Bond Order Formula: Definition, Calculation, Solved Examples
  • Written By Pavithra VG
  • Last Modified 23-06-2022
  • Written By Pavithra VG
  • Last Modified 23-06-2022

Bond Order Formula: Learn Formula, Significance, Examples

Hydrogen \(\left( {{{\rm{H}}_2}} \right)\), oxygen \(\left( {{{\rm{O}}_2}} \right)\), and nitrogen \(\left( {{{\rm{N}}_2}} \right)\) are diatomic molecules. How many bonds are present between each molecule? Is the number of bonds in all three molecules the same? What determines the total number of bonds? All of these questions are answered in the article “Bond order formula.” In this post, you’ll learn how to calculate bond order using molecular orbital theory, Lewis structure, and a variety of formulas and examples.

The bond order describes the bond’s stability. The molecular orbital makes the concept of a chemical bond’s bond order simple to comprehend. It measures the strength of atom-to-atom covalent connections. Bond order is significant in molecular orbital theory for determining bond strength and is also used in valence bond theory. The number of electrons participating in bonding between two atoms during the formation of a molecule is known as bond order. It is used as a measure of a bond’s stability. The concept of bond order and the bond order formula will be discussed in this article.

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Molecular Orbital Theory

Molecular orbitals are formed by the combination of atomic orbitals of the bonded atoms. For example, the formation of the hydrogen molecule. A hydrogen molecule is formed by the combination of atomic orbitals of hydrogen \({\rm{A}}\) and \({\rm{B}}\) Each hydrogen atom has one electron in \({\rm{ 1s }}\) orbital. These atomic orbitals may represent by the wave functions \({\Psi _{\rm{A}}}\) and \({\Psi _{\rm{B}}}\).

Molecular Orbital Theory

The molecular orbital \({\rm{\sigma }}\) formed by the addition of atomic orbitals is called the bonding molecular orbital, and \({{\rm{\sigma }}^ * }\) by subtraction of atomic orbitals is called an antibonding molecular orbital.

The antibonding orbital energy is raised above the parent atomic orbitals energy that has combined. As a result, the energy of the bonding orbital is lower than the parent orbitals. However, the two molecular orbitals’ total energy remains the same as that of the two original atomic orbitals.

What is a Bond Order?

The relative stability of a molecule can be determined on the basis of bond order. Bond order is defined as the number of covalent bonds between two atoms in a molecule. The formula mentioned to measure bond order is equal to one half of the difference between the number of electrons in the bonding and antibonding molecular orbitals.

\({\rm{Bond}}{\mkern 1mu} {\rm{order}} = \frac{{{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}}}{2} = \frac{1}{2}\left( {{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right)\)

Here \({{\rm{N}}_{\rm{b}}}\) is the number of electrons in the bonding molecular orbitals.

\({{\rm{N}}_{\rm{a}}}\) is the number of electrons in antibonding molecular orbitals.

The bond orders of \(1, 2\) or \(3\) correspond to single, double, or triple bonds and bond order may be fractional also.

The sequence of energy levels of molecular orbitals for \({\rm{L}}{{\rm{i}}_2},{\rm{B}}{{\rm{e}}_2},\;{{\rm{B}}_2},{{\rm{C}}_2}\) and \({{\rm{N}}_2}\) is

\({\sigma _{1\;{\rm{s}}}} < {\sigma _{1\;{\rm{s}}}}^* < {\sigma _{2\;{\rm{s}}}} < {\sigma _{2\;{\rm{s}}}}^* < \left( {{{\rm{\pi }}_{{\rm{2Px}}}} = {{\rm{\pi }}_{{\rm{2Py}}}}} \right) < {\sigma _{{\rm{2Pz}}}} < \left( {{{\rm{\pi }}^ * }_{2{\rm{Px}}} = {{\rm{\pi }}^ * }_{2{\rm{Py}}}} \right) < {\sigma _{2{\rm{Pz}}}}^ * \)

The sequence of increasing order of energy of various molecular orbitals for \({{\rm{O}}_2},\;{{\rm{F}}_2}\) is

