A) \[{{I}^{-}}<{{F}^{-}}<H{{S}^{-}}<NH_{2}^{-}\]
B) \[H{{S}^{-}}<NH_{2}^{-}<{{F}^{-}}<{{I}^{-}}\]
C) \[{{F}^{-}}<{{I}^{-}}<\text{ }NH_{2}^{-}<H{{S}^{-}}\]
D) \[NH_{2}^{-}<H{{S}^{-}}<{{I}^{-}}<{{F}^{-}}\]
Correct Answer: A
Solution :
[a] 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, \[NH_{2}^{-}\] must be a very strong base. Therefore the correct order of proton affinity is \[{{I}^{-}}<{{F}^{-}}<H{{S}^{-}}<N{{H}_{2}}^{-}\]
You need to login to perform this action.
You will be redirected in
3 sec
