Why is $\ce{NH_3}$ more basic than $\ce{CH3CN}$?

I tried to reason it out with the hybridization of nitrogen in both the molecules. In $\ce{NH_3}$ nitrogen is $\mathrm{sp^3}$ hybridized while in $\ce{CH3CN}$ nitrogen is $\mathrm{sp}$ hybridized. Hence it follows that nitrogen in $\ce{CH3CN}$ is more electronegative as compared to that in $\ce{NH_3}$, which makes $\ce{NH_3}$ more basic. Is this the main reason for $\ce{NH_3}$ to be more basic or is there any other reason?

  • 1
    $\begingroup$ Electronegativity is a property of atoms, not molecules. $\endgroup$ Jan 16, 2017 at 10:18

1 Answer 1

  1. p orbitals are higher in energy than s orbitals.

  2. sp3 orbitals with 75% p-character are therefore higher in energy than sp orbitals with 50% p-character.

  3. This means that the lone pair in ammonia is higher in energy than that in acetonitrile.

  4. Higher energy lone pair corresponds to greater basicity.

No need to bring electronegativity into the picture.

For a similar question see Relative acidities of alkanes, alkenes, and alkynes


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.