Question

consider the reaction:
bf₃ + ncl₃ → f₃b - ncl₃
part 1 of 2
what changes in hybridization (if any) of the b atom are a result of this reaction? check all that apply.
before the reaction boron is sp² hybridized, and after the reaction it is sp³d² hybridized.
before the reaction boron is sp³ hybridized, and after the reaction it is sp³d hybridized.
before the reaction boron is sp³ hybridized, and after the reaction it is sp³d² hybridized.
none of the above.
part 2 of 2

Explanation:

• For \(BF_3\):
• Boron has 3 valence electrons. It forms 3 \(\sigma\) bonds with \(F\) atoms. Using the formula \(hybridization = number\ of\ \sigma\ bonds+number\ of\ lone\ pairs\), here number of \(\sigma\) bonds \(= 3\) and number of lone pairs \(= 0\). So, the hybridization of \(B\) in \(BF_3\) is \(sp^{2}\) (because \(sp^{2}\) hybridization has \(3\) equivalent hybrid orbitals).
• For \(F_3B - NCl_3\):
• Boron now forms 4 \(\sigma\) bonds (3 with \(F\) and 1 with \(N\)). Using the formula \(hybridization=number\ of\ \sigma\ bonds + number\ of\ lone\ pairs\), number of \(\sigma\) bonds \(= 4\) and number of lone pairs \(= 0\). So, the hybridization of \(B\) in \(F_3B - NCl_3\) is \(sp^{3}\) (because \(sp^{3}\) hybridization has \(4\) equivalent hybrid orbitals). None of the given options (which suggest \(sp^{3}d\) or \(sp^{3}d^{2}\) or incorrect pre - reaction hybridization) are correct.

Answer:

None of the above.