Question

In which of the following pairs, there is greatest difference in the oxidation number of the mentioned element?

• A. $N{O}_{2}and{N}_2{O}_4$ for $N$
• B. $P_2{O}_{5}and{P}_4{O}_{10}$ for $P$
• C. $N_{2}OandNO$ for $N$
• D. $S{O}_{2}andS{O}_3$ for $S$

Answer 1

The correct option is D $S{O}_{2}andS{O}_3$ for $S$

Let O.N. of N in $N{O}_2$ is $x$,
$\Rightarrow x+2\times (-2)=0\Rightarrow x=+4$

Let O.N. of N in $N_2{O}_4$ is $y$,
$\Rightarrow 2y+4\times (-2)=0\Rightarrow x=+4$

$\therefore$ difference is zero.


Similarly, for other cases,

O.N. of P in $P_2{O}_5,$ and $P_4{O}_{10}$ is + 5
$\therefore$ difference is zero.

O.N. of N in $N_{2}O$ is + 1 and in $NO$ is +2.
$\therefore$ difference is 1.

O.N. of S in $S{O}_2$ is +4 and in $S{O}_3,$ is + 6.
$\therefore$ difference is +2.

Clearly, (d) is the correct answer.