Question

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

Answer 1

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.

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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.