Will any hybridized orbital be lower in energy than all unhybridized orbitals?

ORBITAL HYBRIDIZATION EXAMPLE

We know that for example, #2s# orbitals are lower in energy than #2p# orbitals.

So, when one #2s# and three #2p# orbitals hybridize, we get four #sp^3# orbitals (like in methane, ammonia, water, etc).

Each #sp^3# orbital has #25%# #s# character, and #75%# #p# character, and they are all identical to each other. Their energies are automatically similar to the pure orbitals that contributed to their hybridization.

Any #p# orbital started with #100%# #p# character, and any #s# orbital started with #100%# #s# character.

Hence, with #25%# #s# character, the contribution from the #2s# orbital makes the #sp^3# orbitals more similar to the energy of a pure #2s# orbital than they were when they were pure, which means they become lower in energy.

In fact, the hybridized orbitals become intermediate in energy between the lowest-energy pure orbital and the higher-energy pure orbitals.

HYBRIDIZATION IN GENERAL

Of course, that example doesn't prove it in general, but it does get the main idea across. So now, let's put it this way.

For any #(l)(l+1)^m(l+2)^n(l+3)^o# hybrid orbital (where #l# is the angular momentum quantum number; #m#, #n#, and #o# are the number of #(l+1)#, #(l+2)#, or #(l+3)# orbitals contributing to the hybridization, respectively; and #m_"max" = 3#, #n_"max" = 5#, and #o_"max" = 7#):

• No matter what, there will be an orbital or two for a particular #l# that is lowest in energy, whether it's an #ns# orbital, an #(n-1)d# orbital, or an #(n-2)f# orbital, or whatever.
• That lowest-energy pure orbital's contribution to the hybridized orbital makes the hybridized orbital more similar in energy to that lowest-energy orbital.

Therefore, the hybrid orbital will always be lower in energy than the highest-energy pure contributing orbital.