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

##### 1 Answer

**ORBITAL HYBRIDIZATION EXAMPLE**

We know that for example,

So, when **one** **three** **four**

Each **they are all identical to each other**. Their energies are automatically *similar* to the pure orbitals that contributed to their hybridization.

Any

Hence, with **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

- 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.*