A magnetic moment of $1.73 \mathrm{BM}$ will be shown by one among the following

Question

Aicharya · Accepted Answer

Magnetic moment, $\mu$ is related with number of unpaired electrons as $\begin{aligned} \mu & =\sqrt{n(n+2)} \mathrm{BM} \\ (1.73)^2 & =n(n+2) \end{aligned}$ On solving $n=1$ Thus, the complex/compound having one unpaired electron exhibit a magnetic moment of 1.73 $\mathrm{BM}$. (a) $\begin{aligned} & \ln \left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \\ & \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 d^9 \end{aligned}$

(Although in the presence of strong field ligand $\mathrm{NH}_3$, the unpaired electron gets excited to higher energy level but it still remains unpaired). (b) $\begin{aligned} & \ln \left[\mathrm{Nl}(\mathrm{CN})_4\right]^{2-} \\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 30^8 \end{aligned}$

But $\mathrm{CN}^{-}$being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons $=0$. (c) $\ln \left[\mathrm{TiCl}_4\right]$ $\mathrm{Ti}^{4+}=[\mathrm{Ar}]$ No unpaired electron. (D) $\begin{aligned} & \ln \left[\mathrm{CoCl}_6\right]^{4-} \\ & \qquad \mathrm{Co}^{2+}=[\mathrm{Ar}] 3 d^7 \end{aligned}$

It contains three unpaired electrons. Thus, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$ is the complex that exhibits a magnetic moment 1.73 BM.

A magnetic moment of $1.73 \mathrm{BM}$ will be shown by one among the following

Solution