Question

What will be the bond order of the species with electronic configuration $1{s^2}2{s^2}2{p^5}$ ?
(A) One
(B) Two
(C) Three
(D) Zero

Answer 1

Hint: Bond order is usually the difference between the number of bonds and antibonds. The bond number is the number of electron pairs between a pair of atoms. It shows the number of chemical bonds present between a pair of atoms. It also describes the stability of the bond.
Formula used: Bond order $ = \dfrac{1}{2}[{N_b} - {N_a}]$
Where,
${N_b} = $ Number of bonding electrons
${N_a} = $ Number of antibonding electrons

Complete answer:
The species with the electronic configuration $1{s^2}2{s^2}2{p^5}$ is fluorine. To find the bond order of a given species, we should know the number of bonding electrons and the number of antibonding electrons. This can be found using the molecular orbital configuration. The molecular orbital theory helps us to describe the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The molecular orbital configuration of fluorine is:$(\sigma 1{s^2})({\sigma ^ * }1{s^2})(\sigma 2{s^2})({\sigma ^ * }2{s^2})(\sigma 2p_z^2)(\pi 2p_x^2 = \pi 2p_y^2)({\pi ^ * }2p_x^2 = {\pi ^ * }2p_y^2)$
We know that Bond order $ = \dfrac{1}{2}[{N_b} - {N_a}]$
The number of bonding electrons in fluorine $ = 10$
The number of antibonding electrons in fluorine $ = 8$
Bond order $ = \dfrac{1}{2}[{N_b} - {N_a}]$
$ \Rightarrow B.O = \dfrac{1}{2} \times (10 - 8)$
$ \Rightarrow B.O = 1$
The bond order of the species with electronic configuration $1{s^2}2{s^2}2{p^5}$ is $1$ .
Therefore, option A is the correct answer.

Note:
The molecular orbital configuration of a species is derived as follows: First, we have to find the valence electron configuration of each atom in the molecule. Then we have to decide if the molecule is homonuclear or heteronuclear. Then we have to fill molecular orbitals using energy and bonding properties of the overlapping atomic orbitals.