Which of the following metal complexes will show spin only magnetic moment of 4.90 BM?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → [Cr(H2O)6]2+
The spin-only magnetic moment depends only on the number of unpaired electrons (n):
\( \displaystyle \mu=\sqrt{n(n+2)}\ \mathrm{BM} \)
For 4.90 BM:
\( \displaystyle \sqrt{n(n+2)}=4.90 \)
\( \displaystyle n(n+2)=24 \)
\( \displaystyle n=4 \)
\( \ce{H2O} \) is a weak field ligand, so the electrons stay unpaired as far as possible. In \( \ce{[Cr(H2O)6]^{2+}} \), chromium is \( \ce{Cr^{2+}} \) (water is neutral). Cr (Z = 24) is [Ar]3d54s1; losing the 4s electron and one 3d electron leaves 3d4, with four unpaired electrons. So its spin-only moment is 4.90 BM.
Shortcut: 4.90 = √24 means four unpaired electrons, that is a high-spin d4 ion. Only \( \ce{Cr^{2+}} \) among the given ions is d4.
Why the other options are incorrect
Option (1) [V(H2O)6]2+ — Incorrect. V (Z = 23) is [Ar]3d34s2; losing the two 4s electrons gives \( \ce{V^{2+}} \): 3d3, with three unpaired electrons, so μ = √15 = 3.87 BM.
Option (3) [Ni(H2O)6]2+ — Incorrect. Ni (Z = 28) is [Ar]3d84s2; losing the two 4s electrons gives \( \ce{Ni^{2+}} \): 3d8, with two unpaired electrons, so μ = √8 = 2.83 BM.
Option (4) [Ti(H2O)6]2+ — Incorrect. Ti (Z = 22) is [Ar]3d24s2; losing the two 4s electrons gives \( \ce{Ti^{2+}} \): 3d2, with two unpaired electrons, so μ = 2.83 BM.
NCERT (2026-27 edition):
Magnetic Properties – …
