The correct order of the spin-only magnetic moments of the following complexes is: (I) $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}$ (II) $\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$ (III) $\mathrm{Na}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right](\mathrm{Do}>\mathrm{P})$ (IV) $\left(\mathrm{Et}_{4} \mathrm{~N}\right)_{2}\left[\mathrm{CoCl}_{4}\right]$

(A) (I) $>$ (IV) $>$ (III) $>$ (II)

(B) (II) $\gg$ (I) $>$ (IV) $>$ (III)

(C) (III) $>$ (I) $>$ (IV) $>$ (II)

(D) (III) $>$ (I) $>$ (II) $>$ (IV)

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Coordination Compounds

The correct order of the spin-only magnetic moments of the following complexes is:
(I)

[Cr (H_{2} O)_{6}] Br_{2}

(II)

Na_{4} [Fe (CN)_{6}]

(III)

Na_{3} [Fe (C_{2} O_{4})_{3}] (Do > P)

(IV)

(Et_{4} N)_{2} [CoCl_{4}]

Answer: (A)

Using the spin-only moment

μ = n (n + 2)

: in

[Cr (H_{2} O)_{6}] B r_{2}

the

C r^{2 +}

ion is

3 d^{4}

with four unpaired electrons,

μ = 24 \approx 4.90

; in

N a_{4} [Fe (CN)_{6}]

the

F e^{2 +}

ion with the strong-field

C N^{-}

is low-spin

t_{2 g}^{6}

with no unpaired electrons,

μ = 0

; in

N a_{3} [Fe (C_{2} O_{4})_{3}]

, where

Δ_{o} > P

, the

F e^{3 +}

ion is low-spin

t_{2 g}^{5}

with one unpaired electron,

μ = 3 \approx 1.73

; and in

(E t_{4} N)_{2} [CoC l_{4}]

the tetrahedral

C o^{2 +}

has three unpaired electrons,

μ = 15 \approx 3.87

. Ranking the moments gives

[Cr (H_{2} O)_{6}]^{2 +} > [CoC l_{4}]^{2 -} > [Fe (C_{2} O_{4})_{3}]^{3 -} > [Fe (CN)_{6}]^{4 -}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option