What is bond order?
Answer
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Hint: Concept of bond order and its calculation is explained using Molecular Orbital Theory ( MOT). Bond order is the number of electron pairs involved in bonding between a pair of atoms. It is also defined as the difference between the number of bonds and antibonds.
Complete step by step answer:
Molecular Orbital Theory gives a formula to calculate bond order. According to Molecular Orbital Theory, the bond order of a molecule is half the difference between number of electrons in bonding molecular orbital and number of electrons in antibonding molecular orbital. It can be written as,
Bond Order $ = \dfrac{{{N_b} - {N_a}}}{2}$
Where ${N_b}$ is the number of electrons in bonding molecular orbital and ${N_a}$ is the number electrons in antibonding molecular orbital.
So in order to find the bond order first we need to write the electronic configuration of the element. Let us calculate the bond order of nitrogen (${N_2}$) molecules. Total number of electrons in ${N_2}$ is $14$.
Electronic configuration of ${N_2}$ is,
\[\left( {\sigma 1{s^2}} \right)\left( {\sigma *1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {\sigma *2{s^2}} \right)\left( {\pi 2{p^2}_x,\pi 2{p_y}^2} \right)\left( {\sigma 2{p_z}^2} \right)\]
Number of electrons in bonding molecular orbital $ = 10$
Number electrons in antibonding molecular orbital $ = 4$
Bond order $ = \dfrac{{10 - 4}}{2} = \dfrac{6}{2} = 3$
Hence the bond order of nitrogen molecules is three. This means that the bond between the nitrogen atoms in the nitrogen molecule is a triple bond. i.e. nitrogen molecules exist as $N \equiv N$.
In the same way we can calculate the bond order of other molecules like ${C_2},{O_2},{F_2}$ etc. Bond order of ${C_2},{O_2},{F_2}$ are $2,2$ and $1$ respectively. We can also calculate the bond order of ions using the same formula.
Higher the value of bond order, stronger will be bond.
Note:
Bond order can be an integer or fraction. In Molecular Orbital Theory, bond order is used to predict the existence and stability of molecules. But this concept is not acceptable for every molecule.
Complete step by step answer:
Molecular Orbital Theory gives a formula to calculate bond order. According to Molecular Orbital Theory, the bond order of a molecule is half the difference between number of electrons in bonding molecular orbital and number of electrons in antibonding molecular orbital. It can be written as,
Bond Order $ = \dfrac{{{N_b} - {N_a}}}{2}$
Where ${N_b}$ is the number of electrons in bonding molecular orbital and ${N_a}$ is the number electrons in antibonding molecular orbital.
So in order to find the bond order first we need to write the electronic configuration of the element. Let us calculate the bond order of nitrogen (${N_2}$) molecules. Total number of electrons in ${N_2}$ is $14$.
Electronic configuration of ${N_2}$ is,
\[\left( {\sigma 1{s^2}} \right)\left( {\sigma *1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {\sigma *2{s^2}} \right)\left( {\pi 2{p^2}_x,\pi 2{p_y}^2} \right)\left( {\sigma 2{p_z}^2} \right)\]
Number of electrons in bonding molecular orbital $ = 10$
Number electrons in antibonding molecular orbital $ = 4$
Bond order $ = \dfrac{{10 - 4}}{2} = \dfrac{6}{2} = 3$
Hence the bond order of nitrogen molecules is three. This means that the bond between the nitrogen atoms in the nitrogen molecule is a triple bond. i.e. nitrogen molecules exist as $N \equiv N$.
In the same way we can calculate the bond order of other molecules like ${C_2},{O_2},{F_2}$ etc. Bond order of ${C_2},{O_2},{F_2}$ are $2,2$ and $1$ respectively. We can also calculate the bond order of ions using the same formula.
Higher the value of bond order, stronger will be bond.
Note:
Bond order can be an integer or fraction. In Molecular Orbital Theory, bond order is used to predict the existence and stability of molecules. But this concept is not acceptable for every molecule.
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