\({\sigma _{1s}} < {\sigma _{1s}}^* < {\sigma _{2\;{\rm{s}}}} < {\sigma _{2\;{\rm{s}}}}^* < {\sigma _{2{\rm{Pz}}}} < \left( {{{\rm{\pi }}_{2{\rm{Px}}}} = {{\rm{\pi }}_{2{\rm{Py}}}}} \right) < \left( {{{\rm{\pi }}^ * }_{2{\rm{Px}}} = {{\rm{\pi }}^ * }_{2{\rm{Py}}}} \right) < {\sigma _{2{{\rm{P}}_{\rm{y}}}}}^*\)

Significance of Bond Order

The formula of bond order conveys very important information about the molecules and ions. These are briefly discussed.

a) Stability of Molecules or Ions: The stability of the molecules and their ions can be predicted in terms of bond order. If bond order is positive \(\left( {{{\rm{N}}_{\rm{b}}} > {{\rm{N}}_{\rm{a}}}} \right),\) the molecule or ion will be stable. In case it is zero \(\left( {{{\rm{N}}_{\rm{b}}} = {{\rm{N}}_{\rm{a}}}} \right)\) or negative \(\left( {{{\rm{N}}_{\rm{b}}} < {{\rm{N}}_{\rm{a}}}} \right),\) the molecule or ion will be unstable. A fractional value of the bond order suggests that the molecule is unstable. However, it still exists.

b) Bond Dissociation Enthalpy: Bond dissociation enthalpy is directly proportional to the bond order. Thus, the more the value of bond order, the greater will be the bond dissociation enthalpy and vice versa. This is further supported by the following data.

MoleculeBond orderBond dissociation enthalpy \(({\rm{kJ}}/{\rm{mol}})\)
Nitrogen \(({\rm{N}} \equiv {\rm{N}})\)\(3\)\(945\)
Oxygen \(({\rm{O}} = {\rm{O}})\)\(2\)\(498\)
Fluorine \({\rm{ (F – F) }}\)\(1\)\(158\)

c) Bond length: Bond order is inversely proportional to bond length. This means that the greater the bond order, the smaller the bond length.

MoleculeBond orderBond length (pm)
Nitrogen \(({\rm{N}} \equiv {\rm{N}})\)\(3\)\(110\)
Oxygen \(({\rm{O}} = {\rm{O}})\)\(2\)\(121\)
Fluorine \({\rm{ (F – F) }}\)\(1\)\(142\)

d) Number of bonds: The value of bond order also predicts the number of bonds in the molecule. Bond orders \(1, 2, 3\) signifies the presence of single, double, triple bonds, respectively, in a particular molecule. Chemical bonds are always integral, but bond order can be fractional also.

Example for Calculation of Bond Order

The examples of Bond Order calculations are explained below:

Beryllium Molecule \(\left( {{\rm{B}}{{\rm{e}}_2}} \right)\)

The electronic configuration of beryllium is \({\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{,2}}{{\rm{s}}^{\rm{2}}}\). The total number of electrons in \({\rm{B}}{{\rm{e}}_2}\) should be \(8\), four from each beryllium atom. Keeping in view the Aufbau principle and Pauli exclusion principle, the electronic configuration of \({\rm{B}}{{\rm{e}}_2}\) will be \({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}\) or in terms of energy level diagram, it is represented as below. 

Beryllium Molecule

Here, \({{\rm{N}}_{\rm{b}}} = 4,\;{{\rm{N}}_{\rm{a}}} = 4\) So that

Bond order \( = \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {4 – 4} \right) = 0\)

i.e., no bond is formed between two beryllium atoms or in other words, in \({\rm{B}}{{\rm{e}}_2}\) does not exist.

Hydrogen Molecular Positive Ion (H2+)

This is the simplest molecular species containing one electron only. Its existence had been detected spectroscopically when an electric discharge passed through a discharge tube containing hydrogen gas at low pressure. 

