The formation of molecular complex BF_3-NH_3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In $BF_3$,the boron atom is $sp^2$ hybridized and possesses an empty $p_z$ orbital. When $BF_3$ reacts with $NH_3$,the lone pair of electrons from

Vedclass · Accepted Answer

from $sp^2$ to $sp^3$

Vedclass · Answer

(B) In $BF_3$,the boron atom is $sp^2$ hybridized and possesses an empty $p_z$ orbital. When $BF_3$ reacts with $NH_3$,the lone pair of electrons from the nitrogen atom in $NH_3$ is donated into the empty $p_z$ orbital of boron. This coordinate covalent bond formation changes the geometry of boron from trigonal planar to tetrahedral. Consequently,the hybridization of boron changes from $sp^2$ to $sp^3$.

The formation of molecular complex $BF_3-NH_3$ results in a change in hybridization of boron

Similar Questions

The species in which the nitrogen atom is in a state of $sp$ hybridisation is

Sulphur reacts with chlorine in $1:2$ ratio and forms $X$. Hydrolysis of $X$ gives a sulphur compound $Y$. What is the structure and hybridisation of the anion of $Y$?

What is the type of chemical covalent bond between the carbon-carbon atoms in $C_2H_4$?

Which $C$ atom is the most electronegative in the structure $CH_3^{III} - CH_2^{II} - C \equiv CH^I$?

The hybridization of $P$ exhibited in $PF_{5}$ is $sp^{x}d^{y}$. The value of $y$ is $...$

Vedclass Test Series

Exam Paper Generator

Online Exam Module