(a) How would you account for the following:

(i) The chemistry of actinoids is more complicated as compared to lanthanoids.

(ii) Transition metals form complex compounds.

(b) Complete the following equation:

$2 {M n O_{4}}^{-} + 6 H^{+} + {S O_{3}}^{2 -} \to$

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Solution

(a)

(i) The energy difference between the 6d, 7s and 5f orbitals is very small in actinoids. In addition, the 5f orbitals of the actinoids are less shielded than the 4f orbitals of the lanthanoids. Thus, in the lanthanoids, only 4f electrons of Ce participate in chemical bonding, whereas in the lighter actinoids, up to americium, the 5f electrons also take part. This makes the chemistry of actinoids more complicated than that of lanthanoids.

(ii) The transition elements form complexes because they are able to accept pairs of electrons from donor molecules or ions to form dative covalent bonds. This happens because they have vacant orbitals of suitable energy which can accept the non-bonding pairs from the ligands.

$(b) 2 {M n O_{4}}^{-} + 6 H^{+} + 5 {S O_{3}}^{2 -} \to 2 M n^{2 +} + 3 H_{2} O + 5 {S O_{4}}^{2 -}$

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