Question

The magnetic moment of a transition metal compound has been calculated to be $3.87$ B.M. The metal ion is

• A. $T_1^{2+}$
• B. $M{n}^{2+}$
• C. $V^{2+}$
• D. $C{r}^{2+}$

Answer 1

Answer: (C)

$C{r}^{+2}:[Ar],3d^4,4s^{0}n=4,\mu =\sqrt{4(4+2)}=\sqrt{24}$
$=4.89BM$
$M{n}^{+2}:[Ar],3d^5,4s^{0}n=5,\mu =\sqrt{5(5+2)}=\sqrt{35}$
$=5.91BM$
$V^{+2}:[Ar],3d^3,4s^{0}n=3,\mu =\sqrt{3(3+2)}=\sqrt{15}$
$=3.87BM$
$T{i}^{+2}:[Ar],3d^2,4s^{0}n=2,\mu =\sqrt{2(2+2)}=\sqrt{8}$
$=2.82BM$