The only electron present in \({\rm{H}}_2^ + \) ion enters the \(\sigma (1\;{\rm{s}})\) molecular orbital, which has the lowest energy. Thus, the electronic configuration of \({\rm{H}}_2^ + \) is \(\sigma {(1\;{\rm{s}})^1}\).

In terms of energy level diagram, it represents as

Hydrogen Molecular Positive Ion (H2+)

Here, \({{\rm{N}}_{\rm{b}}} = 1,\;{{\rm{N}}_{\rm{a}}} = 0\) so that

Bond order \( = \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {1 – 0} \right) = \frac{1}{2}\)

Bond order is \(\frac{1}{2}\) and is positive. Hence \({\rm{H}}_2^ + \) ions should be stable. However, the less value of bond order indicates that the stability is not very high. 

Bond Order of Oxygen Molecule (O2)

Electronic configuration of the O-atom \(({\rm{Z}} = 8)\) is \(1\;{{\rm{s}}^2},2\;{{\rm{s}}^2},2{\rm{p}}{\rm{x}}^2,2{{\rm{p}}{\rm{y}}}^12{{\rm{p}}_{\rm{z}}}^1\). Total number of electrons in the \({{\rm{O}}_2}\) molecule is \(16\), eight from each oxygen atom. 16 electrons are filled in the various molecular orbitals as follows:

\({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{*}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ <  }}{{\rm{\sigma }}^ * }{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2p}}_{\rm{z}}^{\rm{2}}{\rm{ < }}\left( {{\rm{\pi 2p}}_{\rm{x}}^{\rm{2}}{\rm{ = }}\,{\rm{\pi 2p}}_{\rm{y}}^{\rm{2}}} \right){\rm{ <  }}\left( {{{\rm{\pi }}^ * }{\rm{2p}}_{\rm{x}}^{\rm{1}}{\rm{ =  }}{{\rm{\pi }}^ * }{\rm{2p}}_{\rm{y}}^{\rm{1}}} \right)\)

Here, \({{\rm{N}}_{\rm{b}}} = 10,\;{{\rm{N}}_{\rm{a}}} = 6\) so that

Bond order \(= \frac{1}{2}\left( {{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {10 – 6} \right) = 2\)

The bond order \(2\) justifies the presence of a double bond in the \({{\rm{O}}_2}\) molecule.

Bond Order of Superoxide Ion \(\left( {{\rm{O}}_2^ – } \right)\)

Superoxide ion is formed by the gain of one electron by an \({{\rm{O}}_2}\) molecule. This electron is added up in the \({\pi ^*}2{{\rm{p}}_{\rm{x}}}\) or \({\pi ^*}2{{\rm{p}}_{\rm{y}}}\) molecular orbital. Hence, the electronic configuration of \({\rm{O}}_2^ – \) is

\({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2p}}{\rm{z}}^{\rm{2}}{\rm{ < }}\left( {{\rm{\pi 2p}}{\rm{x}}^{\rm{2}}{\rm{ = \pi 2p}}{\rm{y}}^{\rm{2}}} \right){\rm{ < }}\left( {{{\rm{\pi }}^{\rm{}}}{\rm{2p}}{\rm{x}}^{\rm{2}}{\rm{ = }}{{\rm{\pi }}^{\rm{}}}{\rm{2}}{{\rm{p}}_{\rm{y}}}^{\rm{1}}} \right)\)

Here, \({{\rm{N}}_{\rm{b}}} = 10,\;{{\rm{N}}_{\rm{a}}} = 7\) so that

Bond order \(= \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {10 – 7} \right) = \frac{3}{2} = 1.5\)

Bond Order of Peroxide Ion \(\left( {{{\rm{O}}_2}^{2 – }} \right)\)

Peroxide ion \(\left( {{{\rm{O}}_2}^{2 – }} \right)\) is formed by the gain of two electrons by an \({{\rm{O}}_2}\) molecule. These electrons are added up in the \({\pi ^*}2{{\rm{p}}_{\rm{x}}}\) and \({\pi ^*}2{{\rm{p}}_{\rm{y}}}\) molecular orbital. Hence, the electronic configuration of \({{\rm{O}}_2}^{2 – }\) is

