Answer
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Step 1:I'll solve this problem step by step for both Lewis structures, following the precise LaTeX formatting guidelines:
For $$\mathrm{ICI}_{4}^{-}$$ Lewis Structure:
Step 2:: Count total valence electrons
- Total valence electrons: $$7 + 28 + 1 = 36$$ electrons
- Iodine (I): 7 valence electrons - Chlorine (Cl): 7 valence electrons × 4 = 28 - Negative charge adds 1 electron
Step 3:: Place central atom
- Iodine (I) is the central atom
Step 4:: Distribute electrons around outer atoms
- Place 4 chlorine atoms around central iodine - Connect each chlorine with single bonds - Distribute remaining electrons as lone pairs on chlorines
Step 5:: Complete central atom's electron configuration
For $$\mathrm{SO}_{3}$$ Lewis Structure:
- Remaining electrons go around central iodine as lone pairs
Step 6:: Count total valence electrons
- Total valence electrons: $$6 + 18 = 24$$ electrons
- Sulfur (S): 6 valence electrons - Oxygen (O): 6 valence electrons × 3 = 18
Step 7:: Place central atom
- Sulfur (S) is the central atom
Step 8:: Connect oxygen atoms to sulfur
- Create three single bonds between sulfur and oxygen atoms - Distribute remaining electrons as lone pairs on oxygen atoms
Step 9:: Adjust bonding to satisfy octet rule
- If needed, convert single bonds to double bonds to complete octets
Final Answer
- For \mathrm{ICI}_{4}^{-}: A Lewis structure with iodine at center, four chlorines bonded, with additional lone pairs - For \mathrm{SO}_{3}: A Lewis structure with sulfur at center, three oxygen atoms bonded, potentially with double bonds to satisfy octet rule Note: Precise visual representation would require drawing the actual Lewis structure, which cannot be done in text format.
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