Problem 118 - Chemistry The Molecular Nature of Matter

Question

Use formal charges to establish the preferred Lewis structures for the \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{ClO}_{4}^{-}\) ions. Draw resonance structures for both ions and determine the average \(\mathrm{Cl}-\mathrm{O}\) bond order in each. Which of these ions would be expected to have the shorter \(\mathrm{Cl}-\mathrm{O}\) bond length?

Step-by-Step Solution

Answer

The \(\mathrm{ClO}_{3}^{-}\) ion, with an average bond order of \(\frac{4}{3}\), is expected to have shorter \(\mathrm{Cl}-\mathrm{O}\) bond lengths than the \(\mathrm{ClO}_{4}^{-}\) ion, which has an average bond order of 1.

Step 1: Determine Total Number of Valence Electrons

For both \(\mathrm{ClO}_{3}^{-}\) and \(\mathrm{ClO}_{4}^{-}\), calculate the total number of valence electrons available for bonding. Chlorine (Cl) has 7 valence electrons, each oxygen (O) has 6, and the extra electron comes from the negative charge. For \(\mathrm{ClO}_{3}^{-}\): \(7 \times 1 + 6 \times 3 + 1 = 26\) electrons. For \(\mathrm{ClO}_{4}^{-}\): \(7 \times 1 + 6 \times 4 + 1 = 32\) electrons.

Step 2: Draw the Lewis Structures

Place Cl at the center as it is the least electronegative, and distribute oxygen atoms around it. For \(\mathrm{ClO}_{3}^{-}\), form bonds between Cl and each O and complete octets for the oxygens, leaving two lone pairs on Cl. For \(\mathrm{ClO}_{4}^{-}\), do the same but with four O atoms. Use remaining electrons to satisfy the octet (or expanded octet in case of Cl) rules.

Step 3: Calculate Formal Charges

Calculate the formal charge for each atom to find the most stable structure. For Cl in both ions, calculate based on the equation: \(\text{Formal charge} = \text{valence electrons} - \text{non-bonding electrons} - \frac{\text{bonding electrons}}{2}\). Adjust double bonds accordingly to minimize charges.

Step 4: Draw Resonance Structures

For \(\mathrm{ClO}_{3}^{-}\), draw three resonance structures by rotating the double bond among the O atoms. For \(\mathrm{ClO}_{4}^{-}\), since all oxygens can have a double bond with Cl, draw three additional resonance structures, each showing a double bond with a different O atom.

Step 5: Determine the Average Bond Order

Calculate the average \(\mathrm{Cl}-\mathrm{O}\) bond order by dividing the total number of bonds between Cl and O by the number of Cl-O bonds. For \(\mathrm{ClO}_{3}^{-}\): \(\frac{4}{3} \.\) For \(\mathrm{ClO}_{4}^{-}\): \(\frac{4}{4} = 1\).

Step 6: Compare Bond Lengths

Based on bond order, the higher the bond order, the stronger and shorter the bond. Thus, \(\mathrm{ClO}_{3}^{-}\) with a bond order of \(\frac{4}{3}\) will have shorter Cl-O bonds than the \(\mathrm{ClO}_{4}^{-}\) with a bond order of 1.

Key Concepts

Formal Charge Calculation

Understanding formal charge is crucial when drawing Lewis structures because it helps find the most stable arrangement of electrons. The formal charge on an atom is given by the formula:
\[\begin{equation}\text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{(\text{Bonding electrons})}{2}\end{equation}\]
For instance, in the \[\begin{equation}\mathrm{ClO}_3^- \end{equation}\]
ion, chlorine has 7 valence electrons. If chlorine forms a single bond with each of the three oxygen atoms and retains two lone pairs, it’d be associated with 8 non-bonding electrons and 6 bonding electrons, resulting in a formal charge of +1. Adjusting the structure to have double bonds can reduce this charge, leading to a more stable ion.
When done correctly, formal charge calculation pinpoints that configuration where the charges are minimized across the molecule, crucial for the molecule's stability in a given environment.

Resonance Structures

Resonance structures depict different possible arrangements of electrons that describe the same molecule. These structures are necessary when a single Lewis structure can't accurately portray bonding in a molecule. For the \[\begin{equation}\mathrm{ClO}_3^- \end{equation}\]
and \[\begin{equation}\mathrm{ClO}_4^- \end{equation}\]
ions, their resonance structures showcase the delocalization of electrons among the oxygen atoms. In \[\begin{equation}\mathrm{ClO}_3^- \end{equation}\]
, it's the rotation of the double bonds, while in \[\begin{equation}\mathrm{ClO}_4^- \end{equation}\]
, each oxygen can potentially form double bonds with chlorine.
It's essential to remember that these diagrams are merely representations; the actual molecule is a hybrid of all resonance structures, resonating to stabilize the molecule's electron density distribution.

Average Bond Order

The average bond order is a way of quantifying the strength and length of bonds between two atoms in a molecule that exhibits resonance. Bond order is calculated as the number of chemical bonds between a pair of atoms. For example, calculating the average bond order in \[\begin{equation}\mathrm{ClO}_3^- \end{equation}\]
is done by dividing the total number of Cl-O bonds by the number of Cl-O linkages, which is: \[\begin{equation}\frac{4}{3}. \end{equation}\]
In contrast, for \[\begin{equation}\mathrm{ClO}_4^- \end{equation}\]
the average bond order is 1. Higher average bond order indicates stronger and shorter bonds. Thus, a molecule with a bond order of \[\begin{equation}\frac{4}{3}\end{equation}\]
will have shorter bonds compared to one with a bond order of 1, like in \[\begin{equation}\mathrm{ClO}_4^- \end{equation}\]
. Consequently, this concept serves as a useful predictor for comparing bond lengths in different ions or resonance structures.