Question

Perchlorates are powerful oxidizing agents used in fireworks, flares, and the booster rockets of space shuttles. Lewis structures for the perchlorate ion \(\left(\mathrm{ClO}_{4}^{-}\right)\) can be drawn with all single bonds or with one, two, or three double bonds. Draw each of these possible resonance forms, use formal charges to determine the more important structure, and calculate its average bond order.

Short answer

Draw the Lewis structures, calculate and compare formal charges, identify the most important structure, and then calculate the average bond order.

Step-by-step solution

Drawing the Lewis structures

Draw the possible Lewis structures for the perchlorate ion \(\text{ClO}_{4}^{-}\). This includes one with all single bonds, and others with one, two, or three double bonds.

Calculating formal charges

Calculate the formal charge for each atom in each resonance structure to determine which structure has the minimal and most evenly spread formal charges.

Identifying the most important structure

Identify the resonance structure with the least formal charges on each atom as the most important.

Calculating the average bond order

Determine the bond order by considering all resonance structures. Sum the bond orders and average them to find the average bond order for a \(\text{Cl-O}\) bond in \(\text{ClO}_{4}^{-}\).

Key concepts

Lewis structures

Understanding Lewis structures is crucial for visualizing molecules. They show the arrangement of atoms and valence electrons in a molecule.
Drawing Lewis structures for the perchlorate ion \(\text{ClO}_{4}^{-}\) involves placing chlorine (Cl) at the center with four oxygen (O) atoms around it. Chlorine has 7 valence electrons and each oxygen has 6 valence electrons.
This sums to a total of 32 valence electrons (Chlorine: 7 \( \times 1 \) + Oxygen: 6 \( \times 4 \) + 1 extra for the negative charge).
The structure needs to show all these valence electrons.

• Start with single bonds from Cl to each O.
• Distribute remaining electrons to satisfy the octet rule.

It's possible to draw structures with one, two, or three double bonds because of resonance.

formal charge calculation

Calculating formal charges helps to determine the most stable Lewis structure.
The formal charge of an atom is calculated using the formula:
\[ \text{Formal Charge} = \text{valence electrons} - (\text{non-bonding electrons} + \frac{1}{2} \text{bonding electrons}) \]
For each resonance form of \(\text{ClO}_{4}^{-}\):

• Count bonding and non-bonding electrons for each atom.
• Plug these values into the formula.
• Calculate the formal charge for Cl and each O.

Example: In a form with all single bonds, Cl has 7 valence electrons, 4 bond pairs (8 electrons shared), and no lone pairs.
\[ \text{Formal Charge of Cl} = 7 - (0 + 4) = +3 \] This structure is less stable because of high positive formal charge.

resonance structures

Resonance structures represent different ways to arrange electrons in a molecule. For \(\text{ClO}_{4}^{-}\), we can have structures with different numbers of double bonds.
These structures contribute to the actual form of the molecule.

• Draw each resonance form.
• Calculate formal charges to find the most stable arrangement.
• Remember, the actual structure is a hybrid.

Having resonance means no single Lewis structure can fully describe the ion. The electrons are delocalized, and the true form is a weighted average of all valid structures.

average bond order

The average bond order provides insight into the bonding in resonance structures. It's the average number of bonds between two atoms across all resonance forms.
For \(\text{ClO}_{4}^{-}\):

• Count the total number of bonds in each structure.
• Sum these values.
• Divide by the number of resonance structures.

Example: With four structures (one single, one double bond in various positions):
\[ \text{Total Bonds} = 4 (\text{single}) + 3 (\text{with one double}) + 2 (\text{with two doubles}) + 1 (\text{with three doubles}) = 10 \]
Divide by four structures: \[ \text{Average Bond Order} = \frac{10}{4} = 2.5 \]
The average bond order tells us each \(\text{Cl-O}\) bond is somewhere between a single and double bond.