A) \[{{H}_{2}}S<{{H}_{2}}Se<{{H}_{2}}O<{{H}_{2}}Te\]
B) \[{{H}_{2}}O<{{H}_{2}}S<{{H}_{2}}Se<{{H}_{2}}Te\]
C) \[{{H}_{2}}Te<{{H}_{2}}Se<{{H}_{2}}S<{{H}_{2}}O\]
D) \[{{H}_{2}}O<{{H}_{2}}Se<{{H}_{2}}S<{{H}_{2}}Te\]
Correct Answer: B
Solution :
[b] Assume that each has lost a proton. So we get: \[H{{O}^{+}},\]\[H{{S}^{-}},\]\[HS{{e}^{-}},\]\[HT{{e}^{-}}\] It can be easily seen that the volume available for the negative charge is increasing from \[H{{O}^{-}}\] to \[HT{{e}^{-}},\] therefore (i) volume available for the negative charge is increasing from left to right (ii) charge density is decreasing from left to right (iii) basicity is decreasing from left to right (iv) acidity of conjugate acids is increasing from left to right \[{{H}_{2}}O<{{H}_{2}}S<{{H}_{2}}Se<{{H}_{2}}Te\]
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