Question

The correct order of basicity of oxides of vanadium is

• A. $V _2 O _3> V _2 O _5> V _2 O _4$
• B. $V _2 O _4> V _2 O _3> V _2 O _5$
• C. $V _2 O _3> V _2 O _4> V _2 O _5$
• D. $V _2 O _5> V _2 O _4> V _2 O _3$

Hint:

The basicity of metal oxides decreases as the oxidation state of the metal increases due to the increasing tendency to act as an acid.

Answer 1

The Correct Option is C

The question is about determining the order of basicity of the oxides of vanadium, which is a transition metal. In general, the basicity of an oxide is inversely related to its oxidation state. In other words, oxides with lower oxidation states tend to be more basic compared to those with higher oxidation states.

Let's analyze the given oxides:

• \(V_2O_3\): In vanadium(III) oxide, vanadium is in the +3 oxidation state. This is a relatively low oxidation state, suggesting that \(V_2O_3\) is more basic.
• \(V_2O_4\): In vanadium(IV) oxide, vanadium is in the +4 oxidation state. This is a moderate oxidation state, making \(V_2O_4\) intermediate in basicity.
• \(V_2O_5\): In vanadium(V) oxide, vanadium is in the +5 oxidation state. This represents a higher oxidation state, indicating that \(V_2O_5\) is the least basic and more acidic.

Thus, the order of basicity from most basic to least basic (most acidic) is \(V_2O_3 > V_2O_4 > V_2O_5\).

Hence, the correct answer is:

$V _2 O _3> V _2 O _4> V _2 O _5$