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QJEEM24Ch1638: JEE Main 2024 Chemistry - Class 12 JEEM24Ch1638 Solution: Consider the following test for a group-IV cation. The…

Consider the following test for a group-IV cation.

M2++H2S→A (Black precipitate)+ byproduct \mathrm{M}^{2+}+\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{A} \text { (Black precipitate)+ byproduct }

A+ aqua regia →B+NOCl+S+H2O\mathrm{A}+\text { aqua regia } \rightarrow \mathrm{B}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}

B+KNO2+CH3COOH→C+ byproduct \mathrm{B}+\mathrm{KNO}_2+\mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{C}+\text { byproduct }

The spin-only magnetic moment value of the metal complex C\mathrm{C} is _________ BM\mathrm{BM} (Nearest integer)

Step-by-Step Verified Solution

JEE Main 2024 (Online) 9th April Evening Shift Chemistry - Coordination Compounds Question 44 English Explanation

amp; In K3[Co(NO2)6],Co+3:3 d64 s0amp;Co3+:d2sp3 Hybridisation amp; Number of unpaired e−=0amp; Magnetic moment =n(n+2)=0 B.M amp;\begin{aligned}& \text { In } \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right], \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 4 \mathrm{~s}^0 \\& \mathrm{Co}^{3+}: \mathrm{d}^2 \mathrm{sp}^3 \text { Hybridisation } \\& \text { Number of unpaired } \mathrm{e}^{-}=0 \\& \text { Magnetic moment }=\sqrt{n(n+2)}=0 \text { B.M } \\&\end{aligned}

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