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Question: The bond order of \(O_{2}^{2-},O_{2}^{-},{{O}_{2}}\) are respectively x, y, and z. Then what is th...

The bond order of O22−,O2−,O2O_{2}^{2-},O_{2}^{-},{{O}_{2}} are respectively x, y, and z.
Then what is the value of 2(x + y + z)?

Explanation

Solution

To obtain the final answer, we need to determine the bond order of the given molecules. The formula for calculating bond order is given below:
BO=Nb−Na2BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}
Where, BO = Bond order
Nb{{N}_{b}} = number of bonding molecular orbitals
Na{{N}_{a}} = number of antibonding molecular orbitals

Complete step by steps solution:
- Bond order is the number of bonds present between 2 atoms inside the same molecular species. In other words, bond order is a measure of stability of bonds.
- Higher the bond order between two species, lower is its overall energy, and also lower is the distance between their respective centres.
- As we already know, the molecular orbital is of two types, namely bonding molecular orbitals, which is stable, and antibonding molecular orbitals, which is unstable.
- 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∗2py)<σ∗2pz\sigma 1s<\sigma *1s<\sigma 2s<\sigma *2s<\sigma 2{{p}_{z}}<(\pi 2{{p}_{x}}=\pi 2{{p}_{y}})<(\pi *2{{p}_{x}}=\pi *2{{p}_{y}})<\sigma *2{{p}_{z}}
- The bond order is based upon the relative energies of the molecular orbitals with each other. We will calculate bond order individually for all the given molecular species.
- 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∗2py2)<σ∗2pz\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}=\pi 2{{p}_{y}}^{2})<(\pi *2{{p}_{x}}^{2}=\pi *2{{p}_{y}}^{2})<\sigma *2{{p}_{z}}
So, we get,
BO=Nb−Na2BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}
BO=10−82=1BO=\dfrac{10-8}{2}=1
- 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∗2py1)<σ∗2pz\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}=\pi 2{{p}_{y}}^{2})<(\pi *2{{p}_{x}}^{2}=\pi *2{{p}_{y}}^{1})<\sigma *2{{p}_{z}}
So, we get,
BO=Nb−Na2BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}
BO=10−72=1.5BO=\dfrac{10-7}{2}=1.5
- O2{{O}_{2}}
There are 16 electrons here and the arrangement is as follows:
σ1s2<σ∗1s2<σ2s2<σ∗2s2<σ2pz2<(π2px2=π2py2)<(π∗2px1=π∗2py1)<σ∗2pz\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}=\pi 2{{p}_{y}}^{2})<(\pi *2{{p}_{x}}^{1}=\pi *2{{p}_{y}}^{1})<\sigma *2{{p}_{z}}
So, we get,
BO=Nb−Na2BO=\dfrac{{{N}_{b}}-{{N}_{a}}}{2}
BO=10−62=2BO=\dfrac{10-6}{2}=2
- Therefore, the values of x, y, and z are 1, 1.5, and 2 respectively.
- So, 2(x + y + z) = 2(1 + 1.5 + 2) = 9

Note: The template of arrangement of molecular orbitals should be chosen by calculating the total electrons present in the neutral molecule and adding or subtracting according to the number of charges if any.