A) \[\underline{N}{{O}_{2}}\] and \[{{\underline{N}}_{2}}{{O}_{4}}\]
B) \[{{\underline{P}}_{2}}{{O}_{5}}\] and \[{{\underline{P}}_{4}}{{O}_{10}}\]
C) \[{{\underline{N}}_{2}}O\] and \[\underline{N}O\]
D) \[\underline{S}{{O}_{2}}\] and \[\underline{S}{{O}_{3}}\]
Correct Answer: D
Solution :
[d] O.N. of N in \[N{{O}_{2}}\] and \[{{N}_{2}}{{O}_{4}}\] is +4 \[\therefore \] difference is zero. 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. The difference is 1 O.N. of S in \[S{{O}_{2}}\] is +4 and in \[S{{O}_{3}}\] is +6. The difference is +2.
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