p-Block Elements

Rank these four as Lewis acids, weakest to strongest.

$\mathrm{B}{\mathrm{F}}_3,\mathrm{BC}{\mathrm{l}}_3,\mathrm{BB}{\mathrm{r}}_3,\mathrm{B}{\mathrm{I}}_3$

The reasoning everyone uses: fluorine is the most electronegative element on the table, so it drains boron hardest, so boron is left most electron-deficient, so $\mathrm{B}{\mathrm{F}}_3$ must be the strongest acid.

Every step of that is true. The conclusion is exactly backwards.

$\mathrm{B}{\mathrm{F}}_3<\mathrm{BC}{\mathrm{l}}_3<\mathrm{BB}{\mathrm{r}}_3<\mathrm{B}{\mathrm{I}}_3$

$\mathrm{B}{\mathrm{F}}_3$ is the weakest of the four, and not by a whisker.

Fluorine does drain boron through the sigma bond. But it also gives electron density straight back, sideways, from a filled 2p orbital into boron's empty 2p. Those two orbitals are almost the same size, so the overlap is excellent.

Iodine's 5p orbital is enormous by comparison. It cannot reach boron's compact 2p in any useful way, so nothing flows back and the boron stays genuinely hungry.

Size matching beat electronegativity. That sentence is not a footnote about boron halides. It is most of this chapter.

Two further ideas complete the toolkit. Ordinary periodicity makes atoms larger and less electronegative down a group. And the $n{s}^2$ pair becomes progressively harder to use, which is the inert pair effect and is the p-block's own signature.

Learn those three and most of the individual facts stop needing to be remembered separately, because they can be rebuilt.

1. What the p-Block Is

Groups 13 to 18, with the last electron entering a p orbital, giving valence configurations from $n{s}^{2}n{p}^1$ to $n{s}^{2}n{p}^6$.

It is the only block holding metals, metalloids and non-metals together, and the boundary between them runs diagonally, which is why aluminium is a metal and boron directly above it is not.

The general valence is the group number minus 10, so group 13 tends towards $+3$ and group 15 towards $+5$ or $-3$.

General trends

Two irregularities matter. Gallium is smaller than aluminium, because the ten 3d electrons preceding it shield poorly. And ionisation enthalpies do not fall smoothly down groups 13 and 14 for the same reason, with the d-block and later the f-block interrupting the pattern.

Illustration 1

Melting points down group 13 run Al 933 K, Ga 303 K, In 430 K, Tl 577 K. Gallium melts at 30 degrees Celsius, low enough to liquefy in a closed hand. What has gone wrong with the trend?

Nothing has gone wrong with the periodic trend. Something has gone right with gallium's crystal.

Metals melt when the delocalised bonding across the whole lattice gives way. Solid gallium is not a straightforward metallic lattice: its atoms pair up into $\mathrm{G}{\mathrm{a}}_2$ units, so the solid is closer to a molecular crystal of dimers than to a sea of electrons.

Melting only has to separate those loosely held dimers, which costs very little. Once molten, gallium becomes a normal metal again and needs 2477 K to boil.

The gap between them is the largest liquid range of any element, over 2100 K, which is why gallium fills high-temperature thermometers where mercury would have boiled away long before.

Trap. A trend describes what changes smoothly. It cannot describe a change of structure, and structural anomalies are where the interesting questions live.

2. The Inert Pair Effect

Down each p-block group, the lower oxidation state becomes progressively more stable relative to the higher one.

Why it happens. The $n{s}^2$ pair is held increasingly tightly on descent, partly because poor shielding by intervening d and f electrons raises $Z_{\text{eff}}$, and partly because the energy released by forming two extra bonds falls as bond enthalpies weaken with increasing size.

The pair is not literally inert. It is simply not worth unpairing, because the bonds it would form no longer repay the promotion cost. That is an accounting statement, and accounting statements can be checked with numbers.

