(a) \(E^{\circ}_{(\mathrm{Mn^{2+}/Mn})}\) is \(-1.18\ \mathrm{V}\). Why is this value highly negative in comparison to neighbouring d-block elements?
(b) What is lanthanoid contraction?
(c) Zn, Cd and Hg are soft metals. Why?
(a) \(E^{\circ}_{(\mathrm{Mn^{2+}/Mn})}\) is \(-1.18\ \mathrm{V}\). Why is this value highly negative in comparison to neighbouring d-block elements?
(b) What is lanthanoid contraction?
(c) Zn, Cd and Hg are soft metals. Why?
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Solution and Explanation
Three short theory answers are needed, one mark each. They test the stability of d-configurations, the f-block trend and the metallic bonding of group 12.
Step 2: Part (a) Highly negative \(E^{\circ}\) of Mn.
The value \(E^{\circ}_{\mathrm{Mn^{2+}/Mn}}\) refers to the change \(\mathrm{Mn^{2+} + 2e^{-} \to Mn}\).
The \(\mathrm{Mn^{2+}}\) ion has the configuration \(3d^{5}\), a half-filled d subshell. This is a very stable arrangement.
It is hard to push two electrons into a stable \(3d^{5}\) ion to form the atom. So the reduction is not favourable and the value comes out highly negative. Put simply, \(\mathrm{Mn^{2+}}\) is unusually stable.
Step 3: Part (b) Lanthanoid contraction.
The lanthanoids run from Ce (Z = 58) to Lu (Z = 71), with electrons going into the inner 4f subshell.
The 4f electrons shield each other very poorly because of the diffuse shape of the f orbitals. The nuclear charge goes up by one at each step, so the outer electrons are pulled in more strongly.
As a result, atomic and ionic radii decrease slowly and steadily from La to Lu. This regular decrease in size is called lanthanoid contraction.
Step 4: Part (c) Zn, Cd and Hg are soft.
The configuration of these elements is \((n-1)d^{10}\,ns^{2}\). The d subshell is completely filled, so there are no unpaired d electrons.
Only the two s electrons are available for metallic bonding. Metallic bonding is therefore weak.
These metals have low enthalpies of atomisation and low melting points (Hg is a liquid), and they are soft.
Final Answer:
(a) Stable half-filled \(3d^{5}\) in \(\mathrm{Mn^{2+}}\). (b) Steady decrease in size from La to Lu due to poor 4f shielding. (c) Filled d10, so only weak metallic bonding.
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