\({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2p}}{\rm{z}}^{\rm{2}}{\rm{ < }}\left( {{\rm{\pi 2p}}{\rm{x}}^{\rm{2}}{\rm{ = \pi 2p}}{\rm{y}}^{\rm{2}}} \right){\rm{ < }}\left( {{{\rm{\pi }}^{\rm{}}}{\rm{2p}}{\rm{x}}^{\rm{2}}{\rm{ = }}{{\rm{\pi }}^{\rm{}}}{\rm{2p}}_{\rm{y}}^{\rm{2}}} \right)\)

Here, \({{\rm{N}}_{\rm{b}}} = 10,\;{{\rm{N}}_{\rm{a}}} = 8\) so that

Bond order \( = \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {10 – 8} \right) = \frac{2}{2} = 1\)

Bond Order of \({\rm{O}}_2^ + \) Ion

\({\rm{O}}_2^ + \) is formed by the loss of one electron by an \({{\rm{O}}_2}\) molecule. This electron is lost from the \({\pi ^*}2{{\rm{p}}_{\rm{x}}}\) or \({\pi ^*}2{{\rm{p}}_{\rm{y}}}\) molecular orbital. Hence, the electronic configuration of \({\rm{O}}_2^ + \) is

\({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2p}}{\rm{z}}^{\rm{2}}{\rm{ < }}\left( {{\rm{\pi 2p}}{\rm{x}}^{\rm{2}}{\rm{ = \pi 2p}}{\rm{y}}^{\rm{2}}} \right){\rm{ < }}{{\rm{\pi }}^{\rm{*}}}{\rm{2}}{{\rm{p}}{\rm{x}}}^{\rm{1}}\)

Here, \({{\rm{N}}_{\rm{b}}} = 10,\;{{\rm{N}}_{\rm{a}}} = 5\) so that

Bond order \( = \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {10 – 5} \right) = \frac{5}{2} = 2.5\)

Bond Order of Polyatomic Molecules and Ions

Bond order of \({\rm{NO}}\) Molecule and \({\rm{N}}{{\rm{O}}^ – }\) ion:

The total number of electrons in \({\rm{NO}} = 7\) (from N) \(+ 8\) (from O) \(= 15\)

Order of energy levels and hence the order of filling off molecular orbitals is the same as for homonuclear diatomic molecules with electrons greater than \(14\).

Hence electronic configuration of \({\rm{NO}}\) is

\({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2p}}{\rm{z}}^{\rm{2}}{\rm{ < }}\left( {{\rm{\pi 2p}}{\rm{x}}^{\rm{2}}{\rm{ = \pi 2p}}{\rm{y}}^{\rm{2}}} \right){\rm{ < }}{{\rm{\pi }}^{\rm{*}}}{\rm{2}}{{\rm{p}}{\rm{x}}}^{\rm{1}}\)

Bond order \( = \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {10 – 5} \right) = \frac{5}{2} = 2.5\)

Bond or of NO+ Molecule

However, \({\rm{N}}{{\rm{O}}^ + }\) ion is formed by the removal of one electron from NO. The electronic configuration of \({\rm{N}}{{\rm{O}}^ + }\) is

\({\rm{\sigma 1}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2}}{{\rm{s}}^{\rm{2}}}{\rm{ < }}{{\rm{\sigma }}^{\rm{}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{ < \sigma 2p}}{\rm{z}}^{\rm{2}}{\rm{ < }}\left( {{\rm{\pi 2p}}{\rm{x}}^{\rm{2}}{\rm{ = \pi 2p}}_{\rm{y}}^{\rm{2}}} \right)\)

Bond order \( = \frac{1}{2}\left( {\;{{\rm{N}}_{\rm{b}}} – {{\rm{N}}_{\rm{a}}}} \right) = \frac{1}{2}\left( {10 – 4} \right) = \frac{6}{2} = 3\)

Bond Order Formula Based On Lewis Structure

The bond order is also defined as the number of bonds between two atoms in a molecule.

