Question

The decreasing order of C=C bond length in the following complexes is \[ \text{I: [Cl$_3$Pt(CH$_2$=CH$_2$)]}^- \quad \text{II: [Cl$_3$Pt(C(CN)$_2$=C(CN)$_2$)]}^- \quad \text{III: [Cl$_3$Pt(CF$_2$=CH$_2$)]}^- \quad \text{IV: [Cl$_3$Pt(CF$_2$=CF$_2$)]}^- \]

• A. II>I>III>IV
• B. IV>II>I>III
• C. II>IV>III>I
• D. IV>I>II>III

Hint:

Backbonding from metal to alkene $\pi^*$ orbitals shortens the C=C bond. More electron-withdrawing substituents enhance backbonding, decreasing bond length.

Answer 1

The Correct Option is C

Step 1: Concept.
C=C bond length decreases with increasing $\pi$-backbonding from metal to alkene. Stronger $\pi$-acceptor substituents (like F, CN) increase back-donation, reducing C=C bond length.
Step 2: Analyze substituent effects.
Electron-withdrawing groups stabilize back-donation. Order of $\pi$-acceptor strength: CF$_2$=CF$_2$>CF$_2$=CH$_2$>CH$_2$=CH$_2$>C(CN)$_2$=C(CN)$_2$.
However, in the C(CN)$_2$ case, the $\pi$-system is highly delocalized, leading to longer bond due to weaker localized backbonding.
Step 3: Resulting bond order.
Thus, C=C bond length order: II (longest)>I>III>IV (shortest).
Step 4: Conclusion.
The decreasing order of C=C bond length is II>I>III>IV.