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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

user8167818
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1 Answers1

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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.

Ivan Neretin
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