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)

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

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.