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Question: Which of the following is true for nitrate anion. A.The formal charge on \[N\] is zero B.The bon...

Which of the following is true for nitrate anion.
A.The formal charge on NN is zero
B.The bond order of NONO bond is 43\dfrac{4}{3}
C.The average formal charge on oxygen is 13\dfrac{1}{3}
D.There are 2−π2-\pi bonds in the ion

Explanation

Solution

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: FC=V−N−B2{\text{FC}} = {\text{V}} - {\text{N}} - \dfrac{{\text{B}}}{{\text{2}}}
where VV is the no. of valence electron NN is the no. of non-bonding electrons and BB 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≡NN \equiv N is 3 and bond order between the carbon atoms in H−H≡C−HH - 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: FC=V−N−B2{\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)
FC of N in NO3−=0{\text{FC of N in N}}{{\text{O}}_{\text{3}}}^ - = 0
FC of O in NO3−=−13{\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 =43=1.33 = \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 - 2 . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of −1 - 1 . Example, Na2O2N{a_2}{O_2}
Superoxide- Every oxygen atom is allocated an oxidation number of − 12 - \dfrac{{{\text{ }}1}}{2} . Example, KO2K{O_2}
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of +1 + 1 .