Q. how to calculate hybridization

Hybridization describes how atomic orbitals mix to form new, equivalent hybrid orbitals that explain the geometry of bonds around a central atom. The procedure to calculate hybridization is a sequence of clear steps. I will explain each step in detail and then apply the procedure to several representative examples.

Step 1. Draw the correct Lewis structure for the molecule or ion. Determine a plausible arrangement of atoms. Count the total number of valence electrons available. Form bonds to satisfy connectivity. Place remaining electrons as lone pairs to satisfy the octet rule when possible. If several Lewis structures are possible, choose the one with the lowest formal charges and that satisfies known experimental facts.

Step 2. Identify the central atom for which you want the hybridization. On that central atom, count the number of electron domains, also called regions of electron density. An electron domain is any of the following on the central atom bonded atoms connected by a single, double, or triple bond lone pairs of electrons For multiple bonds, count the entire double or triple bond as one electron domain because it occupies one region of space for sigma bonding.

Step 3. Compute the steric number. The steric number is the count of electron domains. You can think of it as

\( \text{steric number} = \text{number of sigma bonds (bonding domains)} + \text{number of lone pairs} \).

Step 4. Convert the steric number to hybridization using the standard mapping

\( \text{steric number} = 2 \Rightarrow \text{hybridization } = \mathrm{sp} \).

\( \text{steric number} = 3 \Rightarrow \text{hybridization } = \mathrm{sp^2} \).

\( \text{steric number} = 4 \Rightarrow \text{hybridization } = \mathrm{sp^3} \).

\( \text{steric number} = 5 \Rightarrow \text{hybridization } = \mathrm{sp^3d} \).

\( \text{steric number} = 6 \Rightarrow \text{hybridization } = \mathrm{sp^3d^2} \).

Step 5. Consider special cases. For elements in period 3 or lower that can expand their octet, hybridizations such as \(\mathrm{sp^3d}\) and \(\mathrm{sp^3d^2}\) are common. For molecules with extensive resonance or delocalization, hybridization can be described as fractional or the bonding described better by molecular orbitals. For radicals count the unpaired electron as an electron domain. Always use the best Lewis structure when counting domains.

Worked examples. I show each step explicitly for several common molecules. In each example I name the central atom count bonded atoms and lone pairs compute the steric number and state the hybridization.

Example 1. Methane \(\mathrm{CH_4}\). Central atom carbon. Carbon forms four single bonds to hydrogen. There are zero lone pairs on the carbon. Number of electron domains is four. Steric number is

\( 4 = 4 \text{ bonded atoms} + 0 \text{ lone pairs} \Rightarrow \mathrm{sp^3} \text{ hybridization} \).

Example 2. Carbon dioxide \(\mathrm{CO_2}\). Central atom carbon. Carbon forms two double bonds to oxygen. Each double bond counts as one electron domain. There are zero lone pairs on the carbon in the best Lewis structure. Number of electron domains is two. Steric number is

\( 2 = 2 \text{ bonded atoms (two double bonds counted each as one domain)} + 0 \text{ lone pairs} \Rightarrow \mathrm{sp} \text{ hybridization} \).

Example 3. Boron trifluoride \(\mathrm{BF_3}\). Central atom boron. Boron forms three single bonds to fluorine. There are zero lone pairs on boron in the standard Lewis structure. Number of electron domains is three. Steric number is

\( 3 = 3 \text{ bonded atoms} + 0 \text{ lone pairs} \Rightarrow \mathrm{sp^2} \text{ hybridization} \).

Example 4. Ammonia \(\mathrm{NH_3}\). Central atom nitrogen. Nitrogen forms three single bonds to hydrogen and has one lone pair. Number of electron domains is four. Steric number is

\( 4 = 3 \text{ bonded atoms} + 1 \text{ lone pair} \Rightarrow \mathrm{sp^3} \text{ hybridization} \).

Example 5. Water \(\mathrm{H_2O}\). Central atom oxygen. Oxygen forms two single bonds to hydrogen and has two lone pairs. Number of electron domains is four. Steric number is

\( 4 = 2 \text{ bonded atoms} + 2 \text{ lone pairs} \Rightarrow \mathrm{sp^3} \text{ hybridization} \).

Example 6. Acetylene \(\mathrm{C_2H_2}\). Consider one carbon atom. That carbon is bonded to one hydrogen and has a triple bond to the other carbon. The triple bond counts as one electron domain for the central carbon. There are no lone pairs on that carbon. Number of electron domains is two. Steric number is

\( 2 = 1 \text{ bonded hydrogen} + 1 \text{ triple bond to carbon counted as one domain} + 0 \text{ lone pairs} \Rightarrow \mathrm{sp} \text{ hybridization} \).

Example 7. Sulfur hexafluoride \(\mathrm{SF_6}\). Central atom sulfur. Sulfur forms six single bonds to fluorine and in this case expands the octet. There are zero lone pairs on sulfur in the standard Lewis structure. Number of electron domains is six. Steric number is

\( 6 = 6 \text{ bonded atoms} + 0 \text{ lone pairs} \Rightarrow \mathrm{sp^3d^2} \text{ hybridization} \).

Notes on multiple bonds and resonance. For counting electron domains treat a single double bond or triple bond as one domain because hybridization is determined by the sigma bonding framework. Pi bonds occupy unhybridized p orbitals in most simple cases. In resonance stabilized systems, such as benzene, each carbon has three sigma bonds and one delocalized pi electron system. That gives a steric number of three and therefore \(\mathrm{sp^2}\) hybridization for each ring carbon.

Summary quick checklist. 1 Draw the Lewis structure and choose the correct central atom. 2 Count electron domains on the central atom bonded atoms count once each lone pairs count once count each multiple bond as one domain. 3 Map the steric number to hybridization 2 to \(\mathrm{sp}\), 3 to \(\mathrm{sp^2}\), 4 to \(\mathrm{sp^3}\), 5 to \(\mathrm{sp^3d}\), 6 to \(\mathrm{sp^3d^2}\). 4 Watch for special cases such as expanded octets, radicals, and delocalization.