For example, bond order of \({{\rm{H}}_2}\) is 1, i.e., it has a single shared pair of electrons, bond order of \({{\rm{O}}_2}\) is \(2\), i.e., it has two shared pairs of electrons and \({{\rm{N}}_2}\) has three shared pairs of electrons; hence bond order is \(3\).

Bond Order Formula Based On Lewis Structure

The isoelectronic species (molecules and ions) have the same bond order.

Example: \({{\rm{F}}_2},{\rm{O}}_2^{2 – }\) has \(18\) electrons each and have bond order 1. \({{\rm{N}}_2},{\rm{CO}}\) and \({\rm{N}}{{\rm{O}}^ + }\) have \(14\) electrons each and bond order is \(3\).

With the increase in bond order, bond enthalpy increases, and bond length decreases.

Bond Order Formula for Resonance Structures

\({\rm{Bond}}\,{\rm{order}} = \frac{{{\rm{Sum}}\,{\rm{of}}\,{\rm{individual}}\,{\rm{bond}}\,{\rm{orders}}}}{{{\rm{number}}\,{\rm{of}}\,{\rm{bonding}}\,{\rm{groups}}}}\)

Example: Bond order of carbonate ion \(\left( {{\rm{C}}{{\rm{O}}_3}^{2 – }} \right)\)

Lewis dot structure of \({\rm{CO}}_3^{2 – }\) is as follows,

Bond Order Formula for Resonance Structures

Hence, \({\rm{Bond}}\,{\rm{order}} = \frac{{{\rm{Sum}}\,{\rm{of}}\,{\rm{individual}}\,{\rm{bond}}\,{\rm{orders}}}}{{{\rm{number}}\,{\rm{of}}\,{\rm{bonding}}\,{\rm{groups}}}}\)

Bond order \( = \frac{{1 + 1 + 2}}{3}\)

Bond order \({\rm{ = 1}}{\rm{.33}}\)

The bond length of C-O bond is \(142\) pm, C\(=\)O is \(116\) pm whereas, the bond length of the carbonate ion is \(131\) pm.

Average Bond Order Formula

Average Bond Order is given by

\({\rm{Average}}\,{\rm{bond}}\,{\rm{order}} = \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{bonds}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{resonance}}\,{\rm{structures}}}}\)

Example: Bond order of nitrite ion \(\left( {{\rm{NO}}_2^ – } \right)\)

The Lewis structure of nitrite ion is

Average Bond Order Formula

\({\rm{Average}}\,{\rm{bond}}\,{\rm{order}} = \frac{{{\rm{Number}}\,{\rm{of}}\,{\rm{bonds}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{resonance}}\,{\rm{structures}}}}\)

\({\rm{ Average\, bond\, order }} = \frac{{2 + 1}}{2}\)

\({\rm{ Average\, bond\, order }} = 1.5\)

Shortcut for Bond Order Calculation Between 10 to 18 Electrons

Bond order can be calculated by the following formula

\({\rm{ Bond\, order }} = 3 – 0.5{\rm{n}}\)

Here n is the difference between the total number of electrons and \(14\) in the given molecule.

Example 1: Bond order of \({\rm{CO}}\)

Answer: Total number of electrons \(= 6\) (from \({\rm{C}}\)) \(+ 8\) (from \({\rm{O}}\)) \(= 14\)

\({\rm{n = 14 – 14 = 0}}\)

\({\rm{ Bond\, order }} = 3 – 0.5{\rm{n}}\)

\({\rm{ Bond\, order }} = 3 – 0.5 \times 0\)

\({\rm{ Bond\, order }} = 3\)

Example 2: Bond order of \({\rm{O}}_2^{2 – }\)

Answer: Total number of electrons \( = 8 \times 2\) (from each oxygen ) \(+ 2\) (for the -ve charge) \(= 18\)

\({\rm{n = 18 – 14 = 4}}\)

\({\rm{ Bond\, order }} = 3 – 0.5{\rm{n}}\)

\({\rm{ Bond\, order }} = 3 – 0.5 \times 4\)

\({\rm{ Bond\, order }} = 3 – 2 = 1\)

Summary

In the article Bond order formula, you have grasped the formulas to find the bond order based on Molecular Orbital Theory and Lewis structure. You can explain the information conveyed by the bond order, such as stability, number of bonds, etc. The bond order of the oxygen molecule, peroxide ion, superoxide ion, carbonate ion and many more can be calculated by formula or short trick method with the knowledge of this article.