Illustration 2

Put a number on the effect. The standard reduction potentials are $E^{\circ }(\mathrm{S}{\mathrm{n}}^{4+}/S{\mathrm{n}}^{2+})=+0.15$ V and $E^{\circ }(\mathrm{P}{\mathrm{b}}^{4+}/P{\mathrm{b}}^{2+})=+1.67$ V. What do these say about tin and lead?

A more positive potential means the higher oxidation state is a stronger oxidising agent, that is, more eager to fall back down.

For tin, $+0.15$ V is barely positive. $\mathrm{S}{\mathrm{n}}^{2+}$ is therefore easily pushed up to $\mathrm{S}{\mathrm{n}}^{4+}$, which is why tin(II) chloride is a standard laboratory reducing agent.

For lead, $+1.67$ V is enormous, comparable to permanganate. $\mathrm{P}{\mathrm{b}}^{4+}$ is desperate to become $\mathrm{P}{\mathrm{b}}^{2+}$, which is why $\mathrm{Pb}{\mathrm{O}}_2$ is a powerful oxidiser and why the lead accumulator can store useful energy in it.

That gap of 1.52 V is the inert pair effect, measured. One group, one step down, and the preferred oxidation state has flipped completely. Multiply by $nF$ with $n=2$ and the difference is roughly 293 kJ mol${}^{-1}$, which is the size of a chemical bond.

3. The Anomalous First Element

Boron, carbon, nitrogen, oxygen and fluorine each behave unlike the rest of their group, always for the same three reasons.

What follows

Nitrogen forms no pentahalide while phosphorus forms $\mathrm{PC}{\mathrm{l}}_5$. Oxygen reaches covalency 2 and fluorine 1, while sulphur reaches 6 and chlorine 7.

Second-period elements also form strong $p\pi$-$p\pi$ multiple bonds, because their small size lets p orbitals overlap sideways effectively. Heavier elements cannot, and that single difference explains a great deal.

Illustration 3

Nitrogen exists as ${\mathrm{N}}_2$ and phosphorus as ${\mathrm{P}}_4$. Both are group 15. Decide the winner in each case from bond enthalpies alone: $\mathrm{N}\equiv \mathrm{N}$ 941, $\mathrm{N}-\mathrm{N}$ 163, $\mathrm{P}\equiv \mathrm{P}$ 490, $\mathrm{P}-\mathrm{P}$ 201, all in kJ mol${}^{-1}$.

Compare like with like by working out the bonding energy released per two atoms.

A tetrahedral ${\mathrm{E}}_4$ molecule has one bond along each of the six edges, so four atoms carry six single bonds, which is three single bonds per two atoms.

Identical arithmetic, opposite answers.

Nitrogen's triple bond is worth nearly twice three single bonds, so ${\mathrm{N}}_2$ wins comfortably. Phosphorus's triple bond is feeble by comparison, because 3p orbitals are too diffuse to overlap sideways, so three single bonds win instead and phosphorus builds the tetrahedron.

Notice which number does the damage. The single bond enthalpies are similar, 163 against 201. It is the triple bond that collapses, from 941 to 490, and it collapses precisely because sideways overlap needs small orbitals.

Everything else follows from those two structures. ${\mathrm{N}}_2$ has to have a 941 kJ bond broken before nitrogen will react at all, which is why it makes up most of the atmosphere unchanged. ${\mathrm{P}}_4$ has 60 degree bond angles under severe strain, which is why white phosphorus ignites in air and is stored under water.

Diagonal relationships follow from the same anomaly: boron resembles silicon and beryllium resembles aluminium, because moving diagonally roughly cancels the size change.

4. Group 13: The Boron Family

Boron is a metalloid; aluminium, gallium, indium and thallium are metals. The $+3$ state dominates at the top and $+1$ at the bottom.

Boron trifluoride is the classic Lewis acid, with only six electrons around boron. It accepts an electron pair readily, which is why it complexes with ammonia. The acidity order and its cause are the hook this chapter opened on.

