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Chapter 10: Problem 34

Consider the reaction $$ \mathrm{BF}_{3}+\mathrm{NH}_{3} \longrightarrow \mathrm{F}_{3} \mathrm{~B}-\mathrm{NH}_{3} $$ Describe the changes in hybridization (if any) of the \(\mathrm{B}\) and \(\mathrm{N}\) atoms as a result of this reaction.

Short Answer

Expert verified
In the given reaction, the hybridization of Boron (B) changes from sp2 to sp3 and that of Nitrogen (N) changes from sp3 to sp2.

Step by step solution

01

Identify Initial Hybridization States

Initially, \(\mathrm{BF}_{3}\) possesses a boron atom (B) in sp2 hybridization, because it forms 3 sigma bonds with Fluorine atoms and it doesn’t have any lone pairs. On the other hand, \(\mathrm{NH}_{3}\) has a Nitrogen (N) atom in sp3 hybridization as it forms three sigma bonds and has one lone pair.
02

Identify Final Hybridization States

After the reaction, \(\mathrm{F}_{3} \mathrm{B}-\mathrm{NH}_{3}\) forms where the same Nitrogen (N) atom still forms 3 sigma bonds, but now it uses the lone pair to form a coordinate bond with the boron atom which changes its hybrid state to sp2. The Boron (B) atom in this complex still forms 3 sigma bonds with fluorine, but it too participates in the coordinate bond with nitrogen, which adds an additional electron density to the boron, and perceived as a Boron with a lone pair, changing its hybridization from sp2 to sp3.
03

Conclusion

In conclusion, during this reaction, the hybridization of the Boron (B) atom changes from sp2 to sp3, and the hybridization of Nitrogen (N) changes from sp3 to sp2.

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Most popular questions from this chapter

The geometry of \(\mathrm{CH}_{4}\) could be square planar, with the four \(\mathrm{H}\) atoms at the corners of a square and the \(\mathrm{C}\) atom at the center of the square. Sketch this geometry and compare its stability with that of a tetrahedral \(\mathrm{CH}_{4}\) molecule.

The bonds in beryllium hydride \(\left(\mathrm{BeH}_{2}\right)\) molecules are polar, and yet the dipole moment of the molecule is zero. Explain.

Given that the order of molecular orbitals for \(\mathrm{NO}\) is similar to that for \(\mathrm{O}_{2}\), arrange the following species in increasing bond orders: \(\mathrm{NO}^{2-}, \mathrm{NO}^{-}, \mathrm{NO}, \mathrm{NO}^{+}\) \(\mathrm{NO}^{2+}\)

The geometries discussed in this chapter all lend themselves to fairly straightforward elucidation of bond angles. The exception is the tetrahedron, because its bond angles are hard to visualize. Consider the \(\mathrm{CCl}_{4}\) molecule, which has a tetrahedral geometry and is nonpolar. By equating the bond moment of a particular \(\mathrm{C}-\mathrm{Cl}\) bond to the resultant bond \(\mathrm{mo}-\) ments of the other three \(\mathrm{C}-\mathrm{Cl}\) bonds in opposite directions, show that the bond angles are all equal to \(109.5^{\circ}\)

Draw the Lewis structure of ketene \(\left(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}\right)\) and describe the hybridization states of the \(\mathrm{C}\) atoms. The molecule does not contain \(\mathrm{O}-\mathrm{H}\) bonds. On separate diagrams, sketch the formation of sigma and pi bonds.
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