A magnetic moment of 1.73 BM will be shown by one among the following

Magneitc moment, $\mu$μ is related with number of unpaired electrons as $$\begin{aligned} \mu & =\sqrt{n(n+2)} \mathrm{BM} \\ (1.73)^2 & =n(n+2) \end{aligned}$$μ(1.73)2​=n(n+2)​BM=n(n+2)​ On solving $n=1$n=1 Thus, the complex/compound having one unpaired electron exhibit a magnetic moment of 1.73 $\mathrm{BM}$BM. (a) $$\begin{aligned} & \ln \left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \\ & \mathrm{Cu}^{2+}=[\mathrm{Ar}] 3 d^9 \end{aligned}$$​ln[Cu(NH3​)4​]2+Cu2+=[Ar]3d9​

(Although in the presence of strong field ligand $\mathrm{NH}_3$NH3​, the unpaired electron gets excited to higher energy level but it still remains unpaired). (b) $$\begin{aligned} & \ln \left[\mathrm{Nl}(\mathrm{CN})_4\right]^{2-} \\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 30^8 \end{aligned}$$​ln[Nl(CN)4​]2−Ni2+=[Ar]308​

But $\mathrm{CN}^{-}$CN−being strong field ligand pair up the unpaired electrons and hence in this complex, number of unpaired electrons $=0$=0. (c) $\ln \left[\mathrm{TiCl}_4\right]$ln[TiCl4​] $\mathrm{Ti}^{4+}=[\mathrm{Ar}]$Ti4+=[Ar] No unpaired electron. (D) $$\begin{aligned} & \ln \left[\mathrm{CoCl}_6\right]^{4-} \\ & \qquad \mathrm{Co}^{2+}=[\mathrm{Ar}] 3 d^7 \end{aligned}$$​ln[CoCl6​]4−Co2+=[Ar]3d7​

It contains three unpaired electrons. Thus, $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$[Co(NH3​)4​]2+ is the complex that exhibits a magnetic moment 1.73 BM.