Redox Reactions And Electrochemistry Question 695
Question: In which of the following pairs, there is greatest difference in the oxidation number of the underlined elements-
Options:
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 $
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Answer:
Correct Answer: D
Solution:
[d] O.N. of N in $ NO_2 $ and $ N_2O_4 $ is + 4
$ \therefore $ difference is zero. O.N of P in $ P_2O_5, $ and $ P_4O_{10} $ is + 5
$ \therefore $ difference is zero O.N. of N in $ N_2O $ is + 1 and in NO is +2. The difference is 1 O.N. of S in $ SO_2, $ is +4 and in $ SO_3, $ is + 6. The difference is + 2.
BETA
