I've read that nonpolar bonds are stronger than polar bonds. If it is true, why $\ce{C-H}$ bond entalpy ($\pu{413 kJ/mol}$) is more than $\ce{C-C}$ ($\pu{348 kJ/mol}$)? I'm very confused at the moment.
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asked Dec 3, 2014 at 17:08
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2 Answers
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The "strength" of a bond can't be evaluated just based on its enthalpy.
According to this relation
$$ \Delta H = \sum \Delta H_\text{bonds broken} - \sum \Delta H_\text{bonds created} $$
the strength of your bond will depend on the reactive nearby the bond. If the molecule is likely to create bonds with higher enthalpy after the bond's breakdown, then the enthalpy variation will be negative. The process will then be spontaneous and the initial bond will breaks "easily".
There is also some steric consideration, the hydrogen atom is smaller, the bond is then more easy to be "attacked".
Finally, for a polar bond, the electronic cloud is delocalised closer to one atom rather than the other (according to their electronegativity) so you have "less perturbation to add" to fully remove the covalent electron from the less electronegative atom.
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$\begingroup$ Thanks! Could you tell me why in asymmetric alkenes the carbon with more hydrogen is more electronegative that the other? $\endgroup$ Dec 3, 2014 at 17:32
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$\begingroup$ What do you mean by "will depend on the reactive nearby the bond"? What is "reactive"? Did you mean to write something else? $\endgroup$ Mar 17, 2018 at 4:57
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$\begingroup$ Moreover, there are a few anomalies: "If "the hydrogen atom is smaller, the bond is then more easy to be "attacked"", then shouldn't its energy be lower instead, because other atoms are able to attack it and break the bond "easily" as you say? Similarly, "you have "less perturbation to add" to fully remove the covalent electron from the less electronegative atom." This again hints at the C-H bond (the more polar one) being weaker. Both are contrary since $\ce{C-H}$ has a higher bond enthalpy. Could you please explain? Thank you! $\endgroup$ Mar 17, 2018 at 5:02
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I disagree with the currently accepted answer, and believe that there are other reasons at play.
First of all, the difference in the electronegativities of carbon (2.55) and hydrogen (2.2) is roughly 0.35. The $\ce{C-H}$ bond is indeed slightly more polar than the $\ce{C-C}$ bond, but this slightly increased polaity can in no solely way account for the approximately 20% increase in bond energy. In fact, Wikipedia writes: "Because of this small difference in electronegativities, the C−H bond is generally regarded as being non-polar."
I believe the real reason here is atomic size. The hydrogen atom is much smaller than the carbon atom. Smaller bonds lead to higher bond energy, therefore $\ce{C-H}$ bond has higher bond enthalpy than the $\ce{C-C}$ bond.
answered Mar 17, 2018 at 5:23