Hint: In this question, the bond order of N-O in \[NO_3^ - \] is calculated by dividing the number of chemical bonds present between the atoms of the ion divided by the total number of resonating structures.
Complete step by step answer:
The nitrate ion \[NO_3^ - \], can be represented by more than one structure known as resonance structure. The resonance structure is defined as the set of Lewis structure which describes the delocalization of electrons in a polyatomic ion.
The resonance structure of nitrate ion is shown below.
The nitrate ion contains three oxygen atoms attached to the nitrogen atom with a negative charge.
The bond order states the number of chemical bonds present between the atoms in a molecule. The bond order describes the stability of the bond.
For resonating structure, the bond order is calculated by the formula as shown below.
\[B.O = \dfrac{B}{R}\]
Where,
B.O is the bond order
B is total number of chemical bond present
R is the total number of resonating structure
In nitrate ion two oxygen atoms are attached by a single bond and one oxygen atom is attached by a double bond. Total 4 bonds are present.
Total 3 resonating structures are formed by the nitrate ion.
Substitute the values in the above equation.
\[ \Rightarrow B.O = \dfrac{4}{3}\]
\[ \Rightarrow B.O = 1.33\]
Thus, the bond order of N-O bonds in \[NO_3^ - \] is 1.33.
Therefore, the correct option is D.
Note:
The bond order is usually calculated using molecular orbital theory where the number of bonding electrons is subtracted by the number of antibonding electrons divided by 2.
