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Question: The bond order of the species with the orbital configuration \({{(\sigma 1s)}^{2}}\text{ }{{(\sig...

The bond order of the species with the orbital configuration
(σ1s)2 (σ∗1s)2 (σ2s)2 (σ∗2s)2 (σ2pz)1{{(\sigma 1s)}^{2}}\text{ }{{(\sigma *1s)}^{2}}\text{ }{{(\sigma 2s)}^{2}}\text{ }{{(\sigma *2s)}^{2}}\text{ }{{(\sigma 2{{p}_{z}})}^{1}} will be _________

Explanation

Solution

You should know that; each orbital in an orbital configuration occupies two electrons in its orbital. The bond order can be calculated from the number of the bonding electrons and the number of the antibonding electrons.

Complete step by step solution:
-Let us first know about what is the bond order of a species.
-So, bond order is the number of bonds present between two atoms inside the same molecular species. It means that the bond order is actually a measure for stability of the bonds. In short, we can state that the higher the bond order between two species, the higher will be its overall energy and also lower will be the distance between their respective centres.
-Bond order is always defined for a molecule and therefore, the molecular orbitals influence these values. And we know that the molecular orbitals are of two types, bonding and antibonding molecular orbitals.
-The formula of bond order is given by:
Bond order=Number of electrons in (bonding molecular orbitals−antibonding molecular orbitals)2Bond\text{ order}=\dfrac{\text{Number of electrons in (bonding molecular orbitals}-\text{antibonding molecular orbitals)}}{2}
-So, here as per the question the orbital configuration of a species is given as:
(σ1s)2 (σ∗1s)2 (σ2s)2 (σ∗2s)2 (σ2pz)1{{(\sigma 1s)}^{2}}\text{ }{{(\sigma *1s)}^{2}}\text{ }{{(\sigma 2s)}^{2}}\text{ }{{(\sigma *2s)}^{2}}\text{ }{{(\sigma 2{{p}_{z}})}^{1}}
-Here, the bonding molecular orbitals are (σ1s)2{{(\sigma 1s)}^{2}}, (σ2s)2{{(\sigma 2s)}^{2}} and (σ2pz)1{{(\sigma 2{{p}_{z}})}^{1}}.
-So, the number of electrons present in the bonding molecular orbital is (2+2+1)=5(2+2+1)=5.
-While, the number of antibonding molecular orbitals are (σ∗1s)2{{(\sigma *1s)}^{2}} and (σ∗2s)2{{(\sigma *2s)}^{2}} .
-So, the total number of electrons present in the antibonding molecular orbitals is (2+2)=4(2+2)=4.
-Now, by using the formula of bond order, the bond order (consider as BO) of the given species will be:
BO=5−42=12=0.5BO=\dfrac{5-4}{2}=\dfrac{1}{2}=0.5

Hence, the bond order of the given species with orbital configuration (σ1s)2 (σ∗1s)2 (σ2s)2 (σ∗2s)2 (σ2pz)1{{(\sigma 1s)}^{2}}\text{ }{{(\sigma *1s)}^{2}}\text{ }{{(\sigma 2s)}^{2}}\text{ }{{(\sigma *2s)}^{2}}\text{ }{{(\sigma 2{{p}_{z}})}^{1}} is 0.50.5.

Note: Bond is not always an integer for some molecules. So, not all molecules obey the molecular orbital theory. Molecules containing unpaired electrons will show paramagnetism. Antibonding molecular orbitals usually have higher energy. Thus, they are filled after bonding molecular orbitals.