Question

Consider the following ions :
(i) $Ni ^{2+}$
(ii) $Co ^{2+}$
(iii) $Cr ^{2+}$
(iv) $Fe ^{3+}$
(Given, at. no. $Cr = 24,\, Fe = 26,\, Co = 27,\, Ni = 28$)
The correct order of magnetic moment of these ions is

• A. (i) < (ii) < (iii) < (iv)
• B. (iv) < (ii) < (iii) < (i)
• C. (i) < (iii) < (ii) < (iv)
• D. (iii) < (iv) < (ii) < (i)

Answer 1

Answer: (A)

Magnetic moment, $\mu=\sqrt{n(n+2)} BM$

where, $n=$ number of unpaired electrons

Higher the number of unpaired electrons, more the value of $\mu$.

The electronic configuration of given species is

$Ni ^{2+}=[ Ar ] 3 d^{8}$ (two unpaired electrons)

$Co ^{2+}=[ Ar ] 3 d^{7}$ (three unpaired electrons)

$Cr ^{2+}=[ Ar ] 3 d^{4}$ (four unpaired electrons)

$Fe ^{3+}=[ Ar ] 3 d^{5}$ (five unpaired electrons)

Thus, the order of magnetic moment is

$\underset{(i)}{Ni ^{2+}}< \underset{(ii)}{Co ^{2+}}< \underset{(iii)}{Cr ^{2+}}< \underset{(iv)}{Fe ^{3+}}$