Question

Bond order of which among the following molecules is zero?
(A) $ {F_2} $
(B) $ {O_2} $
(C) $ B{e_2} $
(D) $ L{i_2} $

Answer 1

Hint: Molecular orbital theory states that atoms combine together to form molecular orbitals. The total number of molecular orbitals formed will be equal to the atomic orbitals forming molecular orbitals. We solve this sum by applying the relation between molecular orbital theory and bond order. Bond order tells us about the number of bonds between two compounds. A compound with no bond between each other will have a bond order of 0.
 $ {\text{Bond Order}} = \dfrac{{{\text{bonding electrons}} - {\text{antibonding electrons}}}}{{{\text{number of molecules}}}} $

Complete Step by step solution
In molecular orbital theory, there are bonding molecular orbitals and antibonding molecular orbitals. In bonding molecular orbitals, the probability of finding electrons is high. In antibonding molecular orbitals, the probability of finding electrons is low. Also, in antibonding molecular orbitals, there is a node where electron density is zero. The bonding molecular orbitals are formed by the additive effect of the atomic orbitals. The antibonding molecular orbitals are formed by the subtractive effect of the atomic orbitals.
The number of the chemical bonds between the pair of atoms is indicated by the bond order. Bond order indicates the stability of a bond.
Bond order can be calculated by the subtracting bonding electrons with antibonding electrons and dividing by 2.
The common molecular orbitals are sigma orbitals and pi orbitals.
 $ {F_2} $
As the number of electrons in $ {F_2} $ are 18. The pairing of electrons takes place as follows,
 $ {F_2} = \sigma 1{s^2},\sigma 1{s^2} * ,\sigma 2{s^2},\sigma 2{s^2} * ,\sigma 2{p_z}^2,\pi 2{p_x}^2 \approx \pi 2{p_y}^2,\pi 2{p_x}^2 * \approx \pi 2{p_y}^2 * $
Here, the bonding electrons are 10 and the antibonding electrons are 8.
Bond order of $ {F_2} $ molecule = $ \dfrac{{10 - 8}}{2} $ = 1
 $ {O_2} $
As the number of electrons in $ {O_2} $ are 16. The pairing of electrons takes place as follows,
 $ {O_2} = \sigma 1{s^2},\sigma 1{s^2} * ,\sigma 2{s^2},\sigma 2{s^2} * ,\sigma 2{p_z}^2,\pi 2{p_x}^2 \approx \pi 2{p_y}^2,\pi 2{p_x}^1 * \approx \pi 2{p_y}^1 * $
Here, the bonding electrons are 10 and the antibonding electrons are 6.
Bond order of $ {O_2} $ molecule = $ \dfrac{{10 - 6}}{2} $ = 2
 $ B{e_2} $
As the number of electrons in $ B{e_2} $ are 8. The pairing of electrons takes place as follows,
 $ B{e_2} = \sigma 1{s^2},\sigma 1{s^2} * ,\sigma 2{s^2},\sigma 2{s^2} * $
Here, the bonding electrons are 4 and the antibonding electrons are 4.
Bond order of $ B{e_2} $ molecule = $ \dfrac{{4 - 4}}{2} $ = 0
 $ L{i_2} $
As the number of electrons in $ L{i_2} $ are 6. The pairing of electrons takes place as follows,
 $ L{i_2} = \sigma 1{s^2},\sigma 1{s^2} * ,\sigma 2{s^2} $
Here, the bonding electrons are 4 and the antibonding electrons are 2.
Bond order of molecule = $ \dfrac{{4 - 2}}{2} $ = 1
Therefore, the molecule with zero bond order is $ B{e_2} $ .
Hence, the correct answer is option C.

Note
The bond order of a single bond is 1 and the bond order of a double bond is 2. More is the bond order; more is the strength of the bond. The stronger the bond that is the greater the bond order the more the energy required to break the bond. Bond order can also be fractional.