Aluminium is amphoteric, dissolving in both acids and alkalis, which is why aluminium cookware is attacked by both. Aluminium chloride exists as the covalent dimer $\mathrm{A}{\mathrm{l}}_2\mathrm{C}{\mathrm{l}}_6$, an application of Fajans' rules.

Illustration 4

Lewis acidity of the boron trihalides runs $\mathrm{B}{\mathrm{F}}_3<\mathrm{BC}{\mathrm{l}}_3<\mathrm{BB}{\mathrm{r}}_3<\mathrm{B}{\mathrm{I}}_3$. Electronegativity says fluorine should strip boron of electron density hardest and therefore make the strongest acid. Resolve the contradiction.

The inductive argument is real but it is not the dominant one, and here it gives the wrong answer outright.

What decides the order is back-bonding. Boron in $\mathrm{B}{\mathrm{X}}_3$ is $s{p}^2$ with an empty 2p orbital perpendicular to the molecular plane, and each halogen has filled p orbitals. A halogen lone pair can donate sideways into that empty orbital.

Fluorine's 2p is the same size and energy as boron's 2p, so the overlap is excellent and a substantial amount of π donation takes place. The very orbital a Lewis acid needs to keep empty is being partly filled by its own substituents.

Going down the group the halogen's donor orbital becomes 3p, then 4p, then 5p — larger, more diffuse, and progressively worse matched to boron's compact 2p. Back-bonding weakens, boron's vacancy stays open, and the Lewis acidity climbs.

There is a second cost pointing the same way. Accepting a lone pair rehybridises boron from $s{p}^2$ to $s{p}^3$, which destroys the back-bonding entirely. $\mathrm{B}{\mathrm{F}}_3$ has the most to give up, so it is the most reluctant.

Note that this is the size-matching argument from the first-element anomaly, reappearing: 2p overlaps well with 2p and badly with everything larger.

5. Group 14: The Carbon Family

Carbon is a non-metal, silicon and germanium metalloids, tin and lead metals. The $+4$ state dominates at the top, $+2$ at the bottom by the inert pair effect.

Catenation, the ability to form chains of like atoms, falls sharply down the group.

$\mathrm{C}\gg \mathrm{Si}>\mathrm{Ge}\approx \mathrm{Sn}\gg \mathrm{Pb}$

Illustration 5

Silicon is directly below carbon and is the second most abundant element in the Earth's crust, yet there is no silicon-based organic chemistry. The usual reason given is that the $\mathrm{Si}-\mathrm{Si}$ bond at 297 kJ mol${}^{-1}$ is weaker than $\mathrm{C}-\mathrm{C}$ at 348. Is that the whole story?

It is not, and the missing half is more decisive than the half usually quoted.

A chain does not merely have to be strong. It has to survive competition from the alternatives available in its environment, and the environment is full of oxygen.

For carbon, a chain and an oxidised carbon are worth almost the same, so long chains persist indefinitely in an oxygen atmosphere and only burn once something supplies the activation energy.

For silicon, every $\mathrm{Si}-\mathrm{Si}$ bond is 155 kJ mol${}^{-1}$ worse than the $\mathrm{Si}-\mathrm{O}$ bond that could replace it. Silicon chains are not merely fragile; they are thermodynamically doomed anywhere oxygen exists.

That single number is why the Earth's crust is silicates and living things are carbon compounds. The two elements are chemically similar and the environment picked between them.

Allotropy is a group 14 speciality. Diamond is a giant tetrahedral network, extremely hard and non-conducting. Graphite is layers of hexagonal sheets with delocalised electrons, so it conducts and the layers slide, making it a lubricant. Fullerenes are discrete cage molecules, of which ${\mathrm{C}}_{60}$ is the best known.

Silicon's 3p orbitals cannot reach oxygen's 2p sideways, so no double bond forms and silicon satisfies its valency with four single bonds to four different oxygens, each bridging onward. The structure difference is a size-matching difference.

6. Group 15: The Nitrogen Family

Nitrogen and phosphorus are non-metals, arsenic and antimony metalloids, bismuth a metal. Oxidation states run from $-3$ to $+5$.

