Problem 64 - Chemistry The Science in Context

Question

Draw the Lewis structure of the chlorite ion, \(\mathrm{ClO}_{2}^{-}\) which is used as a bleaching agent. Include all resonance structures in which formal charges are closest to zero. What is the shape of the ion? Suggest a hybridization scheme for the central chlorine atom that accounts for the structures you have drawn.

Step-by-Step Solution

Answer

The Lewis structure of the chlorite ion (\(\mathrm{ClO}_{2}^{-}\)) can be drawn with two resonance structures that have formal charges of 0 on each atom. These resonance structures have chlorine double-bonded to both oxygen atoms: $$ \text{Cl}=O-\text{O}^{\cdot} \leftrightarrow \text{Cl}-\text{O}^{\cdot}-O $$ The molecular shape of the chlorite ion is bent or V-shaped, with a central chlorine atom surrounded by two oxygen atoms, due to the one lone pair on the chlorine atom. The hybridization of the central chlorine atom is \(\mathrm{sp^2}\).

Step 1: Calculate the total number of valence electrons

In the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), we have the following valence electrons: - Cl: 7 valence electrons, as it is in Group 7 - O: 6 valence electrons (two O atoms, so 6x2=12) - Additional electron due to the negative charge: 1 Total valence electrons: 7 + 12 + 1 = 20

Step 2: Connect the atoms by single bonds

Place the chlorine atom (Cl) in the center and connect it to the two oxygen (O) atoms with single bonds. This would use 4 of the 20 valence electrons: Cl-O-O

Step 3: Complete the octets for the surrounding atoms

Distribute the remaining 16 electrons (20 - 4 from Step 2) on the two O atoms to complete their octets: $$ \text{Cl}\text{---}\text{O}^{\cdot}\text{---}\text{O} $$

Step 4: Place any remaining electrons on the central atom

No valence electrons are left to place on the central chlorine atom.

Step 5: Evaluate the formal charge for each atom

To calculate the formal charge, the formula is: Formal charge = Valence electrons - Non-bonding electrons - (Bonding electrons/2) - The formal charge of the first O atom (triple bonded): 6 (valence electrons) - 4 (non-bonding electrons) - (6 bonding electrons/2) = +1 - The formal charge of the second O atom (single bonded): 6 (valence electrons) - 6 (non-bonding electrons) - (2 bonding electrons/2) = -1 - The formal charge of the central Cl atom: 7 (valence electrons) - 0 (non-bonding electrons) - (8 bonding electrons/2) = 0

Step 6: Draw resonance structures and minimize formal charges

A way to minimize formal charges is distributing the electrons between the chlorine and the first triple bonded oxygen as a double bond to both oxygen atoms. Now we have two resonance structures with formal charges of 0 on each atom: $$ \text{Cl}=O-\text{O}^{\cdot} \leftrightarrow \text{Cl}-\text{O}^{\cdot}-O $$

Step 7: Calculate the molecular shape and hybridization of the central atom

We have two electron groups around the central chlorine atom: two double bonds (or resonance structures) and one lone pair. This corresponds to a trigonal planar geometry. However, since there is one lone pair, the actual molecular shape is bent or V-shaped. To accommodate the trigonal planar arrangement of the electron groups, the hybridization of the chlorine atom should be \(\mathrm{sp^2}\).

Key Concepts

Resonance Structures

Resonance structures are alternative ways of drawing Lewis structures for a molecule or ion, which can represent the same arrangement of atoms. For the chlorite ion \(\mathrm{ClO}_{2}^{-}\), resonance structures illustrate how electrons can be distributed across different bonds. In this case, the electrons can be shifted between the chlorine atom and the oxygen atoms to create distinct bonding patterns.

• Among several resonance structures, only those configurations that minimize formal charge are considered stable.
• They help give a more accurate depiction of the electron distribution in a molecule than a single static structure.

This means each oxygen can alternatively be double bonded or singly bonded to chlorine, distributing the electrons in slightly different ways.

Formal Charge

Formal charge is a concept used to estimate the charge on an atom in a molecule, aiding in the determination of the most stable resonance structure. It is calculated by using the formula:
\[\text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \left( \frac{\text{Bonding electrons}}{2} \right)\]The goal when assigning formal charges is to have the smallest values possible or zero across the atoms, which indicates higher stability of the molecule or ion.

• In \(\mathrm{ClO}_{2}^{-}\), the structure where each atom has a formal charge as close to zero as possible is favored.
• Each resonance form should preferably have atoms carrying zero formal charge to achieve a stable configuration.
• This stability results from minimizing electron-electron repulsions within the ion or molecule.

Molecular Shape

Molecular shape determines many properties of a molecule, including reactivity, polarity, and biological activity. In the chlorite ion \(\mathrm{ClO}_{2}^{-}\), the molecular shape is affected by the positioning of the electrons around the chlorine atom.

• For the chlorite ion, it assumes a 'bent' or 'V-shaped' geometry.
• This results because of the electron pairs' repulsion, causing the bonds to angle away from each other.
• Although the theoretical geometry based on electron pairs (the two double-bond and lone pair) would be trigonal planar, the lone pair's presence alters this to bent.

The actual shape is critical as it dictates how the ion interacts with other molecules, crucial for understanding chemical reactions and biological processes.

Hybridization

Hybridization is a concept explaining how atomic orbitals fuse to form new hybrid orbitals, which influence the geometry and bonding properties of molecules. In the chlorite ion \(\mathrm{ClO}_{2}^{-}\), hybridization helps account for observed molecular shapes.

• Chlorine in the chlorite ion is \(\mathrm{sp^2}\) hybridized.
• This hybridization arises because the ion includes two double bonds (in resonance), which corresponds to three electron-density regions requiring trigonal planar electron-group geometry.
• The chlorine atom uses one \(s\) orbital and two \(p\) orbitals to create three \(\mathrm{sp^2}\) hybrid orbitals.

The \(\mathrm{sp^2}\) hybridization correctly supports the bent shape due to electron pair repulsion and the delocalization from resonance.