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Chapter 8: Problem 50

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

Short Answer

Expert verified
The dominant Lewis structure for phosphorus trifluoride (PF3) has P as the central atom, single bonded to three F atoms with three lone pairs on each F. In PF3, the oxidation number of P is +3, and the oxidation number of F is -1. The formal charges are +2 for P and 0 for F.

Step by step solution

01

Draw the Lewis Structure

To draw the dominant Lewis structure of phosphorus trifluoride (PF3), first, we need to count the total number of valence electrons in the molecule. Phosphorus (P) has 5 electrons in its outermost shell while fluoride (F) has 7 electrons in its outermost shell. Since there are 3 fluorine atoms, this gives us 5+(3脳7) = 5+21 = 26 valence electrons. Next, we have to find the central atom and arrange the atoms around it. Phosphorus, in this case, is the central atom since it has the lowest electronegativity value. Connect the central atom (P) to the surrounding atoms (F) by single bonds using 2 electrons for each bond. Now, we have 26 鈥� 6 = 20 valence electrons left. We should distribute these remaining electrons as lone pairs to the external atoms (F) to satisfy the octet rule. After giving 3 lone pairs (6 electrons) to each F atom to complete its octet, we used up all 20 remaining electrons, and thus we have the correct Lewis structure for PF3.
02

Determine the Oxidation Numbers

To determine the oxidation numbers of the P and F atoms in PF3, we should use the general rules for assigning oxidation numbers. The oxidation number of fluorine in any compound is always -1. In this case, let's look at PF3. The oxidation number of F is -1, and there are three F atoms, so the total oxidation number contribution by F is -3. Because the sum of oxidation numbers in a neutral molecule is zero, the oxidation number of phosphorus (P) must be +3. So, in PF3, the oxidation number of P is +3, and the oxidation number of F is -1.
03

Determine the Formal Charges

To determine the formal charges of the P and F atoms in PF3, we should use the following formula: Formal Charge = (valence electrons) - (non-bonding electrons) - 1/2(bonding electrons) For P atom: Valence electrons = 5 Non-bonding electrons = 0 (since it has no lone pairs) Bonding electrons = 6 (3 single bonds with F atoms) Formal Charge of P = 5 - 0 - 1/2(6) = 5 - 3 = +2 For F atom: Valence electrons = 7 Non-bonding electrons = 6 (3 lone pairs) Bonding electrons = 2 (1 single bond with P) Formal Charge of F = 7 - 6 - 1/2(2) = 7 - 6 - 1 = 0 Therefore, the formal charges of the P and F atoms in PF3 are +2 for P and 0 for F.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Numbers
Understanding the oxidation numbers of atoms within a molecule is critical in chemistry, as they provide insight into the electron distribution and the likely types of reactions the molecule might undergo.

When determining oxidation numbers, there are some key rules to remember. For instance, the oxidation number of a pure element is always zero, and for monoatomic ions, the oxidation number is the same as the ion's charge. In molecules, the more electronegative atom typically holds a negative oxidation number, while the less electronegative ones have positive values. Also, the overall sum of the oxidation numbers in a non-ionic compound must be zero.

For phosphorus trifluoride (PF3), fluorine, being the more electronegative, always has an oxidation number of -1. Because there are three fluorine atoms, the combined oxidation number for fluorine is -3. To balance this in a neutral molecule, the phosphorus must have an oxidation number of +3.
Formal Charges
Formal charges help identify which atoms within a molecule may have an excess or deficit of electrons relative to their stable state. The calculation of formal charges can guide us in predicting the reactivity of different parts of a molecule and assessing the molecule's most stable structure.

To calculate the formal charge on an atom, one uses the formula:
Formal Charge = (valence electrons) - (non-bonding electrons) - 1/2(bonding electrons).

For example, in phosphorus trifluoride, the phosphorus atom (P) has a formal charge of +2. This is because P has five valence electrons and shares six electrons through three single bonds. The fluorine atoms (F) each have a formal charge of 0, since they have seven valence electrons, six of which are non-bonding and one that is shared in the bond with P.

By determining the formal charges, we can verify that the structure with a +2 formal charge on P and 0 on each F is a plausible Lewis structure for PF3.
Valence Electrons
The concept of valence electrons is foundational in understanding how atoms bond in molecules. Valence electrons are the electrons located in an atom's outermost shell and are the ones involved in forming chemical bonds.

In the case of phosphorus trifluoride, phosphorus (P) has 5 valence electrons, and fluorine (F), being in Group 17 of the periodic table, has 7 valence electrons. Since PF3 is composed of one P atom and three F atoms, we calculate the total valence electrons for the molecule as 5 + (3 x 7) = 26.

The Lewis structure aims to illustrate how these valence electrons are arranged in bonding and as non-bonding pairs, leading to a stable arrangement known as the octet rule. This rule states that atoms tend to bond in a way that each atom ends up with eight electrons in its valence shell, resembling the electron configuration of a noble gas.

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Most popular questions from this chapter

Predict the ordering, from shortest to longest, of the bond lengths in \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}{ }^{2-}\).

(a) State whether the bonding in each compound is likely to be covalent or not: (i) iron, (ii) sodium chloride, (iii) water, (iv) oxygen, (v) argon. (b) A substance XY, formed from two different elements, boils at \(-33{ }^{\circ} \mathrm{C}\). Is XY likely to be a covalent or an ionic substance?

For each of these Lewis symbols, indicate the group in the periodic table in which the element \(X\) belongs: [Section 8.1] (a) \(\dot{X}\) (b) \(\mathrm{X}\) (c) \(\dot{X} \cdot\)

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 \(\mathrm{ClO}^{-}\)?

A major challenge in implementing the "hydrogen economy" is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example, \(\mathrm{NaAlH}_{4}\) can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(s), \mathrm{Al}(s)\), and \(\mathrm{H}_{2}(g) . \mathrm{NaAlH}_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4}\). (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, predict the identity of the polyatomic anion. Draw a Lewis structure for this ion. (d) What is the formal charge on hydrogen in the polyatomic ion?
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