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Chapter 8: Problem 32

Use Lewissymbols and Lewis structures to diagram the formation of \(\mathrm{PF}_{3}\) from \(\mathrm{P}\) and \(\mathrm{F}\) atoms.

Short Answer

Expert verified
To show the formation of \(\mathrm{PF}_{3}\) using Lewis symbols and structures, first determine the valence electrons for \(\mathrm{P}\) and \(\mathrm{F}\) atoms: P has 5 valence electrons and F has 7. Next, draw their Lewis symbols: P: \\[ \cdot P ^{\circ}\cdot \\] F: \\[-] F ^{\bullet\circ} \\] Combine one P atom and three F atoms to form \(\mathrm{PF}_{3}\) by sharing two electrons between each P and F, making single covalent bonds. The final Lewis structure for \(\mathrm{PF}_{3}\) is: \\[ \mathrm{F} : \overset{\circ}{\mathrm{F}} \mathrm{P} - \mathrm{F} : \\] This shows that \(\mathrm{PF}_{3}\) has three single covalent bonds between the central P atom and each of the three F atoms, with three lone pairs on each F atom.

Step by step solution

01

Determine the valence electrons for P and F atoms

To draw the Lewis symbols and structures, we need to know the number of valence electrons for each element. Phosphorus is in group 15 and has 5 valence electrons, while Fluorine is in group 17 and has 7 valence electrons.
02

Draw Lewis symbols for P and F atoms

Using the number of valence electrons, we can draw the Lewis symbols for both phosphorus and fluorine. The phosphorus atom is represented by the symbol 'P' surrounded by 5 dots representing the 5 valence electrons. The fluorine atom is represented by the symbol 'F' surrounded by 7 dots representing the 7 valence electrons. P: \\[ \cdot P ^{\circ}\cdot \\] F: \\[-] F ^{\bullet\circ} \\]
03

Combine P and F atoms to form PF3

To form \(\mathrm{PF}_{3}\), one phosphorus atom bonds with three fluorine atoms. Each bond between the P and F atoms is formed by sharing two electrons, one from P and one from F. This forms a single covalent bond. The Lewis structure of \(\mathrm{PF}_{3}\) will have P as the central atom, surrounded by three F atoms, connected by single covalent bonds.
04

Complete the Lewis structure for PF3

To show the Lewis structure for \(\mathrm{PF}_{3}\), we need to indicate the shared and unshared electrons in the molecule. The shared electrons between the P and F atoms form the single covalent bonds, and the unshared electrons remain with the F atoms as lone pairs. The final Lewis structure for \(\mathrm{PF}_{3}\) is: \\[ \mathrm{F} : \overset{\circ}{\mathrm{F}} \mathrm{P} - \mathrm{F} : \\] This structure shows that \(\mathrm{PF}_{3}\) has three single covalent bonds between the central P atom and each of the three F atoms, with three lone pairs on each F atom.

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Most popular questions from this chapter

Write the Lewis symbol for atoms of each of the following elements: (a) \(\mathrm{Al}\), (b) \(\mathrm{Br}\), (c) \(\mathrm{Ar}\), (d) \(\mathrm{Sr}\).

(a) How does a polar molecule differ from a nonpolar one? (b) Atoms X and Y have different electronegativities. Will the diatomic molecule \(X-Y\) necessarily be polar? Explain. (c) What factors affect the size of the dipole moment of a diatomic molecule?

The iodine monobromide molecule, IBr, has a bond length of \(2.49 \AA\) and a dipole moment of \(1.21 \mathrm{D}\). (a) Which atom of the molecule is expected to have a negative charge? Explain. (b) Calculate the effective charges on the I and Br atoms in IBr, in units of the electronic charge \(e\).

Using only the periodic table as your guide, select the most electronegative atom in each of the following sets: (a) Se, \(\mathrm{Rb}, \mathrm{O}, \mathrm{In} ;\) (b) \(\mathrm{Al}, \mathrm{Ca}, \mathrm{C}, \mathrm{Si} ;\) (c) Ge, As, P, Sn; (d) Li, \(\mathrm{Rb}, \mathrm{Be}, \mathrm{Sr}\)

The following three Lewis structures can be drawn for \(\mathrm{N}_{2} \mathrm{O}:\) \(: \mathrm{N} \equiv \mathrm{N}-\ddot{O}: \longleftrightarrow: \ddot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{O}: \longleftrightarrow: \ddot{\mathrm{N}}=\mathrm{N}=\ddot{\mathrm{O}}:\) (a) Using formal charges, which of these three resonance forms is likely to be the most important? (b) The \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(1.12 \AA\), slightly longer than a typical \(\mathrm{N} \equiv \mathrm{N}\) bond; and the \(\mathrm{N}-\mathrm{O}\) bond length is \(1.19 \AA\), slightly shorter than a typical \(\mathrm{N}=\mathrm{O}\) bond. (See Table 8.5.) Rationalize these observations in terms of the resonance structures shown previously and your conclusion for (a).
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