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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Practical Chemistry

JEE Main 2024 · 9 April · Evening Shift · Chemistry · Practical Chemistry

Consider the following test for a group-IV cation. aligned & M^ 2+...

Consider the following test for a group-IV cation. M2++H2S→A (Black precipitate) + by product A+ aqua regia →B+NOCl+S+H2OB+KNO2+CH3COOH→C+ by product \begin{aligned} & \mathrm{M}^{2+}+\mathrm{H}_2 \mathrm{ S} \rightarrow \mathrm{A} \text{ (Black precipitate) }+ \text{ by product } \\ & \mathrm{A}+\text{ 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}+\text{ by product } \end{aligned})TheThespin-only magnetic moment value of the metal complex (\mathrm{C}) is ______ BM (Nearest integer)$

Correct answer

0

Step-by-step solution

If M+2=Co+2M^{+2}=C o^{+2}thenthenCo+2+H2 S→CoS Black  [A] + byproduct CoS+ aqua regia →CoCl2[B]+NOCl+S+H2O\begin{aligned} & \mathrm{Co}^{+2}+\mathrm{H}_2 \mathrm{~S} \rightarrow \underset{\substack{\text{ Black } \\ \text{ [A] }}}{\mathrm{CoS}}+\text{ byproduct } \\ & \mathrm{CoS}+\text{ aqua regia } \rightarrow \underset{\text{[B]}}{\mathrm{CoCl}_2}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O} \end{aligned})(CoCl2+KNO2→CH3COOHk3[CO(NO2)6]+ byproduct )([C]=Co+3→3d6⇒NO2)isSFLsoCo+3haszerounpaired) (\mathrm{CoCl}_2+\mathrm{KNO}_2 \xrightarrow{\mathrm{CH}_3 \mathrm{COOH}} k_3\left[\mathrm{CO}\left(\mathrm{NO}_2\right)_6\right]+\text{ byproduct }) ([C]=C o^{+3} \rightarrow 3 d^6 \Rightarrow N O_2) is SFL so Co +3 has zero unpairede^{-}(⇒n=0⇒u=n(n+2)=0B.M)(\Rightarrow n=0 \Rightarrow u=\sqrt{n(n+2)}=0 B . M)
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