Question

Although both the \(\mathrm{BF}_{3}\) and \(\mathrm{NF}_{3}\) molecules contain the same number of atoms, the \(\mathrm{BF}_{3}\) molecule is flat, whereas the \(\mathrm{NF}_{3}\) molecule is trigonal pyramidal. Explain.

Short answer

The difference in molecular geometry for \(\mathrm{BF}_{3}\) and \(\mathrm{NF}_{3}\) molecules can be explained using the VSEPR theory. In \(\mathrm{BF}_{3}\), the Boron central atom has 3 bonding pairs with Fluorine atoms and no lone pairs, resulting in a trigonal planar electron domain and molecular geometry. In \(\mathrm{NF}_{3}\), the Nitrogen central atom has 3 bonding pairs with Fluorine atoms and 1 lone pair, resulting in a tetrahedral electron domain geometry and a trigonal pyramidal molecular geometry due to the repulsion caused by the lone pair.

Step-by-step solution

Determine the central atom and its electron domains

For both molecules, the central atom is the first atom in the chemical formula. In \(\mathrm{BF}_{3}\), the central atom is Boron (B), and in \(\mathrm{NF}_{3}\), it is Nitrogen (N). To determine the electron domains, we need to count the number of valence electrons and then distribute them around the central atom in the form of bonding pairs and lone pairs.

Count valence electrons and distribute them

In \(\mathrm{BF}_{3}\), Boron has 3 valence electrons, and each Fluorine atom has 7. So, there is a total of 3 + 3(7) = 24 valence electrons. These electrons will be distributed in the form of 3 bonding pairs between Boron and each Fluorine atom (using 6 electrons) and 9 lone pairs on the Fluorine atoms (using the remaining 18 electrons). In \(\mathrm{NF}_{3}\), Nitrogen has 5 valence electrons, and each Fluorine atom has 7. Therefore, there is a total of 5 + 3(7) = 26 valence electrons. These electrons will be distributed in the form of 3 bonding pairs between Nitrogen and each Fluorine atom (using 6 electrons), 9 lone pairs on the Fluorine atoms (using 18 more electrons), and 1 lone pair on Nitrogen (using the remaining 2 electrons).

Analyze the electron domain geometry using VSEPR Theory

VSEPR (Valence Shell Electron Pair Repulsion) Theory states that the electron pairs, whether bonding or lone pairs, will arrange themselves around the central atom so that they are as far apart from each other as possible to minimize repulsion. In \(\mathrm{BF}_{3}\), Boron has 3 bonding pairs with Fluorine atoms and no lone pairs. Therefore, the electron domain geometry is trigonal planar. As all the bonding pairs are distributed around Boron in the shape of an equilateral triangle, and there are no lone pairs, the molecular geometry will also be trigonal planar. In \(\mathrm{NF}_{3}\), Nitrogen has 3 bonding pairs with Fluorine atoms and 1 lone pair. This results in a tetrahedral electron domain geometry. However, since there is a lone pair, the molecular geometry will be different from the electron domain geometry. The lone pair occupies more space, causing the bonding pairs to be pushed slightly together. As a result, the molecular geometry is trigonal pyramidal.

Conclusion

To summarize, the \(\mathrm{BF}_{3}\) molecule has a trigonal planar electron domain and molecular geometry due to the presence of 3 bonding pairs and no lone pairs on the central atom Boron. In contrast, the \(\mathrm{NF}_{3}\) molecule has a tetrahedral electron domain geometry due to the presence of 3 bonding pairs and 1 lone pair on the central atom Nitrogen, resulting in a trigonal pyramidal molecular geometry.