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# What charge would be needed on ${{F}_{2}}$ to generate an ion with a bond order of 2? Verified
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Hint: As we know that bond order is the difference of number of bonding electrons and antibonding electrons which is divided by two. So by using the concept of bond order we have to tell the charge required by ${{F}_{2}}$ molecule to generate a bond order of 2.
Formula used:
We will use the following formula:-
Bond order = $\dfrac{1}{2}\text{(Bonding electrons}-\text{Anti bonding electrons)}$

-Let us first understand the concept of bond order followed by calculation of charge that would be required by ${{F}_{2}}$ molecule to generate a bond order of 2.
-Bond order can be defined as the number of bonds or electron pairs between two atoms and it is the difference between the number of bonding electrons and antibonding electrons which is divided by two. Bond order can be calculated by using the following formula:-
Bond order = $\dfrac{1}{2}\text{(Bonding electrons}-\text{Anti bonding electrons)}$
-The number of bonding electrons and antibonding electrons can be counted by writing down the molecular orbital configuration according to their increasing order of energy shown as follows:-
$\sigma 1s,{{\sigma }^{*}}1s,\sigma 2s,{{\sigma }^{*}}2s,\sigma 2{{p}_{z}},\pi 2{{p}_{x}}=\pi 2{{p}_{y}},{{\pi }^{*}}2{{p}_{x}}={{\pi }^{*}}2{{p}_{y}},{{\sigma }^{*}}2{{p}_{z}}$ and so on.
Here molecular orbitals with asterisk $(*)$ are anti bonding molecular orbital, so its electrons will be counted as antibonding electrons and rest as bonding electrons.
-Calculation of bond order in ${{F}_{2}}$molecule:-
Total number of electrons in ${{F}_{2}}$molecule is 18. So the molecular orbital configuration of fluorine molecule is: ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{2}}{{({{\pi }^{*}}2{{p}_{y}})}^{2}}$
Number of bonding electrons= 10
Number of bonding electrons= 8
Bond order = $\dfrac{1}{2}\text{(Bonding electrons}-\text{Anti bonding electrons)}$
Bond order = $\dfrac{1}{2}(10-8)$ = 1
-We know that 0.5 bond order is contributed by each electron. In the case of ${{F}_{2}}$molecule, the highest occupied molecular orbital (HOMO) is anti-bonding and the same for the lowest unoccupied molecular orbital (LUMO) as well. So if we add electrons to the molecule, they will enter in antibonding LUMO which will decrease the bond order by 0.5 whereas if we remove the electron from anti-bonding HOMO, it will increase the bond order by 0.5.
-Since the bond order is 1 and we require 2, so to increase the bond order we will remove 2 electrons (each electron makes an increment of 0.5 to the bond order).
So the ion will be ${{F}_{2}}^{2+}$ whose molecular orbital configuration is shown below: ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{1}}{{({{\pi }^{*}}2{{p}_{y}})}^{1}}$
Number of bonding electrons= 10
Number of bonding electrons= 6
Bond order = $\dfrac{1}{2}\text{(Bonding electrons}-\text{Anti bonding electrons)}$
Bond order = $\dfrac{1}{2}(10-6)$ = 2
-Hence +2 charges would be needed on ${{F}_{2}}$ molecule to generate a bond order of 2.

Note:
-Remember that bond order is always positive and can be an integer or fraction.
-The molecule whose bond order is calculated to be zero does not exist because there is no electron pair in the molecule.