Easy Method to Determine Hybridization of Atoms in Molecules, Examples

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

  1. 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
  2. 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
  3. Examples for finding hybridization of atoms
    • Examples of sp hybridization
    • Examples of sp2 hybridization
    • Examples of sp3 hybridization
    • Examples of sp3d hybridization



Easy Method to Determine Hybridization of Atoms in Molecules, Examples

Number of sigma bonds and lone pairs around an atom in a molecule

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.


Count number of sigma bonds and number of lone pairs

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.



Get the summation of sigma bonds and number of lone pairs around the atom

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.

sigma bonds and lone pairs in H2O.jpg

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 sp2 One s orbital and two p orbitals have hybridized
4 sp3 One s orbital and three p orbitals have hybridized
5 sp3d One s orbital, three p orbitals and one d orbital have hybridized


General equation to find hybridization of atoms

We can introduce a general mathematical equation to find hybridization of atoms. Generally, we can represent the hybridization as sxpydz


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



When summation of number of sigma bonds and number of lone pairs around an atom equals to 1

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.



When summation of number of sigma bonds and number of lone pairs around an atom equals to the 2 or 3 or 4

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



Examples for finding hybridization of atoms

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



Examples of sp hybridization

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


Hybridization of Beryllium chloride

  • As the first step, draw the lewis structure of Beryllium chloride (BeCl2).
  • 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


Examples of sp2 hybridization

sp2 hybridization is given when one s orbital and two p orbitals are hybridized. Following examples are illustrated to understand hybridization of sp2.


Hybridization of sulfur dioxide

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 (SO2) 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 = sp2
hybridization of sulfur atom in SO2

Examples for sp3 hybridization


Hybridization of water molecule

  • For H2O 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= sp3
H2O hybridization of oxygen atom

Hybridization of CH4

There are four sigma bonds around the carbon atom in CH4. But, no lone pair exist on carbon atom.

  • For CH4 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 = sp3
hybridization of CH4

Hybridization of chlorine atom of HCl molecule

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 = sp3
HCl hybridization in Cl atom

Examples of sp3d hybridization

In sp3d hybridized orbitals, s orbital, three p orbitals and one d orbital are hybridized. When atom has sp3d 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


Hybridization of phosphorus pentachloride (PCl5)

There are only five sigma bonds around the phosphorus atom in PCl5 (No lone pairs on phosphorus atom).

hybridization of PCl5



Questions





Select which compound contains sp3 hybridized nitrogen atom?

  1. Nitric acid (HNO3)
  2. Nitrous acid (HNO2)
  3. Nitrogen trichloride (NCl3)
  4. 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 sp3.

Answer is 3



Combustion reaction of methane and oxygen is given below. From reactants and products, in which atom, is there a sp3 hybridized atom(s)?

CH4 + O2 → CO2 + H2O

  1. Oxygen atom in O2 molecule
  2. Oxygen atom in H2O molecule
  3. Carbon atom in CH4 and Oxygen atom in H2O molecule
  4. Carbon atom in CO2 and Oxygen atom in H2O molecule

Answer is 3






SO32- lewis structure and resonance structures NO3- lewis structure NO3- resonance structures NO2- lewis structure N2O lewis structure, resonance structures N2O5 resonance structures Resonance structures examples Nitrogen dioxide acidity