# How to tell if a central element in a molecule needs to form hybridized orbitals?

Feb 18, 2016

Simply put, if a central atom has to bond to more than one outer atom, especially if more than one outer atom is different, it has to hybridize.

Diatomic molecules will always point compatible $\sigma$ bonding orbital lobes along the internuclear axis, and be able to pair compatible orbitals, so there is no hybridization in molecules like $\text{HCl}$, ${\text{NO}}^{+}$, ${\text{Cl}}_{2}$, etc.

An easy way to tell when an atom has to hybridize is to count the number of surrounding atoms. I've listed examples below. Essentially:

• Octahedral electron geometry? $s {p}^{3} {d}^{2}$
• Trigonal bipyramidal electron geometry? $s {p}^{3} d$
• Tetrahedral electron geometry? $s {p}^{3}$
• Trigonal planar electron geometry? $s {p}^{2}$
• Linear polyatomic electron geometry? $s p$

HYBRIDIZATION IN WATER

Oxygen in $\text{H"_2"O}$ has to contribute four $s {p}^{3}$-hybridized atomic orbitals to bond because:

1. It is bonding to more than one hydrogen. That tells you that hybridization can occur.
2. It is bonding in a non-horizontal direction to at least one hydrogen. That tells you that hybridization should occur to orient all the orbitals correctly.
3. It is bonding identically to each hydrogen. That tells you that hybridization has occurred to make the orbitals compatible.

It is $s {p}^{3}$ because four electron groups are surrounding oxygen: two bonding to one hydrogen each, and two lone pairs not bonding at all.

Oxygen had to hybridize one $2 s$ and three $2 p$ orbitals together to generate four diagonal-oriented orbitals in three dimensions ($x , y , z$). Two of them could be used, but are not being used.

HYBRIDIZATION IN BH3

Boron in ${\text{BH}}_{3}$ has to contribute three $s {p}^{2}$-hybridized atomic orbitals to bond because:

1. It is bonding to more than one hydrogen. That tells you that hybridization can occur.
2. It is bonding in a non-horizontal direction to at least one hydrogen. That tells you that hybridization should occur to orient all the orbitals correctly.
3. It is bonding identically to each hydrogen. That tells you that hybridization has occurred to make the orbitals compatible.

It is $s {p}^{2}$ because three electron groups are surrounding boron: three bonding to one hydrogen each, and one empty ${p}_{z}$ orbital from boron that isn't compatible with hydrogen's $1 s$ atomic orbital. It isn't using that one to bond at the moment.

Boron had to hybridize one $2 s$ and two $2 p$ orbitals together to generate three diagonal-oriented orbitals in two dimensions ($x , y$).

HYBRIDIZATION IN ACETYLENE

One chosen carbon in $\text{H"-"C"-="C"-"H}$ has to contribute two $s p$-hybridized atomic orbitals to bond because:

1. It is bonding to more than one atom. That tells you that hybridization could occur.
2. It is NOT bonding in a non-horizontal direction with any atoms. This doesn't tell you anything about hybridization.
3. It is NOT bonding identically to each surrounding atom (the other $\text{C}$ and a $\text{H}$). That tells you that hybridization had to occur to make the orbitals of $\text{C}$ and $\text{H}$ compatible. Naturally, $\text{C}$ is compatible with itself, so hybridization is necessary to bond with BOTH $\text{C}$ and $\text{H}$.

It is $s p$ because two electron groups are surrounding one chosen carbon: one bonding to one hydrogen and one bonding to the other carbon.

Carbon had to hybridize one $2 s$ and one $2 p$ orbital together to generate two horizontally-oriented $s p$ hybridized orbitals in one dimension to $\sigma$ bond to two different atoms.

Separately, the remaining two bonds to be made to the other carbon (one triple bond has one $\sigma$ and two $\pi$ bonds) are made using the ${p}_{x}$ and ${p}_{y}$ atomic orbitals of carbon.

So, with acetylene, carbon is using two $s p$ hybridized atomic orbitals and one $2 {p}_{x}$ and one $2 {p}_{y}$ atomic orbital to bond.