3
$\begingroup$

I'm guessing sp hybridization for nitrogen in diatomic nitrogen based on the lone pair on nitrogen and the triple bond but there is no mixing of s and p orbitals so I want to say that there is no hybridization. Are there any chemicals without hybridization? What are some examples of chemicals without hybridization

Why are lone pairs counted in hybridization if they do not contribute to the mixing of the orbitals, a.k.a. hybridization?

$\endgroup$
  • 2
    $\begingroup$ 1. An $sp$ is exactly that, a mix of and $s$ an a $p$ orbital. 2. What about diatomic hydrogen? 3. You are wrong, they do. $\endgroup$ – ringo Oct 4 '16 at 8:13
  • $\begingroup$ s and p orbitals do mix in $\ce{N2}$, regardless of what theory you are using, even in MOT $\mathrm{2s}$-$\mathrm{2p}_z$ mixing is a huge deal in homonuclear diatomics. It would certainly be sp hybridisation. $\endgroup$ – orthocresol Oct 4 '16 at 9:08
3
$\begingroup$

Hybridisation is something you typically deduce from the geometry of the compound you are observing. It is primarily a mathematical concept.

That said, to form the diatomic molecules you asked about, no hybridisation is needed. In fact, you get much better results from assuming unhybridised atoms and then combining their atomic orbitals to molecular ones. Except for hydrogen (which only has 1σ and 2σ), the molecular orbitals of all these diatomic compounds ($\ce{N2, O2, F2, Cl2, Br2, I2}$) are basically identical save the energy levels of the MOs and their populations. Thus, these would all count as unhybridised compounds.

Also, there is no nead to invoke hybridisation to explain the bonding situation in $\ce{HCl}$ either. Instead agan consider a σ bond formed between hydrogen’s 1s orbital and chlorine’s 3p. All of chlorine’s other orbitals will sit there happily filled and unhybridised.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy