Q16 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Chemistry
Match the following

A B C D

A B C D
- A1 2 3 4
- B3 2 4 1
- C3 1 2 4
- Correct answerD3 1 4 2
Solution
(A) $\mathrm{H}_3 \mathrm{BO}_3-$ Hydrogen bonding
Boron is attached to three $\mathrm{O}$-atoms and each oxygen is bonded to one hydrogen, each such unit is $\mathrm{H}_3 \mathrm{BO}_3$.

(B) $\mathrm{AlCl}_3$ - Dimeric form $\mathrm{AlCl}_3$ form dimer and exist as $\mathrm{Al}_2 \mathrm{Cl}_6 \cdot \mathrm{AlCl}_3$ is an electron deficient compound, thus behave as Lewis acid.
(C) $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$ - Inorganic benzene $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$ (Borazine) is a polar inorganic cyclic compound, three $\mathrm{BH}$ units and three $\mathrm{NH}$ units alternate.
(D) $\mathrm{BF}_3-$ Back bonding The exchange of electrons between an atomic orbital on one atom and an antibonding orbital on another atom is known as back bonding.
So, correct match is $\mathrm{A}-3, \mathrm{~B}-1, \mathrm{C}-4, \mathrm{D}-2$

(B) $\mathrm{AlCl}_3$ - Dimeric form $\mathrm{AlCl}_3$ form dimer and exist as $\mathrm{Al}_2 \mathrm{Cl}_6 \cdot \mathrm{AlCl}_3$ is an electron deficient compound, thus behave as Lewis acid.

(C) $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$ - Inorganic benzene $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$ (Borazine) is a polar inorganic cyclic compound, three $\mathrm{BH}$ units and three $\mathrm{NH}$ units alternate.

(D) $\mathrm{BF}_3-$ Back bonding The exchange of electrons between an atomic orbital on one atom and an antibonding orbital on another atom is known as back bonding.

So, correct match is $\mathrm{A}-3, \mathrm{~B}-1, \mathrm{C}-4, \mathrm{D}-2$
Hence, the correct answer is option: 'D'.