How can we explain the shorter carbon-carbon bond length in $\ce{(CCl3)2}$ ($1.49 \,\mathrm{\mathring{A}}$) than that of $\ce{(CH3)2}$ ($1.534 \,\mathrm{\mathring{A}}$)?
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One reasoning would be applying Bent's rule:
- In ethane the $\ce{C-H}$ bonds are approximately composed from $\ce{61\% sp^{3.3} C}$ and $\ce{39\% s H}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{2.3} C}$ orbitals. - In hexachloroethane the $\ce{C-Cl}$ bonds are composed from $\ce{48\% sp^{3.7} C}$ and $\ce{52\% sp^{4.8} Cl}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{1.8} C}$ orbitals.
Higher $\ce{s}$-character in bonds usually means that the optimal overlap of these orbitals is at shorter inter-nuclear distances. Since the $\ce{s}$-character in the $\ce{C-C}$ bond in hexachloroethane is much higher than in ethane, the bond length is expected to be shorter.
(Values obtained from an Natural Bond Orbital Analysis on the DF-BP86/def2-SVP level of theory.)
answered Dec 27, 2016 at 10:02