# NEET Chemistry The p-Block Elements (XII) Questions Solved

BOARD

How would you account for the following:

(i) ${H}_{2}S$ is more acidic than ${H}_{2}O.$

(ii) The N-O bond in $N{O}_{2}^{-}$ is shorter than the N-O bond in $N{O}_{3}^{-}.$

(iii) Both  stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine.

(i) Since the size of sulphur is more than oxygen, S-H bond length increases and hence bond dissociation energy of S-H is less than O-H. Therefore S-H easily loses ${H}^{+}$ and thus is more acidic than ${H}_{2}O.$

(ii) The resonating structure of $N{O}_{2}^{-}$ and $N{O}_{3}^{-}$ show that in $N{O}_{2}^{-}$ two bonds are sharing a double bond while in $N{O}_{3}^{-},$ 3 bonds are sharing a double bond. That's why $N{O}_{2}^{-}$ has shorter bond than that of $N{O}_{3}^{-}.$

(iii) Oxygen stabilizes the highest oxidation state even more than fluorine.

Example : Highest fluoride of Mn is $Mn{F}_{4}$ whereas highest oxide is $M{n}_{2}{O}_{7}$. It is due to ability of oxygen to form multiple bonds with the metal atoms.

Difficulty Level:

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