JEE Chemistry 2023 Question with Solution

Given: The question asks for the correct order of basicity of oxides of vanadium.

Find: The decreasing order of basic character among $V_2O_3$, $V_2O_4$ and $V_2O_5$.

As the oxidation state of a metal increases, the oxide becomes more acidic and less basic.

For vanadium oxides:

$$V_2O_3 : V = +3$$ $$V_2O_4 : V = +4$$ $$V_2O_5 : V = +5$$

With increase in the percentage of oxygen, the acidic nature of oxides increases and basic nature decreases.

Therefore, the order of basicity is:

$$V_2O_3 > V_2O_4 > V_2O_5$$

So, the correct option corresponds to A. The solution labels option C, but the worked conclusion matches option A in the given options.

Given: Basicity of metal oxides depends on the oxidation state of the metal.

Find: Which vanadium oxide is most basic and which is least basic.

In $V_2O_3$, vanadium is in the lower oxidation state $+3$, so the oxide is more ionic and more basic.

In $V_2O_4$, vanadium is in oxidation state $+4$, so its basicity is lower than $V_2O_3$.

In $V_2O_5$, vanadium is in oxidation state $+5$, so acidic character dominates and the basic nature is minimum.

Hence,

$$V_2O_3 > V_2O_4 > V_2O_5$$

Therefore, the correct answer is A.