Question

What is the order of stability of ${N_2}$ and its ions?

$$A.\;\;\;\;\;{N_2} > {\text{ }}{N_2}^ + = {\text{ }}{N_2}^ - > {\text{ }}{N_2}^{2 - } \\\ B.\;\;\;\;\;{N_2}^ + > {\text{ }}{N_2}^ - > {\text{ }}{N_2} > {\text{ }}{N_2}^{2 - } \\\ C.\;\;\;\;\;{N_2}^ - > {\text{ }}{N_2}^ + > {\text{ }}{N_2} > {\text{ }}{N_2}^{2 - } \\\ D.\;\;\;\;\;{N_2}^{2 - } > {\text{ }}{N_2}^ - = {\text{ }}{N_2}^ + > {\text{ }}{N_2} \\\$$

Answer 1

Hint : We need to calculate the bond order of${N_2}$. If higher the bond orders of the molecule/ion, higher the stability. Therefore, we can find the order of stability of ${N_2}$and its ions.
Formula used:
Bond order =$\dfrac{{{\text{Number of electrons in Bond orbitals - Number of electron in non - bond orbitals}}}}{2}$

Complete step by step solution :
In this question first we need to calculate the no. of bonding orbitals and the no. of non-bonding orbitals. For that we need the electronic configuration of each molecule.
Electronic configuration of nitrogen (N) is $1{s^2}2{s^2}2{p^3}$ as N is having only 7 electrons.
Similarly, the electronic configuration for N- will be $1{s^2}2{s^2}2{p^4}$ as ${N^ - }$is having 8 electrons in total due to borrowing 1 electron.
Electronic configuration of N+ will be $1{s^2}2{s^2}2{p^2}$ as ${N^ + }$has donated 1 electron.
Electronic configuration of N2- will be $1{s^2}2{s^2}2{p^5}$ as ${N^{2 - }}$has borrowed 2 electrons.
So the bond for unfilled orbitals is being counted for bond and non-bond orbitals i.e. on 2p orbital will be counted.
In case of${N_2}$, 2 N are being bonded together, also the bonding orbitals are being filled first and after that non-bonding orbitals are being filled.
So for ${N_2}$ Since only 6 electrons are there in unfilled orbitals then only bond orbitals are being filled hence, Its bond order will be 6/2 = 3.
In the case of${N^{2 - }}$, one N and one N- will be there. The number of electrons in an unfilled orbital of N is 3 and that of ${N^ - }$is 4 so the total number of electrons is 7. But the bond orbital can carry up to 6 electrons and one electron in a non-bond orbital. So, its bond order will be 5/2 = 2.5.
In case of${N_2}^ +$, one N and one ${N^ + }$will be there. The number of electrons in an unfilled orbital of N is 3 and that of ${N^ + }$ is 2 so the total number of electrons is 5. Since it does not contain more than 6 electrons so its bond order will be 5/2 = 2.5.
In the case of${N_2}^{2 - }$, one N and one N2- will be there. The number of electrons in an unfilled orbital of N is 3 and that of ${N^{2 - }}$is 5 so the total number of electrons is 8. Since, it contains more than 8 electrons so 6 electrons will be in bond orbital and 2 in non-bond orbital. So, its bond order will be 4/2 = 2.
So, the answer to this question is$A.{\text{ }}{N_2} > {\text{ }}{N_2}^ + = {\text{ }}{N_2}^ - > {\text{ }}{N_2}^{2 - }$.

Note : While answering this question we should know the concept of stability and bond order along with bond and non-bond orbital. The relevance of bond order is that it tells us about how strongly the two molecules are bonded together. For example, the bond order for $N_2$ is 3 and for $O_2$ is 2, so the $N_2$ molecule is more tightly bonded as compared to $O_2$ molecule.