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# Give an example of ion with minimum internuclear distance?

Last updated date: 13th Jul 2024
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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. The distance between two nuclei of a molecule is termed as internuclear distance which is inversely proportional to bond order. So here we have to give an example of an ion with minimum internuclear distance.
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 and internuclear distance followed by giving an example of ion with minimum internuclear distance:-
-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.
-Internuclear distance is the distance between two nuclei of a molecule and it is inversely proportional to bond order. This means greater the bond order, less will be the internuclear distance due to more number of bonds or electron pairs between the two nuclei of a molecule.
-An example of ion with minimum internuclear distance is as follows:-
Let's take the case of the oxygen molecule. The total electrons in ${{O}_{2}}$ is 16. So the molecular orbital configuration of oxygen molecule and its ions are shown below:-
i) ${{O}_{2}}$ = ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{1}}{{({{\pi }^{*}}2{{p}_{y}})}^{1}}$
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
ii)${{O}_{2}}^{-}$ = ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{2}}{{({{\pi }^{*}}2{{p}_{y}})}^{1}}$
Number of bonding electrons= 10
Number of bonding electrons= 7
Bond order = $\dfrac{1}{2}\text{(Bonding electrons}-\text{Anti bonding electrons)}$
Bond order = $\dfrac{1}{2}(10-7)$ = 1.5
iii)${{O}_{2}}^{+}$ = ${{(\sigma 1s)}^{2}}{{({{\sigma }^{*}}1s)}^{2}}{{(\sigma 2s)}^{2}}{{({{\sigma }^{*}}2s)}^{2}}{{(\sigma 2{{p}_{z}})}^{2}}{{(\pi 2{{p}_{x}})}^{2}}{{(\pi 2{{p}_{y}})}^{2}}{{({{\pi }^{*}}2{{p}_{x}})}^{1}}$
Number of bonding electrons= 10
Number of bonding electrons= 5
Bond order = $\dfrac{1}{2}\text{(Bonding electrons}-\text{Anti bonding electrons)}$
Bond order = $\dfrac{1}{2}(10-5)$ = 2.5
-Since the bond order of ${{O}_{2}}^{+}$ is the highest, therefore its internuclear distance will be minimum.

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.