JEE Main 2023 — Chemistry Inorganic Chemistry

${V}_{2}{O}_{3}$V2​O3​: In, ${V}_{2}{O}_{3}$V2​O3​, the oxidation state of $\text{V}$V is as follows

$$2(x)+3(-2)=0$$2(x)+3(−2)=0

$$\Rightarrow 2x-6=0$$⇒2x−6=0

$$\Rightarrow x=\frac{6}{2}=3$$⇒x=26​=3

The oxidation state of $V$V in ${V}_{2}{O}_{3}$V2​O3​ 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.

${V}_{2}{O}_{5}$V2​O5​: In ${V}_{2}{O}_{5}$V2​O5​, the oxidation state of $V$V is as follows

$$2(x)+5(-2)=0$$2(x)+5(−2)=0

$$\Rightarrow 2(x)-10=0$$⇒2(x)−10=0

$$\Rightarrow x=\frac{10}{2}=5$$⇒x=210​=5

The oxidation state of $V$V in ${V}_{2}{O}_{5}$V2​O5​ is +5. The Vanadium tends to donate and accept the pair of electrons. It is amphoteric oxide.

${V}_{2}{O}_{4}$V2​O4​: In ${V}_{2}{O}_{4}$V2​O4​, the oxidation state of $V$V is as follows

$$2(x)+4(-2)=0$$2(x)+4(−2)=0

$$\Rightarrow 2(x)-8=0$$⇒2(x)−8=0

$$\Rightarrow x=\frac{8}{2}=4$$⇒x=28​=4

The oxidation state of $V$V in ${V}_{2}{O}_{4}$V2​O4​ 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 ${V}_{2}{O}_{3}$V2​O3​, more basic than ${V}_{2}{O}_{5}$V2​O5​.

So, the correct order of basicity of oxides of vanadium is ${V}_{2}{O}_{3}>{V}_{2}{O}_{4}>{V}_{2}{O}_{5}$V2​O3​>V2​O4​>V2​O5​.