Question

Borazine is called ‘inorganic benzene’ in view of its ring structure with alternate BH and NH groups. Which of the following statements is correct about benzene?
A. Each B and N atom is $s{p^2}$ hybridised
B. Borazine satisfies the $\left( {\left( {4n + 2} \right)\pi } \right)$ Huckel’s rule
C. Like organic benzene, borazine doesn’t give addition product with $HCl$
D. Borazine contains dative $p\pi - p\pi$ bond

Answer 1

Hint- Borazine is a dreary fluid with a fragrant smell. In water it hydrolyses to boric acid, ammonia, and hydrogen. Borazine has a standard enthalpy change of formation ΔHf of −531 kJ/mol and is thermally entirely steady.

Complete answer:
Borazine has always been a polar inorganic compound with the chemical formula ${B_3}{H_6}{N_3}$. In this cyclic compound, the three BH units and three NH units substitute. The compound is isoelectronic and isostructural with benzene. Therefore, borazine is once in a while alluded to as "inorganic benzene". Like benzene, borazine is a colourless fluid.
Borazine is isostructural with benzene. The six B-N bonds have a length of 1.436. The carbon-carbon bond in benzene has a shorter length at 1.397. The boron-nitrogen bond length is between that of the boron-nitrogen single bond with 0.151 nm and the boron-nitrogen twofold bond which is 0.131 nm. This recommends partial delocalisation of nitrogen lone-pair electrons.
Borazine fulfils the $\left( {\left( {4n + 2} \right)\pi } \right)$ Huckel's standard and every B and N molecule is $s{p^2}$ hybridized.

Hence, option A and B are correct options.

Note: The electronegativity of boron (2.04 on the Pauling scale) contrasted with that of nitrogen (3.04) and furthermore the electron inadequacy on the boron particle and the lone pair on nitrogen favour elective mesomer structures for borazine.