# How to Calculate Bond Order

Bond order is defined in simple words as the number of bonds formed between two atoms. This article covers the basic concept, application, and calculation of this order.

Bond order is a term used to determine the number of bonds in a covalent bond formation. For example, when nitrogen forms a triple bond with another nitrogen atom, its value is 3. Similarly, between a carbon and hydrogen atom (C-H), it is 1. However, this value is not always a whole number, and it also can be a fraction. For resonating structures like benzene, the value is a fraction. This value also determines the stability of a bond. It is directly proportional to the energy required to break the bond, i.e., more the value, higher is the energy required to break it.

**Bond Order Formula**

This concept is commonly used in molecular orbital theory. According to this concept, bond order is half of the difference between the number of bonding electrons and the number of antibonding electrons.

B.O. = ½ (number of bonding electrons - number of antibonding electrons)

Before calculating, you must know about both these types of electrons. Bonding and antibonding orbitals are formed when two atoms combine to form a molecule. The bonding is governed by Pauli's Exclusion Principle, which states that no two electrons can have the same quantum numbers, i.e.,

*n, l, m, s*, and they must be different for the electrons to have opposite spins. The antibonding orbitals, as the name suggests, are at a higher energy level, are unstable, and do not support bond formation. The ones present in this orbital are the antibonding electrons. Similarly, bonding electrons, are the ones present in the binding orbitals, which are at a lower and stable energy state, and hence, support binding. Knowing the structures and placing the electrons in correct orbitals can help to determine the electrons in the bonding or antibonding orbitals.

**Calculations**

*For C*

_{2}-You must know the number of electrons in the bonding and antibonding orbitals. A clue to find this out is to draw an orbital structure, and determine the number of electrons in each orbit. In the case of C

_{2}, there is an extra electron in the bonding orbital. Thus, after applying the formula provided above,

B.O. = ½ (9-4) = 2.5

*For N*

_{2}+ and N_{2}-The value for both of these is 2.5, as in both cases there is an unpaired electron either in the bonding or antibonding orbital, reducing the value by half.

*For Benzene*

Benzene is an aromatic cyclic compound, and hence, has resonance structures. In this compound, 6 pi electrons are shared by 6 carbon atoms; every carbon has half a pi bond and one sigma bond. Thus, the bond order of benzene is 1.5 for each bond.

**For a Diatomic Molecule**

Linus Pauling derived a formula to determine the bond order, and it was later realized that this formula can only be applied to diatomic molecules. It is given as,

B.O. (S

_{ij}) = exp[R

_{ij}- d

_{ij}/ b]

Here, R

_{ij}stands for experimentally measured bond length,

d

_{ij}stands for the single bond length, and

'b' is a constant, calculated by Pauling as 0.353.

It is important to remember that this subject covers a very small part of chemistry, but at the same time, a very important one as well. It is necessary to know all the basics of orbital structures like their properties and principles, without which understanding bond order is not possible.