Hybridization is a section of bonding in general chemistry. In examinations, you may have some questions such as identifying hybridization of atoms and determine which hybridized orbitals are attached to make bonds. At the end of this tutorial, you will have the ability of determine the hybridization of atoms in a molecule in a short time period.

**Content**

**Number of sigma bonds and lone pairs around an atom in a molecule**- Count number of sigma bonds and number of lone pairs
- Get the summation of sigma bonds and number of lone pairs around the atom

**General equation to find hybridization of atoms**- When summation of number of sigma bonds and number of lone pairs around an atom equals to 1
- When summation of number of sigma bonds and number of lone pairs around an atom equals to the 2 or 3 or 4

**Examples for finding hybridization of atoms**- Examples of sp hybridization
- Examples of sp
^{2}hybridization - Examples of sp
^{3}hybridization - Examples of sp
^{3}d hybridization

To determine the hybridization of atom, we **only consider number of sigma bonds and number of lone pairs** around that
atom. We are **not consider pi bonds and unpaired electrons**. Steps are explained below.

In a covalent bond, an atom has sigma bonds and lone pairs. Sometimes, lone pairs does not exist on atoms such as sulfur atom in sulfur trioxide molecule.

When you add sigma bonds and lone pairs, most occasions you will get values like 1, 2, 3, 4, 5, 6. Lets take water molecule as an example.

In water molecule, there are two sigma bonds and two lone pairs around oxygen atom. Therefore, summation of number of sigma bonds and number of lone pairs around oxygen atom is 4.

According to the **summation of number of sigma bonds and number of lone pairs around an atom**, hybridization can be decided as below.

Summation of number of sigma bonds and number of lone pairs | Hybridization | Hybridized orbitals |
---|---|---|

1 | s | Not hybridized |

2 | sp | One s orbital and only one p orbital have hybridized |

3 | sp^{2} |
One s orbital and two p orbitals have hybridized |

4 | sp^{3} |
One s orbital and three p orbitals have hybridized |

5 | sp^{3}d |
One s orbital, three p orbitals and one d orbital have hybridized |

We can introduce a general mathematical equation to find hybridization of atoms. Generally, we can represent the
**hybridization as s ^{x}p^{y}d^{z}**

**Summation of number of sigma bonds and number of lone pairs around an atom = x + y + z**

Because , there is only one s orbital in a period, always x = 1. Otherwise, we can say, for hybridization only one s orbital is contributed.

Summation of number of sigma bonds and number of lone pairs around an atom equals to the 2 or 3 or 4, only s and p orbitals have contributed for hybridization. Then our equation is simplified as below.

- Summation of number of sigma bonds and number of lone pairs around an atom = x + y
- Summation of number of sigma bonds and number of lone pairs around an atom = 1 + y (x=1)

Therefore y value can be found by subtracting 1 from summation of number of sigma bonds and number of lone pairs.

**y = summation of number of sigma bonds and number of lone pairs - 1**

You can study following examples to understand how hybridization is determined from the summation of sigma bonds and lone pairs around a specific atom.

sp hybridization is given when one s orbital and one p orbital are hybridized. Following examples are illustrated to understand the hybridization of sp.

- As the first step, draw the lewis structure of Beryllium chloride (BeCl
_{2}). - Because, we want to know the hybridization of Beryllium atom, calculate the summation of sigma bonds and lone pairs around the Beryllium atom.
- There are two sigma bonds and no lone pairs around the Beryllium atom. Therefore summation of sigma bonds and lone pairs around the Beryllium atom is two.
- Then use the algebraic equation as following. Because summation of sigma bonds and lone pairs
around the Beryllium atom is 3 or under 3,
**x+y = summation of sigma bonds and lone pairs**is used.- x+ y = 2
- 1 + y = 2
- y = 1

sp^{2} hybridization is given when one s orbital and two p orbitals are hybridized. Following examples are
illustrated to understand hybridization of sp^{2}.

There are two sigma bonds and a lone pair around the sulfur atom in sulfur dioxide molecule. For oxygen atom, there are one sigma bond and two lone pairs.

- In sulfur dioxide
(SO
_{2}) molecule, summation of number of sigma bonds and number of lone pairs around sulfur atom = 2 + 1 = 3 - Therefore, y = 3 - 1
- y = 2
**Hybridization = sp**^{2}

- For H
_{2}O molecule, summation of number of sigma bonds and number of lone pairs around oxygen atom = 2 +2 = 4 - Therefore, y = 4 - 1
- y = 3
**Hybridization of oxygen atom= sp**^{3}

There are four sigma bonds around the carbon atom in CH_{4}. But, no lone pair exist on carbon atom.

- For CH
_{4}molecule, summation of number of sigma bonds and number of lone pairs = 4 +0 = 4 - Therefore, y = 4 - 1
- y = 3
**Hybridization of carbon atom = sp**^{3}

There are only one sigma bond around the carbon atom in HCl. But, there are three lone pairs exist on chlorine atom.

- For HCl molecule, summation of number of sigma bonds and number of lone pairs = 1 + 3 = 4
- Therefore, y = 4 - 1
- y = 3
**Hybridization of chlorine atom in HCl = sp**^{3}

In sp^{3}d hybridized orbitals, s orbital, three p orbitals and one d orbital are hybridized.
When atom has sp^{3}d hybridization, summation of number of sigma bonds and number of lone pairs
around that atom should be 5.

Because all p orbitals are hybridized, y = 3. Therefore we can find the z from our equation.

- Summation of number of sigma bonds and number of lone pairs around an atom = x + y + z = 5
- 1 + 3 + z = 5
- z = 1

There are only five sigma bonds around the phosphorus atom in PCl_{5} (No lone pairs on
phosphorus atom).

Questions

- Nitric acid (HNO
_{3}) - Nitrous acid (HNO
_{2}) - Nitrogen trichloride (NCl
_{3}) - Hydrogen cynide (HCN)

In nitrogen trichloride, there are one lone pair and three N-Cl bonds around nitrogen bonds. Therefore, summation of number of lone pairs and sigma bonds around is four. Therefore, hybridization of nitrogen atom should be sp^{3}.

Answer is 3

CH_{4} + O_{2} → CO_{2} + H_{2}O

- Oxygen atom in O
_{2}molecule - Oxygen atom in H
_{2}O molecule - Carbon atom in CH
_{4}and Oxygen atom in H_{2}O molecule - Carbon atom in CO
_{2}and Oxygen atom in H_{2}O molecule

Answer is 3

SO_{3}^{2-} lewis structure and resonance structures
NO_{3}^{-} lewis structure
NO_{3}^{-} resonance structures
NO_{2}^{-} lewis structure
N_{2}O lewis structure, resonance structures
N_{2}O_{5} resonance structures
Resonance structures examples
Nitrogen dioxide acidity