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Chapter 8: Problem 59

What is the most common exception to the octet rule? Give two examples.

Short Answer

Expert verified
The most common exception to the octet rule is the presence of atoms with an incomplete octet, such as elements from Group 13 (Boron, Aluminum, etc.) that usually have only six valence electrons. Two examples of this exception are: 1. Boron Trifluoride (BF3): In this molecule, boron forms three covalent bonds with fluorine atoms and has only six valence electrons in its outer shell. 2. Aluminum Chloride (AlCl3): In this molecule, aluminum shares an electron with each chlorine atom, forming three covalent bonds, and also has only six valence electrons in its outer shell.

Step by step solution

01

Identify the most common exception to the octet rule

The most common exception to the octet rule is the presence of atoms with an incomplete octet, which means they have fewer than eight electrons in their valence shell. This usually occurs with elements from Group 13 of the periodic table (Boron, Aluminum, etc.), as they have three valence electrons and often form compounds by sharing only three electrons with other atoms, resulting in only six valence electrons.
02

Example 1: Boron Trifluoride (BF3)

In BF3, boron (B) is the central atom and is bonded to three fluorine (F) atoms. It forms three covalent bonds with fluorine atoms, sharing one electron from each fluorine. As a result, boron has only six valence electrons in its outer shell rather than the typical eight. The electron configuration of BF3 is as follows: Boron: 1s^2 2s^2 2p^1 Fluorine: 1s^2 2s^2 2p^5 In the BF3 molecule, Boron: 1s^2 2s^2 2p^3 Fluorine: 1s^2 2s^2 2p^6
03

Example 2: Aluminum Chloride (AlCl3)

In AlCl3, aluminum (Al) is the central atom and is bonded to three chlorine (Cl) atoms. Aluminum shares one electron with each chlorine atom, forming three covalent bonds. As a result, aluminum only has six valence electrons in its outer shell. The electron configuration of AlCl3 is as follows: Aluminum: 1s^2 2s^2 2p^6 3s^2 3p^1 Chlorine: 1s^2 2s^2 2p^6 3s^2 3p^5 In the AlCl3 molecule, Aluminum: 1s^2 2s^2 2p^6 3s^2 3p^3 Chlorine: 1s^2 2s^2 2p^6 3s^2 3p^6

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Most popular questions from this chapter

(a) State the octet rule. (b) Does the octet rule apply to ionic as well as to covalent compounds? Explain, using examples as appropriate.

Predict the ordering of the \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{CO}\), \(\mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}^{2-}\)

One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity \(=k(\mathrm{IE}-\mathrm{EA})\), where \(k\) is a proportionality constant. (a) How does this definition explain why the electronegativity of \(\mathrm{F}\) is greater than that of \(C l\) even though \(C l\) has the greater electron affinity? (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity? (c) By using data in Chapter 7 , determine the value of \(k\) that would lead to an electronegativity of \(4.0\) for \(\mathrm{F}\) under this definition. (d) Use your result from part (c) to determine the electronegativities of \(\mathrm{Cl}\) and \(\mathrm{O}\) using this scale. Do these values follow the trend shown in Figure \(8.6\) ?

The lattice energies of \(\mathrm{KBr}\) and \(\mathrm{CsCl}\) are nearly equal (Table 8.2). What can you conclude from this observation?

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