Question

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

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

Answer 1

Answer: (B)

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

Same magnetic moment = Same number of unpaired electrons
$\mu =\sqrt{n(n+2)}$
where n = number of unpaired electrons
$C{o}^{2+}=3d^7$, 3 unpaired electrons;
$C{r}^{2+}=3d^4$, 4 unpaired electrons;
$M{n}^{2+}=3d^5$, 5 unpaired electrons;
$F{e}^{2+}=3d^6$, 4 unpaired electrons.