three equatorial lone pairs, ${\text{XeF}}_4$ square planar with two axial lone pairs, and ${\text{XeF}}_6$ a distorted octahedron because its single lone pair still occupies space.

Their hydrolysis is examined constantly:

${\text{XeF}}_6+3{\text{H}}_2\text{O}\to {\text{XeO}}_3+6\text{HF},{\text{XeF}}_6+{\text{H}}_2\text{O}\to {\text{XeOF}}_4+2\text{HF}$

so complete hydrolysis gives the trioxide and partial hydrolysis the oxofluoride.

Illustration 10

Predict the shapes of ${\text{XeO}}_3$ and ${\text{XeO}}_4$, and explain the difference.

${\text{XeO}}_3$: three bonding domains and one lone pair, four in all, giving a trigonal pyramidal shape.

${\text{XeO}}_4$: four bonding domains and no lone pair, giving a perfect tetrahedron.

The lone pair in the trioxide compresses its angles below $109.5^{\circ }$, to about $103^{\circ }$.

Counting domains works exactly as it does for ordinary molecules. Nothing about noble gas compounds requires special rules once the compounds exist at all.

Illustration 11

Explain why xenon forms fluorides but helium and neon do not.

Compound formation requires the noble gas's ionisation enthalpy to be low enough that bonding to fluorine repays the cost.

Xenon's is $1170$ kJ mol${}^{-1}$, comparable with oxygen's $1314$, and oxygen forms fluorides readily.

Helium and neon have ionisation enthalpies of $2372$ and $2081$ kJ mol${}^{-1}$, far too high, and are also too small to accommodate the fluorines.

Krypton sits at the boundary, forming ${\text{KrF}}_2$ only under forcing conditions, exactly as the ionisation trend predicts.

Illustration 12

A compound of xenon has a formula ${\text{XeF}}_x$ and on complete hydrolysis gives a xenon compound in which xenon is at $+6$. Identify $x$ and write the equation.

Xenon at $+6$ in an oxide means ${\text{XeO}}_3$.

Since hydrolysis does not change the oxidation state, xenon must be $+6$ in the fluoride too, giving $x=6$.

${\text{XeF}}_6+3{\text{H}}_2\text{O}\to {\text{XeO}}_3+6\text{HF}$

Hydrolysis conserves oxidation state. Reading it off the product and working backwards identifies the starting fluoride without any other information.

Illustration 13

Explain why perchloric acid is the strongest of the chlorine oxyacids but the weakest oxidant among them.

Acidity depends on how well the anion is stabilised. Perchlorate has four oxygens over which the negative charge delocalises, the largest number in the series, so the proton leaves most readily.

Oxidising power depends on how easily the acid gives up oxygen. The same extensive delocalisation makes perchlorate exceptionally stable and reluctant to be reduced.

The two properties therefore run in opposite directions across the series.

Hypochlorous acid, the weakest acid, is the strongest bleach. This is the reason bleaching powder is based on hypochlorite rather than on any of the higher oxyacids.

Summary

• Basicity equals the number of $\text{X}-\text{OH}$ groups. ${\text{H}}_3{\text{PO}}_3$ is dibasic and ${\text{H}}_3{\text{PO}}_2$ monobasic, because the remaining hydrogens sit on phosphorus.
• Those same $\text{P}-\text{H}$ bonds make the lower phosphorus acids strong reducing agents.
• Diborane's $12$ electrons make four terminal bonds and two three-centre two-electron bridges; the bridge bonds are longer and weaker.
• Boric acid is a Lewis acid, accepting hydroxide from water, and is monobasic despite three $\text{OH}$ groups.
• ${\text{CO}}_2$ is a gas of discrete molecules; ${\text{SiO}}_2$ is a giant network, because silicon cannot form effective $p\pi$-$p\pi$ bonds.
• ${\text{CCl}}_4$ resists hydrolysis for lack of $d$ orbitals and space — it is unstable thermodynamically but inert kinetically.
• Silicates classify by corners shared per tetrahedron: $0$ discrete, $2$ chains, $3$ sheets, $4$ network — and cleavage follows directly.
• Silicone architecture is set by the chlorine count: ${\text{R}}_3\text{SiCl}$ caps, ${\text{R}}_2{\text{SiCl}}_2$ extends, ${\text{RSiCl}}_3$ cross-links.
• ${\text{N}}_2$ is inert through its $941$ kJ mol${}^{-1}$ triple bond; phosphorus instead forms strained ${\text{P}}_4$ tetrahedra.
• Ammonia is out of place on any boiling-point trend because of hydrogen bonding, but in place on every basicity trend.
• Ozone is bent at $117^{\circ }$ with two equal intermediate-length bonds, and oxidises by donating one oxygen atom.
• The average oxidation-number rule fails for peroxo and thio acids; draw the structure instead.
• $\text{HF}$ is weak because its bond enthalpy is $567$ kJ mol${}^{-1}$; acidity rises down the group as that falls.
• Chlorine oxyacids: acidity rises with oxidation state while oxidising power falls, so the weakest acid is the best bleach.
• Interhalogens always place the larger halogen at the centre, since fluorine cannot expand its octet.
• ${\text{XeF}}_2$ linear, ${\text{XeF}}_4$ square planar, ${\text{XeF}}_6$ distorted octahedral; complete hydrolysis gives ${\text{XeO}}_3$ and partial gives ${\text{XeOF}}_4$.