CBSE Class 12 · 2005 · Foreign · d and f Block ElementsShort Answer

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The outer electronic configurations of atoms of two members of the lanthanoid series are given below:
i) 4f¹5d¹6s²
ii) 4f⁷5d⁰6s²
Find their atomic numbers. What oxidation states will possibly be exhibited by these elements?

Answer: (i) Z = 58, cerium — states +3 and +4 (f⁰). (ii) Z = 63, europium — states +3 and +2 (f⁷).

Solution

(i) 4f

^{1}

^{1}

^{2}

: count = 54 (Xe core) + 4 = Z = 58 → cerium.
Oxidation states: +3 (normal) and +4 — Ce

^{4 +}

attains the stable 4f

^{0}

configuration.
(ii) 4f

^{7}

^{0}

^{2}

: 54 + 9 = Z = 63 → europium.
Oxidation states: +3 (normal) and +2 — Eu

^{2 +}

keeps the stable half-filled 4f

^{7}

CBSE Class 12 Board · 2005 · Foreign · Subjective

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