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In diatomic nitrogen
There are two electron configurations I'd like to compare:

  1. 1s2 2s2(lone pair) 2p3(the bonding electrons sigma,pi,pi)
  2. 1s2 2sp(lone pair) 2sp(sigma bonding) 2p2(pi,pi bonding)

Why is option 2 favored over option 1?
I thought having the lone pair in a 2s orbital would be favored over 2sp.

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I thought having the lone pair in a 2s orbital would be favored over 2sp

That's true, placing the lone pair in a 2s orbital would be put the lone pair electrons at lower energy then if they were in a 2sp orbital. However you're only looking at a part of the puzzle. As you note, in this case the sigma bonding electrons would be in a pure p orbital and they would then be higher in energy then if they were in an sp orbital. You need to optimize the total energy of the molecule, so you need to consider all electrons and find the hybridization that gives the lowest possible energy to both the bonding and non-bonding electrons.

Which hybridization scheme a molecule adopts is the result of a delicate balance between electron stabilization and other factors such as bond strengths, steric and electrostatic effects. The balance of these factors can shift and that is why some molecules hybridize one way and other molecules hybridize differently.

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