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Column-I (Compounds)	Column-II (Characteristics)	Column-III (Characteristics)
(P) $Na_{2} B_{2} O_{4} (OH)_{4}$	(1) Contains even number of $π$ -bonds	(i) Contains X-O-X linkage
(Q) $Na_{2} B_{4} O_{5} (OH)_{4}$	(2) Contains odd number of $π$ -bonds	(ii) Has cyclic structure
(R) $C_{3} N_{3} (NH_{2})_{3}$	(3) No $π$ -bonds are present	(iii) Has $\geq 2 sp^{2}$ hybridised atoms
(S) $Na_{6} [(PO_{3})_{6}]$	(4) Has ionic bond	(iv) Has $\leq 3 sp^{3}$ hybridised atoms

.

Select the CORRECT combination.

Answer:

C

Solution:

Compound (R) is Melamine, C₃N₃(NH₂)₃, which contains a six-membered aromatic ring with alternating single and double bonds.

It possesses exactly 3 $π$ -bonds within the ring, which represents an odd number of $π$ -bonds, matching perfectly with Column-II (2).

All three carbon atoms and three nitrogen atoms in the aromatic ring are sp² hybridized (total of 6 atoms).

This fulfills the condition of having $\geq 2$ sp² hybridized atoms, matching with Column-III (iii).

Thus, the correct combination is R — 2 — (iii).

Stream:JEESubject:ChemistryTopic:CHEMICAL BONDINGSubtopic:VBT(overlapping & Hybridisation)

⏱ 2mℹ️ Source: QB

Chemistry - CHEMICAL BONDING Question with Solution | TestHub