Question

Magnetic moment of $C{r}^{2+}$ is nearest to

• A. $F{e}^{2+}$
• B. $M{n}^{2+}$
• C. $C{o}^{2+}$
• D. $N{i}^{2+}$

Answer 1

Answer: (A)

Cr²⁺ = 3d⁴, No. of unpaired electrons (n) = 4
Magnetic moment $=\sqrt{n\left(n+2\right)}$ BM
$=\sqrt{4\left(4+2\right)}=\sqrt{24}=4.89$ BM
Fe²⁺ = 3d⁶, No. of unpaired electrons (n) = 4
Magnetic moment $=\sqrt{4\left(4+2\right)}$ BM
$=\sqrt{24}=4.89$ BM
Mn²⁺ = 3d⁵, No. of unpaired electrons (n) = 5
Magnetic moment $=\sqrt{5\left(5+2\right)}$ BM
$=\sqrt{35}=5.91$ BM
Co²⁺ = 3d⁷, No. of unpaired electrons (n) = 3
Magnetic moment $=\sqrt{3\left(3+2\right)}$ BM
$=\sqrt{15}=3.87$ BM
Ni²⁺ = 3d⁸, No. of unpaired electrons (n) = 2
Magnetic moment $=\sqrt{2\left(2+2\right)}$ BM
$=\sqrt{8}=2.82$ BM