Question

Calculate the spin only magnetic moment of $F{e}^{2+}$.[ At. no. = 26]

Answer 1

Electronic configuration of ${Fe}^{+2}=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$

It has $4$ unpired electrons.

As we know that

Magnetic moment $\left(\mu \right)=\sqrt{n(n+2)}$

whereas $n$ is the no. of unpaired electrons

$\mu =\sqrt{4(4+2)}$

$\Rightarrow \mu =\sqrt{4\times 6}\approx 5BM$