A) \[{{I}^{-}}<{{F}^{-}}<H{{S}^{-}}<NH_{2}^{-}\]
B) \[{{\operatorname{HS}}^{-}}<\operatorname{NH}_{2}^{-}>{{\operatorname{F}}^{-}}<{{\operatorname{I}}^{-}}\]
C) \[{{\operatorname{F}}^{-}}<{{\operatorname{I}}^{-}}<\operatorname{NH}_{2}^{-}<{{\operatorname{HS}}^{-}}\]
D) \[NH_{2}^{1}<H{{S}^{-}}<{{I}^{-}}<{{F}^{-}}\]
Correct Answer: A
Solution :
The species with the greatest proton affinity will be the strongest base, and .its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of\[{{K}_{a}}.\] Since HI is a stronger acid than HF which is a stronger acid than \[{{H}_{2}}S,\] a partial order of proton affinity is \[{{I}^{-}}<{{F}^{-}}<H{{S}^{-}}\] Since \[N{{H}_{3}}\] is a very weak acid, \[N{{H}_{2}}^{-}\]must be a very strong base. Therefore the correct order of proton affinity is\[{{I}^{-}}<{{F}^{-}}<H{{S}^{-}}<NH_{2}^{-}\]
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