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Which of the following is true for nitrate anion.
A.The formal charge on \[N\] is zero
B.The bond order of \[NO\] bond is $\dfrac{4}{3}$
C.The average formal charge on oxygen is \[\dfrac{1}{3}\]
D.There are \[2-\pi \] bonds in the ion

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Last updated date: 21st Jun 2024
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Answer
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Hint: To answer this question, you should recall the concept of formal charge and bond order. It’s a theoretical charge over an individual atom of an ion as the real charge over a polyatomic molecule or ion is distributed on an ion as a whole and not over a single atom.
The formula used: \[{\text{FC}} = {\text{V}} - {\text{N}} - \dfrac{{\text{B}}}{{\text{2}}}\]
where $V$ is the no. of valence electron \[N\] is the no. of non-bonding electrons and \[B\] is the no. of electrons in the covalent bond

Complete Step by step solution:
The bond order shows the number of chemical bonds present between a pair of atoms. For instance, the bond order of diatomic nitrogen \[N \equiv N\] is 3 and bond order between the carbon atoms in \[H - H \equiv C - H\] is also three. The bond order describes the stability of the bond. The molecular orbital provides an easy understanding of the concept of the bond order of a chemical bond. It quantifies the degree of covalent bonds between the atoms. The formal charge of nitrogen can be calculated using: \[{\text{FC}} = {\text{V}} - {\text{N}} - \dfrac{{\text{B}}}{{\text{2}}}\]
 (V = no. of valence electron N = No. of nonbonding electrons B = No. of electrons in covalent bond)
\[{\text{FC of N in N}}{{\text{O}}_{\text{3}}}^ - = 0\]
\[{\text{FC of O in N}}{{\text{O}}_{\text{3}}}^ - = - \dfrac{1}{3}\]
The Lewis structure of nitrate ion can be drawn as:
seo images

We can see that the total number of bonds = 4.
The number of bond groups between individual atoms = 3 and bond order \[ = \dfrac{{4}}{3} = 1.33\]. There is one pi bond.

Therefore, we can conclude that the correct answer to this question is option D.

Note: In most of the compounds, the oxidation number of oxygen is $ - 2$ . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $ - 1$ . Example, \[N{a_2}{O_2}\]
Superoxide- Every oxygen atom is allocated an oxidation number of \[ - \dfrac{{{\text{ }}1}}{2}\] . Example, \[K{O_2}\]
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $ + 1$ .