How would you account for the following situations?
i) The transition metals generally form coloured compounds.
ii) With a 3d⁴ configuration, Cr²⁺ acts as a reducing agent but Mn³⁺ acts as an oxidising agent (Z: Cr = 24, Mn = 25).
iii) The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
i) The transition metals generally form coloured compounds.
ii) With a 3d⁴ configuration, Cr²⁺ acts as a reducing agent but Mn³⁺ acts as an oxidising agent (Z: Cr = 24, Mn = 25).
iii) The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
Solution
(i) Transition-metal ions have incompletely filled d-orbitals. The ligand field splits these orbitals, and electrons jump between the split levels (d–d transitions) by absorbing particular visible wavelengths — the transmitted light gives the compound its colour.
(ii) Both ions are d, but stability lies in opposite directions:
Cr → Cr + e attains the stable t (d) — Cr donates electrons, i.e. reduces.
Mn + e → Mn attains the extra-stable half-filled d — Mn accepts electrons, i.e. oxidises.
(iii) In the actinoids the 5f, 6d and 7s orbitals are nearly degenerate, so differing numbers of electrons can engage in bonding, producing states up to +7; lanthanoid 4f electrons are buried, restricting them essentially to +3.
(ii) Both ions are d, but stability lies in opposite directions:
Cr → Cr + e attains the stable t (d) — Cr donates electrons, i.e. reduces.
Mn + e → Mn attains the extra-stable half-filled d — Mn accepts electrons, i.e. oxidises.
(iii) In the actinoids the 5f, 6d and 7s orbitals are nearly degenerate, so differing numbers of electrons can engage in bonding, producing states up to +7; lanthanoid 4f electrons are buried, restricting them essentially to +3.
CBSE Class 12 Board · 2008 · All India · Subjective