JEE MAIN - Chemistry (2024 - 9th April Evening Shift - No. 28)
Consider the following test for a group-IV cation.
$$\mathrm{M}^{2+}+\mathrm{H}_2 \mathrm{S} \rightarrow \mathrm{A} \text { (Black precipitate)+ byproduct }$$
$$\mathrm{A}+\text { aqua regia } \rightarrow \mathrm{B}+\mathrm{NOCl}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}$$
$$\mathrm{B}+\mathrm{KNO}_2+\mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{C}+\text { byproduct }$$
The spin-only magnetic moment value of the metal complex $$\mathrm{C}$$ is _________ $$\mathrm{BM}$$ (Nearest integer)
Explanation
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$$\begin{aligned} & \text { In } \mathrm{K}_3\left[\mathrm{Co}\left(\mathrm{NO}_2\right)_6\right], \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 4 \mathrm{~s}^0 \\ & \mathrm{Co}^{3+}: \mathrm{d}^2 \mathrm{sp}^3 \text { Hybridisation } \\ & \text { Number of unpaired } \mathrm{e}^{-}=0 \\ & \text { Magnetic moment }=\sqrt{n(n+2)}=0 \text { B.M } \\ & \end{aligned}$$
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