Question

Spin only magnetic moment $2.84 \, \text{B} \text{M}$ is given by:

(Atomic Number of $\text{Ni} = 28 , \text{Ti} = 22 , \text{Cr} = 24 , \text{Co} = 27$ )

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

Answer 1

Answer: (A)

$\text{μ} = \sqrt{\text{n} \left(\right. \text{n} + 2 \left.\right)} = 2.84 \Rightarrow \text{n} = 2$

electronic configuration of $\text{Ni}^{+ 2} = 1 \text{s}^{2} 2 \text{s}^{2} 2 \text{p}^{6} 3 \text{s}^{2} 3 \text{p}^{6} 3 \text{d}^{8} \, \text{e}^{-}$ having two unpaired electrons

electron configuration of $\text{T} \text{i}^{3 +} = 1 \text{s}^{2} 2 \text{s}^{2} 2 \text{p}^{6} 3 \text{s}^{2} 3 \text{p}^{6} 3 \text{d}^{1}$ only one unpaired electron

electron configuration of $\text{C} \text{r}^{2 +} = 1 \text{s}^{2} 2 \text{s}^{2} 2 \text{p}^{6} 3 \text{s}^{2} 3 \text{p}^{6} 3 \text{d}^{4}$ having four unpaired electrons

electron configuration of $\text{C} \text{o}^{+ 2} = 1 \text{s}^{2} 2 \text{s}^{2} 2 \text{p}^{6} 3 \text{s}^{2} 3 \text{p}^{6} 3 \text{d}^{7}$ having three unpaired electrons.