Question

The pair in which both the species have the same magnetic moment (spin only) is:

• A. $[Co{\left(OH\right)}_4{]}^{2-}and[Fe{\left(N{H}_3\right)}_6{]}^{2+}$
• B. $[Mn{\left(H_{2}O\right)}_6{]}^{2+}and{\left[Cr\right(H_{2}O\left)\right]}^{2+}$
• C. $[Cr{\left(H_{2}O\right)}_6{]}^{2+}and{\left[CoCl4\right]}^{2-}$
• D. ${\left[Cr{\left(H_{2}O\right)}_6\right]}^{2+}and{\left[Fe{\left(H_{2}O\right)}_6\right]}^{2+}$

Answer 1

The correct option is D

${\left[Cr{\left(H_{2}O\right)}_6\right]}^{2+}and{\left[Fe{\left(H_{2}O\right)}_6\right]}^{2+}$

The explanation for the correct answer:

In case of option D:

1. Spin only magnetic moment, $U_s=\sqrt{n(n+2)}$ where, $n=$ number of unpaired electrons.
2. Number of unpaired electrons in $C{r}^2+\left(\right[Ar]3d4)$ is $4$, in $C{O}^2+\left(\right[Ar\left]3d7\right)$ is $3$, in $F{e}^{2+}\left(\right[Ar]3d6)$ is 4, in $M{n}^{2+}\left(\right[Ar]3d5)$ is $5$.
3. As the number of unpaired electrons in $C{r}^{2+}andF{e}^{2+}$ are same, hence $[Cr{\left(H_{2}O\right)}_6{]}^{2+}and[Fe{\left(H_{2}O\right)}_6{]}^{2+}$ will have same magnetic moment.

Hence option D is the correct answer.