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QuestionChemistry

Draw the Lewis dot structure for the following ions: Al³⁺, Mg²⁺, Se²⁻, N³⁻
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Step 1:
I'll solve this Lewis dot structure problem step by step, following the specified formatting guidelines:

Step 2:
: Understanding Lewis Dot Structures

- Lewis dot structures represent the valence electrons of atoms - For ions, we need to account for the charge by adding or removing electrons - The goal is to show the electron configuration around each ion

Step 3:
: Al³⁺ (Aluminum Ion)

- Aluminum's ground state electron configuration: $$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{1}
- Valence electrons: 3 - Loses 3 electrons to become Al³⁺ - Lewis dot structure: No dots (all valence electrons removed)

Step 4:
: Mg²⁺ (Magnesium Ion)

- Magnesium's ground state electron configuration: $$1s^{2} 2s^{2} 2p^{6} 3s^{2}
- Valence electrons: 2 - Loses 2 electrons to become Mg²⁺ - Lewis dot structure: No dots (all valence electrons removed)

Step 5:
: Se²⁻ (Selenide Ion)

\ddot{Se}\ddot{}
- Valence electrons: 6 - Gains 2 electrons to become Se²⁻ - Lewis dot structure: 8 dots around Se (full octet)

Step 6:
: N³⁻ (Nitride Ion)

\ddot{N}\ddot{}
- Valence electrons: 5 - Gains 3 electrons to become N³⁻ - Lewis dot structure: 8 dots around N (full octet)

Final Answer

- Al³⁺: No dots - Mg²⁺: No dots - Se²⁻: \ddot{Se}\ddot{} (8 dots) - N³⁻: \ddot{N}\ddot{} (8 dots)