Chemistry - Complete d-block Question with Solution | TestHub

${\mathrm{V}}_2{\mathrm{O}}_3$: In, ${\mathrm{V}}_2{\mathrm{O}}_3$, the oxidation state of $\text{V}$ is as follows

$2\left(\mathrm{x}\right) + 3(-2) = 0$

$\Rightarrow 2\mathrm{x} - 6 = 0$

$\Rightarrow \mathrm{x} = \frac{6}{2} = 3$

The oxidation state of $\mathrm{V}$ in ${\mathrm{V}}_2{\mathrm{O}}_3$ is +3. The vanadium is present in its lowest oxidation state. The vanadium tends to donate the pair of electrons. Hence, it is basic oxide.

${\mathrm{V}}_2{\mathrm{O}}_5$: In ${\mathrm{V}}_2{\mathrm{O}}_5$, the oxidation state of $\mathrm{V}$ is as follows

$2\left(\mathrm{x}\right) + 5(-2) = 0$

$\Rightarrow 2\left(\mathrm{x}\right) - 10 = 0$

$\Rightarrow \mathrm{x} = \frac{10}{2} = 5$

The oxidation state of $\mathrm{V}$ in ${\mathrm{V}}_2{\mathrm{O}}_5$ is +5. The Vanadium tends to donate and accept the pair of electrons. It is amphoteric oxide.

${\mathrm{V}}_2{\mathrm{O}}_4$: In ${\mathrm{V}}_2{\mathrm{O}}_4$, the oxidation state of $\mathrm{V}$ is as follows

$2\left(\mathrm{x}\right) + 4(-2) = 0$

$\Rightarrow 2\left(\mathrm{x}\right) - 8 = 0$

$\Rightarrow \mathrm{x} = \frac{8}{2} = 4$

The oxidation state of $\mathrm{V}$ in ${\mathrm{V}}_2{\mathrm{O}}_4$ is +4. The Vanadium is present in its other low oxidation state. The Vanadium tends to donate and accept the pair of electrons. It is amphoteric oxide and less basic than ${\mathrm{V}}_2{\mathrm{O}}_3$, more basic than ${\mathrm{V}}_2{\mathrm{O}}_5$.

So, the correct order of basicity of oxides of vanadium is ${\mathrm{V}}_2{\mathrm{O}}_3>{\mathrm{V}}_2{\mathrm{O}}_4>{\mathrm{V}}_2{\mathrm{O}}_5$.