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Q.

Hybrid orbitals of central atom are involved in bonding in

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a

H2S

b

H2O

c

H2Te

d

H2Se

answer is B.

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Detailed Solution

Down the group as the size of the atom increases affinity of hydrogen towards central atom decreases. Therefore

a) X-H bond enthalpy decreases

b) Thermal stability decreases….H2O and H2S are exothermic products,remaining three hydrides are endothermic products.

c) Reducing power increases

d) acidic strength (Ka) increases

e) pKa decreases

Down the group as vanderwaal's forces increases

Melting and boiling points increases

Expected BP and MP….H2O<H2S<H2Se<H2Te

Due to H-bonding in H2O, it has higher BP and MP than expected.

BP trend…...H2S<H2Se<H2Te<H2O

MP trend….H2S<H2Se<H2Te<H2O

Central atom in these hydrides contain two lone pairs, according to Bent rules

As the Electronegativity of central atom decreases, bond angle decreases

VSEPR theory is required to explain the shape of water molecule, VBT can explain the shape of other hydrides;

Bond angle in water is about 1050 whereas it is 900 in other hydrides;

Oxygen in water molecule is sp3 hybridized whereas pure p-orbitals of other chalcogens are involved in bonding in their hydrides;

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