Question

The calculated spin only magnetic moment of $C{r}^{2+}$ ion is :

• A. 4.90 BM
• B. 5.92 BM
• C. 3.87 BM
• D. 2.84 BM

Answer 1

The correct option is A 4.90 BM

$C{r}^{2+}\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^4$
$\therefore \text{n(number of unpaired electrons)}=4$$\therefore \mu =\sqrt{n(n+2)}$
$=\sqrt{\left(4\right(4+2)}$
$=4.8989$
$=4.9B.M$