The incorrect statement regarding the BF_2NH_ | vedclass

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(D) $1$. The molecule $BF_2NH_2$ exhibits back bonding from $N$ to $B$,which gives the $B-N$ bond partial double bond character.$2$. Both boron and ni

Vedclass · Accepted Answer

Hybridization of $N$ atom is $sp^3$

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(D) $1$. The molecule $BF_2NH_2$ exhibits back bonding from $N$ to $B$,which gives the $B-N$ bond partial double bond character.$2$. Both boron and nitrogen atoms are $sp^2$ hybridized to maintain planarity and facilitate back bonding.$3$. The $FBF$ bond angle is less than $120^{\circ}$ due to the presence of the $B=N$ double bond character,which exerts more repulsion than single bonds.$4$. The $HNH$ bond angle is greater than $109^{\circ}28'$ because the nitrogen atom is $sp^2$ hybridized (ideal angle $120^{\circ}$).$5$. Since the nitrogen atom is $sp^2$ hybridized,option $D$ is incorrect.$6$. The molecule contains $N-H$ bonds,which allows for intermolecular hydrogen bonding.

The incorrect statement regarding the $BF_2NH_2$ molecule is:

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