Question 10.8: Using Formal Charges in Writing Lewis Structures Write the m...

Regardless of the skeletal structure chosen, the number of valence electrons (dots) that must appear in the final Lewis structure is

5 from N + 6 from O + 7 from Cl = 18

When we apply the four steps listed below to the two possible skeletal structures, we obtain a total of four Lewis structuresــــtwo for each skeletal structure. This doubling occurs because in step 4, there are two ways to complete the octets of the central atoms. The final Lewis structures obtained are labeled (a_1), (a_2), (b_1), and (b_2).

\begin{matrix} (a) & & (b) \\ \begin{array}{r c} \begin{matrix}O-Cl-N \end{matrix} \end{array} & \text{1. Assign four electrons.} & \begin{array}{r c} \begin{matrix}O-N-Cl \end{matrix} \end{array} \\ \begin{array}{r c} \begin{matrix} :\underset{\cdot \cdot}{\ddot{O} }-Cl-\underset{\cdot \cdot}{\ddot{N} } : \end{matrix} \end{array} & \text{2. Assign twelve more electrons.} & \begin{array}{r c} \begin{matrix} :\underset{\cdot \cdot}{\ddot{O} }-N-\underset{\cdot \cdot}{\ddot{Cl} } : \end{matrix} \end{array} \\ \begin{array}{r c} \begin{matrix} :\underset{\cdot \cdot}{\ddot{O} }-\ddot{Cl} -\underset{\cdot \cdot}{\ddot{N} } : \end{matrix} \end{array} & \text{3. Assign the last two electrons.} & \begin{array}{r c} \begin{matrix} :\underset{\cdot \cdot}{\ddot{O} }-\ddot{N} -\underset{\cdot \cdot}{\ddot{Cl} } : \end{matrix} \end{array} \\ & \text{4. Complete the octet on the central atom.} & \end{matrix}

\begin{matrix} (a_{1}) & (a_{2}) & (b_{1}) & (b_{2}) \\ \begin{array}{r c}\begin{matrix}:\ddot{O} =\ddot{Cl}-\underset{\cdot \cdot}{\ddot{N} } : \end{matrix} \end{array} & \begin{array}{r c}\begin{matrix}:\underset{\cdot \cdot}{\ddot{O} } -\ddot{Cl}=\ddot{N} : \end{matrix} \end{array} & \begin{array}{r c}\begin{matrix}:\ddot{O} =\ddot{N}-\underset{\cdot \cdot}{\ddot{Cl} } : \end{matrix} \end{array} & \begin{array}{r c}\begin{matrix}:\underset{\cdot \cdot}{\ddot{O} } -\ddot{N}=\ddot{Cl} : \end{matrix} \end{array} \end{matrix}

Evaluate formal charges by using equation (10.16). In structure (a_{1}),

FC = \text{number valence } e^- \text{in free atom } –  \text{number lone-pair } e^- -\frac{1}{2} \text{number bond-pair } e^-                                        (10.16)

for the N atom,

F C = 5  –  6  –  \frac{1}{2}(2) = -2

for the O atom,

F C = 6  –  4  –  \frac{1}{2}(4) = 0

for the Cl atom,

F C = 7  –  2  –  \frac{1}{2}(6) = +2

Proceed in a similar manner for the other three structures. Summarize the formal charges for the four structures.

\begin{matrix} & (a_{1}) & (a_{2}) & (b_{1}) & (b_{2}) \\ \text{N:} & -2 & -1 & 0 & 0 \\ \text{O:} & 0 & -1 & 0 & -1 \\ \text{Cl:} & +2 & +2 & 0 & +1 \end{matrix}

Select the best Lewis structure in terms of the formal-charge rules. First, note that all four structures obey the requirement that formal charges of a neutral molecule add up to zero. In structure (a_{1}), the formal charges are large (+2 on Cl and -2 on N) and the negative formal charge is not on the most electronegative atom. Structure (a_{2}) has formal charges on all atoms, one of them large (+2 on Cl). Structure (b_{1}) is the ideal we seekــــno formal charges. In structure (b_{2}), we again have formal charges. The best Lewis structure of nitrosyl chloride is

\begin{array}{r c} \begin{matrix} :\ddot{O} =\ddot{N} -\underset{\cdot \cdot}{\ddot{Cl} }: \end{matrix}\end{array}

Assess
Based on structure (b_{1}), ONCl is a better way to write the formula of nitrosyl chloride.