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Consider the following reaction $\mathrm{BCl}_3+\mathrm{NH}_3 \longrightarrow \mathrm{BCl}_3 \cdot \mathrm{NH}_3$
The geometries of $\mathrm{BCl}_3$ and $\mathrm{BCl}_3 \cdot \mathrm{NH}_3$ respectively are
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The geometries of $\mathrm{BCl}_3$ and $\mathrm{BCl}_3 \cdot \mathrm{NH}_3$ respectively are
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trigonal planar and tetrahedral
The molecular geometry of $\mathrm{BCl}_3$ is trigonal planar.
In $\mathrm{BCl}_3 \cdot \mathrm{NH}_3$, the central atom $\mathrm{B}$ has 6 valence electrons that makes it an electron deficient molecule. $\mathrm{BCl}_3$ acts as a Lewis acid as it needs 2 electrons to complete its octet. $\mathrm{NH}_3$ can donate its lone pair to boron as it acts as Lewis acid. The resulting structure is tetrahedral.
In $\mathrm{BCl}_3 \cdot \mathrm{NH}_3$, the central atom $\mathrm{B}$ has 6 valence electrons that makes it an electron deficient molecule. $\mathrm{BCl}_3$ acts as a Lewis acid as it needs 2 electrons to complete its octet. $\mathrm{NH}_3$ can donate its lone pair to boron as it acts as Lewis acid. The resulting structure is tetrahedral.
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