Question

In which of the following pairs, there is greatest difference in the oxidation number of the underlined elements ?

• A. $\underline{N}O_2$ and $\underline{N}_2O_4$
• B. $\underline{P}_2O_5$ and $\underline{P}_4O_{10}$
• C. $\underline{N}_2O$ and $\underline{N}O$
• D. $\underline{S}O_2$ and $\underline{S}O_3$

Answer 1

Answer: (D)

The oxidation number of elements have been mentioned in brackets as shown below:
a. $NO _{2}( x =4)$ and $N _{2} O _{4}( x =4)$
Difference in oxidation state of $N =0$.
b. $P _{2} O _{5}( x =+5)$ and $P _{4} O _{10}( x =+5)$
Difference in oxidation state of $P$ is zero.
c. $N _{2} O ( x =1)$ and $NO ( x =2)$
Difference in oxidation state of $N =1$.
d. $SO _{2}( x =4)$ and $SO _{3}( x =6)$
Difference in oxidation state of $S =2$.
Hence, the difference in oxidation states of underlined elements is greatest in $SO _{2}$.