Let's solve part **12. BONUS** as requested, in **2 steps** with explanations and a final answer. --- ## **Step 1: Draw the electron configuration of oxygen in the water molecule** **Explanation:** - **Ground-state electron configuration of oxygen (O):** - Atomic number = 8 - Electron configuration: \(1s^2 \ 2s^2 \ 2p^4\) - In a **water molecule (H\(_2\)O)**, oxygen forms two sigma bonds with hydrogen atoms using its 2p and 2s orbitals, which are **sp\(^3\) hybridized**. This means: - The 2s and three 2p orbitals mix to form four equivalent sp\(^3\) hybrid orbitals. - Two of these sp\(^3\) orbitals form sigma bonds with hydrogens, and the other two hold lone pairs. **Drawing:** - **Ground-state O:** ``` 1s ↑↓ 2s ↑↓ 2p ↑↓ ↑ ↑ ``` - **In H\(_2\)O (sp\(^3\) hybridized):** - Four sp\(^3\) orbitals, each can hold two electrons. - Two with lone pairs (each has 2 electrons). - Two with unpaired electrons (each bonds with an H atom). --- ## **Step 2: Does the ground-state electron configuration explain the shape of the water molecule?** **Explanation:** - **Shape explanation:** - The ground-state configuration (\(2s^2 2p^4\)) predicts two unpaired electrons, which would suggest two bonds. - In the molecule, oxygen's orbitals hybridize to form four sp\(^3\) orbitals, with two forming bonds and two holding lone pairs. - The four hybrid orbitals arrange themselves to minimize electron repulsion, resulting in a **tetrahedral electron geometry**. - Because two vertices are occupied by lone pairs, the molecular shape is **bent (angular)**, with a bond angle of about 104.5°. --- ## **Final Answer** **1.** The ground-state electron configuration of oxygen is \(1s^2 \ 2s^2 \ 2p^4\). In water, oxygen's orbitals hybridize to form four sp\(^3\) orbitals; two form O-H bonds and two hold lone pairs. **2.** Yes, the ground-state configuration (and subsequent hybridization) explains the shape of the water molecule. The four sp\(^3\) orbitals arrange tetrahedrally, but since two contain lone pairs, the shape of H\(_2\)O is bent, not linear, with a bond angle of 104.5°.

# Solving the Problem in 2 Steps --- ## **Step 1: Understand the Electron Configuration of Oxygen in Water** **Explanation:** - **Ground-state electron configuration of oxygen (O):** - Atomic number = 8, so it has 8 electrons. - The electron configuration can be expressed as: \[ 1s^2 \ 2s^2 \ 2p^4 \] - In a **water molecule (H\(_2\)O)**: - Oxygen utilizes its 2s and 2p orbitals to form bonds. - These orbitals undergo **sp\(^3\) hybridization**, creating four equivalent hybrid orbitals. - Two of these sp\(^3\) orbitals form sigma bonds with hydrogen (H) atoms, while the other two contain lone pairs of electrons. **Drawing:** - **Ground-state O:** ``` 1s ↑↓ 2s ↑↓ 2p ↑↓ ↑ ↑ ``` - **In H\(_2\)O (sp\(^3\) hybridized):** - Four sp\(^3\) orbitals: - Two hold lone pairs. - Two bond with H atoms. --- ## **Step 2: Relate Electron Configuration to the Shape of Water** **Explanation:** - **Shape Explanation:** - The ground-state configuration suggests there are two unpaired electrons, which indicates the formation of two bonds. - The hybridization leads to four sp\(^3\) orbitals positioned to minimize repulsion, resulting in a **tetrahedral electron geometry**. - Since two of these orbitals are occupied by lone pairs, the resulting molecular shape is **bent** (angular), with a bond angle of approximately 104.5°. --- ## **Final Answer** 1. The ground-state electron configuration of oxygen is \(1s^2 \ 2s^2 \ 2p^4\). In water, oxygen’s orbitals hybridize to form four sp\(^3\) orbitals; two form O-H bonds, and two hold lone pairs. 2. Yes, the ground-state electron configuration and hybridization explain the shape of the water molecule. The four sp\(^3\) orbitals arrange tetrahedrally, but with two containing lone pairs, the shape of H\(_2\)O is bent, with a bond angle of 104.5°.