Question

Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is $25$ .

Answer 1

$\text{Magnetic moment}$

Given, Atomic number $=25$. So, the metal is manganese $\left(Mn\right)$ with valence shell electronic configuration $3d^{5}4s^2$.
So, the divalent ion $M{n}^{2+}$ will have valence shell electronic configuration $3d^{5}4s^0$. Hence, number of unpaired electrons, $n=5$

We know, spin only magnetic moment is given by $\sqrt{n(n+2)}$.

Put $n=5$, we get,

$\sqrt{n(n+2)}=\sqrt{5(5+2)}=\sqrt{35}=5.92BM$

$\text{Final answer}:$

Spin only magnetic moment is$5.92BM$.