QQuestionChemistry
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}
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