Question

What do you mean by term bond order? Calculate the bond order of ${N_{2,}}{C_{2,}}{H_2}.$

Answer 1

The difference between the number of bonds and antibonds, as established by Linus Pauling, is known as bond order.The bond number is the total number of electron pairs (bonds) that exist between two atoms. For example, the bond number in diatomic nitrogen $N \equiv N$ is 3, the bond number between the two carbon atoms in ethyne $H - C \equiv C - H$ is $3$ , and the $C - H$ bond order is $1$ . A bond's stability is determined by its bond number. The bond number of isoelectronic species is the same.

Formula Used:
The bond order is given by:
Bond Order $= \dfrac{1}{2}[a - b]$
Where,
$a =$ Number of electrons in bonding molecular orbitals.
$b =$ Number of electrons in antibonding molecular orbitals.

Complete answer:
The bond order indicates how many chemical bonds exist between two atoms. Half of the difference between the number of electrons in bonding orbitals and antibonding orbitals is the Bond Order Formula.
Let us first know about antibonding and bonding orbitals.
An antibonding molecular orbital in chemical bonding theory, is a type of molecular orbital (MO) that weakens the chemical bond between two atoms and helps to raise the energy of the molecule relative to separate atoms. In the bonding region between the nuclei, such an orbital has one or more nodes. The orbital electron density is concentrated outside the bonding region, pulling one nucleus away from the other and causing mutual repulsion between the two atoms.
Chemical bonding theory describes the attractive contact between the atomic orbitals of two or more atoms in a molecule as bonding Molecular Orbital. Electrons are depicted as moving in waves in this idea. When more than one of these waves comes near together, their in-phase interaction creates an interaction that results in a species that is highly stabilised. The constructive interference of the waves leads the electron density to be found inside the binding zone, resulting in a stable link between the two species.
Let us see ${N_2}$ structure:
Electronic configuration of ${N_2}$ $= {\text{ }}KK'{\left( {\sigma 2s} \right)^{2\;}}{\left( {\sigma *2s} \right)^2}\;{\left( {\pi 2Px} \right)^2}\;{\left( {\pi 2py} \right)^{2\;}}{\left( {\sigma 2pz} \right)^2}$
$a = 10$
$b = 4$
Bond Order $= \dfrac{{10 - 4}}{2} = 3$
Similarly, Bondorder of ${C_2}$ $= \dfrac{{8 - 4}}{2} = 2$
Bond Order of ${H_2} = \dfrac{{2 - 0}}{2} = 1$

Note:
In order to solve this problem there are some very important points which we should keep on our fingertips. We should know Molecular Orbital theory regarding bonding and antibonding orbital and the bond order formula . And this will be achieved by only practicing multiple problems.