Problem 86 - Chemistry: The Central Science

Question

The substance chlorine monoxide, \(\mathrm{ClO}(\mathrm{g})\), 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 \(\mathrm{Cl}\) - \(\mathrm{O}\) bond length is \(1.60 \hat{A}\). (a) Determine the magnitude of the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) 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 \(\mathrm{ClO}\) 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 \(\mathrm{Cl}\) for the best Lewis structure for \(\mathrm{ClO}^{-}\)?

Step-by-Step Solution

Answer

In summary, for the chlorine monoxide (ClO) molecule: a) The magnitude of the charges on the Cl and O atoms is approximately \(1.61\,e\). b) Oxygen (O) has a partial negative charge due to its higher electronegativity value. c) The dominant Lewis structure for ClO has formal charges of +1 on Cl and -1 on O. d) The formal charge on Cl in the ClO\(^{-}\) ion is 0.

Step 1: Recall or find the given information

We have been given that the molecule has a dipole moment of 1.24 D, and the Cl-O bond length is 1.60 Å. We need to determine the magnitude of the charges on the Cl and O atoms in units of the electronic charge, e.

Step 2: Convert bond length from Å to meters

We must first convert the bond length from Å (angstrom) to meters (m) since the unit of charge is related to units of meters. Recall that 1 Å = 10\(^{-10}\) m. Therefore, the bond length of 1.60 Å is equal to 1.60 x 10\(^{-10}\) m.

Step 3: Calculate the magnitude of the charges

In a polar molecule, the dipole moment can be calculated using the formula: \[ \mu = q \cdot d,\] where \( \mu\) is the dipole moment, \(q\) is the magnitude of the charges on the atoms (in Coulombs), and \(d\) is the bond length (in meters). We can re-arrange the formula to solve for the magnitude of the charges: \[ q = \frac{\mu}{d}\] Now, substitute the values of \( \mu\) and \(d\) in the equation to find the charge magnitude: \[ q = \frac{1.24 D}{1.60 × 10^{-10} m}\] Keep in mind that 1 Debye (D) equals 3.336 x 10\(^{-30}\) Cm, so convert the dipole moment to Coulomb-meters by multiplying by this conversion factor: \[ q = \frac{1.24 \times 3.336 \times 10^{-30} Cm}{1.60 × 10^{-10} m}\] Finally, calculate the charge magnitude in Coulombs: \[ q \approx 2.58 × 10^{-19} C\]

Step 4: Convert charge magnitude to electronic charge

The electronic charge, e, is equal to 1.602 × 10\(^{-19}\) C. To express the charge magnitude in units of e, divide the charge magnitude in Coulombs by the electronic charge value: \[ q_e = \frac{2.58 × 10^{-19} C}{1.602 × 10^{-19} C}\] Thus, the magnitude of the charges on the Cl and O atoms is: \[ q_e \approx 1.61 \,e\] #b) Determine which atom has a partial negative charge#

Step 1: Recall the electronegativity values for Cl and O

The electronegativities of Cl and O are 3.16 and 3.44 respectively on the Pauling scale.

Step 2: Identify the atom with a higher electronegativity value

Since the electronegativity value of O (3.44) is higher than that of Cl (3.16), O has a greater tendency to attract electrons and therefore will have a partial negative charge. #c) Propose the dominant Lewis structure for the ClO molecule#

Step 1: Count the total number of valence electrons

Chlorine has 7 valence electrons, and oxygen has 6 valence electrons. Their total combined valence electrons for ClO are 7+6 = 13 electrons.

Step 2: Draw the skeletal structure and distribute the electrons

Connect Cl to O, using a single bond which uses 2 electrons. Distribute the remaining 11 electrons as lone pairs, starting with the outer atoms, following the octet rule (except for hydrogen). In this case, both atoms are outer atoms. Oxygen will have 3 lone pairs (6 electrons), and chlorine will have 2 lone pairs (4 electrons) and one unpaired electron.

