Question

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

Short answer

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

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.

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

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.