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a
H2O > H2Te > H2Se > H2S - M.P & B.P
b
H2O> H2S > H2Se > H2Te - Ka value
c
H2O> H2S > H2Se > H2Te --X-H bond enthalpy
d
O-H < S-H < Se-H < Te-H - Bond length

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detailed solution

Correct option is B

Down the group in the hydrides of 15th , 16th , 17th groups

A) Affinity towards Hydrogen decreases

B) X-H bond enthalpy decreases

C) X-H bond length increases

D) Thermal stability decreases

E) Acidic strength (Ka) increases

F) Reducing power increases

Thus 'Ka' of H2O< H2S < H2Se < H2Te

HB.D.E  of  H2O> H2S > H2Se > H2Te  (BDE=Bond dissociation enthalpy)

X-H bond length O-H < S-H < Se-H < Te-H

from H2O to H2Te molecular weight and molecular size increases , vanderwaals forces become stronger

 Expected  MP & BP - H2O< H2S < H2Se < H2Te

But H2O molecules are associated through intermolecular H- bonding.

It has abnormally high MP & BP

MP & BP order will be  H2S< H2Se < H2Te< H2O  (or) H2O > H2Te> H2Se > H2S

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