Step 3: Calculate formal charges

For the Cl atom, we have: \[ FC(Cl) = Valence(Cl) - (Nonbonding\, e^{-} + \frac{ Bonding\, e^{-}}{2})\] \[ FC(Cl) = 7 - (5 + \frac{2}{2}) \] \[ FC(Cl) = +1 \] For the O atom, we have: \[ FC(O) = Valence(O) - (Nonbonding\, e^{-} + \frac{ Bonding\, e^{-}}{2})\] \[ FC(O) = 6 - (6 + \frac{2}{2}) \] \[ FC(O) = -1 \] Since these formal charges are relatively small, this proposed Lewis structure is likely dominant. #d) Determine the formal charge on Cl for the best Lewis structure for ClO\(^-\)#

Step 1: Account for the additional electron in ClO\(^-\)

The ClO\(^-\) ion has one extra electron compared to the neutral ClO molecule. This electron gets added to the unpaired electron on the Cl atom in the molecule.

Step 2: Calculate the formal charge on Cl

For the Cl atom in the ClO\(^-\) ion, we have: \[ FC(Cl) = Valence(Cl) - (Nonbonding\, e^{-} + \frac{ Bonding\, e^{-}}{2})\] \[ FC(Cl) = 7 - (6 + \frac{2}{2}) \] \[ FC(Cl) = 0 \] The formal charge on Cl for the best Lewis structure for ClO\(^-\) is 0.

Key Concepts

Ozone layer depletion

The depletion of the ozone layer is a significant environmental issue that affects Earth's atmosphere. Chlorine monoxide (\(\mathrm{ClO}\)) plays a crucial role in this process. When chlorine gas is released into the atmosphere, it reacts with ozone (\(\mathrm{O_3}\)) to form chlorine monoxide and oxygen. This reaction is part of a catalytic cycle where one chlorine atom can destroy many ozone molecules. The ozone layer, situated in the stratosphere, acts as a shield, absorbing the majority of the Sun's harmful ultraviolet radiation. Ozone depletion leads to increased UV exposure on Earth's surface, which can result in health issues like skin cancer and cataracts, as well as negative impacts on ecosystems.

Dipole moment calculation

The dipole moment of a molecule is a measure of the separation of positive and negative charges and indicates the molecule's polarity. Given the bond length and experimental dipole moment of chlorine monoxide, we can determine the magnitude of the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms. To calculate the magnitude of the charges (\(q\)), we use the formula:\[\mu = q \cdot d,\]where \(\mu\) is the dipole moment, \(q\) is the charge, and \(d\) is the bond length. Substituting the known values and converting to the appropriate units allows us to find \(q\). Once calculated in Coulombs, this can be converted to electronic charge units (\(e\)), providing us insight into the molecule's polar characteristics.

Lewis structure

The Lewis structure of a molecule is a diagram that represents the atomic structure, showing valence electrons and bonds between atoms. For chlorine monoxide (\(\mathrm{ClO}\)), we need to account for the total valence electrons. Chlorine has 7, and oxygen has 6, making a total of 13 valence electrons for the molecule. In the optimal Lewis structure:

• At least one bond connects chlorine and oxygen, using 2 electrons.
• The remaining electrons are distributed as lone pairs to satisfy the octet rule as much as possible.
• Oxygen will have 3 lone pairs, and chlorine will have 2 lone pairs and one unpaired electron.

This configuration gives formal charges that are reasonably close to zero, suggesting a stabile structure.

Formal charge

Formal charge is a concept used to determine the most plausible Lewis structure by assessing charge distribution on atoms in a molecule. It's calculated using the formula \(FC = Valence\,e^{-} - (Nonbonding\,e^{-} + \frac{Bonding\,e^{-}}{2})\). This determines the electron distribution compared to neutral atoms.For chlorine in \(\mathrm{ClO}\), chlorine may gain an extra electron in the case of \(\mathrm{ClO}^{-}\), relieving the lone electron to form a complete octet. The formal charges for \(\mathrm{ClO}^-\) can be calculated, resulting in a formal charge of 0 for chlorine, which is more stable. Understanding formal charges is crucial, as it helps predict which resonance structures most likely represent the molecule's actual arrangement of atoms.