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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?

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    $\begingroup$ Electronegativity is a property of atoms, not molecules. $\endgroup$ – Ivan Neretin Jan 16 '17 at 10:18
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  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

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