Question

The $d$ electron 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?
(At. Nos. $Cr=24,Mn=25,Fe=26,Co=27$ )

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

Answer 1

Answer: (B)

As the number of unpaired electron increases, the magnetic moment increases and hence, the paramagnetic behaviour increases.
So, $C{r}^{2+}(22)=3d^4$ (4 unpaired electrons)
$M{n}^{2+}(23)=3d^5$(5 unpaired electrons)
$F{e}^{2+}(24)=3d^6$(4 unpaired electrons)
$N{i}^{2+}(26)=3d^8$(2 unpaired electrons)
So, ${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}$ exhibit minimum paramagnetic behaviour.