Unhybridized P Orbitals

[sanguinee]

How many unhybridized P orbitals are on the central N of azide (N3- ion)?

The answer is 2 unhybridized orbitals to participate in 2 π bonds, but I can't seem to wrap my head around this. I understand the bonds are sp, but was hoping someone could explain this further?

Thanks!

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[aldol16]

What part of it don't you understand?

If you hybridize the central N to sp, that means you have two mutually orthogonal p orbitals left. So say you use px to hybridize. That leaves you with py and pz. Each of those can overlap with the corresponding p orbital on the adjacent nitrogen atom to form a pi bond. Note that these pi bonds are not in the same plane relative to each other - they only appear to be in the same plane because it's a 2D representation.

 

[betterfuture]

First, you find the steric # which is # of sigma bonds - # of lone pairs. The central N has 2 sigma bonds and 0 lone pairs, hence 2-0= 2.

2 corresponds to the hybridized sp orbital because you have 1 s orbital and 1 p orbital, so altogether you have 2 orbitals. Then, you find that there is only one p orbital used while there are 2 remaining p orbitals which ARE NOT hybridized and are empty.

One p orbital can participate in a pi bond and so can the other p orbital because they were not used and thus empty. Therefore, you find that there are indeed 2 pi bonds; 1 on the left N and the other on the right N.

 

[sanguinee]

@betterfuture @aldol16 this made a lot of sense; thank you!

 

[betterfuture]

I apologize. There should be a correction.

Steric # is formulated by taking the # of sigma bonds +# of lone pairs. In the above post, I wrote minus instead of plus. It should be PLUS.