Question

How do you determine which orbitals are used in hybridization?

Answer 1

Yes. For general chemistry, we just count the number of electron groups around the central atom, and assume that the orbitals used are in order of angular momentum `l`.

"NORMAL", GENERAL CHEMISTRY WAY

We assume an ordering of

`s, p, p, p, d, d, d, d, d, f, f, f, f, f, f, f`

or

`sp^3d^5f^7`.

For example, in `"NH"_3`, three `"N"-"H"` bonds are made and one lone pair of electrons is leftover, so we assume that the hybridization is `s+p+p+p -> color(blue)(sp^3)`.

(You can think of it as the central atom "anticipating" future bonds it can possibly make.)

Or, in `"BH"_3`, three `"B"-"H"` bonds are made, and there are no other lone pairs or bonds. So, we assume the hybridization is `s+p+p -> color(blue)(sp^2)`.

In short, count the orbitals from left to right in the list above until you have the same number of orbitals as the number of electron groups.

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MORE RIGOROUS WAY

There is a more rigorous way to do it using Group Theory. You do not have to know this for general chemistry, but I am just providing it for the interested reader.

Here it is in a scratchpad:
https://socratic.org/scratchpad/25f1d95969583d1e1c9d

Answer 2

Here's a method using the VSEPR Theory to determine number of hybrid orbitals needed when using the Valence Bond Theory.

Explanation:

Here's a method that uses the VSEPR theory to determine how many hybrid orbitals would be needed for the central element of a given binary compound.
Need:
1. number of bonded electron pairs
2. number of non-bonded electron pairs
3. (NBPr + BPr)/2 => Parent Geometry => Number of Hybrids needed from central element's valence electrons.

Note: The detailing of this procedure will appear to be extensive, however, once the sequence is understood & with a bit of practice, these can almost be determined by inspection. I will finish with several examples of how this is a 'short cut' system.

DETAILED VERSION:
Example: Determine the geometry of the following:
`BeCl_2`

(BPr) = Bonded Pr (BPr) = 2 => That is, 2 substrates attached to central element)

(NBPr) = NonBonded Pr (NBPr)
= `("Valence""Number"(V) - "Substrate""Number"(S))/2`

Valence Number = Number of Valence Electrons in given compound = Be + 2Cl = 1(2) + 2(7) = 16

Substrate Number = Number of Octet electrons associated with bonded substrates in given compound = 2Cl = 2(8) = 16 (Duet if working with Hydrogen as a substrate)

`(NBPr) = (V - S)/2` = `(16-16)/2=0`

Total Electron Pairs Associated with Central Element of Parent Geometry = BPr + NBPr = 2 + 0 = 2 => `AX_2` Geometry => Linear.

The TTL `e^-` pairs associated with the central element also is the number of hybrid orbital needed for the VB Theory.

That is, 2 sp hybrid orbitals. sp hybrids => linear geometry

ABREVIATED VERSION:
Determine geometry of `SnCl_2`.
(Is this linear of a derived structure?)

`BPr = 2`

`NBPr = (V-S)/2 = (18-16)/2 = 2/2 = 1`

`V = 1Sn + 2Cl = 1(4) + 2(7) = 18`
`S = 2Cl = 2(8) = 16`

Parent = BPr + NBPr = 2 + 1 = 3 => Trigonal Planar Parent `(AX_3)`

=> Derived Geometry `(AX_2E)` => Bent Angular Geometry with 2 bonded pair and 2 nonbonded pair.

=> This is also the number of hybrid orbitals needed. => `3(sp^2)` hybrids (2 bonded pr and 1 with nonbonded pr)

Determine the Geometry of `NH_3`.
(Is this Trigonal Planar or a Derived Structure?)

`BPr = 3` (number substrates attached to central element)
`NBPr = (V-S)/2 = (8-6)/2 = 2/2 = 1`

`V = 1N + 3H = 1(5) + 3(1) = 8`
`S = 3H = 3(2) = 6`

BPr + NBPr = 3 + 1 = 4 => Tetrahedral Parent `(AX_4)` => Pyrimidal Geometry `(AX_3E)` => 4`(sp^3)` Hybrid Orbitals ( 3 with bonded pairs & 1 with nonbonded pair).

Determine the Geometry of `ICl_4^-`.
(Is this a Tetrahedral Geometry or a Derived Structure?)

`BPr = 4`
`NBPr = (V-S)/2 = (36-32)/2 = 4/2 = 2`

`V = 1I + 4Cl + 1e^- = 1(7) + 4(7) + 1 = 36`
`S = 4Cl = 4(8) = 32`

BPr + NBPr = 4 + 2 = 6 => Octahedral Parent `(AX_6)` => Square Planar Geometry `(AX_4E_2)` => 6`(sp^3d^2)` Hybrid Orbitals ( 4 with bonded pairs & 2 with nonbonded pair).

STURCTURES WITH MULTIPLE BONDS
Determine the Geometry of `CH_2=O`.
(The Lewis Structure shows this to be `H_2C=O`. Carbon is the central element and assume the double bond to be a single bond for defining geometry.)

BPr = 3
NBPr = 0
BPr + NBPr = 3 => Trigonal Planar Geometry about Carbon. This means that the structure is composed of `3sp^2` hybrid orbitals and 1 unhybridized `p`-orbital at the carbon valence level.