Question

1. draw the lewis structure for the no3^- ion.

a) draw all resonance structures.
b) indicate the formal charges on atoms in one of the resonance structures.
c) determine the average charge on o atoms and n - o bond order.

Explanation:

Step1: Calculate the total number of valence electrons

The valence electrons of \(N\) is \(5\), of \(O\) is \(6\), and there is an extra electron due to the \(- 1\) charge. So the total number of valence electrons \(=5 + 3\times6+1=24\).

Step2: Draw the basic structure

Start by connecting the \(N\) atom to three \(O\) atoms. Assume single bonds first. Each single bond uses \(2\) electrons. So far, \(3\times2 = 6\) electrons are used. Then we have \(24 - 6=18\) electrons left for lone - pairs.

Step3: Distribute the remaining electrons

Each \(O\) atom needs \(6\) electrons (to complete the octet). If we give \(3\) lone - pairs (\(6\) electrons) to each \(O\) atom, we use \(3\times6 = 18\) electrons. But \(N\) has only \(6\) electrons (from three single bonds). So we need to form double bonds.

Step4: Draw resonance structures

• Structure 1: \(N\) forms a double bond with one \(O\) atom and single bonds with the other two \(O\) atoms.
• Structure 2: \(N\) forms a double bond with a different \(O\) atom and single bonds with the remaining two \(O\) atoms.
• Structure 3: \(N\) forms a double bond with the third \(O\) atom and single bonds with the other two \(O\) atoms.

Step5: Calculate formal charges (for one resonance structure, say the first one)

The formula for formal charge \(FC=V - N - \frac{B}{2}\), where \(V\) is the number of valence electrons in the free atom, \(N\) is the number of non - bonding electrons, and \(B\) is the number of bonding electrons.

• For \(N\): \(V = 5\), \(N = 0\), \(B = 8\). \(FC=5-0-\frac{8}{2}=+1\)
• For the double - bonded \(O\): \(V = 6\), \(N = 4\), \(B = 4\). \(FC=6 - 4-\frac{4}{2}=0\)
• For the single - bonded \(O\)s: \(V = 6\), \(N = 6\), \(B = 2\). \(FC=6-6-\frac{2}{2}=-1\)

Step6: Calculate the average charge on \(O\) atoms

There are three \(O\) atoms. The sum of charges on \(O\) atoms in one resonance structure is \((- 1)+(-1)+0=-2\). The average charge on \(O\) atoms \(=\frac{-2}{3}\approx - 0.67\)

Step7: Calculate the \(N - O\) bond order

In the resonance hybrid, there are a total of \(4\) bonding pairs (\(3\) single bonds and \(1\) double bond in the resonance description) over \(3\) \(N - O\) bonds. The bond order \(=\frac{4}{3}\approx1.33\)

Answer:

a) The resonance structures of \(NO_{3}^{-}\) are as follows (showing the movement of double bond among the three \(O\) atoms around \(N\)).
b) In one resonance structure (e.g., \(N\) double - bonded to one \(O\) and single - bonded to two \(O\)s), formal charges: \(N:+1\), double - bonded \(O:0\), single - bonded \(O:-1\) (for each single - bonded \(O\)).
c) Average charge on \(O\) atoms: \(-\frac{2}{3}\), \(N - O\) bond order: \(\frac{4}{3}\)