Answer
See the explanation
Work Step by Step
For the molecules or ions in Exercises 81 and 87 from Chapter 3:
Exercise 81: BF3 (Boron trifluoride)
The central atom in BF3 is boron (B). The expected hybridization of the boron atom is sp2.
Explanation:
Boron has three valence electrons, and in BF3, it forms three covalent bonds with the three fluorine atoms. To accommodate these three bonds, the boron atom undergoes sp2 hybridization, where one 2s orbital and two 2p orbitals combine to form three equivalent sp2 hybrid orbitals. This allows the boron atom to form the three B-F bonds in a trigonal planar geometry.
Exercise 87: NH4+ (Ammonium ion)
The central atom in the ammonium ion (NH4+) is nitrogen (N). The expected hybridization of the nitrogen atom is sp3.
Explanation:
Nitrogen has five valence electrons, and in the ammonium ion, it forms four covalent bonds with the four hydrogen atoms. To accommodate these four bonds, the nitrogen atom undergoes sp3 hybridization, where one 2s orbital and three 2p orbitals combine to form four equivalent sp3 hybrid orbitals. This allows the nitrogen atom to form the four N-H bonds in a tetrahedral geometry.
