Question

What is the correct order of spin only magnetic moment (in BM) of $M{n}^{2+},C{r}^{2+}$ and $V^{2+}$ ?

• A. $M{n}^{2+}>V^{2+}>C{r}^{2+}$
• B. $V^{2+}>C{r}^{2+}>M{n}^{2+}$
• C. $M{n}^{2+}>C{r}^{2+}>V^{2+}$
• D. $C{r}^{2+}>V^{2+}>M{n}^{2+}$

Answer 1

Answer: (C)

Spin only magnetic moment depends upon die number of unpaired electrons, more the number of unpaired electrons, greater will be the spin only magnetic moment.
${}_{25}Mn=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^2$
$M{n}^{2+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^o$
Number of unpaired electrons =5
${}_{24}Cr=1s^2,2s^{2}2p^6,3s^{2}3d^5,4s^1$
$C{r}^{2+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^4,4s^0$
Number of unpaired electrons =4
Number of unpaired electrons =3
So, the correct order of spin only magnetic moment is
$M{n}^{2+}>C{r}^{2+}>V^{2+}$