The correct order of the spin only magnetic moment of metal ions in the following low spin complexes, $\left[\mathrm{V}(\mathrm{CN})_{6}\right]^{4-},\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-},\left[\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$, and $\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$, is:

(B) $\mathrm{V}^{2+}>\mathrm{Cr}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Fe}^{2+}$

(A) $\mathrm{Cr}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Fe}^{2+}>\mathrm{V}^{2+}$

(C) $\mathrm{Cr}^{2+}>\mathrm{V}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Fe}^{2+}$

(D) $\mathrm{V}^{2+}>\mathrm{Ru}^{3+}>\mathrm{Cr}^{2+}>\mathrm{Fe}^{2+}$

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Coordination Compounds

The correct order of the spin only magnetic moment of metal ions in the following low spin complexes,

[V (CN)_{6}]^{4 -}, [Fe (CN)_{6}]^{4 -}, [Ru (NH_{3})_{6}]^{3 +}

, and

[Cr (NH_{3})_{6}]^{2 +}

, is:

Answer: (B)

All four complexes are low-spin, so count the

t_{2 g}

electrons.

[V (CN)_{6}]^{4 -}

has

V^{2 +}

d^{3}

with three unpaired electrons;

[Cr (N H_{3})_{6}]^{2 +}

has

Cr^{2 +}

as low-spin

d^{4}

with two;

[Ru (N H_{3})_{6}]^{3 +}

has

Ru^{3 +}

as low-spin

d^{5}

with one; and

[Fe (CN)_{6}]^{4 -}

has

Fe^{2 +}

as low-spin

d^{6}

with none. Since the spin-only moment grows with the number of unpaired electrons, the order is

V^{2 +} > Cr^{2 +} > Ru^{3 +} > Fe^{2 +}

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option