Question

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

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

Answer 1

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

a.
Let Oxidation number (O.N.) of "P" in $P_2{O}_5$ is $x$.
So , $2\times \left(x\right)+5\times (-2)=02x=10x=+5$

Let Oxidation number (O.N.) of "P" in $P_4{O}_{10}$ is $y$.
$4\times \left(y\right)+10\times (-2)=04y=20y=+5$

So, O.N. of N in $P_2{O}_5$ is + 5 and in $P_4{O}_{10}$ is +5.
$\therefore$ difference is 0.

b.
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 $y$,
$\Rightarrow 2y+4\times (-2)=0\Rightarrow y=+4$

So, O.N. of N in $N{O}_2$ is +4 and in $N_2{O}_4$ is +4.
$\therefore$ difference is 0.

c.
Let Oxidation number (O.N.) of "N" in $N_{2}O$ is $x$.
So , $2\times \left(x\right)+1\times (-2)=02x=2x=+1$

Let Oxidation number (O.N.) of "N" in $NO$ is $y$.
$y+1\times (-2)=0y=+2$

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

d.
Let Oxidation number (O.N.) of "S" in $S{O}_2$ is $x$.
So , $x+2\times (-2)=0x=+4$

Let Oxidation number (O.N.) of "S" in $S{O}_3$ is $y$.
$y+3\times (-2)=0y=+6$
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.