Which one of the following is an outer orbital complex and exhibits...

Outer orbital complex utilises nd orbitals for bonding and exhibit paramagnetic behaviour, only if there present unpaired electrons. $$\begin{aligned} & \text{(a) } \operatorname{In}\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}: \\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 4 s^0 \end{aligned}$$ So, this is an outer orbital complex having paramagnetic character. (b) In $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$ : $\mathrm{Zn}^{2+}=[\mathrm{Ar}] 3 d^{10}$ Thus, it is also an outer orbital complex but it is diamagnetic as all the electrons are paired. $\begin{gathered} \text{(c) } \operatorname{In}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} \\ \mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 d^3 \\ {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}=} \end{gathered}$ Because of the involvement of $(n-1) d$, i.e., $3 d$ orbital in hybridisation, it is an inner orbital complex. Its nature is paramagnetic because of the presence of three unpaired electrons. $$\begin{aligned} & \text{(d) } \operatorname{In}\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \text{ : } \\ & \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 \end{aligned}$$ As all the electrons are paired, it is a diamagnetic complex.