Question

Arrange the following ions in order of their magnetic moment:

$\left(i\right)V^{4+}$ $\left(ii\right){Mn}^{4+}$ $\left(iii\right){Fe}^{3+}$ $\left(iv\right){Ni}^{2+}$

[Atomic number of $V=23,Mn=25,Fe=26,Ni=28$]

• A. $\left(ii\right)>\left(iii\right)>\left(i\right)>\left(iv\right)$
• B. $\left(iii\right)>\left(ii\right)>\left(iv\right)>\left(i\right)$
• C. $\left(i\right)>\left(iv\right)>\left(iii\right)>\left(ii\right)$
• D. $\left(iv\right)>\left(iii\right)>\left(i\right)>\left(ii\right)$

Answer 1

Answer: (B)

$\left(iii\right)>\left(ii\right)>\left(iv\right)>\left(i\right)$

The transition metal with more unpaired electrons corresponds to the higher magnetic moment.

$V^{4+}\to 4s^{0}3d^1$, 1 unpaired electron;

$M{n}^{4+}\to 4s^{0}3d^3$, 3 unpaired electron;

$F{e}^{3+}\to 4s^{0}3d^5$, 5 unpaired electron and

$N{i}^{2+}\to 4s^{0}3d^8$, 2 unpaired electron.

Hence, option B is correct.