Question

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

• A. $N{i}^{2+}$
• B. $T{i}^{3+}$
• C. $C{r}^{2+}$
• D. $C{o}^{2+}$

Answer 1

Answer: (A)

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