Question

Select the correct statement(s) regarding $B{F}_{2}N{H}_2$ molecule :

• A. $FBF$ bond angle < ${120}^0$
• B. $HNH$ bond angle > ${109}^0{28}^{\prime }$
• C. $HNH$ bond angle < ${109}^{0}28$
• D. $FBF$ bond angle > ${120}^0$

Answer 1

Answer: (A), (B)

The correct options are
A $FBF$ bond angle < ${120}^0$
B $HNH$ bond angle > ${109}^0{28}^{\prime }$

According to Bent's rule there is more p-character in B-F bond than in B-N bond; hence FBF is less than ${120}^0$
As size of $-N{H}_2$ group is larger than F-atom hence repulsion between $-N{H}_2$ group and F-atoms is higher than between two F-atoms. As lone pair of N-atom in $N{H}_2$ group participate in back bonding hence H-N-H is greater than ${109}^0{28}^{\prime }$', as hybridization of N is changing from spa towards close to $s{p}^2$