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Chapter 8: Problem 50

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

Short Answer

Expert verified
The dominant Lewis structure of PF$_3$ is: F | P - F - F | F The oxidation numbers of the P and F atoms are -1 and +1, respectively. The formal charges of both P and F atoms are 0.

Step by step solution

01

Determine the valence electrons for P and F atoms

To draw the Lewis structure of PF3, first determine the number of valence electrons for each atom. Phosphorus (P) is in Group 15 (or Group 5A) and has 5 valence electrons. Fluorine (F) is in Group 17 (Group 7A) and has 7 valence electrons.
02

Connect the atoms and distribute the valence electrons

Connect the P atom to three F atoms with a single bond (two valence electrons) to each F atom. This will use 6 of the 5 valence electrons of P. Leave the remaining 3 F atoms' valence electrons as lone pairs. Each F atom will now have two lone pairs (four valence electrons), and then we need to place the remaining two electrons on the P atom to complete its octet.
03

Complete the Lewis structure

Draw the additional lone pair on the P atom. The final Lewis structure for PF3 is as follows: F | P - F - F | F
04

Determine the oxidation numbers of P and F atoms

To calculate the oxidation numbers, use the following rules: 1. The oxidation number of an atom in a free element is 0. 2. For monoatomic ions, the oxidation number is equal to the charge of the ion. 3. The sum of the oxidation numbers in a neutral species is 0. In this case, since P is more electronegative than F, P will have a -1 oxidation number while F will have a +1 oxidation number. Oxidation numbers: P = -1, F = +1
05

Determine the formal charges of P and F atoms

To calculate the formal charges, follow the formula: Formal Charge = (Valence Electrons) - (0.5 x Bonding Electrons) - (Lone-pair Electrons) For the P atom: Formal Charge = (5) - (0.5 x 6) - (2) = 0 For the F (Any single) atom: Formal Charge = (7) - (0.5 x 2) - (6) = 0 Formal charges: P = 0; F = 0

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Most popular questions from this chapter

Consider the element silicon, Si. (a) Write its electron configuration. (b) How many valence electrons does a silicon atom have? (c) Which subshells hold the valence electrons?

A carbene is a compound that has a carbon bonded to two atoms and a lone pair remaining on the carbon. Many carbenes are very reactive. (a) Draw the Lewis structure for the simplest carbene, \(\mathrm{H}_{2} \mathrm{C}\) . ( b) Predict the length of the carbon-carbon bond you would expect if two \(\mathrm{H}_{2} \mathrm{C}\) molecules reacted with each other by a combination reaction.

(a) Triazine, \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}_{3},\) is like benzene except that in triazine every other \(\mathrm{C}-\mathrm{H}\) group is replaced by a nitrogen atom. Draw the Lewis structure(s) for the triazine molecule. (b) Estimate the carbon-nitrogen bond distances in the ring.

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) Al and F, (b) K and \(S,(\mathbf{c}) \mathrm{Y}\) and \(\mathrm{O},(\mathbf{d}) \mathrm{Mg}\) and \(\mathrm{N} .\)

(a) Using Lewis symbols, diagram the reaction between magnesium and oxygen atoms to give the ionic substance MgO. (b) How many electrons are transferred? (c) Which atom loses electrons in the reaction?
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