Question

In which case change of oxidation number of V is maximum?

• A. Equal in each case
• B. $V{O}^{2+}\to V{O}_2^{+}$
• C. $V{O}_2^{+}\to V{O}^{+}$
• D. $V{O}^{2+}\to V^{3+}$

Answer 1

The correct option is C $V{O}_2^{+}\to V{O}^{+}$

Analysing the options
Option (A): Vanadium gets oxidised from (+4) to (+5).
$V{O}^{2+}\to V{O}_2^{+}$
Oxidation number of V in $V{O}^{2+}=+4$
Oxidation number of V in $V{O}_2^{+}=+5$
$\Delta O.N.=5-4=1$

Option (B):
The vanadium gets reduced from (+5) to (+3).
$V{O}_2^{+}\to V{O}^{+}$
Oxidation number of V in $V{O}_2^{+}=+5$
Oxidation number of V in $V{O}^{+}\left(V\right)=+3$
$\Delta O.N.=5-3=2$

Option (C):
The vanadium gets reduced from (+4) to (+3).
$V{O}^{2+}\to V^{3+}$
Oxidation number of V in $V{O}^{2+}=+4$
Oxidation number of V in $V^{3+}=+3$
$\Delta O.N=4-3=1$
Hence, change in oxidation no. of V in option B is maximum.
So,the correct answer is option (B).