Looking for classes? Ksquare Career Institute, Bengaluru →
The calculated spin only magnetic moment of $Cr^{2+}$ ion is:
A
4.90 BM
B
5.92 BM
C
2.84 BM
D
3.87 BM
Detailed Solution
Cr: [Ar] $3d^54s^1$ ; $Cr^{2+}$: [Ar] $3d^4$ → 4 unpaired electrons
$\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} = 4.90$ BM