Problem 50 - Chemistry The Central Science In Si Units

Question

(a) Draw the dominant Lewis structure for the phosphorus trifluoride molecule, \(\mathrm{PF}_{3}\). (b) Determine the oxidation numbers of the \(\mathrm{P}\) and \(\mathrm{F}\) atoms. \((\mathbf{c})\) Determine the formal charges of the \(\mathrm{P}\) and \(\mathrm{F}\) atoms.

Step-by-Step Solution

Answer

PF₃ has a central P atom bonded to three F atoms; P has an oxidation number of +3, F atoms have -1. Both P and F have a formal charge of 0.

Step 1: Drawing the Lewis Structure

To draw the Lewis structure of PF₃, identify the number of valence electrons. Phosphorus (P) has 5 valence electrons, and each fluorine (F) has 7 valence electrons. Since PF₃ consists of one P atom and three F atoms, the total valence electrons for PF₃ are calculated as follows: 5 (from P) + 3×7 (from each F) = 26 valence electrons. Place the P atom in the center and connect each of the three F atoms with single bonds to P. Distribute the remaining electrons to satisfy the octet rule.

Step 2: Determining the Oxidation Numbers

Oxidation numbers can be determined by assuming that in a polar bond, electrons are assigned to the more electronegative atom. For PF₃, fluorine is more electronegative than phosphorus. Assign an oxidation number of -1 to each F, as fluorine usually has an oxidation number of -1. Phosphorus will have an oxidation number of +3 to balance the three F atoms, each contributing -1.

Step 3: Calculating Formal Charges

Formal charges can be determined by the formula: \( \text{Formal Charge} = \text{Valence Electrons} - (\text{Bonding Electrons}/2 + \text{Non-Bonding Electrons}) \). For P in PF₃, the formal charge is calculated using 5 valence electrons and 3 bonds: \( 5 - (6/2 + 2) = 0 \). Each F atom has 7 valence electrons, 1 bond, and 6 non-bonding electrons, giving a formal charge of \( 7 - (2/2 + 6) = 0 \). Both P and F atoms have a formal charge of 0.

Key Concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a crucial role in bond formation. For phosphorus trifluoride (\( \mathrm{PF}_{3} \)), identifying the number of valence electrons is the first step in drawing its Lewis structure.
Phosphorus belongs to Group 15 of the periodic table and has 5 valence electrons. Each fluorine atom, belonging to Group 17, has 7 valence electrons. To determine the total number of valence electrons in \( \mathrm{PF}_{3} \), you add the valence electrons from the phosphorus atom with those from the three fluorine atoms:

• Phosphorus (P): 5 valence electrons
• Fluorine (F) for three atoms: \(3 \times 7 = 21\) valence electrons
• Total: \(5 + 21 = 26\) valence electrons

These 26 valence electrons are used to draw the Lewis structure by placing phosphorus in the center and forming single bonds with each fluorine atom. The remaining electrons are distributed to satisfy the octet rule, ensuring that each atom achieves a stable electronic configuration akin to the noble gases.

Oxidation Numbers

Oxidation numbers provide insight into the charge distribution in a compound, assuming that bonds are purely ionic. This assumption simplifies the identification of oxidation numbers, especially in polar bonds where electrons are considered to be transferred to the more electronegative atom.
In \( \mathrm{PF}_{3} \), fluorine is more electronegative than phosphorus, so you assign electrons from the phosphorus-fluorine bonds entirely to fluorine. This gives each fluorine atom an oxidation number of \(-1\), its typical value in compounds, and the phosphorus must balance this with an oxidation number of \(+3\) because:

• 3 fluorine atoms \( \times (-1) = -3 \)
• Phosphorus must have \(+3\) for the sum to be zero, balancing the molecule

Thus, oxidation numbers help understand electron transfer and provide insight into a molecule's stability and reactivity. They are vital for balancing reactions and understanding redox processes.

Formal Charges

Formal charges are a bookkeeping tool that helps chemists determine the ideal distribution of electrons in a Lewis structure. Unlike oxidation numbers, they assume electrons in bonds are shared equally, which helps in assessing the most stable structure for a molecule.
To calculate formal charges in \( \mathrm{PF}_{3} \), we use the formula:\[\text{Formal Charge} = \text{Valence Electrons} - \left( \frac{\text{Bonding Electrons}}{2} + \text{Non-Bonding Electrons} \right)\]Applying this to phosphorus (P) and each fluorine (F) atom:

• **Phosphorus**: 5 valence electrons, 3 bonds (6 bonding electrons), and 2 non-bonding electrons: \ \(5 - \left(\frac{6}{2} + 2 \right) = 0\)
• **Fluorine**: 7 valence electrons, 1 bond (2 bonding electrons), and 6 non-bonding electrons: \ \(7 - \left(\frac{2}{2} + 6 \right) = 0\)

Both phosphorus and fluorine atoms have a formal charge of 0, indicating that the Lewis structure represents a stable and minimal energy configuration. Using formal charges is essential in verifying that a molecule is depicted accurately and that no atom is unnecessarily charged.