Why are there FOUR $sp^3$ hybridized orbitals in methane?

(This may be a stupid question, but I've only learnt about hybridization yesterday.)


I will take the example of methane.
The valence shell electron configuration of ground state of carbon is $2s^2 2p^2$. There are 2 unpaired electrons, both in the $2p$ orbitals. But we need 4 unpaired electrons to form 4 bonds. So one of the electrons from the $1s$ orbital is excited to the 3rd $2p$ orbital. So, there is an $s$ orbital with an unpaired electron, and 3 $p$ orbitals with unpaired electrons, which can all participate in bonding, so now we can form 4 bonds. In the end, there will be **4** $sp^3$ hybridized orbitals.


My question is, why are there going to be **4** hybridized orbitals? I imagine the hybridization as "there is an $s$ orbital and 3 $p$ orbitals, hence $sp^3$." I thought a single $sp^3$ hybridized orbital is the combination of a single $s$ orbital and 3 $p$ orbitals. So how can there be 4 $sp^3$ orbitals in the end if the valence shell of carbon only *has* a single $s$ orbital and three $p$ orbitals? Intuitively it seems to me that 4 $sp^3$ orbitals would require 4 $s$ orbitals and 12 $p$ orbitals.




Hybrid orbitals are *linear combinations* of unhybridised orbitals:


$$\mathrm{sp^3} = a \mathrm{s} + b\mathrm{p}\_x + c\mathrm{p}\_y + d\mathrm{p}\_z$$


Now, you seem to have picked up the notion that $a = b = c = d = 1$. This is partly true in that for a perfectly tetrahedral molecule like methane you have that $a = b = c = d$; but recall that each orbital only has space for 2 electrons,\* one spin-up and one spin-down. When you add up four orbitals like that, you'll find that you have space for way more electrons, which is not quite right. An orbital should be *normalised*, in that it only has space for 2 electrons, one spin-up and one spin-down. Or, mathematically, we require that


$$|a|^2 + |b|^2 + |c|^2 + |d|^2 = 1,$$


(the modulus is required because in general these numbers may be complex, and it's $1$ not $2$ because we don't count the electrons of opposite spin). Now, if you set $a = b = c = d = 1$ you see that you've completely overshot the target of $1$. Instead of doing that, you need to scale down of the coefficients so in fact we have that $a = b = c = d = 1/2$.


Now, that looks like it should only be enough to make two hybrid orbitals, but it's a curious fact of quantum mechanics that only $|a|^2$ of the s-orbital (i.e., $(1/2)^2 = 1/4$) goes into each of the hybrid orbitals. So you *do* have enough for four hybrid orbitals.


The other $\mathrm{sp^3}$ orbitals also have that $|a| = |b| = |c| = |d| = 1/2$, but they are combined with different signs. See e.g. <https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/05%3A_Bonding_in_polyatomic_molecules/5.2%3A_Valence_Bond_Theory_-_Hybridization_of_Atomic_Orbitals/5.2D%3A_sp3_Hybridization>.




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\* OK, if you're really clever at chemistry you know this should really be $1$, and that what I'm talking about here is really a *spatial orbital*, and that's why the equation below says $1$ and not $2$.


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The sum of the exponents in $\mathrm{sp^3}$ is $1 + 3 = 4$.


The first $\mathrm{sp^3}$ is made with $(1/2)^2 = 1/4$ of the $\mathrm{2s}$ orbital, one quarter of the $\mathrm{2p\_x}$ orbital, one quarter of the $\mathrm{2p\_y}$ orbital and one quarter of the $\mathrm{2p\_z}$ orbital. Total $4$ quarters = $1$ hybrid orbital. This process is repeated four times, taking care to change twice the sign + to minus according to the character table. The definition of the four hybrid orbitals are :


$(1/2)[\mathrm{2s + 2p\_x + 2p\_y + 2p\_z}$]


$(1/2)[\mathrm{2s - 2p\_x - 2p\_y + 2p\_z}$]


$(1/2)[\mathrm{2s - 2p\_x + 2p\_y - 2p\_z}$]


$(1/2)[\mathrm{2s + 2p\_x - 2p\_y - 2p\_z}$]


This choice of plus or minus signs allows the hybrid orbitals to have a maximum density in the four directions of the tetrahedron.


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