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Chapter 8: Problem 87

The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Short Answer

Expert verified
The Lewis structures for the given compounds or ions are as follows: PF5 has Phosphorus as the central atom bonded to five Fluorine atoms; SF4 has Sulfur as the central atom bonded to four Fluorine atoms; ClF3 has Chlorine as the central atom bonded to three Fluorine atoms; and Br3- has a central Bromine atom bonded to two other Bromine atoms. Elements like Phosphorus, Sulfur, Chlorine, and Bromine, which are in the third period and beyond, can have more than eight electrons around them due to the availability of d-orbitals. This allows them to form stronger and more stable molecules by expanding their octet using d-orbitals, in addition to the s and p orbitals.

Step by step solution

01

1. Drawing the Lewis structure for PF5 (phosphorus pentafluoride)

: 1. Count valence electrons: Phosphorus has 5 valence electrons and Fluorine has 7. Since there are 5 fluorine atoms, the total number of electrons is 5(Phosphorus) + 5x7(Fluorine) = 40. 2. Place the central atom, Phosphorus, and add fluorine atoms around it. 3. Create bonds between each fluorine and phosphorus by sharing a pair of electrons. 4. Distribute the remaining electrons to outer atoms (fluorine) to fill their octets. 5. As there are no additional electrons left, the Lewis Structure for PF5 is complete.
02

2. Drawing the Lewis structure for SF4 (sulfur tetrafluoride)

: 1. Count valence electrons: Sulfur has 6 valence electrons and Fluorine has 7. Since there are 4 fluorine atoms, the total number of electrons is 6(Sulfur) + 4x7(Fluorine) = 34. 2. Place the central atom, Sulfur, and add fluorine atoms around it. 3. Create bonds between each fluorine and sulfur by sharing a pair of electrons. 4. Distribute the remaining electrons to outer atoms (fluorine) and sulfur to fill their octets or expanded octets. 5. The Lewis Structure for SF4 is complete.
03

3. Drawing the Lewis structure for ClF3 (chlorine trifluoride)

: 1. Count valence electrons: Chlorine has 7 valence electrons and Fluorine has 7 as well. Since there are 3 fluorine atoms, the total number of electrons is 7(Chlorine) + 3x7(Fluorine) = 28. 2. Place the central atom, Chlorine, and add fluorine atoms around it. 3. Create bonds between each fluorine and chlorine by sharing a pair of electrons. 4. Distribute the remaining electrons to outer atoms (fluorine) and chlorine to fill their octets or expanded octets. 5. The Lewis Structure for ClF3 is complete.
04

4. Drawing the Lewis structure for Br3- (tribromide ion)

: 1. Count valence electrons: Bromine has 7 valence electrons. Since there are 3 bromine atoms and the ion has an extra electron, the total number of electrons is 3x7(Bromine) +1 = 22. 2. Place one Bromine atom as the central atom and other two Bromine atoms around it. 3. Create bonds between central bromine and surrounding bromines by sharing a pair of electrons. 4. Distribute the remaining electrons to outer atoms (bromine) and the central bromine to satisfy their octets or expanded octets. 5. The Lewis Structure for Br3- is complete.
05

5. Identifying elements that can have more than eight electrons around them and rationalization

: From the drawn structures, the central atoms in the molecules PF5, SF4, and ClF3, as well as the ion Br3-, have more than eight electrons. These elements are Phosphorus, Sulfur, Chlorine, and Bromine. The central atoms of these compounds belong to the third period and beyond, which have a d-orbital available for electron occupation. They can accommodate more than eight electrons in their valence shell, called expanded octets, by using their d-orbital in addition to the s and p orbitals. As a result, they can form stronger and more stable molecules by having more than 8 valence electrons in their outer shell.

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: Given the following bond lengths, $$ \begin{array}{llll} \mathrm{N}-\mathrm{N} & 167 \mathrm{pm} & \mathrm{N}=\mathrm{O} & 115 \mathrm{pm} \\ \mathrm{N}=\mathrm{N} & 120 \mathrm{pm} & \mathrm{N}-\mathrm{O} & 147 \mathrm{pm} \\ \mathrm{N} \equiv \mathrm{N} & 110 \mathrm{pm} & & \end{array} $$ rationalize the observations that the \(\mathrm{N}-\mathrm{N}\) bond length in \(\mathrm{N}_{2} \mathrm{O}\) is \(112 \mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is \(119 \mathrm{pm}\). Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\). Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3}\) are examples of this type of exception. Draw the Lewis structures for \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3}\).

Two different compounds have the formula \(\mathrm{XeF}_{2} \mathrm{Cl}_{2} .\) Write Lewis structures for these two compounds, and describe how measurement of dipole moments might be used to distinguish between them.

Consider the following energy changes: Magnesium oxide exists as \(\mathrm{Mg}^{2+} \mathrm{O}^{2-}\) and not as \(\mathrm{Mg}^{+} \mathrm{O}^{-}\). Explain.

Some plant fertilizer compounds are \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}, \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), \(\mathrm{K}_{2} \mathrm{O}, \mathrm{P}_{2} \mathrm{O}_{5}\), and \(\mathrm{KCl}\). Which of these compounds contain both ionic and covalent bonds?
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