The decreasing order of C=C bond length in the following complexes is · ABC26IN0771
Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Organometallics – General · Exam: JAM 2005-2022 · Marks: 2 · Difficulty: Hard
The decreasing order of $\mathrm{C}=\mathrm{C}$ bond length in the following complexes is | $\left[\mathrm{Cl}_{3} \mathrm{Pt}\left(\mathrm{CH}_{2}=\mathrm{CH}_{2}\right)\right]^{-}$ | $\left[\mathrm{Cl}_{3} \mathrm{Pt}\left(\mathrm{C}(\mathrm{CN})_{2}=\mathrm{C}(\mathrm{CN})_{2}\right)\right]^{-}$ | $\left[\mathrm{Cl}_{3} \mathrm{Pt}\left(\mathrm{CF}_{2}=\mathrm{CH}_{2}\right)\right]^{-}$ | $\left[\mathrm{Cl}_{3} \mathrm{Pt}\left(\mathrm{CF}_{2}=\mathrm{CF}_{2}\right)\right]^{-}$ |
|---|
| I | II | III | IV |
(a)II $>$ III $>$ IV $>$ I
(b)IV $>$ II $>$ III $>$ I
(c)II $>$ IV $>$ III $>$ I
(d)IV $>$ II $>$ I $>$ III
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
Electron-poor alkenes accept more back-donation, lengthening C=C: TCNE (II) > C₂F₄ (IV) > CF₂=CH₂ (III) > C₂H₄ (I) (c).
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