Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour?

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.