QQuestionChemistry
QuestionChemistry
Draw a Lewis structure for H^2O that obeys the octet rule if possible and answer the following questions based on your drawing.
For the central oxygen atom:
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Answer
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Step 1:I'll solve this step by step, focusing on drawing the Lewis structure for H^2O and analyzing the central oxygen atom.
Step 2:: Determine the total number of valence electrons
- Total valence electrons: $$6 + (2 \times 1) = 8$$ electrons
- Oxygen (O) has 6 valence electrons - Hydrogen (H) has 1 valence electron
Step 3:: Arrange atoms in the molecule
- Oxygen will be the central atom - Two hydrogen atoms will bond to the oxygen
Step 4:: Draw initial Lewis structure
- Connect hydrogen atoms to oxygen with single bonds - Each single bond uses 2 electrons \begin{array}{c} H-O-H \end{array}
Step 5:: Add remaining electrons as lone pairs
- Total valence electrons: 8 - Electrons used in bonds: 4 (2 single bonds) - Remaining electrons: 4 - Place these 4 electrons as 2 lone pairs on oxygen Final Lewis Structure: \begin{array}{c} H-O: \end{array} Answers about the central oxygen atom:
Step 6:
Number of lone pairs: 2
Step 7:
Number of bonding electron pairs: 2
Step 8:
Electron geometry: Tetrahedral
Step 9:
Molecular geometry: Bent (due to lone pairs)
Final Answer
The Lewis structure for H^2O shows a central oxygen atom with 2 lone pairs and 2 single bonds to hydrogen atoms, satisfying the octet rule.
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