Consider the following test for a group-IV cation. $\mathrm{M}^{2+}+\mathrm{H}_{2} \mathrm{~S} \rightarrow \mathrm{~A}$ (Black precipitate) + byproduct $\mathrm{A}+$ aqua regia $\rightarrow \mathrm{B}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_{2} \mathrm{O}$ $\mathrm{B}+\mathrm{KNO}_{2}+\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{C}+$ byproduct The spin only magnetic moment value of the metal complex C is ____ BM. (Nearest integer)

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Coordination Compounds

Consider the following test for a group-IV cation.

M^{2 +} + H_{2} S \to A

(Black precipitate) + byproduct

A +

aqua regia

\to B + NOCl + S + H_{2} O

B + KNO_{2} + CH_{3} COOH \to C +

byproduct
The spin only magnetic moment value of the metal complex C is ____ BM.
(Nearest integer)

Answer: 0

Solution

Co^{2 +} + H_{2} S \to CoS ↓ (black, A)

CoS + aqua regia \to Co^{2 +} (a q) (B) + NOCl + S + H_{2} O

Co^{2 +} + KNO_{2} + CH_{3} COOH \to K_{3} [Co(NO_{2})_{6}] (C)

, in which cobalt is oxidised to Co

^{3 +}

(

3 d^{6}

), and with NO

_{2}^{-}

as a strong-field ligand the complex is low-spin,

d^{2} s p^{3}

, with all electrons paired.

Number of unpaired electrons in

C

= 0, so

μ = n (n + 2) = 0 BM

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Integer answer

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