# Is each hybridized orbital made up of multiple complete orbitals?

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.

• Because $s^{0.25}p^{0.75}$ is a mouthful.
– Zhe
Dec 4, 2017 at 19:39
• 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. Dec 4, 2017 at 19:53
• @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.
– Zhe
Dec 4, 2017 at 21:53
• (1s + 3p) is in parentheses. It reads four (----) orbitals. 1 + 3 = 4. Dec 4, 2017 at 21:56

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.