Q. no3- lewis structure formal charge

Step 1. Identify the species and count total valence electrons.

The species is the nitrate ion, \(\mathrm{NO_3}^{-}\). Count valence electrons for each atom and add one extra electron for the negative charge. Nitrogen has 5 valence electrons. Each oxygen has 6 valence electrons, and there are three oxygens. Add 1 electron for the overall \(-1\) charge. So the total number of valence electrons is

\[ 5 + 3\times 6 + 1 = 5 + 18 + 1 = 24. \]

Step 2. Draw the skeletal structure and place bonding electrons.

Nitrogen is the less electronegative atom, so place N in the center with three O atoms bonded to it. Form three single N–O bonds initially. Each single bond uses 2 electrons. Electrons used in three single bonds:

\[ 3 \times 2 = 6. \]

Remaining electrons after making the single bonds:

\[ 24 – 6 = 18. \]

Step 3. Distribute remaining electrons to satisfy octets, starting with the outer atoms (the oxygens).

Place the remaining 18 electrons as lone pairs on the three oxygens to fill their octets. Each oxygen already has 1 bonding pair (2 electrons) from the N–O single bond, so each oxygen needs 6 more electrons to complete an octet. That uses exactly 18 electrons (3 oxygens × 6 electrons each). After this placement, each oxygen has an octet, and nitrogen has only 6 electrons around it (three single bonds). Nitrogen does not yet have an octet.

Step 4. Form a multiple bond to give nitrogen an octet.

To complete nitrogen’s octet, convert one lone pair from one oxygen into a bonding pair between that oxygen and nitrogen, forming an N=O double bond. This yields a structure with one N=O double bond and two N–O single bonds. At this point every atom has an octet.

Step 5. Calculate formal charges to check the best Lewis structure. Use the formal charge formula:

\[ \text{Formal charge} = V – N – \frac{B}{2}, \]

where \(V\) is the number of valence electrons for the neutral atom, \(N\) is the number of nonbonding (lone pair) electrons on the atom in the structure, and \(B\) is the total number of electrons in bonds (count bonding electrons, not pairs). Compute formal charges for each atom in the structure with one double-bonded oxygen and two single-bonded oxygens.

Compute the formal charge on the double-bonded oxygen.

\[ V(\mathrm{O}) = 6, \quad N = 4 \text{ (two lone pairs, 4 electrons)}, \quad B = 4 \text{ (two bonds, 4 electrons)}. \]

\[ \text{Formal charge on double-bonded O} = 6 – 4 – \frac{4}{2} = 6 – 4 – 2 = 0. \]

Compute the formal charge on a single-bonded oxygen.

\[ V(\mathrm{O}) = 6, \quad N = 6 \text{ (three lone pairs, 6 electrons)}, \quad B = 2 \text{ (one bond, 2 electrons)}. \]

\[ \text{Formal charge on single-bonded O} = 6 – 6 – \frac{2}{2} = 6 – 6 – 1 = -1. \]

Compute the formal charge on nitrogen.

\[ V(\mathrm{N}) = 5, \quad N = 0 \text{ (no lone pairs on N in this structure)}, \quad B = 8 \text{ (four bonds total, 8 electrons)}. \]

\[ \text{Formal charge on N} = 5 – 0 – \frac{8}{2} = 5 – 0 – 4 = +1. \]

Step 6. Summarize the formal charges and overall charge.

In this Lewis structure there is one oxygen with formal charge \(0\), two oxygens each with formal charge \(-1\), and nitrogen with formal charge \(+1\). The sum of the formal charges is

\[ (+1) + 0 + (-1) + (-1) = -1, \]

which matches the overall charge of the ion, \(-1\).

Step 7. Account for resonance.

The double bond can be placed between nitrogen and any one of the three oxygens, producing three equivalent resonance structures. In each resonance structure, nitrogen carries a formal charge of \(+1\), one oxygen is neutral \(0\), and the other two oxygens carry \(-1\) each. The true structure of the nitrate ion is a resonance hybrid in which the negative charge is delocalized equally over the three oxygens, and all N–O bond distances are equivalent and intermediate between a single and a double bond.

Final statement of the Lewis structure and formal charges.

The nitrate ion, \(\mathrm{NO_3}^{-}\), is represented by three equivalent resonance structures, each having formal charges: nitrogen \(+1\), one oxygen \(0\), and the other two oxygens \(-1\) each. The overall charge is \(-1\), and the negative charge is delocalized over the three oxygens in the resonance hybrid.