Which is the correct Lewis structure for boron trifluoride

Part A Which is the correct Lewis structure for boron trifluoride? #-B-#-B #-B-F: #-B-E; #-B-F: Submit Request Answer

The Correct Answer and Explanation is:

The correct answer is the second option.

Explanation:

To determine the correct Lewis structure for boron trifluoride (BF₃), we can follow these steps:

1. Count Total Valence Electrons: First, we find the total number of valence electrons for the molecule. Boron (B) is in Group 13, so it has 3 valence electrons. Fluorine (F) is in Group 17, so it has 7 valence electrons. Since there are three fluorine atoms, the total count is:
   3 (from B) + 3 × 7 (from F) = 3 + 21 = 24 valence electrons.
2. Identify the Central Atom: Boron is less electronegative than fluorine, so it will be the central atom. The three fluorine atoms will be bonded to it.
3. Draw the Skeletal Structure: Connect the central boron atom to each of the three fluorine atoms with a single bond. Each bond uses 2 electrons, so we have used 3 × 2 = 6 electrons so far.
   Remaining electrons = 24 – 6 = 18.
4. Complete the Octets of Terminal Atoms: Distribute the remaining 18 electrons as lone pairs on the terminal fluorine atoms. Each fluorine atom needs 6 more electrons to complete its octet (2 from the bond + 6 lone pair electrons = 8). We place 3 lone pairs (6 electrons) on each of the three fluorine atoms.
   3 fluorine atoms × 6 electrons/atom = 18 electrons.
   This uses all the remaining electrons.
5. Check the Central Atom and Formal Charges: At this point, each fluorine atom has a complete octet. The central boron atom, however, is only surrounded by 6 electrons (from the three single bonds). While this is an incomplete octet, it is a common exception for boron.To verify this is the best structure, we check the formal charges:
   • Formal Charge of Boron: 3 (valence e⁻) – 0 (lone pair e⁻) – 3 (bonds) = 0
   • Formal Charge of Fluorine: 7 (valence e⁻) – 6 (lone pair e⁻) – 1 (bond) = 0

Since all atoms have a formal charge of zero, this is the most stable and widely accepted Lewis structure for BF₃, even with an electron-deficient boron atom. The other options are incorrect because they either have the wrong atomic arrangement, an incorrect number of electrons, or result in unfavorable formal charges (like the structure with a double bond).|
:F-B-F:
¨ ¨Generated code

*(Note: Each fluorine atom should have 3 lone pairs of electrons, for a total of 6 dots around each F)*

### Explanation

To determine the correct Lewis structure for boron trifluoride (BF₃), we follow a systematic process based on valence electrons, the octet rule, and formal charges.

**1. Count Total Valence Electrons:**
First, we calculate the total number of valence electrons for the molecule. Boron (B) is in Group 13 of the periodic table, so it has 3 valence electrons. Fluorine (F) is in Group 17, so each atom has 7 valence electrons.
Total valence electrons = (1 × electrons from B) + (3 × electrons from F)
Total valence electrons = 3 + (3 × 7) = 3 + 21 = 24 electrons.

**2. Determine the Central Atom and Skeletal Structure:**
Boron is less electronegative than fluorine, so it serves as the central atom. The three fluorine atoms are bonded to the central boron atom. We start by drawing single bonds between the central boron and each fluorine atom. Each single bond uses 2 electrons, so we have used 3 × 2 = 6 electrons.

**3. Distribute Remaining Electrons:**
We have 24 - 6 = 18 electrons remaining. These are distributed as lone pairs around the terminal atoms (the fluorines) to satisfy their octets. Each fluorine atom needs 6 more electrons (3 lone pairs) to have a full octet. Since there are three fluorine atoms, this uses exactly 3 × 6 = 18 electrons.

**4. Check Octets and Formal Charges:**
At this point, each fluorine atom has a complete octet (one bond and three lone pairs). However, the central boron atom only has three single bonds, giving it a total of 6 valence electrons, which is an incomplete octet.

While we could form a double bond by moving a lone pair from a fluorine atom to create a B=F bond, this would give boron an octet but would also create unfavorable formal charges. The fluorine in the double bond would have a formal charge of +1, and the boron would have a formal charge of -1. Placing a positive charge on fluorine, the most electronegative element, is highly destabilizing.

The structure with only single bonds results in a formal charge of zero for all atoms. This is the most stable and preferred arrangement. Boron is a known exception to the octet rule and is stable with an electron-deficient configuration of only six valence electrons. Therefore, the second option, which shows boron with three single bonds and no lone pairs, is the correct Lewis structure.