Hybridization and concept of sigma and pi bond

Hybridization

The structures of different molecules can be explained on the basis of hybridization. For e.g., in case of carbon, the ground state electronic

To explain the tetravalency of carbon, it was proposed that one of the electrons from 2s filled orbital is promoted to the 2p empty orbital (2p_z), which is in a higher energy state. Thus, four half-filled orbitals form in the valence shell this accounts for the bonding capacity of four carbon atoms. This state is known as excited state and the configuration of carbon in the excited state is:

The above configuration reveals that all the four bonds formed by carbon will not be identical. For e.g., in the formation of CH₄ molecule, one C-H bond will be formed by the overlapping of 2s-orbital of C and 1s-orbital of H whereas the other three C-H bonds will be formed by the overlapping of 2p-orbitals of C and 1s-orbital of H. Therefore, all the bonds will not be equivalent.

But actually, in most of the carbon compounds, such as methane (CH₄), carbon tetrachloride (CCl₄) etc., all the four bonds of carbon atom are equivalent. The equivalent character of the bonds can be explained with the help of hybridisation.

Hybridisation may be defined as the phenomenon of intermixing of the orbitals of slightly different energies so as to redistribute their energies and to give new set of orbitals of equivalent energy and shape. The new orbitals formed as a result of hybridization are called hybrid or hybridized orbitals. Thus, to form four equivalent bonds, one 2s and three 2p-orbitals of carbon hybridize and form four new orbitals.such orbitals are called sp³ hybrid orbitals.

The important characteristics of hybridisation are listed below:

(i) The number of hybridized orbitals formed is equal to the number of orbitals that get hybridized.

(ii) The hybridized orbitals are always equivalent in energy and shape.

(iii) The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.

(iv) The hybrid orbitals are directed in space in some preferred directions to have stable arrangements.

Therefore, the type of hybridization gives the geometry of the molecule. Depending upon the different combinations of s- and three p-orbitals, three types of hybridizations are known.

sp hybridization

This involves the mixing of one s- and one p-orbital forming two sp-hybrid orbitals. The two sp-hybrid orbitals are oriented in a linear arrangement and bond angle is 180°. For e.g., BeF₂ involves sp-hybridization and is, therefore, linear.2-hybridization">

sp² hybridization

In this case, one s- and two p-orbitals hybridize to form three sp² hybrid orbitals. These three sp² hybrid orbitals are oriented in a trigonal planar arrangement. For e.g., in BH₃ boron atom undergoes sp²hybridization and therefore, BH₃ has trigonal planar geometry and HBH bond angle is 120^o.3-hybridization">

sp³ hybridization

In this case, one s- and three p-orbitals hybridize to form four sp³ hybrid orbitals. These four sp³-hybrid orbitals are oriented in a tetrahedral arrangement. The common example of molecule involving sp³-hybridisation is methane (CH₄). Therefore, CH₄ has tetrahedral geometry and HCH bond angle is 109.5^o.

Hybrid orbitals and molecular shapes involving s and p-orbitals

Sigma and pi bonds

Two types of covalent bonds are formed depending on the way the two atomic orbitals overlap with each other.s)">

Sigma Bond (s)

When the overlap of orbitals of two atoms takes place along the line joining the two nuclei (orbital axis) then the covalent bond formed is called sigma (s) bond. These bonds can be formed due to 's-s', 's-p' or 'p-p' overlap along the orbital axis. Free rotation around a sigma bond is always possible.

Fig: 6.4 - Formation of sigma bond due to various overlapping

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Pi bond (p)

When the two atoms overlap due to the sideways overlap of their 'p' orbitals, the covalent bond is called as pi(p) bond. In a pi bond the electron density is concentrated in the region perpendicular to the bond axis.

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Characteristics of the pi (p) bonds

• The pi bonds are weak because the orbtial overlap is partial.

• For a complete side ways overlap the 'p' orbitals should be parallel to each other. This is possible when all the atoms of the molecule are in the same plane i.e., there is no rotation of one part of the molecule relative to the other about the pi (p) bond.

Fig: 6.5 - Rotation about a double bond

• In molecules containing double bond, which has a pi bond, there is no free rotation and such molecules exist in isomeric forms of 'cis' and 'trans'. In the 'cis' form the similar atoms lie on the same side of a plane placed along the internuclear axis while in the 'trans' form the similar atoms lie on the opposite sides.

• The electrons in the pi (p) bond are placed above and below the plane of the bonding atoms and so they are more exposed. They are more susceptible to attack by electron seeking or oxidizing agents. Hence, they are the most reactive centers in unsaturated (multiple bonded) compounds.

Comparative properties of sigma and pi bonds

Sigma (σ) bond	Pi (π) bond
Formed due to the axial overlap of two orbitals (‘s-s’, ‘s-p’or’p-p’).	Formed by the lateral (sideways) overlap of two ‘p’ orbitals.
Only one sigma bond exists between two atoms.	There can be more than one pi bonds between the two atoms.
The electron density is maximum and cylindrically symmetrical about the bond axis.	The electron density is high along the direction at right angles to the bond axis.
Free rotation about the sigma bond is possible.	Free rotation about the pi bond is not possible.
This bond can be independently formed, i.e., without the formation of a pi bond.	The pi bond is formed after the sigma bond has been formed,
Sigma bond is relatively strong.	Pi bond is a weak bond.

Problems

11. Explain how the valence bond theory accounts for the existence of cis- trans isomers.

Solution

A double bond between carbon-carbon atoms (C=C) consists of a

sp²-sp² s bond and a p bond. These bonds are formed between unhybridized '2p' orbitals of each C atom at right angle to the plane of sigma bond. Since the p bond prevents the free rotation about the carbon-carbon axis, a di-substituted alkene can exist in two forms: cis-isomer and trans-isomer.

In cis-isomer, both the like groups are on the same side of the plane whereas in trans-isomer these groups are on the opposite side of the plane.

For example 1, 2-dichloroethene.

In order to transform one isomer into the other, one end of the molecule must be rotated as the other remains fixed. The bond must be broken in order to do this. Breaking the p bond requires considerable energy, so the cis and trans compounds are not easily inter converted.

12. How many and bonds are present in naphthalene?

Solution

Naphthalene is

pbonds = 19, p bonds = 5