Question

The stable oxidation states of Cr are:

• A. + 3, + 6
• B. + 3, + 4
• C. + 1, + 4
• D. + 2, + 5

Answer 1

The correct option is A + 3, + 6

Cr- $\left[Ar\right]3d^{5}4s^1$ After it has attained a configuration of +3 its electronic configuration becomes $\left[Ar\right]3d^3$ this makes it half filled stable at the $t_{2g}$ level which means three sub-d orbitals i.e. $d_{xy}$, $d_{yz}$ and $d_{xz}$ are half filled. Half filled $t_{2g}$ state is more stable than $d_5$ configuration so ,Chromium is more stable in its $d^3$ state than its $d^5$ state. $C{r}^{+6}$ has stable gas configuration so it is stable.