Question

Which ion has magnetic moment $5.90$ B.M?

• A. $F{e}^{2+}$
• B. $F{e}^{3+}$
• C. $C{o}^{2+}$
• D. $M{n}^{4+}$

Answer 1

Answer: (B)

$F{e}^{3+}$

$\mu =\sqrt{n(n+2)}=5.9$ B.M.
$n(n+2)=5.9\times 5.9=34.8$

$n^2+2n=34.8$

$n^2+2n-34.8=0$

$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$n=\frac{-2\pm \sqrt{2^2-4\left(1\right)(-34.8)}}{2\left(1\right)}$

$n=\frac{-2\pm \sqrt{4+139}}{2}$

$n=\frac{-2\pm \sqrt{143}}{2}$

$n=\frac{-2\pm 12}{2}$

$n=\frac{10}{2}$ or $n=\frac{-14}{2}$

$n=5$ or $n=-7$

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

$\therefore n=5$

$F{e}^{3+}$ ion (with valence shell electronic configuration $3d^{5}4s^0$ has magnetic moment 5.90 B.M. It has 5 unpaired electrons.

Hence, option $B$ is correct.