NEET UG 2012 — Chemistry Inorganic Chemistry

Outer orbital complex utilises nd orbitals for bonding and exhibit paramagnetic behaviour, only if there present unpaired electrons. $$\begin{aligned} & \text {(a) } \operatorname{In}\left[\mathrm{Ni}\left(\mathrm{NH}_3\right)_6\right]^{2+}: \\ & \mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8 4 s^0 \end{aligned}$$​(a) In[Ni(NH3​)6​]2+:Ni2+=[Ar]3d84s0​ So, this is an outer orbital complex having paramagnetic character. (b) In $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_6\right]^{2+}$[Zn(NH3​)6​]2+ : $\mathrm{Zn}^{2+}=[\mathrm{Ar}] 3 d^{10}$Zn2+=[Ar]3d10 Thus, it is also an outer orbital complex but it is diamagnetic as all the electrons are paired. $$\begin{gathered} \text {(c) } \operatorname{In}\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+} \\ \mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 d^3 \\ {\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]^{3+}=} \end{gathered}$$(c) In[Cr(NH3​)6​]3+Cr3+=[Ar]3d3[Cr(NH3​)6​]3+=​ Because of the involvement of $(n-1) d$(n−1)d, i.e., $3 d$3d orbital in hybridisation, it is an inner orbital complex. Its nature is paramagnetic because of the presence of three unpaired electrons. $$\begin{aligned} & \text{(d) } \operatorname{In}\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \text { : } \\ & \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 d^6 \end{aligned}$$​(d) In[Co(NH3​)6​]3+ : Co3+=[Ar]3d6​ As all the electrons are paired, it is a diamagnetic complex.