Problem 50 - Chemistry: The Central Science in SI Units, Global Edition

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

The dominant Lewis structure for $\mathrm{PF}_{3}$ is: P - single bond - F | F - single bond - P - single bond - F With 6 lone pair electrons around each fluorine atom. The oxidation number of P is +3, and the oxidation number of each F is -1. The formal charge for the P atom is -1 and the formal charge for each F atom is 0.

Step 1: Identify the valence electrons in P and F atoms

First, let's determine the number of valence electrons in phosphorus (P) and fluorine (F) atoms. Phosphorus, located in group 15, has 5 valence electrons while fluorine, located in group 17, has 7 valence electrons.

Step 2: Calculate the total number of valence electrons

Now, we need to calculate the total number of valence electrons in PF3. Total valence electrons = 1 (Number of P atoms) x 5 (Valence electrons of P) + 3 (Number of F atoms) x 7 (Valence electrons of F) Total valence electrons = 5 + 21 = 26

Step 3: Draw the base Lewis structure

Place the phosphorus (P) atom in the center and form single covalent bonds with the three fluorine (F) atoms. Since 3 single bonds are formed, 6 valence electrons are used up, leaving 20 valence electrons for putting on the surrounding fluorine atoms.

Step 4: Distribute the remaining valence electrons

We need to distribute the remaining valence electrons in order to complete the octet for all the atoms. Since each fluorine atom already has 2 valence electrons from the single bond, we need to add 6 more valence electrons to each fluorine atom to complete the octet.

Step 5: Draw the dominant Lewis structure

Now, the Lewis structure for PF3 is as follows: P - single bond - F | F - single bond - P - single bond - F With 6 lone pair electrons around each fluorine atom.

Step 6: Determine the oxidation numbers

To find the oxidation numbers for the P and F atoms, we must follow some rules: - The oxidation number of F is always -1. - The sum of oxidation numbers of all atoms in a molecule is equal to the net charge on the molecule. Let x be the oxidation number of P. x + (-1) × 3 = 0 x - 3 = 0 x = 3 The oxidation number of P is +3, and the oxidation number of each F is -1.

Step 7: Determine the formal charges

To calculate the formal charge of an atom, we will use the formula: Formal charge = Valence electrons (neutral atom) - Non-bonded electrons - (1/2) × Bonding electrons Formal charge of P: = 5 (Valence electrons) - 0 (Non-bonded electrons) - (1/2 × 6) (Bonding electrons) = 5 - 0 - 3 = 2 - 3 = -1 Formal charge of each F: = 7 (Valence electrons) - 6 (Non-bonded electrons) - (1/2 × 2) (bonding electrons) = 7 - 6 - 1 = 0 Thus, the formal charge for P is -1, and the formal charge for each F is 0.

Key Concepts

Phosphorus Trifluoride

Phosphorus trifluoride is a chemical compound with the formula \( \text{PF}_3 \). It consists of a central phosphorus atom bonded to three fluorine atoms. This molecule is a type of phosphorus halide, where phosphorus is commonly found bonded with halogen atoms such as fluorine.
In PF\(_3\), phosphorus is surrounded by three fluorine atoms in a trigonal pyramidal shape. This geometry arises due to the lone pair of electrons on the phosphorus, causing a distortion from a perfect tetrahedral shape.
PF\(_3\) acts as a polar molecule due to the difference in electronegativity between phosphorus and fluorine. Fluorine is highly electronegative, drawing electron density towards itself and away from phosphorus, leading to a dipole moment within the molecule. This makes phosphorus trifluoride interesting for various applications such as in industry and in chemical synthesis.

Oxidation Numbers

Oxidation numbers are a way to keep track of electron distribution in molecules, especially for the purpose of balancing redox reactions.
In phosphorus trifluoride, we need to determine the oxidation states of phosphorus (P) and fluorine (F). Fluorine is highly electronegative, consistently having an oxidation number of -1 in compounds. This fact is a key rule when calculating oxidation states.
The PF\(_3\) molecule is neutral, so the sum of oxidation numbers must equate to zero. Given that each of the three fluorine atoms contributes an oxidation number of -1, phosphorus must have an oxidation number that balances this negative charge. Let's consider phosphorus's oxidation number as \( x \). The sum is expressed as:
\[ x + 3(-1) = 0 \]
Hence, solving this equation shows that the phosphorus in PF\(_3\) has an oxidation number of +3, maintaining the neutrality of the molecule.

Formal Charges

Formal charges help us understand the distribution of electrons within a molecule—implying whether an atom is gaining or losing electron density as compared to its neutral atom state.
To calculate the formal charge, use the formula:
Formal charge = Valence electrons - Non-bonded electrons - \( \frac{1}{2} \times \text{Bonding electrons} \)
For the phosphorus atom in PF\(_3\), we start with 5 valence electrons (matching its group number). No non-bonded electrons are present on phosphorus itself, and it shares 6 electrons in the three covalent bonds with fluorine.
Substituting these values into the formula gives a formal charge of:
5 - 0 - \( \frac{1}{2} \times 6 \) = 5 - 3 = +2.
However, the solution in the step-by-step is incorrect here: the formal charge should actually calculate correctly, and ensure it's consistent with the net charge distribution in PF\(_3\). Importantly, each fluorine in the molecule has:

• 7 valence electrons minus 6 non-bonded electrons and
• shared electrons amounting to = \( \frac{1}{2} \times 2 = 1 \) bonding electron.

This, indeed, results in a formal charge of 0 for each fluorine atom, as they are not gaining or losing electrons relative to their neutral state.