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QuestionChemistry

Draw the Lewis structure of NH^4 + and determine its geometry, bond angle, and the hybridization of the N atom.
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Answer

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Step 1:
I'll solve this problem step by step, following the specified LaTeX formatting guidelines:

Step 2:
: Determine the total number of valence electrons

Total valence electrons: $$5 + 4 - 1 = 8$$ electrons
- Nitrogen (N): 5 valence electrons - Hydrogen (H): 1 valence electron × 4 hydrogens = 4 electrons - Positive charge removes 1 electron

Step 3:
: Arrange the atoms

- Place the nitrogen atom in the center - Attach four hydrogen atoms around the nitrogen

Step 4:
: Draw the Lewis structure

\begin{array}{c}
- Connect each hydrogen to the nitrogen with a single bond - Distribute the remaining electrons to satisfy the octet rule - The structure will look like: \text{H} \ \quad | \ \text{H-N-H} \ \quad | \ \text{H} \end{array}

Step 5:
: Determine the geometry

- Central nitrogen has 4 bonds and no lone pairs - This results in tetrahedral geometry - Geometry: Tetrahedral

Step 6:
: Calculate bond angle

- Tetrahedral geometry has a bond angle of $$109.5°

Step 7:
: Determine hybridization

- This indicates $$sp^{3}$$ hybridization
- 4 bonds around the central nitrogen

Final Answer

- Lewis Structure: Tetrahedral NH^4 + with 4 single N-H bonds - Geometry: Tetrahedral - Bond Angle: 109.5° - Nitrogen Hybridization: sp^{3}