Q50 · CSIR-NET Chemistry, June 2013

Paper: CSIR-NET June 2013 · Subject: Inorganic Chemistry · Chapter: Lanthanides and Actinides · Topic: Lanthanides & Actinides – General · Marks: 2 · Difficulty: Medium

The ground state term symbol for Nb (atomic number 41) is $^{6}\mathrm{D}$. The electronic configuration corresponding to this term symbol is
(a)$[\mathrm{Kr}]4d^{3}5s^{2}$
(b)$[\mathrm{Kr}]4d^{4}5s^{1}$
(c)$[\mathrm{Kr}]4d^{5}5s^{0}$
(d)$[\mathrm{Kr}]4d^{3}5s^{1}5p^{1}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
⁶D needs S = 5/2 (five unpaired electrons) and L = 2: [Kr]4d⁴5s¹, the actual ground configuration of Nb.

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