Hydride trends worth knowing

Nitrogen is small, and its 2s and 2p orbitals are close in energy, so hybridising into four $s{p}^3$ orbitals is cheap. Heavier atoms have a larger s-to-p gap, so hybridisation stops being worth it and the bonds use nearly pure p orbitals at close to 90 degrees.

Read the numbers rather than the arrow. Almost the entire change happens in one step, from 107 to 93.6. After phosphorus there is nothing left to lose, because the bonding is already as close to pure p as it can get.

Illustration 6

${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_2$, ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3$ and ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ each contain three hydrogens, yet their basicities are 1, 2 and 3 respectively. Account for this.

Only a hydrogen attached to oxygen can ionise. Count P–OH groups, never hydrogens.

${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4:3\times \mathrm{P}-\mathrm{OH}\Rightarrow \text{tribasic}$ ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3:2\times \mathrm{P}-\mathrm{OH}+1\times \mathrm{P}-\mathrm{H}\Rightarrow \text{dibasic}$ ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_2:1\times \mathrm{P}-\mathrm{OH}+2\times \mathrm{P}-\mathrm{H}\Rightarrow \text{monobasic}$

A hydrogen bonded straight to phosphorus does not come off as a proton, because phosphorus and hydrogen have nearly the same electronegativity. That bond is barely polar, and if it breaks at all the hydrogen leaves carrying the electrons rather than abandoning them.

Which is the second half of the story. Those same P–H bonds make ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_2$ and ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3$ good reducing agents — hypophosphorous acid will reduce silver salts to the metal — while ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$, having none, is not a reducing agent at all.

One structural feature therefore settles two apparently unrelated properties, and a question asking about either is really asking you to draw the structure.

7. Group 16: The Oxygen Family

Oxygen and sulphur are non-metals, selenium and tellurium metalloids, polonium a metal. Oxidation states run from $-2$ to $+6$.

Sulphur's ability to catenate is second only to carbon's, which is why ${\mathrm{S}}_8$ rings exist and ${\mathrm{O}}_8$ does not.

Illustration 7

Water is famous for its high boiling point. It is also the weakest acid of the group 16 hydrides. Reconcile the two.

Water is the outlier at the top of one row and the bottom of the other, which looks contradictory until you notice the two rows are asking different questions.

Boiling point asks how strongly one whole molecule sticks to another. Water hydrogen bonds; the others do not, because sulphur, selenium and tellurium are neither small nor electronegative enough. Hence the 160 K jump, and hence the smooth rise afterwards that is just molecules getting heavier.

Acidity asks how easily one $\mathrm{O}-\mathrm{H}$ bond breaks and how comfortable the anion is afterwards. Down the group the bond weakens and the anion grows, spreading its charge more thinly. Both favour ionisation, and ${\mathrm{H}}_2\mathrm{Te}$ is roughly $10^{13}$ times the acid that water is.

Trap. Electronegativity is the tempting variable and it gets acidity exactly backwards here. Bond strength and anion size decide acidity; hydrogen bonding decides boiling point. Neither is answered by pointing at the periodic table.

8. Group 17 and Group 18

Halogens are the most reactive non-metals, all diatomic, all needing one electron to complete an octet.

Fluorine breaks the pattern twice. Its bond dissociation enthalpy is only 155 kJ mol${}^{-1}$, below chlorine's 242, because the two atoms are so small that their lone pairs repel strongly across the short bond. Yet it is the most reactive halogen, because that weak bond is cheap to break and the bonds it then forms are exceptionally strong.

Oxidising power falls down the group, ${\mathrm{F}}_2>\mathrm{C}{\mathrm{l}}_2>\mathrm{B}{\mathrm{r}}_2>{\mathrm{I}}_2$, following the electrode potentials directly, which is why fluorine displaces every other halogen from its salts and iodine displaces none.

Interhalogen compounds form between halogens of different sizes and are more reactive than the parent halogens, because the mixed bond is weaker than either homonuclear bond.

