Question

A transition metal cation $M^{3+}$ has magnetic moment $\sqrt{35}BM$. The number of unpaired electrons in ns orbital of $M^{3+}$ is :
(Note: transition metal exhibit ferromagnetic property)

• A. 5
• B. 3
• C. 2
• D. 0.0

Answer 1

The correct option is D 0.0

We know spin only magnetic momentum $\left(\mu \right)=\sqrt{n(n+2)}$
According to given data
$\mu =\sqrt{n(n+2)}=\sqrt{35}BM$
$n=5$
i.e. 5 unpaired $e^{-}$ are present in (n-1) d-shell of $M^{3+}$ transition metal cation so it is $F{e}^{3+}$
${}_{26}Fe\to \left[Ar\right]3d^{6}4s^2$
$F{e}^{3+}\to \left[Ar\right]3d^{5}4s^0$