Question

The magnetic moment of a transition metal ion is $\sqrt{15}BM$. Therefore, the number of unpaired electrons present in it is:

• A. $4$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

The magnetic moment of a transition metal ion is $\sqrt{15}BM$. Therefore, the number of unpaired electrons present in it is $3$.
If n is the number of unpaired electrons, then the magnetic moment is given by the expression:

$\mu =\sqrt{n(n+2)}$
$\sqrt{15}=\sqrt{n(n+2)}$
$15=n(n+2)$
$15=n^2+2n$
$n^2+2n-15=0$
This is the quadratic equation with solution

$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$n=\frac{-2\pm \sqrt{2^2-4\left(1\right)(-15)}}{2\left(1\right)}$
$n=\frac{-2\pm 8}{2\left(1\right)}$
$n=3$ or $n=-5$.

The negative value is discarded as the number of unpaired electrons cannot be negative.

Hence, $n=3$ .