Illustration 8

Predict the shapes of $\mathrm{Cl}{\mathrm{F}}_3$, $\mathrm{Br}{\mathrm{F}}_5$ and $\mathrm{I}{\mathrm{F}}_7$ from VSEPR.

Count valence electrons on the central atom, subtract one for each bond, and pair up what remains.

For $\mathrm{Cl}{\mathrm{F}}_3$, chlorine has 7 valence electrons and uses 3, leaving 4 as two lone pairs. Both lone pairs take equatorial positions, where they have only two neighbours at 90 degrees instead of three, and the three fluorines are left in a T.

Note what the series says about size. $\mathrm{I}{\mathrm{F}}_7$ exists but $\mathrm{Cl}{\mathrm{F}}_7$ does not, because iodine is large enough to hold seven fluorines around it and chlorine is not. The formula that forms is a packing question as much as an electronic one, which is the same size argument the chapter opened with, arriving from a different direction.

Noble gases

Group 18 has complete octets, the highest ionisation enthalpies in each period, and positive electron gain enthalpies.

Illustration 9

For sixty years every textbook said group 18 formed no compounds. In 1962 Neil Bartlett made one in a few minutes. What did he notice?

He had prepared ${\mathrm{O}}_2^{+}[\mathrm{Pt}{\mathrm{F}}_6{]}^{-}$, which meant $\mathrm{Pt}{\mathrm{F}}_6$ was aggressive enough to tear an electron off an oxygen molecule. Then he looked up two numbers.

$\text{IE of }{\mathrm{O}}_2=1175\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1},\text{IE of }\mathrm{Xe}=1170\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

Xenon's is lower. If $\mathrm{Pt}{\mathrm{F}}_6$ could take an electron from oxygen, nothing stood in the way of it taking one from xenon.

He mixed the two gases and got a yellow-orange solid immediately. The inertness of group 18 had never been a law of nature; it was an unchecked assumption.

The same two numbers say why the lighter members hold out. Helium needs 2372 kJ mol${}^{-1}$ and neon 2081, and no reagent is that aggressive. Radon at 1037 is easier still than xenon, but it is radioactive and decays too fast to study comfortably.

9. Allotropes and Oxide Character

Phosphorus has white, red and black forms. White is ${\mathrm{P}}_4$ tetrahedra with 60 degree bond angles, severely strained and so reactive it ignites in air. Red is a polymeric chain, far less strained and stable in air. Black is the most stable and most dense.

Sulphur exists mainly as ${\mathrm{S}}_8$ crown rings, in rhombic and monoclinic forms that interconvert at 369 K.

Oxide character follows metallic character exactly. Metal oxides are basic, non-metal oxides acidic, and elements at the metalloid boundary give amphoteric oxides. Across period 3 the sequence runs from basic $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$ through amphoteric $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$ to strongly acidic $\mathrm{C}{\mathrm{l}}_2{\mathrm{O}}_7$; down a group the oxides become more basic, so group 15 runs from acidic ${\mathrm{N}}_2{\mathrm{O}}_5$ to basic $\mathrm{B}{\mathrm{i}}_2{\mathrm{O}}_3$.

Where an element shows several oxidation states, the higher the oxidation state, the more acidic the oxide.

Illustration 10

Chlorine forms four oxoacids. Their $\mathrm{p}{K}_a$ values are $\mathrm{HOCl}$ 7.5, $\mathrm{HCl}{\mathrm{O}}_2$ 2.0, $\mathrm{HCl}{\mathrm{O}}_3$ $-1$, $\mathrm{HCl}{\mathrm{O}}_4$ $-10$. The chlorine is the same atom in all four. What is doing the work?

Each added oxygen drops the $\mathrm{p}{K}_a$ by roughly five units, which is a factor of $10^5$ in acid strength each time.

Acidity is decided by how comfortable the anion is once the proton has gone. In $\mathrm{Cl}{\mathrm{O}}^{-}$ the negative charge sits on a single oxygen. In $\mathrm{Cl}{\mathrm{O}}_4^{-}$ it is shared equally across four equivalent oxygens by resonance, so no single atom carries much of it.

