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


1 Answer 1


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.

  • 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
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    $\begingroup$ Great minds think alike, they say. $\endgroup$ Jun 22, 2017 at 13:20
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    $\begingroup$ but what will happen if s has two signs ( positive and negative ) . Thanks $\endgroup$ Jun 22, 2017 at 13:20
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    $\begingroup$ 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). $\endgroup$ Jun 22, 2017 at 13:22
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    $\begingroup$ @user8167818 en.wikipedia.org/wiki/… is a good way of trying to understand orbitals, IMO. $\endgroup$
    – Stian
    Jun 22, 2017 at 13:30

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