Question

Arrange the following in order of decreasing $N - O$ bond length:
$N{O_2}^ + ,N{O_2}^ - ,N{O_3}^ - $.
A.$N{O_3}^ - > N{O_2}^ + > N{O_2}^ - $
B.$N{O_3}^ - > N{O_2}^ - > N{O_2}^ + $
C.$N{O_2}^ + > N{O_3}^ - > N{O_2}^ - $
D.$N{O_2}^ - > N{O_3}^ - > N{O_2}^ + $

Answer 1

Hint: Bond order: It is defined as the difference between the number of bonds and number of anti-bonds. Or we can say bond order is the number of electron pairs between a pair of atoms.
Bond length: It is basically defined as the average distance between the nuclei of two bonded atoms in a molecule. The relation between the bond order and bond length is as: Bond length is inversely proportional to the bond order.

Complete step by step solution:
When atoms collide with each to form the molecules then two types of orbitals are formed: bonding orbitals and antibonding orbitals.
Bonding orbitals: The orbitals which have lower energies than the orbitals of separate atoms, are known as bonding orbitals.
Non-bonding orbitals: The orbitals which have higher energies than the orbitals of separate atoms, are known as anti-bonding orbitals. The number of bonding orbitals is always higher or equal to the number of antibonding orbitals.
Let us first talk about bond order and bond length and the relation between bond order and bond length.
Bond order: It is defined as the difference between the number of bonds and number of anti-bonds. Or we can say bond order is the number of electron pairs between a pair of atoms.
 For example: In NO molecule there are total $7 + 8 = 15$ electrons because there are $7$ electrons in a nitrogen atom and $8$ electrons in an oxygen atom.
Now when NO molecule is formed then there are total $6$ electrons in bonding $2p$ orbitals and $1$ electrons in anti-bonding $2p$ orbitals. So its bond order will be $\dfrac{{6 - 1}}{2} = 2.5$.
Now the molecule $N{O_2}^ - $ the total number of electrons is two more than the number of electrons in NO. So the bond order will be $\dfrac{{6 - 3}}{2} = 1.5$.
Similarly the bond order of $N{O_2}^ + $ bond order is $\dfrac{{6 - 0}}{2} = 3$ and the bond order of $N{O_3}^ - $ is $\dfrac{{6 - 4}}{2} = 1$.
Hence the order of bond order is as : $N{O_3}^ - < N{O_2}^ - < N{O_2}^ + $.
Bond length: It is basically defined as the average distance between the nuclei of two bonded atoms in a molecule.
The relation between the bond order and bond length is as: Bond length is inversely proportional to the bond order.
So the order of bond length will be $N{O_3}^ - > N{O_2}^ - > N{O_2}^ + $.

Hence, option B is correct.

Note: The $1s$ and $2s$ bonding and antibonding electrons cancel each other’s effect because in $1s$ and $2s$ the number of electrons in bonding and antibonding orbitals is $two$. So we can neglect these two orbitals electrons.