Calculate the spin only magnetic moment of a \( Ni^{2+} \) in \( [Ni(CN)_4]^{2-} \) (atomic number of Ni is 28).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
The spin-only magnetic moment depends only on the number of unpaired electrons, n:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \text{BM} \)
Ni (Z = 28) is [Ar]3d84s2. In \( \ce{[Ni(CN)4]^2-} \), x + 4(−1) = −2 gives x = +2, and losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8. In the free ion these eight electrons leave two unpaired.
\( \ce{CN^{-}} \) is a strong field ligand: it pairs up the two unpaired electrons, so no unpaired electron is left:
\( \displaystyle n = 0 \Rightarrow \mu = \sqrt{0 \times 2} = 0 \)
So the complex is diamagnetic and the spin-only magnetic moment is 0.
NCERT (2026-27 edition):
Valence Bond Theory – …
