What is the B-F bond order in BF_3? | vedclass

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(D) In $BF_3$,the boron atom is electron-deficient and forms three $B-F$ sigma bonds. Due to the back-bonding from the filled $2p$ orbital of fluorine

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(D) In $BF_3$,the boron atom is electron-deficient and forms three $B-F$ sigma bonds. Due to the back-bonding from the filled $2p$ orbital of fluorine to the empty $2p$ orbital of boron,the $B-F$ bond acquires partial double bond character. There are $3$ equivalent resonance structures for $BF_3$. In each structure,there is one double bond and two single bonds among the three $B-F$ positions. The total number of bonds is $1 + 1 + 2 = 4$. The bond order is calculated as $\frac{	ext{Total number of bonds}}{	ext{Number of resonating structures}} = \frac{4}{3} \approx 1.33$.

What is the $B-F$ bond order in $BF_3$?

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