Q.

Statement (I) : BP of H2S < H2Se < H2Te

Statement (II) : Down the group vanderwaal's forces become strong

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a

Both (I) & (II) are true

b

Both (I) & (II) are false

c

(I) is false & (II) is true

d

(I) is true & (II) is false

answer is A.

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Detailed Solution

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

bond dissociation enthalpy  of  H2O> H2S > H2Se > H2Te

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

 Except 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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