Oxidation states and the inert pair effect

A battery, a stain remover and a poison

The brown lead(IV) oxide in a car battery is a strong oxidiser. Tin(II) chloride is a mild reducing agent that removes stains of iron. Thallium compounds were once used as rat poison, and they are poisonous to people too. These substances share a pattern. The heavy members of groups 13 and 14 do not like to give up all their outer electrons.

The pattern of oxidation states

The group 13 atoms have the outer pattern $ns^2\,np^1$ and the group 14 atoms have $ns^2\,np^2$. They can show the maximum state of $+3$ or $+4$, when all outer electrons are used. They can also show a state lower by two, $+1$ or $+2$, when only the $p$ electrons are used and the $ns^2$ pair stays at home.

• Group 13: boron and aluminium are $+3$ only. Gallium is almost always $+3$. Indium has $+3$ and a weak $+1$. Thallium is more stable as $+1$ than as $+3$.
• Group 14: carbon and silicon are $+4$ in nearly all compounds. Germanium is mostly $+4$. Tin has both, and tin(II) is a reducing agent. Lead is more stable as $+2$, and lead(IV) is an oxidiser.

The reluctance of the $ns^2$ pair to take part in bonding is called the inert pair effect.

Why the pair is inert

Two causes work together.

1. Poor shielding. Down the group the $d$ and $f$ electrons are added between the $s$ electrons and the nucleus. They shield badly. So the $ns$ electrons feel a strong pull from the nucleus and are held tightly.
2. Weaker bonds. The bonds that the large atoms form are longer and weaker. The energy that two more bonds give out does not pay for the energy needed to take the $ns$ electrons off.

Heavy atoms have a third effect. Near a heavy nucleus the $6s$ electrons of Tl and Pb move at a large fraction of the speed of light, and this holds them more tightly. We do not need this to use the rule.

Evidence from potentials. A strong oxidiser has a high standard potential for being reduced to the lower state.

• Tl$^{3+}$ to Tl$^+$: $+1.25\ \text{V}$. Thallium(III) is a strong oxidiser.
• PbO$_2$ to Pb$^{2+}$ in acid: $+1.46\ \text{V}$. Lead(IV) oxide is a strong oxidiser.
• Sn$^{4+}$ to Sn$^{2+}$: $+0.15\ \text{V}$. Tin(IV) is hardly an oxidiser, and tin(II) is a reducing agent.

NCERT prints a value near this, $+1.26\ \text{V}$, in its table of group 13. The standard value for Tl$^{3+}$ changing to Tl$^+$ is $+1.25\ \text{V}$, and this is the one used here.

Keep the reducing couple at $0.15\ \text{V}$ (Sn$^{4+}$/Sn$^{2+}$, so tin(II) is the reducer). Slide the oxidiser from 0 to 1.8. A reaction with a log K over 10 or so is complete. Tl(III) and Pb(IV) are well above this. Sn(IV) gives zero, which means that neither side is favoured.

What this means for compounds

• Thallium. TlCl is stable and TlCl$_3$ loses chlorine on heating. Thallium(I) is large and its compounds are ionic, like those of potassium. TlOH is a strong base, and Tl$_2$CO$_3$ dissolves in water. The compound written TlI$_3$ is really Tl$^+$ I$_3^-$, because Tl$^{3+}$ would oxidise iodide.
• Tin. Tin(II) chloride is a reducing agent: $\text{Sn}^{2+} + 2\text{Fe}^{3+} \to \text{Sn}^{4+} + 2\text{Fe}^{2+}$. With mercury(II) chloride it forms white calomel, and then grey mercury with more tin(II): $2\text{HgCl}_{2} + \text{SnCl}_{2} \to \text{Hg}_{2}\text{Cl}_{2} + \text{SnCl}_{4}$ and $\text{Hg}_{2}\text{Cl}_{2} + \text{SnCl}_{2} \to 2\text{Hg} + \text{SnCl}_{4}$.
• Lead. Lead(IV) compounds oxidise. $\text{PbO}_{2} + 4\text{HCl} \to \text{PbCl}_{2} + \text{Cl}_{2} + 2\text{H}_{2}\text{O}$. The chloride $\text{PbCl}_{4}$ is an unstable yellow liquid that loses chlorine, $\text{PbCl}_{4} \to \text{PbCl}_{2} + \text{Cl}_{2}$. $\text{PbBr}_{4}$ and $\text{PbI}_{4}$ do not exist, since Pb(IV) would oxidise bromide and iodide.
• Bond type. The lower state is more ionic (larger, lower charge on the cation). The higher state is more covalent. So $\text{SnCl}_{2}$ and $\text{PbCl}_{2}$ are solids with ionic character, and $\text{SnCl}_{4}$ is a covalent liquid.

Practise with new numbers

Try it in Python

The program takes some couples and tests which oxidations go, and how far.

Common mistakes

• Saying that the inert pair stays in the inner shell. It is the outer $ns^2$ pair. It is only reluctant to bond.
• Using $+3$ for thallium in every compound. It is $+1$ in most.
• Saying that Sn(IV) is a strong oxidiser. It is Pb(IV) that is.
• Counting one electron in a reaction that transfers two.

Remember these

• Heavy elements of groups 13 and 14 prefer the state that is two lower. The $+1$ state of Tl and the $+2$ state of Pb are stable.
• Cause: poor shielding by $d$ and $f$ electrons, and weaker bonds.
• Tl(III), Pb(IV) oxidise. Sn(II), Ge(II) reduce.
• The lower state is more ionic. The higher state is more covalent.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which is the stronger oxidiser, Sn(IV) or Pb(IV), and why?

2. Why does $\text{PbI}_{4}$ not exist?

3. TlI$_3$ is described as Tl$^+$ I$_3^-$. What does it tell us about thallium?

4. Give the two causes of the inert pair effect.

References

• OpenStax, Chemistry 2e, Chapter 18 (Representative Metals, Metalloids and Nonmetals), the section on periodicity.
• NCERT Chemistry for Class 11, the chapter The p-Block Elements (older edition), the sections on oxidation states and the inert pair effect.
• Atkins and Shriver, Inorganic Chemistry, the chapters on the group 13 and group 14 elements.