The correct Lewis structure for BF3 would have exactly: no double bonds. O2 double bonds. 1 triple bond. 1 double bond
The Correct Answer and Explanation is :
The correct answer is “no double bonds.”
Explanation:
The Lewis structure of ( BF_3 ) (boron trifluoride) is drawn by distributing the valence electrons around boron and the three fluorine atoms. Here’s how it works:
- Determine total valence electrons:
- Boron (( B )) has 3 valence electrons.
- Fluorine (( F )) has 7 valence electrons, and there are 3 fluorine atoms.
- Total valence electrons: ( 3 + (7 \times 3) = 24 ).
- Form bonds:
- Boron forms single covalent bonds with each of the three fluorine atoms. Each bond uses 2 electrons.
- Total electrons used in bonds: ( 3 \times 2 = 6 ).
- Distribute remaining electrons:
- After forming bonds, ( 24 – 6 = 18 ) electrons remain.
- These electrons are placed as lone pairs on the fluorine atoms, with each fluorine receiving 6 electrons (3 lone pairs), completing their octets.
- Check boron’s octet:
- Boron has only 6 electrons around it (from three single bonds), which does not complete an octet.
- However, boron is an exception to the octet rule and is stable with only 6 valence electrons in this molecule.
- No double or triple bonds:
- Fluorine is highly electronegative and does not share additional electrons to form double bonds.
- Boron lacks the necessary electrons to form multiple bonds.
Thus, the correct Lewis structure for ( BF_3 ) has three single bonds and no double or triple bonds. Each fluorine atom has a full octet, and boron is stable despite having fewer than 8 electrons. The absence of double or triple bonds in ( BF_3 ) aligns with its actual molecular geometry, which is trigonal planar.