Pi Bonds: Definition, Formula, Examples, Questions

Kinds of Bonding: P$\pi$-P$\pi$ and P$\pi$-D$\pi$

The major types of $\pi$ bonding include $\pi$ (p-p) and $\pi$ (p-d) bonding.

(p$\pi$-p$\pi$) Bonding

$\pi$(p-p) bonding occurs between the two 'p' orbitals of the adjacent atoms. This is so in ethylene, C₂H₄.

The carbon atoms in this molecule are sp² hybridized with one unhybridized p orbital on each carbon atom. These p orbitals overlap laterally and form the pi bond—which is above and below the plane of the molecule—to a large degree. This kind of bonding is frequent in alkenes and generally accounts for their typical reactivity, for instance, in electrophilic addition.

(p$\pi$-d$\pi$) Bonding

The other is $\pi$ (p-d) bonding, where p orbitals are mixed with d orbitals, which is usually observed in transition metals. This kind of bonding is quite important for the creation of metal-metal bonds in compounds like metal carbonyls and metal complexes. $\pi$ (p-d) bonds can exert a strong influence on the electronic properties of these complexes, which means their stability and reactivity.

The two kinds of $\pi$ bonding serve to illustrate just how versatile these bonds can be in different chemical settings and, therefore, their importance both in organic chemistry and inorganic.

p$\pi$-p$\pi$ Bonding

When one p-orbital of one atom contains one electron and the other atom has one electron in the p orbital, both perpendicular to the plane of the molecule, the type of bond formed is known as pπ−pπ bonding.

When one p-orbital of one atom contains one electron and the other atom has one electron, the type of bond formed is known as pπ−pπ bonding. In other words, pπ−dπ bond is formed when the p orbital of one atom and the d orbital of another atom overlap laterally.

For example, PO₄^3- has a pπ−dπ bonding because the lateral overlap of the p-orbital of an oxygen atom with the d-orbital of a sulfur atom occurs.

NCERT Chemistry Notes:

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$\pi$ Bonds: Real-Life Applications

$\pi$ bonds have several applications that stretch beyond theoretical chemistry into real life.

$\pi$ bonds are central in organic chemistry to not only the structure but also the reactivity of unsaturated hydrocarbons. Take alkenes or alkynes as examples; these are unsaturated hydrocarbons with at least one or more $\pi$ bonds. They form very essential intermediates for many synthetic reactions. The $\pi$ bond can undergo electrophilic addition that empowers the chemist to take simple molecules and then transform them into complex compounds. Hence, they become very important in the synthesis of drugs or even materials science.

Materials Science

$\pi$ bonds are important in determining the properties of various polymers. For instance, under conjugated systems where there are alternating single and double bonds, special electronic properties realize that are utilized in organic electronics and photovoltaic devices. In such materials, pi bonds allow the delocalization of electrons, hence enhancing conductivity and light absorption.

Biochemistry

The functions associated with π bonds within the structure of biomolecules are of immense importance in biochemistry. For instance, double bonds in fatty acids and the aromatic rings of nucleotides are of fundamental importance for the functions associated with the respective lipids and nucleic acids. Thus, the physical properties of such molecules result from the presence of π bonds that modulate their interactions and biological activities.

Related Topics:

• Hybridisation
• Vsepr Theory

Some Solved Examples

Example 1:

What is the number and type of bonds in C₂^2- in the CaC₂?

1) one $\sigma$ and 2 $\pi$

2) two $\sigma$ and one $\pi$

3) one $\sigma$ and one $\pi$

4) 2 $\sigma$ and 2 $\pi$

Solution: The structure of C₂^2- is

$
[\ddot{c} \equiv \ddot{c}]^{2-}
$

In this ion, there is one sigma bond and two pi bonds. Therefore, correct option is 1.

Example 2:

Which one of the following does not have a pyramidal shape?

1) (SiH₃)₃ N

2) (CH₃)₃N

3) (SiH₃)₃

4) (CH₃)₃

Solution: The molecule (SiH₃)₃ N has a trigonal planar shape due to sp² hybridization, while(CH₃)₃N does have a pyramidal shape.

Thus, the correct answer is option (1).

Example 3:

How strong is the p$\pi$-p$\pi$ bond compared to p$\pi$-d$\pi$ and d$\pi$-d$\pi$ bonds?

1) More stronger than p$\pi$ -d$\pi$ and d$\pi$-d$\pi$ bond

2) Less strong than p$\pi$-d$\pi$ and d$\pi$-d$\pi$ bond

3) More stronger than p$\pi$-p$\pi$ and d$\pi$-d$\pi$ bond

4) none of the above

Solution: The p$\pi$-p$\pi$ bond is stronger than both p$\pi$-d$\pi$ and d$\pi$-d$\pi$ bonds.

Therefore, the correct answer is option (1).

Example 4:
How many p$\pi$-p$\pi$ and p$\pi$-d$\pi$ bonds are present in SO₃?

1) 2, 2

2) 2, 1

3) 1, 2

4) 1, 0

Solution: In the structure of SO₃, there is 1 p$\pi$-p$\pi$ bond and 2 p$\pi$-d$\pi$ bonds.

Thus, the correct answer is option (3).

Example 5:
What is the number of p$\pi$-p$\pi$ and p$\pi$-d$\pi$ bonds respectively in P₄O₁₀?

1) 2,1

2) 1,2

3) 1, 4

4) 0, 4

Solution: In P₄O₁₀, all four π bonds (P=O) are p$\pi$-d$\pi$ bonds, as phosphorus is sp³ hybridized.

Therefore, the correct answer is option (4).

Example 6:

In the given structure, number of sp and $\mathrm{sp}^2$ hybridized carbon atoms present respectively are :

[JEE Main 2025]

1) 3 and 6

2) 3 and 5

3) 4 and 6

4) 4 and 5

Solution:

Number of sp and $\mathrm{sp}^2$ hybridised carbon atom are 3 and 5.

Hence, the correct answer is option (2).

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Summary

The paper treated $\pi$ bonds, focusing on their definitions, types, and applications in real life. $\pi$ bonds form as a result of a lateral overlap of p orbitals. Therefore, it is of prime importance in the explanation of molecular structure and reactivity. There were two notable types of bonding: p$\pi$-p$\pi$, found in alkenes and alkynes, and p$\pi$-d$\pi$, bonding seen in transition metal complexes.

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