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Chapter 8: Problem 103

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

Short Answer

Expert verified
By analyzing the formal charges of BF3, we find that the structure with the central Boron atom having only 6 valence electrons has zero formal charges for all atoms, leading to a stable electron distribution. Obeying the octet rule for Boron would result in higher formal charges and an unstable electron distribution. Thus, the formal charge analysis rationalizes why BF3 does not follow the octet rule.

Step by step solution

01

Understanding the Octet Rule

The octet rule states that atoms prefer to have eight electrons in their valence shell, forming stable, full-filled energy levels. This usually leads to the formation of stable covalent bonds between atoms. However, some molecules do not follow the octet rule due to the nature of their atoms or their electron configurations.
02

Understanding Formal Charge

Formal charge is a method used to determine the charge distribution within a molecule. It is calculated for each atom in the molecule by comparing the number of valence electrons of the free atom with the number of electrons assigned to the atom in the molecule. The formal charge of an atom is given by the formula: Formal Charge = Valence Electrons - (Non-bonding Electrons + 1/2 Bonding Electrons) In most stable compounds, formal charges are minimized, and they can be used to predict the most likely electron distribution in the molecule.
03

Calculate Formal Charges of BF3

Let's determine the structure of BF3. Boron has 3 valence electrons, while each Fluorine atom has 7 valence electrons. In BF3, Boron forms single covalent bonds with each Fluorine atom. For Boron: Valence electrons = 3 Nonbonding electrons = 0 (Boron does not have lone pairs in the molecule) Bonding electrons = 6 (3 single covalent bonds with 3 Fluorine atoms) Formal Charge (Boron) = 3 - (0 + 1/2 * 6) = 0 For each Fluorine atom: Valence electrons = 7 Nonbonding electrons = 6 (each Fluorine has 3 lone pairs in the molecule) Bonding electrons = 2 (1 single covalent bond with Boron) Formal Charge (each Fluorine) = 7 - (6 + 1/2 * 2) = 0
04

Analyze Formal Charges and the Octet Rule

In the structure of BF3, all formal charges are zero, which creates a stable electron distribution. However, the central Boron atom only has 6 valence electrons instead of the complete octet (8 electrons). In this case, obeying the octet rule for Boron would lead to higher formal charges and an unstable electron distribution.
05

Conclusion

By analyzing the formal charges, we can rationalize why BF3 does not follow the octet rule. The structure of BF3 with incomplete octets for the central Boron atom has the lowest formal charges, thus resulting in a more stable electron distribution.

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Most popular questions from this chapter

\(\mathrm{SF}_{6}, \mathrm{ClF}_{5},\) and \(\mathrm{XeF}_{4}\) are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}\)

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.

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{F}_{2} \quad\left(\mathrm{O}_{2} \mathrm{F}_{2}\right.\) exists as \(\mathrm{F}-\mathrm{O}-\mathrm{O}-\mathrm{F} )\). Assign oxidation states and formal charges to the atoms in O2F2. This compound is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of \(\mathrm{O}_{2} \mathrm{F}_{2}\)?

Place the species below in order of the shortest to the longest nitrogen鈥搊xygen bond. $$\mathrm{H}_{2} \mathrm{NOH}, \quad \mathrm{N}_{2} \mathrm{O}, \quad \mathrm{NO}^{+}, \quad \mathrm{NO}_{2}^{-}, \quad \mathrm{NO}_{3}^{-}$$ \(\left(\mathrm{H}_{2} \mathrm{NOH} \text { exists as } \mathrm{H}_{2} \mathrm{N}-\mathrm{OH} .\right)\)
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