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

Question

(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.

Answer 1

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 $2$.

Answer 2

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.