Question

If the electronic configuration of a transition metal is $\left[Ar\right]3d^{3}4s^2$, then its possible oxidation states are:

• A. $+I,+II,+III$
• B. $+II,+III$
• C. $+II,+III,+IV,+V$
• D. $+I,+II,+III,+IV$

Answer 1

The correct option is C $+II,+III,+IV,+V$

Given that the electronic configuration of a transition metal is $\left[Xe\right]3d^{3}4s^2$. The possible oxidation states shown by the metal are $+II,+III,+IV,+V$.

$+II$ oxidation state is obtained when $2$ electrons from $4s$ orbital are lost to give an electronic configuration $\left[Xe\right]3d^{3}4s^0$. $+V$ oxidation state is obtained when all the five electrons are lost to give the electronic configuration $\left[Xe\right]3d^{0}4s^0$. Similarly, loss of $3$ and $4$ electrons will give $+III$ and $+IV$ oxidation states with electronic configurations, $\left[Xe\right]3d^{2}4s^0$ and $\left[Xe\right]3d^{1}4s^0$ respectively.