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Chapter 8: Problem 95

Use the formal charge arguments to rationalize why \(\mathrm{BF}_{3}\) would not follow the octet rule.

Short Answer

Expert verified
The Lewis structure of BF3 has boron (B) in the center, connected to three fluorine atoms (F) with single bonds and no lone pairs on boron. Calculating the formal charges of B and F following this structure, we find that they are both 0. Despite having only 6 electrons around it, the central boron atom has a formal charge of 0, indicating that the structure is stable. Similarly, each fluorine atom follows the octet rule and has a formal charge of 0. Therefore, using formal charge arguments, we can rationalize that BF3 does not follow the octet rule because the current structure is the most stable one, with all atoms having a formal charge of 0.

Step by step solution

01

Determine the Lewis structure of BF3

First, draw the Lewis structure of the BF3 molecule. Boron (B) will be in the center, connected to three fluorine (F) atoms. Boron has 3 valence electrons, and each fluorine atom has 7 valence electrons. So, the Lewis structure will have single bonds between boron and each fluorine atom, and three lone pairs of electrons on each fluorine atom.
02

Calculate the formal charges of the atoms

To calculate the formal charges, use the formula: Formal charge = (Valence electrons) - (Non-bonding electrons + 1/2 * Bonding electrons) For boron (B): Valence electrons = 3 Non-bonding electrons = 0 (since boron has no lone pairs) Bonding electrons = 6 (since there are three single bonds) Formal charge of B = 3 - (0 + 1/2 * 6) = 0 For fluorine (F): Valence electrons = 7 Non-bonding electrons = 6 (since each fluorine atom has three lone pairs) Bonding electrons = 2 (since there is one single bond between B and each F) Formal charge of each F = 7 - (6 + 1/2 * 2) = 0
03

Analyze the formal charges to explain why the octet rule is not followed

Despite not having a complete octet (only 6 electrons around it), the central boron atom has a formal charge of 0. Meanwhile, each fluorine atom follows the octet rule and also has a formal charge of 0. Since the most stable structure is the one with the least formal charges, the current structure (single bonds between boron and each fluorine atom) is the most stable one for BF3. Thus, using formal charge arguments, it can be rationalized that BF3 does not follow the octet rule because the current structure, which does not have a complete octet for boron, is the most stable one, as all the atoms have a formal charge of 0.

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Most popular questions from this chapter

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 on ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Look up the energies for the bonds in \(\mathrm{CO}\) and \(\mathrm{N}_{2}\). Although the bond in \(\mathrm{CO}\) is stronger, \(\mathrm{CO}\) is considerably more reactive than \(\mathrm{N}_{2}\). Give a possible explanation.

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part \(\mathrm{c}\), why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about \(\mathrm{MgO}\) compared to \(\mathrm{MgO}_{2} ?\) \(\mathrm{Mg}_{2} \mathrm{O} ?\)

Give an example of an ionic compound where both the anion and the cation are isoelectronic with each of the following noble gases. a. Ne c. \(\mathrm{Kr}\) b. \(\mathrm{Ar}\) d. Xe

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl}\) c. \(\mathrm{Sr}\) and \(\mathrm{F}\) b. \(\mathrm{Na}\) and \(\mathrm{O}\) d. Ca and Se
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