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

The octet rule states that most atoms in a molecule will achieve stability by having eight electrons in their valence shell. This means they will share, lose, or gain electrons to achieve a stable electron configuration. To determine whether a molecule follows the octet rule, we must consider the formal charges of its atoms. The formal charge of an atom can be calculated using the following formula: Formal charge = Valence electrons - (Number of bonds + Lone pair electrons). We will use this formula to calculate the formal charge of the atoms in the \(\mathrm{BF}_{3}\) molecule.

The atomic number of Boron (B) is 5, so it has 5 electrons in total. Since it is in group 3 (or 13 depending on the notation used) in the periodic table, it has 3 valence electrons. In the \(\mathrm{BF}_{3}\) molecule, Boron is bonded to 3 Fluorine atoms, so it shares 3 electrons. Since it has no lone pairs of electrons, we can calculate the formal charge of Boron as follows: Formal charge of B = Valence electrons - (Number of bonds + Lone pair electrons) Formal charge of B = 3 - (3 + 0) Formal charge of B = 0 As a result, the formal charge of Boron in \(\mathrm{BF}_{3}\) is 0.

Fluorine (F) has an atomic number of 9 and 7 valence electrons since it is in group 7 (or 17 depending on the notation used) in the periodic table. In the \(\mathrm{BF}_{3}\) molecule, each Fluorine atom is bonded to the central Boron atom. Therefore, each Fluorine atom shares 1 electron with Boron and has 6 lone pair electrons. We can calculate the formal charge of a Fluorine atom as follows: Formal charge of F = Valence electrons - (Number of bonds + Lone pair electrons) Formal charge of F = 7 - (1 + 6) Formal charge of F = 0 Hence, the formal charge of each Fluorine atom in \(\mathrm{BF}_{3}\) is 0.

Based on our calculations above, both Boron and Fluorine atoms in the \(\mathrm{BF}_{3}\) molecule have formal charges of 0, which indicate that they are in their most stable state. However, let's examine if the octet rule holds for the Boron atom. When accounting for the shared electrons in the bonds formed with the three Fluorine atoms, the Boron atom has a total of 6 electrons in its valence shell. As a result, the octet rule is not followed by Boron in the \(\mathrm{BF}_{3}\) molecule. But, \(\mathrm{BF}_{3}\) is still a stable molecule because it has formal charges of 0 for all its atoms. In this case, having a full octet is not necessary for stability. This demonstrates that some molecules, like \(\mathrm{BF}_{3}\), can be stable even if they do not strictly follow the octet rule. The formal charge argument rationalizes that stability can be achieved without a complete octet for certain molecules.