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JEE Main 2024
09-04-2024 S2
Question
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)
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Solution
$ \mathrm{Co}^{2+}+\mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{CoS} \downarrow \text { (Black) }
$
$ \begin{equation*} \mathrm{CoS}+\text { Aqua-regia } \rightarrow \mathrm{Co}^{2+}(\mathrm{aq})+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O} \tag{A} \end{equation*} $
(A)
(B)
$ \mathrm{Co}^{2+}(\mathrm{aq})+\mathrm{KNO}_2+\mathrm{CH}_3 \mathrm{COOH} $
$ \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right]+\mathrm{NO}+\mathrm{S}+\mathrm{H}_2 \mathrm{O} $
$ \mathrm{Co}^{3+}: \mathrm{d}^2 \mathrm{sp}^3 \text { Hybridisation } $
Number of unpaired $\mathrm{e}^{-}=0$
$ \text { Magnetic moment }=\sqrt{n(n+2)}=0 \text { B.M } $
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