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 the number of unpaired electrons, more the number of unpaired electrons, greater will be the spin only magnetic moment.
${}_{25}Mn=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^2$
$M{n}^{2+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^0$

Number of unpaired electrons $=5$
${}_{24}Cr=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^1$
$C{r}^{2+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^4,4s^0$

Number of unpaired electrons $=4$
${}_{23}V=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^3,4s^2$
$V^{2+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^3,4s^0$

Number of unpaired electrons $=3$
So, the correct order of spin only magnetic moment is
$M{n}^{2+}>C{r}^{2+}>V^{2+}$