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# Which of the following is true for nitrate anion.A.The formal charge on $N$ is zeroB.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

Last updated date: 21st Jun 2024
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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:

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 $+ 1$ .