Question

The calculated magnetic moment of $F{e}^{n+}$ ion is 4.90 BM. The sum of the number of unpaired electrons present in $3d$ subshell of $F{e}^{n+}$ ion and the value of n is

Answer 1

We know that, $\mu =\sqrt{n(n+2)}=4.90BM$
`\therefore` `n=4`
The ground state electronic configuration of iron is $\left[Ar\right]3d^{6}4s^2$
The electronic configuration of $F{e}^{2+}$is $\left[Ar\right]3d^{6}4s^0$
The number of unpaired electrons in the $3d$ subshell is `4`
So, the sum of unpaired electrons in the $3d$ subshell and the oxidation state is = 4+4= 8