So, I was wondering that how can a methane molecule form four $\mathrm{sp^3}$ hybrid orbitals, with each hybrid orbital containing one $\mathrm s$ and three $\mathrm p$ orbitals.

In four of such $\mathrm{sp^3}$ hybrid orbitals, it makes a total of four individual $\mathrm s$ orbitals and twelve individual $\mathrm p$ orbitals; whereas, we only have, numerically, one $\mathrm s$ orbital and three $\mathrm p$ orbitals available that constitute in hybridization.

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    $\begingroup$ Because $s^{0.25}p^{0.75}$ is a mouthful. $\endgroup$
    – Zhe
    Dec 4, 2017 at 19:39
  • $\begingroup$ The first rule of converting atomic orbitals to hybrid orbitals is that the number of the former must equal the number of the latter. If you start with four (1s + 3p) orbitals, then you wind up with four sp^3 orbitals. Saying sp^3 is not a mouthful. $\endgroup$
    – user55119
    Dec 4, 2017 at 19:53
  • $\begingroup$ @user55119 I don't understand what you mean by four (1s + 3p) orbitals. That implies you have 4 s orbitals, which is the exactly the issue that the OP is confused about. $\endgroup$
    – Zhe
    Dec 4, 2017 at 21:53
  • $\begingroup$ (1s + 3p) is in parentheses. It reads four (----) orbitals. 1 + 3 = 4. $\endgroup$
    – user55119
    Dec 4, 2017 at 21:56

1 Answer 1


As has been pointed out in the comments, an $\mathrm{sp}^3$ orbital is a linear combination not of one s orbital and three p orbitals, but rather of one part s orbital and three parts p orbital. Stated differently, the numbers in $sp^3$ denotes the ratio and not the number of orbitals we combine.

Oddities such as the $\mathrm{sp^5}$ orbitals in cyclopropane can therefore be rationalised. It does not mean that there are five p orbitals which form hybrids, which is obviously impossible. It simply means that that particular orbital has ~1/6 s orbital character and ~5/6 p orbital character.


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