Question

Which one of the following is an inner orbital complex as well as diamagnetic in behaviour?

[Atomic number of Zn = 30,Cr = 24, Co = 27, Ni = 28]

• A. $\left[Zn\right(N{H}_3{)}_6{]}^{2+}$
• B. $[CrN{H}_3{)}_6{]}^{3+}$
• C. $\left[Co\right(N{H}_3{)}_6{]}^{3+}$
• D. $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$

Answer 1

The correct option is C $\left[Co\right(N{H}_3{)}_6{]}^{3+}$

In $\left[Co\right(N{H}_3{)}_6{]}^{3+}$ oxidation state of $Co=+3$ and its coordination number is six.

The electronic configuration of ${}_{27}Co=1s^2,2s^2,2p^6,3s^{2}3p^{6}3d^7,4s^2$

$C{o}^{3+}=1s^{2}2s^{2}2p^6,3s^{2}3p^{6}3d^6$

Thus, $\left[Co\right(N{H}_3{)}_6{)}^{3+}$ shows inner orbital complex as well as diamagnetic in behaviour (due to absence of an unpaired electron).

$\left[Zn\right(N{H}_3{)}_6{]}^{2+}\to s{p}^3{d}^2$ hybridisation (outer) and diamagnetic.

$\left[Cr\right(N{H}_3{)}_6{]}^{3+}\to d^{2}s{p}^3$ hybridisation (inner) and paramagnetic.

Hence, option C is correct.