FAQs

Q.1: How do you calculate bond order quickly? Write the formula which helps to calculate the bond order.
Ans:
Bond order can be calculated by the following formula
\({\rm{ Bond\, order }} = 3 – 0.5{\rm{n}}\)
Here n is the difference between the total number of electrons and \(14\) in the given molecule.
Example: Bond order of CO
Total number of electrons \(= 6\) (from \({\rm{C}}\)) \(+ 8\) (from \({\rm{O}}\)) \({\rm{ = 14, n = 14 – 14 = 0}}\)
\({\rm{ Bond\, order }} = 3 – 0.5{\rm{n}}\)
\({\rm{ Bond\, order }} = 3 – 0.5 \times 0 = 3\)

Q.2: What is the bond order of CO2?
Ans:
The Lewis structure of \({\rm{C}}{{\rm{O}}_2}\) is

\({\rm{Bond}}{\mkern 1mu} {\rm{order}} = \frac{{{\rm{Sum}}\,{\rm{of}}\,{\rm{individual}}\,{\rm{bond}}\,{\rm{orders}}}}{{{\rm{number}}\,{\rm{of}}\,{\rm{bonding}}\,{\rm{groups}}}}\)
\({\rm{ Bond\, order }} = \frac{{2 + 2}}{2} = 2\)

Q.3: What is the average bond order for the \({\rm{SO}}\) bond in \({\rm{SO}}_4^{2 – }\) ion?
Ans:
The Lewis structure of \({\rm{SO}}_4^{2 – }\) ion is

\({\rm{Bond}}{\mkern 1mu} {\rm{order}} = \frac{{{\rm{Sum}}\,{\rm{of}}\,{\rm{individual}}\,{\rm{bond}}\,{\rm{orders}}}}{{{\rm{number}}\,{\rm{of}}\,{\rm{bonding}}\,{\rm{groups}}}}\)
\({\rm{ Bond\, order }} = \frac{{2 + 2 + 1 + 1}}{4} = 1.5\)

Q.4: Write the electron configuration of \({{\rm{C}}_2}\) molecule.
Ans:
Total number of electrons in \({{\rm{C}}_2}\) molecule is \(12\), six from each carbon atom. \(12\) electrons are filled in the various molecular orbitals as follows:
\(\sigma {\rm{1}}{{\rm{s}}^{\rm{2}}} < {\sigma ^ * }{\rm{1}}{{\rm{s}}^{\rm{2}}} < \sigma {\rm{2}}{{\rm{s}}^{\rm{2}}} < {\sigma ^ * }{\rm{2}}{{\rm{s}}^{\rm{2}}} < \left( {\pi {\rm{2p}}{{\rm{x}}^{\rm{2}}}{\rm{ = }}\pi {\rm{2p}}{{\rm{y}}^{\rm{2}}}} \right)\)

Q.5: What is the shortest trick to calculate bond order?
Ans:
Bond order can be calculated by the short cut method by the following formula
\({\rm{ Bond\, order }} = 3 – 0.5{\rm{n}}\)
Here n is the difference between the total number of electrons and \(14\) in the given molecule.

Q.6: Write the formula to calculate bond order of molecule.
Ans: ({\rm{Bond}}{\mkern 1mu} {\rm{order}} = \frac{{{{\rm{N}}{\rm{b}}} – {{\rm{N}}{\rm{a}}}}}{2} = \frac{1}{2}\left( {{{\rm{N}}{\rm{b}}} – {{\rm{N}}{\rm{a}}}} \right))

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