Exams › JEE Advanced › Chemistry
Correct answer: (P)-(3), (Q)-(1), (R)-(4), (S)-(2)
(P) NO3-: N is in period 2, no d-orbitals. The negative charge is delocalized through p-pi to p-pi bonds (resonance involving 2p orbitals of N and O). Only p-pi to p-pi. -> (3). (Q) (CH3)2O: Oxygen has lone pairs but is bonded only to carbon atoms; C has no d-orbitals and there is no pi bond framework. However, the oxygen lone pairs can donate into empty d-orbitals... In (CH3)2O, there is p-d pi backbonding from O lone pairs to the d-orbitals of S... wait, this is an ether, not a sulfur compound. In dimethyl ether, O is bonded to two sp3 carbons. C has no d-orbitals and O forms no pi bonds here. So no p-p pi and no p-d pi. -> (2). (R) SO3: S is in period 3 with d-orbitals. SO3 has both delocalized p-pi to p-pi bonds (resonance) AND p-pi to d-pi bonding from O lone pairs into empty d-orbitals of S. -> (4). (S) SO4²-: All four S-O bonds are equivalent. The formal charges and geometry lead to p-d pi bonding (O lone pairs into S d-orbitals), but the negative charge and tetrahedral geometry reduce p-p pi contribution significantly - primarily only p-d pi bonding. -> (1). So: P-3, Q-2, R-4, S-1 = option (C).