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Why is the hydbridization of the $N$ atom (rightmost) in diazomethane $sp$ and not $sp^2$ ?

Most books claim that the negative charge (electron pair) of nitrogen can participate in resonance and is placed in a pure $p$ orbital. However, according to the general rule the hydribization number (number of sigma bonds + number of lone pairs) of the $N$ comes out to be $3$ implying ($sp^2$). However, according to several books the hybrdization is $sp$ and not $sp^2$. Why is it so?

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    As always in these questions, draw the resonance structure. – Zhe Mar 03 '17 at 13:01
  • @Zhe I don't agree with you. We can draw a resonance structure only if we know beforehand that the electron pair is in pure p orbital. But, unless we know that the hybridization is $sp$ we cannot do so. If both the electron pairs were in hybrid sp2 orbitals then no resonance would have taken place. –  Mar 03 '17 at 17:22
  • You don't have to agree with me, but it does. That's why diazomethane behaves like a carbene and is nucleophilic at the carbon.

    In addition, the logic here is backwards. You should be thinking about resonance and whether the electron should go into a $p$ orbital to begin with, not the other way around.

     – Zhe Mar 03 '17 at 19:16

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