Question

Magnetic moment $2.84BM$ is given by :

(Atomic numbers : $Ni=28,Ti=22,Cr=24,Co=27)$

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

Answer 1

The correct option is C $N{i}^{2+}$

The relationship between the spin only magnetic moment ${\mu }_{eff}$ and the number of unpaired electrons 'n' is :

${\mu }_{eff}=\sqrt{n(n+2)}$ B.M.

But ${\mu }_{eff}=2.84$ B.M.

Hence, $2.84=\sqrt{n(n+2)}$

$n(n+2)=8=2\times 4$

$n=2$

$N{i}^{2+}$ has $3d^8$ electronic configuration which can also be written as $t_{2g}^6{e}_g^2$.

Thus, it has $2$ unpaired electrons.

Thus, magnetic moment $2.84$ BM is given by $N{i}^{2+}$.

Therefore, the correct option is C.