# How do you determine which orbitals are used in hybridization?

May 30, 2017

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

$s {p}^{3} {d}^{5} {f}^{7}$.

For example, in ${\text{NH}}_{3}$, three $\text{N"-"H}$ bonds are made and one lone pair of electrons is leftover, so we assume that the hybridization is $s + p + p + p \to \textcolor{b l u e}{s {p}^{3}}$.

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

Or, in ${\text{BH}}_{3}$, three $\text{B"-"H}$ bonds are made, and there are no other lone pairs or bonds. So, we assume the hybridization is $s + p + p \to \textcolor{b l u e}{s {p}^{2}}$.

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.

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:

May 31, 2017

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:
$B e C {l}_{2}$

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

(NBPr) = NonBonded Pr (NBPr)
= $\frac{\text{Valence""Number"(V) - "Substrate""Number} \left(S\right)}{2}$

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)

$\left(N B P r\right) = \frac{V - S}{2}$ = $\frac{16 - 16}{2} = 0$

Total Electron Pairs Associated with Central Element of Parent Geometry = BPr + NBPr = 2 + 0 = 2 => $A {X}_{2}$ 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 $S n C {l}_{2}$.
(Is this linear of a derived structure?)

$B P r = 2$

$N B P r = \frac{V - S}{2} = \frac{18 - 16}{2} = \frac{2}{2} = 1$

$V = 1 S n + 2 C l = 1 \left(4\right) + 2 \left(7\right) = 18$
$S = 2 C l = 2 \left(8\right) = 16$

Parent = BPr + NBPr = 2 + 1 = 3 => Trigonal Planar Parent $\left(A {X}_{3}\right)$

=> Derived Geometry $\left(A {X}_{2} E\right)$ => Bent Angular Geometry with 2 bonded pair and 2 nonbonded pair.

=> This is also the number of hybrid orbitals needed. => $3 \left(s {p}^{2}\right)$ hybrids (2 bonded pr and 1 with nonbonded pr)

Determine the Geometry of $N {H}_{3}$.
(Is this Trigonal Planar or a Derived Structure?)

$B P r = 3$ (number substrates attached to central element)
$N B P r = \frac{V - S}{2} = \frac{8 - 6}{2} = \frac{2}{2} = 1$

$V = 1 N + 3 H = 1 \left(5\right) + 3 \left(1\right) = 8$
$S = 3 H = 3 \left(2\right) = 6$

BPr + NBPr = 3 + 1 = 4 => Tetrahedral Parent $\left(A {X}_{4}\right)$ => Pyrimidal Geometry $\left(A {X}_{3} E\right)$ => 4$\left(s {p}^{3}\right)$ Hybrid Orbitals ( 3 with bonded pairs & 1 with nonbonded pair).

Determine the Geometry of $I C {l}_{4}^{-}$.
(Is this a Tetrahedral Geometry or a Derived Structure?)

$B P r = 4$
$N B P r = \frac{V - S}{2} = \frac{36 - 32}{2} = \frac{4}{2} = 2$

$V = 1 I + 4 C l + 1 {e}^{-} = 1 \left(7\right) + 4 \left(7\right) + 1 = 36$
$S = 4 C l = 4 \left(8\right) = 32$

BPr + NBPr = 4 + 2 = 6 => Octahedral Parent $\left(A {X}_{6}\right)$ => Square Planar Geometry $\left(A {X}_{4} {E}_{2}\right)$ => 6$\left(s {p}^{3} {d}^{2}\right)$ Hybrid Orbitals ( 4 with bonded pairs & 2 with nonbonded pair).

STURCTURES WITH MULTIPLE BONDS
Determine the Geometry of $C {H}_{2} = O$.
(The Lewis Structure shows this to be ${H}_{2} C = O$. 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 $3 s {p}^{2}$ hybrid orbitals and 1 unhybridized $p$-orbital at the carbon valence level.