NEET 2016 Phase 1 · ChemistryLanthanoidsPrevious Year Question

The electronic configurations of Eu (Atomic no. 63), Gd (Atomic No. 64) and Tb (Atomic No. 65) are:

Answer: (D) (d) $[Xe] 4 f X^{7} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{9} 6 s X^{2}$ . \textbf{Answer:} (d) $[Xe] 4 f X^{7} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{9} 6 s X^{2}$ .

A.(a) $[Xe] 4 f X^{7} 6 s X^{2}$ , $[Xe] 4 f X^{8} 6 s X^{2}$ and $[Xe] 4 f X^{8} 5 d X^{1} 6 s X^{2}$
B.(b) $[Xe] 4 f X^{6} 5 d X^{1} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{9} 6 s X^{2}$
C.(c) $[Xe] 4 f X^{6} 5 d X^{1} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{8} 5 d X^{1} 6 s X^{2}$
D.(d) $[Xe] 4 f X^{7} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{9} 6 s X^{2}$ ✓

Correct Answer

(D) (d) $[Xe] 4 f X^{7} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{9} 6 s X^{2}$

Solution & Explanation

\textbf{Answer:} (d) $[Xe] 4 f X^{7} 6 s X^{2}$ , $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ and $[Xe] 4 f X^{9} 6 s X^{2}$ . \textbf{Solution:} For lanthanoids the $4 f$ subshell is filled progressively. Eu (Z = 63): $[Xe] 4 f X^{7} 6 s X^{2}$ — a stable half-filled $4 f X^{7}$ . Gd (Z = 64): the extra electron enters $5 d$ (not $4 f$ ) to retain the stable half-filled $4 f X^{7}$ , giving $[Xe] 4 f X^{7} 5 d X^{1} 6 s X^{2}$ . Tb (Z = 65): the configuration becomes $[Xe] 4 f X^{9} 6 s X^{2}$ . This matches option (d).

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