Question

Magnetic moment of $(N{H}_4{)}_{2}MnB{r}_4$ (in BM is):

Answer 1

Oxidation state of $Mn$ in the complex is $+2$. Hence it is a $d^5$ system.
Since the ligands are weak field, it will not cause pairing in the inner $d$ orbital. Hence the number of unpaired $e^{-}$ are 5.
Using spin only formula,
$\sqrt{n(n+2)}=\sqrt{5(5+2)}=5.916\approx 5.92$