p-Block elementsHardBloom L3

Question

When an inorganic compound (X) having 3e-2e as well as 2e-2e bonds reacts with ammonia gas at a certain temperature, gives a compound (Y) iso-structural with benzene. Compound (X) with ammonia at a high temperature, produces a hard substance (Z). Then

Options

A.(X) is B2H6
B.(Z) is known as inorganic graphite
C.(Z) having structure similar to graphite
D.(Z) having structure similar to (X)

Solution

Sol. X is B2​H6​ having 3c-2e as well as 2c-2e bonds. B2​H6​ reacts with NH3​ to form borazine or inorganic benzene. Z is (BN)x​ is called inorganic graphite and has similar structure to graphite.

B2​H6​ + 2NH3 ​→ B2​H6​.2NH3​

When the addition product is heated at 200∘C a volatile compound borazole or inorganic benzene is formed. 3B2​H6​.2NH3​→2B3​N3​H6​+12H2​ (X) is B2​H6​

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