Question

Spin only magnetic moment $2.84 \, BM$ is given by:
(Atomic Number of $Ni=28,Ti=22,Cr=24,Co=27$ )

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

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.