Question

Match coloum - I with coloum - II and select the correct answer using the codes given below the coloums.
$\begin{array}{cccc}& & & \\ & Column-I& & Column-II\\ & \left(Metalion\right)& & \left(Magneticmoment\right(BM\left)\right)\\ \left(a\right)& C{r}^{3+}& \left(p\right)& \sqrt{35}\\ \left(b\right)& F{e}^{2+}& \left(q\right)& \sqrt{30}\\ \left(c\right)& N{i}^{2+}& \left(r\right)& \sqrt{24}\\ \left(d\right)& M{n}^{2+}& \left(s\right)& \sqrt{15}\\ & & \left(t\right)& \sqrt{8}\end{array}$

• A. (a) s; (b) r; (c) t; (d) p
• B. (a) r; (b) s; (c) t; (d) p
• C. (a) s; (b) r; (c) p; (d) t
• D. (a) p; (b) r; (c) t; (d) s

Answer 1

The correct option is A (a) s; (b) r; (c) t; (d) p

$C{r}^{3+}$ has 3 unpaired electrons in the $\left[Ar\right]3d^{3}4s^0$ configuration after losing 3 electrons from the $\left[Ar\right]3d^{5}4s^1$ configuration.

The magnetic moment in (BM) is calculated as = $\sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.

For $C{r}^{3+}$ it comes out to $\sqrt{15}$.
Likewise, for $F{e}^{2+}$ we get $\sqrt{24}$ as n =4 in the $3d^4$ configuration.

$N{i}^{2+}$ only has n = 2 unpaired electrons in the $\left[Ar\right]3d^{8}4s^0$ configuration and thus it has a magnetic moment of $\sqrt{8}$.

Lastly, $M{n}^{2+}$ has n = 5 unpaired electrons and thus it has trhe highest magnetic moment of $\sqrt{35}$