Answer
See the explanation
Work Step by Step
For the molecules or ions in Exercises 82 and 88 from Chapter 3:
Exercise 82: CH4 (Methane)
The central atom in methane (CH4) is carbon (C). The expected hybridization of the carbon atom is sp3.
Explanation:
Carbon has four valence electrons, and in methane, it forms four covalent bonds with the four hydrogen atoms. To accommodate these four bonds, the carbon atom undergoes sp3 hybridization, where one 2s orbital and three 2p orbitals combine to form four equivalent sp3 hybrid orbitals. This allows the carbon atom to form the four C-H bonds in a tetrahedral geometry.
Exercise 88: H2O (Water)
The central atom in water (H2O) is oxygen (O). The expected hybridization of the oxygen atom is sp3.
Explanation:
Oxygen has six valence electrons, and in water, it forms two covalent bonds with the two hydrogen atoms. To accommodate these two bonds and the two lone pairs of electrons, the oxygen atom undergoes sp3 hybridization, where one 2s orbital and three 2p orbitals combine to form four equivalent sp3 hybrid orbitals. This allows the oxygen atom to form the two O-H bonds and accommodate the two lone pairs in a tetrahedral geometry.
