Trends in Chemical Properties of Group 14 Elements and Their Compounds

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Senior 4 · Chemistry

Trends in Chemical Properties of Group 14 Elements and Their Compounds

5 lessons

Lesson 2 of 5

Reactions of the Group 14 Elements and the Inert Pair Effect

25 min

Why tin(II) gives electrons away and lead(IV) takes them

Tin(II) chloride is sold as a reducing agent: it gives up electrons and becomes tin(IV). Lead(IV) oxide, the dark brown solid on the positive plate of a car battery, does the opposite: it is a strong oxidising agent that takes electrons and becomes lead(II). Tin and lead are neighbours in the same group, so why do their favourite oxidation states differ? The answer, called the inert pair effect, runs through the chemistry of the heavier p-block elements. This lesson first surveys how the Group 14 elements react, then explains the effect.

Definition

Inert pair effect: the growing reluctance, down a p-block group, of the outer $n s^{2}$ electrons to be lost or used in bonding. As a result the oxidation state two below the group maximum becomes more stable. In Group 14, +2 becomes more stable than +4 at lead. Example: $\mathrm{PbCl_2}$ is stable, while $\mathrm{PbCl_4}$ decomposes above about 50 °C to $\mathrm{PbCl_2}$ and chlorine.

Reducing agent: a species that gives electrons and is itself oxidised. Oxidising agent: a species that accepts electrons and is itself reduced.

Reactions with oxygen

All the elements burn or are oxidised when heated in air or oxygen. Carbon gives carbon dioxide in excess oxygen and carbon monoxide when oxygen is limited; silicon, germanium and tin give the dioxides; lead gives lead(II) oxide:

\mathrm{C + O_2 \rightarrow CO_2} \qquad \mathrm{2C + O_2 \rightarrow 2CO} \qquad \mathrm{Sn + O_2 \rightarrow SnO_2} \qquad \mathrm{2Pb + O_2 \rightarrow 2PbO}

Notice the switch at lead: even with plenty of oxygen it stops at the +2 oxide. (Heating PbO in air at about 450–480 °C gives the mixed oxide "red lead", $\mathrm{Pb_3O_4}$ , which contains both lead(II) and lead(IV).)

Reactions with water and steam

Red-hot carbon reacts with steam to give "water gas": $\mathrm{C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)}$ .
Silicon and tin react with steam only at red heat: $\mathrm{Sn(s) + 2H_2O(g) \rightarrow SnO_2(s) + 2H_2(g)}$ .
Lead does not react with steam. It is protected by a film of oxide and basic carbonate, although soft, acidic water slowly dissolves traces of lead from old pipes.

Reactions with acids

Carbon and silicon do not react with dilute acids. Hot concentrated nitric acid oxidises carbon to carbon dioxide. Silicon dissolves only in hydrofluoric acid.
Tin dissolves in hydrochloric acid to give tin(II), but an oxidising acid takes it to tin(IV): $\mathrm{Sn(s) + 2HCl(aq) \rightarrow SnCl_2(aq) + H_2(g)}$ With concentrated nitric acid a white solid of hydrated tin(IV) oxide forms.
Lead hardly reacts with dilute hydrochloric or sulphuric acid, because insoluble $\mathrm{PbCl_2}$ or $\mathrm{PbSO_4}$ coats the metal. It dissolves in dilute nitric acid, which gives the soluble nitrate: $\mathrm{3Pb(s) + 8HNO_3(aq) \rightarrow 3Pb(NO_3)_2(aq) + 2NO(g) + 4H_2O(l)}$

Reactions with alkalis

Carbon does not react. Silicon dissolves in hot sodium hydroxide solution, giving hydrogen and sodium silicate. Tin and lead also dissolve slowly in hot concentrated alkali, forming stannate and plumbate ions, which shows that they, like aluminium, are on the borderline between metal and non-metal:

\mathrm{Si(s) + 2NaOH(aq) + H_2O(l) \rightarrow Na_2SiO_3(aq) + 2H_2(g)}

\mathrm{Sn(s) + 2OH^-(aq) + 4H_2O(l) \rightarrow [Sn(OH)_6]^{2-}(aq) + 2H_2(g)}

Reactions with chlorine

Heated silicon, germanium and tin combine with chlorine to give the tetrachlorides, $\mathrm{MCl_4}$ , which are volatile covalent liquids. Lead, even with excess chlorine, gives lead(II) chloride, a white solid:

\mathrm{Sn + 2Cl_2 \rightarrow SnCl_4} \qquad \mathrm{Pb + Cl_2 \rightarrow PbCl_2}

Carbon does not combine directly with chlorine; tetrachloromethane, $\mathrm{CCl_4}$ , is made in other ways.

The inert pair effect

Every Group 14 atom has two s and two p outer electrons. Using all four gives the +4 state; using only the p pair gives +2. Down the group, +2 becomes steadily more important (Figure 9.4), until for lead it is the more stable state.

Figure 9.4: The more stable oxidation state for each element.

Two factors explain the trend:

The s electrons are held unusually tightly in the heavy elements. In tin and lead the outer s electrons penetrate close to the nucleus, where the filled 4d, 4f and 5d subshells shield them poorly; in lead they are also pulled in by relativistic effects (very fast-moving inner electrons). Removing the s pair from lead costs more than from tin (Figure 9.3), even though lead's atom is larger.
Bonds get weaker down the group. Forming two extra bonds is what pays for using the s electrons. For carbon and silicon the bonds formed are strong, so the +4 state wins easily. Bonds to the large lead atom are weak, so the energy released by forming two more bonds no longer repays the cost, and lead stays at +2.

Figure 9.3: The energy to remove the s pair ( $I E_{3} + I E_{4}$ ) is about three times the energy to remove the p electrons, and it rises from tin to lead.

Consequences. Compounds in the less stable state react to reach the more stable one. Tin(II) compounds are reducing agents ( $\mathrm{Sn^{2+} \rightarrow Sn^{4+} + 2e^-}$ , $E^\circ = +0.15\ \mathrm{V}$ for the reverse reduction), while lead(IV) compounds are strong oxidising agents ( $\mathrm{PbO_2 + 4H^+ + 2e^- \rightarrow Pb^{2+} + 2H_2O}$ , $E^\circ = +1.46\ \mathrm{V}$ ). Lead(IV) oxide even oxidises chloride ions in concentrated hydrochloric acid to chlorine:

\mathrm{PbO_2(s) + 4HCl(aq) \rightarrow PbCl_2(s) + Cl_2(g) + 2H_2O(l)}

Lead(IV) oxide, PbO₂, a dark brown powder. Because lead(IV) readily gains two electrons to become lead(II), this solid is a strong oxidising agent; it forms the positive plates of lead–acid car batteries. Photo: Walkerma, public domain, via Wikimedia Commons

Worked example 1: tin(II) as a reducing agent

Tin(II) chloride solution is added to iron(III) chloride solution, and the yellow-brown colour of $\mathrm{Fe^{3+}}$ fades to the pale green of $\mathrm{Fe^{2+}}$ . (a) Write the equation and give the oxidation number changes. (b) Use $E^\circ(\mathrm{Fe^{3+}/Fe^{2+}}) = +0.77\ \mathrm{V}$ and $E^\circ(\mathrm{Sn^{4+}/Sn^{2+}}) = +0.15\ \mathrm{V}$ to show the reaction is feasible.

