Question 4 of 43 What are the group numbers and the generic outer electron configurations for a neutral atom with the given Lewis symbols? O Macmillan Learnin - \( { }_{\circ}^{\circ} \). Select the group number. 1A (1) 3A (13) 5A (15) 7A (17) 2A (2) 4A (14) 6A (16) 8A (18) Select the generic outer electron configuration. \( n s^{2} n p^{1} \) \( n s^{2} n p^{2} \) \( n s^{2} n p^{3} \) \( n s^{2} n p^{4} \) \( n s^{2} n p^{5} \) \( n s^{2} n p^{6} \) - Select the group number. 1A (1) 3A (13) 5A (15) 2A (2) 4A (14) 6A (16) Select the generic outer electron configuration. \( n s^{2} n p^{1} \) \( n s^{2} n p^{3} \) \( n s^{2} n p^{4} \) \( n s^{2} n p^{5} \) \( n s^{2} n p^{6} \)

To determine the **group number** and the **generic outer electron configuration** for a neutral oxygen atom based on its Lewis symbol, let's analyze the provided information. ### Lewis Symbol Analysis: The Lewis symbol for oxygen can be represented as: ``` . : O : . ``` This indicates that oxygen has **6 valence electrons** (2 dots above and below, and 2 on each side). ### Determining the Group Number: The group number in the periodic table corresponds to the number of **valence electrons** for main-group elements. Since oxygen has **6 valence electrons**, it belongs to: - **Group 6A (16)** ### Determining the Generic Outer Electron Configuration: The generic outer electron configuration for elements in Group 6A (16) can be represented as: - **\( n s^{2} n p^{4} \)** This configuration indicates: - **2 electrons** in the **s-orbital** - **4 electrons** in the **p-orbital** ### Summary: - **Group Number:** 6A (16) - **Generic Outer Electron Configuration:** \( n s^{2} n p^{4} \) --- **Final Selections:** - **Group Number:** 6A (16) - **Generic Outer Electron Configuration:** \( n s^{2} n p^{4} \)