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  • What will be the distance in <img alt="\mathrm{\mathrm{BF}_3}" src="https://entrancecorner.oncode

What will be the \mathrm{B.F} distance in \mathrm{\mathrm{BF}_3} molecule if you are given with \mathrm{B-F} distance as \mathrm{139 \mathrm{pm}} in \mathrm{\left(\mathrm{CH}_3\right)_3 \mathrm{~N}-\mathrm{BF}_3} molecule?
 

Option: 1

Because of \mathrm{p \pi-p \pi } back bonding in \mathrm{ B F_3}, it will be less.
 


Option: 2

because of \mathrm{p \pi-p \pi} back bonding, it will be more in \mathrm{\mathrm{BF}_3.}
 


Option: 3

same as in \mathrm{\left(\mathrm{CH}_3\right)_3 \mathrm{~N}-\mathrm{BF}_3}
 


Option: 4

Because \mathrm{BF}_3 exists as dimer, it will be more.


Answers (1)

best_answer

Because of \mathrm{p \pi-p \pi } back bonding in \mathrm{ B F_3}, it will be less.
 

Posted by

vinayak

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