Question

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

• A. $\left[Cr\right(H_{2}O{)}_6{]}^{2+},[CoC{l}_4{]}^{2-}$
• B. $\left[Cr\right(H_{2}O{)}_6{]}^{2+},\left[Fe\right(H_{2}O{)}_6{]}^{2+}$
• C. $\left[Mn\right(H_{2}O{)}_6{]}^{2+},\left[Cr\right(H_{2}O{)}_6{]}^{2+}$
• D. $[CoC{l}_4{]}^{2-},[Fe(H_{2}O{)}_6{]}^{2+}$

Answer 1

The correct option is B $\left[Cr\right(H_{2}O{)}_6{]}^{2+},\left[Fe\right(H_{2}O{)}_6{]}^{2+}$

Magnetic moment $=\sqrt{n(n+2)}$ BM
where $n=$ number of unpaired electrons
The pair that has the same number of unpaired electrons will have the same magnetic moments.

(a) $C{r}^{2+}=3d^4;$ $4$ unpaired electrons
$C{o}^{2+}=3d^7;$ $3$ unpaired electrons

(b) $C{r}^{2+}=3d^4;$ $4$ unpaired electrons
$F{e}^{2+}=3d^6;$ $4$ unpaired electrons

(c) $M{n}^{2+}=3d^5;$ $5$ unpaired electrons
$C{r}^{2+}=3d^4;$ $4$ unpaired electrons

(d) $C{o}^{2+}=3d^7;$ $3$ unpaired electrons
$F{e}^{2+}=3d^6;$ $4$ unpaired electrons