6
$\begingroup$

I have a doubt regarding the shape of the $\mathrm{sp^3}$ hybridised orbital. I looked up for it on stackexchange and found a post where the guy said that where ever the orbitals interfere with opposite signs of the wavefunction they cancel out and hence we get a small lobe . He further said that when the right side of the p orbital overlaps with s orbital( i have linked the post) they both add up.

But the $\mathrm{s}$ orbital wavefunction can also have negative sign (in the post the guy has assumed that s orbital wavefunction is always positive) . So in all four interference combinations are possible : ++ , +- , -+ , - - . Hence the shape should be different but its not.

I am really confused about this. Any help is appreciated.

Thanks in Advance. Link to original post : https://chemistry.stackexchange.com/a/15909

Improve this question
$\endgroup$

1 Answer 1

Reset to default
17
$\begingroup$

Don't put too much trust in the absolute signs of wavefunction, for they all are arbitrary anyway. Look at it this way: an s orbital has one sign (*). One lobe of p orbital has the same sign, so when they add up, it grows huge. Another lobe inevitably has the opposite sign, so when they interfere, it is reduced to a tiny pig tail.

(*) This is not quite true if we consider radial nodes. That's when things get really hairy.

Improve this answer
$\endgroup$
5
  • 3
    $\begingroup$ I think this answer says what I meant to say. I'll delete mine, you were 33 seconds earlier, after all =D $\endgroup$
    – Stian
    Jun 22, 2017 at 13:17
  • 5
    $\begingroup$ Great minds think alike, they say. $\endgroup$
    – Ivan Neretin
    Jun 22, 2017 at 13:20
  • 1
    $\begingroup$ but what will happen if s has two signs ( positive and negative ) . Thanks $\endgroup$
    – user8167818
    Jun 22, 2017 at 13:20
  • 2
    $\begingroup$ That can't happen. Every s orbital looks from the outside exactly the same: as a nice furry ball of one color (one sign, that is). $\endgroup$
    – Ivan Neretin
    Jun 22, 2017 at 13:22
  • 2
    $\begingroup$ @user8167818 en.wikipedia.org/wiki/… is a good way of trying to understand orbitals, IMO. $\endgroup$
    – Stian
    Jun 22, 2017 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.