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Question: Calculate the bond order in \({{O}_{2}}\), \({{O}_{2}}^{-}\), \(O_{2}^{2-}\) and \(O_{2}^{+}\)molecu...

Calculate the bond order in O2{{O}_{2}}, O2−{{O}_{2}}^{-}, O22−O_{2}^{2-} and O2+O_{2}^{+}molecule.

Explanation

Solution

The formula for determining bond order is as given below:
BO=Nb−Na2BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}
Where, “BO” is bond order; Nb{{N}_{b}}is the number of bonding molecular orbitals; Na{{N}_{a}}is the number of antibonding molecular orbitals.

Complete step-by-step answer:
Bond order is the number of bonds present between two atoms inside the same molecular species. What we mean to say is, that bond order is actually a measure of stability of bonds. We can also state that the higher the bond order between two species, the lower is its overall energy and also the lower is the distance between their respective centres.
Bond order is always defined for a molecule and therefore, the molecular orbitals influence these values. As you already know, the molecular orbital is of two types, namely bonding molecular orbitals and antibonding molecular orbitals. Here the former is more stable and therefore contains less energy than the latter.
For molecular species with less than or equal to 14 electrons, the following arrangement of molecular orbitals is taken into consideration-
σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<σ2pz<(π∗2px=π∗2pz)<σ∗2pz\sigma 1s<\sigma *1s<\sigma 2s<\sigma *2s<\left( \pi 2{{p}_{x}}=\pi 2{{p}_{y}} \right)<\sigma 2{{p}_{z}}<\left( \pi *2{{p}_{x}}=\pi *2{{p}_{z}} \right)<\sigma *2{{p}_{z}}
For molecular species with more than 14 electrons, the arrangement of molecular orbitals is as follows:
σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2pz)<σ∗2pz\sigma 1s<\sigma *1s<\sigma 2s<\sigma *2s<\sigma 2{{p}_{z}}<\left( \pi 2{{p}_{x}}=\pi 2{{p}_{y}} \right)<\left( \pi *2{{p}_{x}}=\pi *2{{p}_{z}} \right)<\sigma *2{{p}_{z}}
The arrangement given above is a pure experimental data and the order is based upon the relative energies of the molecular orbitals with each other. This is therefore treated as a constant. As we are dealing with oxygen here, we will use the second arrangement mentioned above because it has a total of sixteen electrons in its neutral molecule.
We will do it individually for all the molecular species given in the question.
- O2{{O}_{2}}
It is worth mentioning here, that the molecular orbitals follow all the rules of atomic orbitals. So filling them up should be done taking this piece of information into account. We have done it as follows:
σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<(π2px2=π2py2)<(π∗2px1=π∗2pz1)<σ∗2pz\sigma 1{{s}^{2}}<\sigma *1{{s}^{2}}<\sigma 2{{s}^{2}}<\sigma *2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\left( \pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2} \right)<\left( \pi *2{{p}_{x}}^{1}=\pi *2{{p}_{z}}^{1} \right)<\sigma *2{{p}_{z}}
Nb=5×2=10{{N}_{b}}=5\times 2=10
This is the number of electrons inside molecular orbitals which do not have an asterisk over their head, in other words, this is the number of electrons which occupy the bonding molecular orbitals.
Na=3×2=6{{N}_{a}}=3\times 2=6
This is the number of electrons inside molecular orbitals which have an asterisk mark over them, in other words, these are the anti-bonding molecular orbitals.
The formula of bond order is given by-
BO=Nb−Na2BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}
Putting the respective values, we get-
BO=10−62=2BO=\dfrac{10-6}{2}=2
So the bond order of dioxygen is “2”.
- O2−{{O}_{2}}^{-}
There are 17 electrons here and the arrangement is as follows:
σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<(π2px2=π2py2)<(π∗2px2=π∗2pz1)<σ∗2pz\sigma 1{{s}^{2}}<\sigma *1{{s}^{2}}<\sigma 2{{s}^{2}}<\sigma *2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\left( \pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2} \right)<\left( \pi *2{{p}_{x}}^{2}=\pi *2{{p}_{z}}^{1} \right)<\sigma *2{{p}_{z}}
So we get:
BO=Nb−Na2=10−72=1.5BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}=\dfrac{10-7}{2}=1.5
- O22−O_{2}^{2-}
There are 18 electrons here and the arrangement is as follows:
σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<(π2px2=π2py2)<(π∗2px2=π∗2pz2)<σ∗2pz\sigma 1{{s}^{2}}<\sigma *1{{s}^{2}}<\sigma 2{{s}^{2}}<\sigma *2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\left( \pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2} \right)<\left( \pi *2{{p}_{x}}^{2}=\pi *2{{p}_{z}}^{2} \right)<\sigma *2{{p}_{z}}
So we get:
BO=Nb−Na2=10−82=1BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}=\dfrac{10-8}{2}=1
- O2+O_{2}^{+}
There are 15 electrons here and the arrangement is as follows:
σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<(π2px2=π2py2)<(π∗2px1=π∗2pz)<σ∗2pz\sigma 1{{s}^{2}}<\sigma *1{{s}^{2}}<\sigma 2{{s}^{2}}<\sigma *2{{s}^{2}}<\sigma 2{{p}_{z}}^{2}<\left( \pi 2{{p}_{x}}^{2}=\pi 2{{p}_{y}}^{2} \right)<\left( \pi *2{{p}_{x}}^{1}=\pi *2{{p}_{z}} \right)<\sigma *2{{p}_{z}}
So we get:
BO=Nb−Na2=10−52=2.5BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}=\dfrac{10-5}{2}=2.5
As you can see, we successfully calculated the bond order of all the molecules that we were given.

Note: Care should be taken while choosing the template of arrangement of molecular orbitals. This should be done by calculating the total electrons present in the neutral molecule and adding or subtracting according to the number of charges if any. In the arrangement of molecular orbitals mentioned in the solution above, an “is equal to” sign means both the orbitals have the same energy. This should be kept in mind, as it will significantly influence the arrangement of electrons.