Question

Oxidation of $Cr$ to $C{r}^{2+}$ ion involves removal of valence electrons from:

• A. 4s and 3d orbitals
• B. 4s and 4p orbitals
• C. 3d orbital only
• D. 4s orbital only

Answer 1

The correct option is A 4s and 3d orbitals

Oxidation of $Cr$ to $C{r}^{2+}$ ion involves removal of valence electrons from 4s and 3d orbitals.
The ground state electronic configuration of Cr is $[Ar{]}^{18}3d^{5}4s^1$.
The electronic configuration of Cr (II) is $[Ar{]}^{18}3d^{4}4s^0$.
It can also be expressed as $\left[Ar{]}^{18}3d^{5}4s^1\underrightarrow{-2e^{-}}\right[Ar{]}^{18}3d^{4}4s^0$.
Hence, electrons are removed from $4s$ and $3d$ orbitals.