Solution. (a) $\mathrm{SnCl_2 + 2FeCl_3 \rightarrow SnCl_4 + 2FeCl_2}$ , or ionically $\mathrm{Sn^{2+} + 2Fe^{3+} \rightarrow Sn^{4+} + 2Fe^{2+}}$ . Tin goes from +2 to +4 (oxidised: it is the reducing agent); each iron goes from +3 to +2 (reduced).
(b) $E^\circ_{\text{cell}} = E^\circ(\text{reduced}) - E^\circ(\text{oxidised}) = 0.77 - 0.15 = +0.62\ \mathrm{V}$ . A positive value means the reaction is feasible.

Worked example 2: chlorine from lead(IV) oxide

2.39 g of lead(IV) oxide is warmed with excess concentrated hydrochloric acid. Calculate the volume of chlorine formed at room temperature and pressure (24.0 dm³ mol⁻¹). $A_{r}$ : Pb 207.2, O 16.00.

Solution. $\mathrm{PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O}$ : 1 mol $\mathrm{PbO_2}$ gives 1 mol $\mathrm{Cl_2}$ . $M(\mathrm{PbO_2}) = 207.2 + 32.00 = 239.2\ \mathrm{g\,mol^{-1}} \qquad n = \frac{2.39}{239.2} = 9.99 \times 10^{-3}\ \mathrm{mol}$ $V(\mathrm{Cl_2}) = 9.99 \times 10^{-3} \times 24.0 = 0.240\ \mathrm{dm^3} = 240\ \mathrm{cm^3}$ Lead goes from +4 to +2 and two chloride ions are oxidised to chlorine. $\mathrm{SnO_2}$ does not do this: tin(IV) is the stable state of tin.

Warning

Common mistakes. (1) Saying the s electrons are "inert" because they cannot be removed: they can be (Sn and Pb do form +4 compounds); the point is that using them is less often repaid by bond energy. (2) Writing $\mathrm{PbCl_4}$ as the product of lead and chlorine: it is $\mathrm{PbCl_2}$ . (3) Expecting tin to give $\mathrm{SnCl_4}$ with hydrochloric acid: it gives $\mathrm{SnCl_2}$ and hydrogen. (4) Mixing up the roles: tin(II) is the reducing agent, lead(IV) the oxidising agent.

Safety. The reaction of lead(IV) oxide with concentrated hydrochloric acid releases chlorine, a toxic gas: it is done only as a small-scale teacher demonstration in a fume cupboard. Lead compounds are toxic: wear gloves and wash hands afterwards. Concentrated acids cause burns; if any splashes on skin or into the eyes, rinse at once with plenty of running water and get medical help.

Tip

Down Group 14, "+4 at the top, +2 at the bottom". A compound in the "wrong" state for its element is reactive: Sn(II) reduces, Pb(IV) oxidises, CO reduces. Learn the two equations $\mathrm{Sn^{2+} + 2Fe^{3+} \rightarrow Sn^{4+} + 2Fe^{2+}}$ and $\mathrm{PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O}$ .

The Organic Chemistry Tutor: Inert Pair Effect (12 min). Watch how the video uses electron configurations to explain why tin forms Sn²⁺ and Sn⁴⁺, and why lower oxidation states become more stable down the p-block groups.

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Practice

Write equations for the reaction of (a) carbon with limited oxygen; (b) tin with dilute hydrochloric acid; (c) lead with chlorine.
Explain why lead does not appear to react with dilute sulphuric acid, but dissolves in dilute nitric acid.
Write an equation for silicon dissolving in sodium hydroxide solution.
Define the inert pair effect and give two pieces of evidence for it from the chemistry of lead.
Use Figure 9.3 to explain why removing four electrons from lead costs more energy than from tin, although lead's atom is larger.
Predict, with a reason, whether tin(IV) oxide will release chlorine from concentrated hydrochloric acid.
Lead(IV) chloride decomposes when warmed. Write the equation, and calculate the mass of lead(II) chloride formed from 3.49 g of $\mathrm{PbCl_4}$ . $A_{r}$ : Pb 207.2, Cl 35.45.

