Question

Magnetic moment $3.87 \,B.M$. is given by
(At. nos. $Ni = 28, Ti = 22, Cr = 24, Co = 27)$

• A. $Ti^{3+}$
• B. $Co^{2+}$
• C. $Ni^{2+}$
• D. $Cr^{2+}$

Answer 1

Answer: (B)

Magnetic moment $(\mu) = \sqrt{n(n + 2)}$
$3.87 \,B.M$. corresponds to $3$ unpaired electrons.
$Cr^{2+} - 3d^4, 4$ unpaired electrons
$Co^{2+} - 3d^7, 3$ unpaired electrons
$Ni^{2+} - 3d^8, 2$ unpaired electrons
$Ti^{3+} - 3d^1, 1$ unpaired electron