Question

State the most stable oxidation state of Cr?
A. +3
B. +4
C. +2
D. +6

Answer 1

Chromium has an atomic number of 24 and the common oxidation state of chromium is +6, +3 and +2. The oxidation state of an atom is the total no. of the electron which can be lost or gained by an atom to form a bond with other molecules.

Complete answer:
- In the given question, we have to find the most stable oxidation state of chromium from the given options.
- Firstly, we have to write the electronic configuration of the chromium atom i.e.
$(\text{Ar) 3}{{\text{d}}^{5}}\text{ 4}{{\text{s}}^{1}}$
- So, firstly the electron will be lost from the outermost shell of the atom i.e. 4s and then from the 3d shell.
- Now, as we know that chromium shows a wide range of oxidation states among which +3 is the most stable oxidation state.
- Whereas +1, +4 and +5 oxidation states of chromium are rare.
- When chromium loses 3 electrons then the electronic configuration becomes $(\text{Ar) 3}{{\text{d}}^{3}}$ which is most stable because at the ${{\text{t}}_{2\text{g}}}$ level the ${{\text{d}}_{\text{xy}}}$, ${{\text{d}}_{\text{yz}}}$ and ${{\text{d}}_{\text{xz}}}$orbitals are half-filled.
- It means that one electron is present in each orbital due to which it is more stable than the ${{\text{d}}^{5}}$ configuration.
- That’s why the oxidation state of +3 is the most stable oxidation state than the +4, +2 and +6 oxidation states.
- Some compounds show an oxidation state of +2 such as $\text{CrO and Cr(OH}{{\text{)}}_{2}}$. Whereas the compounds such as ${{\text{H}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$ showing +6 oxidation state are strong oxidiser.
Therefore, the most stable oxidation state is +3 due to the stable $(\text{Ar) 3}{{\text{d}}^{3}}$ configuration.

Hence option A is the correct one.

Note: In${{\text{t}}_{2\text{g}}}$, 't' means the triply degenerate orbitals and g stands for gerade which consist of three orbitals of d subshell i.e. ${{\text{d}}_{\text{xy}}}$, ${{\text{d}}_{\text{yz}}}$ and ${{\text{d}}_{\text{xz}}}$orbitals. When there is a presence of an unsymmetrical ligand, then undergo splitting of d orbitals into ${{\text{t}}_{2\text{g}}}$ and e.g. levels take place.