Question

The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of \(1.24 \mathrm{D},\) and the \(\mathrm{Cl}-\) O bond length is 1.60 \(\mathrm{A}\) . (a) Determine the magnitude of the charges on the Cl and O atoms in units of the electronic charge, \(e\) (b) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the Clo molecule? (c) Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (d) The anion \(\mathrm{ClO}^{-}\) exists. What is the formal charge on the Cl for the best Lewis structure for \(\mathrm{ClO}^{-}\) ?

Short answer

The magnitude of the charges on Cl and O atoms in ClO is approximately \(0.16e\). Oxygen is expected to have a partial negative charge due to its higher electronegativity. The dominant Lewis structure for ClO molecule has one single bond between O and Cl, with 3 lone pairs on O and 1 lone pair on Cl. Formal charge on Cl is -1, and on O is 0. For the best Lewis structure of ClO^-, there is one single bond between O and Cl, with 4 lone pairs on O and 1 lone pair on Cl. The formal charge on Cl in ClO^- is -1.

Step-by-step solution

Determine the magnitude of charges on Cl and O atoms.

To determine the magnitude of the charges on the Cl and O atoms, we can use the dipole moment formula: \( \mu = q \times d \) Here, \( \mu \) is the dipole moment, \( q \) is the magnitude of the charge on either atom, and \( d \) is the bond length between the atoms. We are given \( \mu = 1.24 \mathrm{D} \) and \( d = 1.60 \mathrm{A}\) (here 1 Å = 100 pm). We need to convert the units here to be consistent. Dipole moments in Debyes (D) can be converted to Coulomb-meters (Cm) using the conversion factor: 1 D = 3.336 x 10^{-30} Cm Bond lengths in Angstroms (A) can be converted to meters (m) with 1 A = 10^{-10} m Now, plug the values into the formula: \( 1.24 \mathrm{D} = q \times 1.60 \mathrm{A} \)

Convert units and solve for the magnitude of charges on the atoms.

First, we will convert the units for \( \mu \) and \( d \): \( \mu = 1.24 \mathrm{D} \times \frac{3.336 \times 10^{-30} \mathrm{Cm}}{1 \mathrm{D}} = 4.137 \times 10^{-30} \mathrm{Cm} \) \( d = 1.60 \mathrm{A} \times \frac{10^{-10} \mathrm{m}}{1 \mathrm{A}} = 1.60 \times 10^{-10} \mathrm{m} \) Now we will replace the values in the equation: \( 4.137 \times 10^{-30} \mathrm{Cm} = q \times 1.60 \times 10^{-10} \mathrm{m} \) Solving for \( q \): \( q = \frac{4.137 \times 10^{-30} \mathrm{C}}{1.60 \times 10^{-10} \mathrm{m}} = 2.59 \times 10^{-20} \mathrm{C} \) To express the magnitude of the charges in units of the electronic charge \(e\), divide the charge magnitude by \(e = 1.602 \times 10^{-19} \mathrm{C}\): \( \frac{2.59 \times 10^{-20} \mathrm{C}}{1.602 \times 10^{-19} \mathrm{C}} \approx 0.16e \)

Identify which atom has a partial negative charge based on electronegativity.

In general, the atom with higher electronegativity will have the partial negative charge. Oxygen has an electronegativity of 3.44, while chlorine has an electronegativity of 3.16. Since oxygen is more electronegative than chlorine, we can expect that oxygen will have a partial negative charge in the ClO molecule.

Propose the dominant Lewis structure.

Based on formal charges, we want to minimize formal charge discrepancies in a molecule. In ClO, chlorine has 7 valence electrons and oxygen has 6 valence electrons. We can create a single covalent bond between the two atoms, sharing one electron pair. After forming the bond, there are 6 non-bonding electrons on chlorine and 8 non-bonding electrons (4 lone pairs) on oxygen. Cl: Non-bonding electrons = 6; Groups = 2 (1 bond + 1 lone pair); Formal charge = Valence - non-bonding - (groups) = 7 - 6 - 2 = -1 O: Non-bonding electrons = 8; Groups = 4 (1 bond + 3 lone pairs); Formal charge = Valence - non-bonding - (groups) = 6 - 8 - 4 = 0. The dominant Lewis structure has one single bond between O and Cl, with 3 lone pairs on O, and 1 lone pair on Cl. The formal charge on Cl is -1, and the formal charge on O is 0.

Determine the formal charge on Cl for the best Lewis structure of ClO^-.

In ClO^-, we have an additional electron in the system. Now, chlorine has 7 valence electrons and oxygen has 7 valence electrons in this molecule. We can create a single covalent bond between the two atoms, sharing one electron pair. After forming the bond, there are 6 non-bonding electrons on chlorine and 10 non-bonding electrons (five lone pairs) on oxygen. Cl: Non-bonding electrons = 6; Groups = 2 (1 bond + 1 lone pair); Formal charge = Valence - non-bonding - (groups) = 7 - 6 - 2 = -1 O: Non-bonding electrons = 10; Groups = 4 (1 bond + 4 lone pairs); Formal charge = Valence - non-bonding - (groups) = 7 - 10 - 4 = 0. The best Lewis structure for ClO^- has one single bond between O and Cl, with 4 lone pairs on O, and 1 lone pair on Cl. The formal charge on Cl in ClO^- is -1.

Key concepts

Dipole Moment Calculation

The dipole moment is a vector quantity that represents the separation of charge within a molecule. For a diatomic molecule like chlorine monoxide (ClO), the dipole moment can be calculated using the equation
\[ \mu = q \times d \]
where \( \mu \) is the dipole moment, \( q \) is the magnitude of the charge on the atoms, and \( d \) is the bond length between the atoms. In the given exercise, the conversion of units is crucial to get the answer in the required units of electronic charge, \(e\).
After conversion and calculation, the magnitude of the charges on the chlorine and oxygen in ClO comes out to be approximately \(0.16e\), which is a fraction of the elementary charge. Understanding how to calculate this provides insight not just for homework but also for understanding molecular properties in chemistry.

Electronegativity and Partial Charges

Electronegativity is a measure of an atom's ability to attract and hold onto electrons. When bonding, atoms with higher electronegativity will pull more strongly on the shared electrons, creating a partial negative charge.
In ClO, oxygen is more electronegative than chlorine, hence it holds a partial negative charge. With this knowledge, we can predict the physical properties and reactivity of the molecule. For example, the direction of the dipole moment in ClO points towards oxygen, indicating oxygen's greater share of electron density. Understanding how electronegativity affects the distribution of charge within a molecule is key to grasping the concept of polarity and interactions with other molecules.

Lewis Structures and Formal Charges

The Lewis structure is a representation of molecules that show all valence electrons and their arrangement. Each dot represents an electron, while lines represent shared pairs forming covalent bonds. Formal charges help to identify the most stable Lewis structure, particularly in molecules where multiple structures are possible.

Calculating Formal Charges

The formal charge on an atom in a molecule is calculated using the formula:
\[ \text{Formal charge} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \frac{1}{2}(\text{Bonding electrons}) \]
For ClO, based on the formal charges calculated in the exercise (Cl with -1 and O with 0), the proposed dominant Lewis structure has one single bond and three lone pairs on O and one lone pair on Cl. This arrangement minimizes formal charge discrepancies, leading to a more stable molecule.
In ClO^-, the additional electron results in the formal charge on Cl being -1, while it remains 0 on O. Through such analysis, we can hypothesize about the likely structure of related molecules.