Show answersHide answers

(a) $\mathrm{2C + O_2 \rightarrow 2CO}$ ; (b) $\mathrm{Sn + 2HCl \rightarrow SnCl_2 + H_2}$ ; (c) $\mathrm{Pb + Cl_2 \rightarrow PbCl_2}$ .
Insoluble lead(II) sulphate forms on the surface and stops further reaction. Lead(II) nitrate is soluble, so nitric acid keeps attacking fresh metal.
$\mathrm{Si + 2NaOH + H_2O \rightarrow Na_2SiO_3 + 2H_2}$ .
The reluctance of the outer $n s^{2}$ pair to be lost or used in bonding, making the state two below the group maximum more stable down the group. Evidence: lead burns to PbO, not $\mathrm{PbO_2}$ ; lead and chlorine give $\mathrm{PbCl_2}$ ; $\mathrm{PbO_2}$ is a strong oxidising agent; $\mathrm{PbCl_4}$ decomposes to $\mathrm{PbCl_2}$ .
$I E_{3} + I E_{4}$ is 7164.5 kJ mol⁻¹ for lead and 6873.3 for tin. Lead's 6s electrons are poorly shielded by the 4f and 5d electrons and are pulled in by relativistic effects, which outweighs the larger size.
No: tin(IV) is the more stable state of tin, so $\mathrm{SnO_2}$ has no tendency to gain electrons; $E^\circ(\mathrm{Sn^{4+}/Sn^{2+}}) = +0.15\ \mathrm{V}$ is far below that for chlorine.
$\mathrm{PbCl_4 \rightarrow PbCl_2 + Cl_2}$ . $M(\mathrm{PbCl_4}) = 207.2 + 4(35.45) = 349.0$ ; $n = \frac{3.49}{349.0} = 0.0100$ mol; $M(\mathrm{PbCl_2}) = 278.1$ ; mass $= 0.0100 \times 278.1 = 2.78\ \mathrm{g}$ .

Quick check

Lead is heated in excess chlorine. What is the main product?

Key points

With oxygen, C, Si, Ge and Sn give $\mathrm{MO_2}$ , but lead gives PbO; with chlorine, Si, Ge and Sn give $\mathrm{MCl_4}$ , but lead gives $\mathrm{PbCl_2}$ .
C and Si resist dilute acids; Sn gives $\mathrm{SnCl_2}$ with HCl; Pb is protected by insoluble $\mathrm{PbCl_2}$ / $\mathrm{PbSO_4}$ but dissolves in dilute $\mathrm{HNO_3}$ . Si, Sn and Pb dissolve in hot alkali.
Inert pair effect: down the group the $n s^{2}$ pair is held more tightly and bonds get weaker, so +2 grows in stability and becomes the more stable state for lead.
So Sn(II) is a reducing agent ( $\mathrm{Sn^{2+} + 2Fe^{3+} \rightarrow Sn^{4+} + 2Fe^{2+}}$ ) and Pb(IV) an oxidising agent ( $\mathrm{PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O}$ ).

Sources and credits

Image: PbO.jpg by Hubertus Giefers, via Wikimedia Commons, public domain.
Image: Lead dioxide.jpg by Walkerma, via Wikimedia Commons, public domain.
Data: ionisation energies from Wikipedia's ionisation energies data page; standard electrode potentials from the standard electrode potential data page; decomposition of lead(IV) chloride from Lead(IV) chloride; preparation of red lead from Lead(II,IV) oxide; all CC BY-SA 4.0, adapted by Ganzaa.
First aid for chemical splashes: NIOSH Pocket Guide, first aid, CDC.
Video: Inert Pair Effect, The Organic Chemistry Tutor, embedded from YouTube.
The text, charts, diagram, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Why tin(II) gives electrons away and lead(IV) takes them

2Reactions with oxygen

3Reactions with water and steam

4Reactions with acids

5Reactions with alkalis

6Reactions with chlorine

7The inert pair effect