Question 10.7: Writing a Lewis Structure for a Polyatomic Ion Write the Lew...

Step 1. Determine the total number of valence electrons. The N atom (group 15) has five valence electrons, and each of the two O atoms (group 16) has six. However, one valence electron must be removed to produce the charge of 1+.
The total number of valence electrons is 5 + 6 + 6 – 1 = 16

Step 2. Identify the central atom(s) and terminal atoms. The N atom has a lower electronegativity (3.0) than the O atoms (3.5). N is the central atom, and the O atoms are the terminal atoms.

Step 3. Write a plausible skeletal structure by joining atoms through single covalent bonds.

\begin{array}{r c} \begin{matrix} O-N-O \end{matrix} \end{array}

Step 4. Subtract two electrons for each bond in the skeletal structure. The two bonds in this structure account for 4 of the 16 valence electrons. This leaves 12 valence electrons to be assigned.

Step 5. Complete octets for the terminal O atoms, and to the extent possible, the central N atom. The remaining 12 valence electrons are sufficient only to complete the octets of the O atoms.

\left[\begin{array}{r c} \begin{matrix} :\underset{\cdot \cdot}{\ddot{O} }-N-\underset{\cdot \cdot}{\ddot{O} }: \end{matrix} \end{array} \right]^+

Step 6. Move lone pairs of electrons from the terminal O atoms to form multiple bonds to the central N atom. The N atom has only four electrons in its valence shell and needs four more to complete an octet. Thus, the N atom requires two additional pairs of electrons, which it acquires if we move one lone pair from each O atom into its bond with the N atom, as shown below.

Assess
After drawing a Lewis structure, and before moving on to the next step of a problem or to the next exercise, check the structure. Each atom is surrounded by 8 electrons (each atom has an octet), and the structure has a total valence of 16 (we have not inadvertently added or dropped electrons). In assessing the structure, we must remember that each line represents two electrons (a bonding pair).