Answer
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Step 1:I'll solve this Lewis structure problem step by step, following the specified formatting guidelines:
Step 2:: Determine the total number of valence electrons
• Total valence electrons: $$7 + (7 \times 5) = 42$$ electrons
• Iodine (I): 7 valence electrons • Chlorine (Cl): 7 valence electrons × 5 = 35 valence electrons
Step 3:: Place the central atom
• Iodine (I) will be the central atom due to its lower electronegativity
Step 4:: Connect the chlorine atoms to the central iodine
• Remaining electrons: $$42 - 10 = 32$$ electrons
• Draw single bonds between I and each Cl
Step 5:: Distribute remaining electrons as lone pairs
• Remaining electrons: $$2$$ electrons
• Place 6 electrons (3 pairs) around each chlorine atom
Step 6:: Place remaining electrons on central iodine
• The final 2 electrons become a lone pair on the central iodine
Step 7:: Check the formal charges
• Verify that the structure minimizes formal charges • Confirm the octet rule for chlorines • Allow iodine to expand its octet (exceeding 8 electrons) Final Lewis Structure: • Central iodine with 5 single bonds to chlorines • 3 lone pairs on each chlorine • 1 lone pair on central iodine • Molecular geometry: Square pyramidal
Final Answer
The Lewis structure for ICl^5 consists of an iodine atom with five chlorine atoms arranged around it, with one lone pair on the central iodine and three lone pairs on each chlorine atom.
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