Problem 148 - Chemistry The Science in Context

Question

A compound consisting of chlorine and oxygen, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) decomposes by the following reaction: $$ \mathrm{Cl}_{2} \mathrm{O}_{7} \rightarrow \mathrm{ClO}_{4}+\mathrm{ClO}_{3} $$ a. Draw two Lewis structures for \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) : one with a chlorine-chlorine bond and one with a \(\mathrm{Cl}-\mathrm{O}-\mathrm{Cl}\) arrangement of atoms. b. Draw a Lewis structure for \(\mathrm{ClO}_{3}\)

Step-by-Step Solution

Answer

Question: Draw two Lewis structures for \(\mathrm{Cl}_2 \mathrm{O}_7\) based on the given arrangements, and draw the Lewis structure for \(\mathrm{ClO}_3\). Answer: The two Lewis structures for \(\mathrm{Cl}_2 \mathrm{O}_7\) are: a. Chlorine-chlorine bond: Cl(Cl)-O-O(Cl)-O-O-O b. \(\mathrm{Cl}-\mathrm{O}-\mathrm{Cl}\) arrangement: \(\mathrm{ClO}_{2} -\mathrm{O}-\mathrm{ClO}_{2}\) The Lewis structure for \(\mathrm{ClO}_{3}\) is \(\mathrm{Cl}(\mathrm{O})_{\mathrm{O}} \mathrm{O}\).

Step 1: Part a: Drawing Lewis structures for \(\mathrm{Cl}_2 \mathrm{O}_7\)

1. Calculate the total number of valence electrons for the given molecule: - Chlorine (Cl) has 7 valence electrons - Oxygen (O) has 6 valence electrons Total valence electrons: (2 x 7) + (7 x 6) = 14 + 42 = 56 valence electrons.

Step 2: 2. Draw the skeleton structure for both given arrangements: a. Chlorine-chlorine bond: Cl - Cl-O-O-O-O-O-O b. \(\mathrm{Cl}-\mathrm{O}-\mathrm{Cl}\) arrangement: Cl-O-Cl-O-O-O-O

3. Distribute the remaining valence electrons in pairs around the atoms to form octets: - For each chlorine atom, add 3 lone pairs (6 electrons). - For each oxygen atom (except the central one in the \(\mathrm{Cl}-\mathrm{O}-\mathrm{Cl}\) arrangement), add 2 lone pairs (4 electrons). a. Chlorine-chlorine bond: Cl(Cl)-O-O(Cl)-O-O-O b. \(\mathrm{Cl}-\mathrm{O}-\mathrm{Cl}\) arrangement: \(\mathrm{ClO}_{2} -\mathrm{O}-\mathrm{ClO}_{2}\) Note: In case of the central oxygen (arrangement b), add 3 lone pairs (6 electrons).

Step 3: Part b: Drawing Lewis structure for \(\mathrm{ClO}_{3}\)

1. Calculate the total number of valence electrons for the given molecule: - Chlorine (Cl) has 7 valence electrons - Oxygen (O) has 6 valence electrons Total valence electrons: 7 + (3 x 6) = 7 + 18 = 25 valence electrons.

Step 4: 2. Draw the skeleton structure: Cl - O - O - O

3. Distribute the remaining valence electrons in pairs around the atoms to form octets: - For each oxygen atom, add 2 lone pairs (4 electrons). - For the chlorine atom, add 2 lone pairs (4 electrons) and one unpaired electron. Cl(Cl) - O - O - O The resulting Lewis structure for \(\mathrm{ClO}_{3}\) is \(\mathrm{Cl}(\mathrm{O})_{\mathrm{O}} \mathrm{O}\).

Key Concepts

Valence Electrons

Valence electrons play a crucial role in determining the chemical properties of an element. These are the electrons located in the outermost shell of an atom and are involved in chemical bonding. For instance, chlorine has 7 valence electrons, while oxygen has 6. To find the total number of valence electrons in a molecule, you sum the valence electrons of all the atoms present. This calculation helps in constructing Lewis structures, where valence electrons are shown as dots around the element symbols.

• For Chlorine ( Cl ), it has 7 valence electrons.
• For Oxygen ( O ), it has 6 valence electrons.

To construct the Lewis structure for Cl_2O_7 , you need to account for all valence electrons: two chlorines contribute 14, and seven oxygens contribute 42, giving a total of 56 electrons. These electrons are then assigned to bonds and lone pairs, respecting the octet rule, to accurately represent the molecule's structure.

Chlorine-Oxygen Compounds

Chlorine-oxygen compounds, like Cl_2O_7 , are significant for their reactivity and various applications, ranging from bleaching agents to disinfectants. These compounds consist of chlorine and oxygen atoms bonded in different configurations, sometimes forming powerful oxidizers.
In the context of Lewis structures, understanding the bonding in chlorine-oxygen compounds is essential. For Cl_2O_7 , you can represent it two ways:

• One structure with a direct chlorine-chlorine bond and oxygen atoms attached.
• Another arrangement where the chlorines are bonded through an oxygen atom ( Cl-O-Cl ), creating an ether-like bridge.

When drawing these structures, it's crucial to fulfill the octet rule for all the involved atoms, ensuring that chlorine and oxygen have complete outer electron shells. This helps in predicting the compound's stability and reactivity.

Molecular Decomposition

Molecular decomposition is the process where a chemical compound breaks down into simpler substances. In the decomposition of Cl_2O_7 , it breaks down to form ClO_4 and ClO_3 . Understanding this reaction is essential for grasping how complex molecules transform and react.
For ClO_3 , we need to consider its total valence electrons: chlorine provides 7, while each of the three oxygens provide 6, accumulating to 25 electrons. The Lewis structure reflects these valence electrons used in bonds and lone pairs, maintaining the octet rule where possible. During decomposition, energy dynamics can influence which bonds break and which new ones form.

• Lewis structures offer a visual insight into bond formation and breaking in decomposition reactions.
• Analytical methods like these help predict the products of decomposition reactions, essential in chemical synthesis and analysis.

Grasping molecular decomposition is vital for understanding chemical processes in industries as well as biological systems.