A) \[{{N}_{2}}{{O}_{5}}>S{{O}_{2}}>CO>C{{O}_{2}}\]
B) \[{{N}_{2}}{{O}_{5}}>S{{O}_{2}}>C{{O}_{2}}>CO\]
C) \[S{{O}_{2}}>C{{O}_{2}}>CO>{{N}_{2}}{{O}_{5}}\]
D) \[S{{O}_{2}}>{{N}_{2}}{{O}_{5}}>CO>C{{O}_{2}}\]
Correct Answer: B
Solution :
Key Idea Acidic nature of oxide \[\propto \] Number of electron outermost shell \[\propto \frac{1}{atomic\text{ }size}\] \[\propto \] electronegative character In a period from left to right the acidic nature of oxide of element increases. Therefore, \[{{N}_{2}}{{O}_{5}}\] is more acidic than \[C{{O}_{2}}\]. Further, when the oxidation number increases, the acidic character is also increases. Hence, \[C{{O}_{2}}\] is more acidic than CO. In a group acidic nature of oxides decreases. So, \[S{{O}_{2}}\] is more acidic than \[C{{O}_{2}}\] but less than\[{{N}_{2}}{{O}_{5}}\]. Therefore, the correct order of acidic character is as\[{{N}_{2}}{{O}_{5}}>S{{O}_{2}}>C{{O}_{2}}>CO\]
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