Consider the following reaction BCl_3 + NH_3 | vedclass

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(A) The molecular geometry of $BCl_3$ is trigonal planar because boron is $sp^2$ hybridized with three bond pairs and no lone pairs.In the adduct $BCl

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trigonal planar and tetrahedral

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(A) The molecular geometry of $BCl_3$ is trigonal planar because boron is $sp^2$ hybridized with three bond pairs and no lone pairs.In the adduct $BCl_3 \cdot NH_3$,the boron atom forms a coordinate bond with the nitrogen atom of $NH_3$.After the formation of the coordinate bond,the boron atom is surrounded by four atoms (three $Cl$ and one $N$),resulting in $sp^3$ hybridization.Therefore,the geometry of $BCl_3 \cdot NH_3$ is tetrahedral.

Consider the following reaction $BCl_3 + NH_3 \longrightarrow BCl_3 \cdot NH_3$. The geometries of $BCl_3$ and $BCl_3 \cdot NH_3$ respectively are

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