The pair of metal ions that can give a spin-only magnetic moment of 3.9 BM for the complex $\left[\mathrm{M}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ is

(A) $\mathrm{V}^{2+}$ and $\mathrm{Co}^{2+}$

(B) $\mathrm{V}^{2+}$ and $\mathrm{Fe}^{2+}$

(C) $\mathrm{Co}^{2+}$ and $\mathrm{Fe}^{2+}$

(D) $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{2+}$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · d and f Block Elements

The pair of metal ions that can give a spin-only magnetic moment of 3.9 BM for the complex

[M (H_{2} O)_{6}] Cl_{2}

Answer: (A)

A spin-only moment of

3.9 BM

comes from

μ = n (n + 2)

with

n = 3

unpaired electrons. Water is a weak-field ligand, so the

d

-electrons stay maximally unpaired.

V X^{2 +}

d^{3}

and plainly has three unpaired electrons;

Co X^{2 +}

d^{7}

, which in the high-spin case fills as

t_{2 g}^{5} e_{g}^{2}

and also leaves three unpaired.

Fe X^{2 +}

(

d^{6}

) would give four. So the pair that both give

3.9 BM

V X^{2 +}

and

Co X^{2 +}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option