QQuestionChemistry
QuestionChemistry
Draw two Lewis structures for a compound with the formula C^4H^10. No atom bears a charge, and all carbon atoms have complete octets.
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Answer
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Step 1:I'll solve this Lewis structure problem step by step:
Step 2:: Determine the total number of valence electrons
- Total valence electrons: $$(4 \times 4) + (10 \times 1) = 16 + 10 = 26$$ valence electrons
- Carbon (C) has 4 valence electrons - Hydrogen (H) has 1 valence electron
Step 3:: Identify possible carbon skeleton arrangements
- C^4H^10 can have two possible structural isomers:
Step 4:
Straight-chain (n-butane)
Step 5:
Branched-chain (isobutane)
Step 6:: Draw Lewis structure for n-butane
\begin{array}{c}
- Connect carbons in a linear chain - Attach hydrogens to fill octets - Distribute 26 valence electrons H-C-C-C-C-H \ | | | | \ H H H H \end{array}
Step 7:: Draw Lewis structure for isobutane
\begin{array}{c}
- Create a branched carbon skeleton - Attach hydrogens to fill octets - Distribute 26 valence electrons \quad H \ \quad | \ H-C-C-C-H \ \quad | \quad | \ \quad H \quad H \end{array}
Final Answer
Two valid Lewis structures for C^4H^10 are n-butane and isobutane, each satisfying the requirements of no charged atoms and complete octets for carbon atoms.
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