Concept: Bond length depends on bond order. • Higher bond order → shorter bond length
• Lower bond order → longer bond length
Resonance can reduce bond order and increase bond length. [width=0.5]18c ans.png
Step 1: Compound (a) \[ CH_3{-}CO{-}CH_3 \] This contains a normal carbonyl bond \(C=O\). Bond order \(=2\). Hence bond length is relatively shorter.
Step 2: Compound (c) \[ CH_3{-}COO^- \] Carboxylate ion has resonance between two oxygen atoms. \[ R{-}C=O \leftrightarrow R{-}C{-}O^- \] Thus both C–O bonds become equivalent with bond order \(1.5\). Hence bond length is intermediate.
Step 3: Compound (b) In this structure, resonance with the benzene ring gives a quasi-aromatic resonance form. \[ \text{Resonance structures (RS1 and RS2)} \] This delocalization decreases the effective bond order of the C–O bond. Therefore C–O bond length becomes maximum.
Step 4: Final order \[ b > c > a \] Thus, \[ \boxed{\text{Option (2) is correct}} \]
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
