Question

The d electron configuration of $C{r}^{2+},M{n}^{2+},F{e}^{2+}andN{i}^{2+}are3d^4,3d^5,3d^{6}and3d^8$ respectively. Which one of the following aqua complex will exhibit the minimum paramagnetic behaviour? (At. No. $Cr=24,Mn=25,Fe=26,Ni=28$)

• A. ${\left[Cr\left(H_{2}O\right)6\right]}^{2+}$
• B. ${\left[Mn\left(H_{2}O\right)6\right]}^{2+}$
• C. ${\left[Fe\left(H_{2}O\right)6\right]}^{2+}$
• D. ${\left[Ni\left(H_{2}O\right)6\right]}^{2+}$

Answer 1

The correct option is D ${\left[Ni\left(H_{2}O\right)6\right]}^{2+}$

Since, in all the complexes of $Cr,Mn,Fe$ and $Ni$, ligand is water so $6$ electrons would be pairing with the central metal atom. In case of Nickel the d electronic configuration is $3d^8$.

i.e. (Refer to Image)

Since, $Ni$ has only two unpaired electrons so it shows minimum paramagnetic behaviour. Other have more than two unpaired electrons so they will show more paramagnetic behaviour.