Question

The $d-$ electronic configuration of $C{r}^{2+},M{n}^{2+},F{e}^{2+}$ and $N{i}^{2+}$ are $3d^4,3d^5,3d^6$ and $3d^8$ respectively, which one of the following aqua-complex will exhibit the minimum paramagnetic behaviour?

• A. $\left[Cr\right(H_{2}O{)}_6{]}^{2+}$
• B. $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$
• C. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• D. $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$

Answer 1

The correct option is D $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$

As $H_{2}O$ is weak ligand, pairing of electrons will not take place until each orbital is singly occupied.
a. $\left[Cr\right(H_{2}O{)}_6{]}^{2+}=3d^4=t_{2g}^3{e}_g^1$
Number of unpaired electrons $=4$
b. $\left[Mn\right(H_{2}O{)}_6{]}^{2+}=3d^5=t_{2g}^3{e}_g^2$
Number of unpaired electrons $=5$
c. $\left[Fe\right(H_{2}O{)}_6{]}^{2+}=3d^6=t_{2g}^4{e}_g^2$
Number of unpaired electrons $=4$
d. $\left[Ni\right(H_{2}O{)}_6{]}^{2+}\to N{i}^{2+}=3d^8\to \left[Ni\right(H_{2}O{)}_6{]}^{2+}$
has $t_{2g}^6{e}_g^2$ configuration in which only two unpaired electrons are present which is minimum in all.
More the number of unpaired electrons, more will be the paramagnetic behaviour.