Which one of the following arrangements represents the correct order of the proton affinity of the given…

Which one of the following arrangements represents the correct order of the proton affinity of the given species :
  1. $\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-} < \mathrm{NH}_2^{-}$
  2. $\mathrm{HS}^{-} < \mathrm{NH}_2^{-} < \mathrm{F}^{-} < \mathrm{I}^{-}$
  3. $\mathrm{F}^{-} < \mathrm{I}^{-} < \mathrm{NH}_2^{-} < \mathrm{HS}^{-}$
  4. $\mathrm{NH}_2^{-} < \mathrm{HS}^{-} < \mathrm{I}^{-} < \mathrm{F}^{-}$

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 $\mathrm{K}_a$. Since $\mathrm{HI}$ is a stronger acid than $\mathrm{HF}$ which is a stronger acid than $\mathrm{H}_2 \mathrm{S}$, a partial order of proton affinity is $\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-}$ Since $\mathrm{NH}_3$ is a very weak acid, $\mathrm{NH}_2^{-}$must be a very strong base. Therefore the correct order of proton affinity is $\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-} < \mathrm{NH}_2^{-}$

Asked in: JEE Main 2013 (25 Apr Online)

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