The oxidation state of chlorine climbs $+1$, $+3$, $+5$, $+7$ across the same series, and it climbs for the same reason: each extra oxygen pulls more density off the chlorine.

So "higher oxidation state means more acidic oxide" is not a separate rule to memorise. It is charge delocalisation, counted in oxygen atoms, and $\mathrm{HCl}{\mathrm{O}}_4$ being one of the strongest acids known is where the counting ends.

A Note on Syllabus Emphasis

The JEE Main unit for this chapter is worded as a general introduction: electronic configuration, general trends in physical and chemical properties across periods and down groups, and the unique behaviour of the first element in each group.

That wording is worth taking seriously. It means the highest-yield preparation is exactly the reasoning in sections 2 and 3, applied to whichever group a question happens to name, rather than a memorised inventory of individual compounds.

Group-specific facts are still asked, and the ones in this chapter are the recurring ones. But if revision time is short, the trends and the first-element anomaly repay it best, because they let unfamiliar facts be reconstructed rather than recalled.

Summary

The p-block spans groups 13 to 18 and is the only block containing metals, metalloids and non-metals together, with the boundary running diagonally.

Size matching beats electronegativity, again and again. It reverses the boron halide Lewis acidity order to $\mathrm{B}{\mathrm{F}}_3<\mathrm{BC}{\mathrm{l}}_3<\mathrm{BB}{\mathrm{r}}_3<\mathrm{B}{\mathrm{I}}_3$, it decides that nitrogen is ${\mathrm{N}}_2$ while phosphorus is ${\mathrm{P}}_4$, that $\mathrm{C}{\mathrm{O}}_2$ is a gas while $\mathrm{Si}{\mathrm{O}}_2$ is a rock, and that $\mathrm{I}{\mathrm{F}}_7$ exists while $\mathrm{Cl}{\mathrm{F}}_7$ does not.

The inert pair effect makes the lower oxidation state progressively more stable down a group, giving $\mathrm{T}{\mathrm{l}}^{+}$, $\mathrm{P}{\mathrm{b}}^{2+}$ and $\mathrm{B}{\mathrm{i}}^{3+}$. It is an accounting balance, not an inertness, and it is measurable: $E^{\circ }$ for the $+4$ to $+2$ couple jumps from $+0.15$ V in tin to $+1.67$ V in lead.

The first element of each group is anomalous for three reasons: very small size, the highest electronegativity in the group, and no valence d orbitals, so its maximum covalency is four.

Catenation falls steeply down group 14, but the decisive number is not $\mathrm{Si}-\mathrm{Si}$ against $\mathrm{C}-\mathrm{C}$; it is that $\mathrm{Si}-\mathrm{O}$ beats $\mathrm{Si}-\mathrm{Si}$ by 155 kJ mol${}^{-1}$ while $\mathrm{C}-\mathrm{O}$ beats $\mathrm{C}-\mathrm{C}$ by only 10, which is why the crust is silicates and life is carbon.

Group 15 hydride bond angles collapse from 107 to 93.6 degrees in one step and barely move afterwards, as hybridisation gives way to nearly pure p bonding, and basicity falls with them.

Group 16 hydrides have water highest in boiling point and lowest in acidity at once, because hydrogen bonding decides one and bond enthalpy with anion size decides the other.

Fluorine has an anomalously low bond dissociation enthalpy from lone pair repulsion and is still the most reactive halogen, because the bonds it forms are exceptionally strong.

Noble gases are not inert. Xenon's ionisation enthalpy of 1170 kJ mol${}^{-1}$ sits just below oxygen's 1175, which is the entire reason its chemistry exists, and its compound shapes follow from VSEPR with $\mathrm{Xe}{\mathrm{F}}_2$ linear, $\mathrm{Xe}{\mathrm{F}}_4$ square planar and $\mathrm{Xe}{\mathrm{O}}_3$ pyramidal.