Question

(i) Which element in 3d series of transition elements does not exhibit variable oxidation states ? Give reason.

(ii) Explain why \( \mathrm{Cu^{+}} \) is colourless in aqueous solution, whereas \( \mathrm{Cu^{2+}} \) is coloured.

Hint:

Scandium gives only the +3 state. For copper ions, count d electrons: colour needs an unpaired d electron for a d-d transition.

Answer 1

Step 1: Understanding the Concept:
Variable oxidation states come from the small energy gap between 3d and 4s electrons. Colour comes from d-d transitions, which need unpaired d electrons.

Step 2: Part (i) - Identify the element:
Scandium (Z = 21, \( 3d^1 4s^2 \)) does not show variable oxidation states.
The reason is that it loses all three outer electrons (two 4s and one 3d) and shows only the +3 state.
Sc(III) then has the stable noble gas configuration \( [\mathrm{Ar}] \). A fourth electron cannot be removed easily.

Step 3: Part (ii) - Configuration of the two ions:
\( \mathrm{Cu^{+}} \) (Z = 29) is \( 3d^{10} \). All five d orbitals are filled, so there is no unpaired electron.
\( \mathrm{Cu^{2+}} \) is \( 3d^9 \), so it has one unpaired electron.

Step 4: Part (ii) - Link to colour:
In water, the d orbitals split into two energy sets. A coloured ion absorbs visible light to lift an electron from the lower set to the higher set (d-d transition).
\( \mathrm{Cu^{2+}} \) has a vacancy in the higher set, so the transition is possible. It absorbs red light and appears blue.
\( \mathrm{Cu^{+}} \) has no vacancy in the d orbitals, so no d-d transition can occur. It absorbs no visible light and is colourless.

Final Answer:
(i) Scandium, because it shows only the +3 state with a stable noble gas core. (ii) \( \mathrm{Cu^{+}} \) is \( 3d^{10} \) with no d-d transition, while \( \mathrm{Cu^{2+}} \) is \( 3d^9 \) with one unpaired electron and shows d-d transition.