Draw the Lewis Structure of CIO3 (OR CI O 31 minus) with the lowest possible formal charge on Cl by expanding its octet.. How many lone pairs (non-bonding pairs) are present?

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**Topic: Drawing Lewis Structures**

**Task:** Draw the Lewis Structure of ClO₃⁻ (OR Cl O 3 1 minus) with the lowest possible formal charge on Cl by expanding its octet. How many lone pairs (non-bonding pairs) are present?

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**Detailed Explanation:**

When drawing the Lewis structure for ClO₃⁻, the goal is to arrange the valence electrons around the atoms such that the octet rule is satisfied with an emphasis on minimal formal charges. The central chlorine (Cl) atom can expand its octet to accommodate extra electrons if necessary.

1. **Count Total Valence Electrons:**
   - Chlorine (Cl): 7 valence electrons
   - Oxygen (O): 6 valence electrons × 3 = 18
   - Additional electron for the negative charge: 1
   - Total: 7 + 18 + 1 = 26 valence electrons

2. **Determine Structure:**
   - Place chlorine as the central atom, surrounded by three oxygen atoms.
   - Form single bonds between Cl and each O atom.
   - Distribute remaining electrons to satisfy the octet rule, prioritizing terminal oxygen atoms first.
   - If needed, form double bonds with an oxygen atom to reduce formal charges or accommodate remaining electrons on chlorine.

3. **Calculate Formal Charges:**
   - Adjust the electron placement such that the formal charge on chlorine is minimized using expanded octets if possible.

4. **Check and Count Lone Pairs:**
   - Calculate and verify the number of lone pairs on each atom, considering any double bonds or expanded electrons on Cl.

**Outcome:** 
Ensure no atom exceeds or is deficient in the traditional octet (other than the expanded octet on Cl), and calculate final lone pairs for clarity in non-bonding electron arrangements.

**Conclusion:** 
Final structural evaluation should confirm that the formal charges are minimized, ideally resulting in the chlorine having the lowest possible formal charge, with precise counting of non-bonding electron pairs on each atom.
Transcribed Image Text:**Topic: Drawing Lewis Structures** **Task:** Draw the Lewis Structure of ClO₃⁻ (OR Cl O 3 1 minus) with the lowest possible formal charge on Cl by expanding its octet. How many lone pairs (non-bonding pairs) are present? --- **Detailed Explanation:** When drawing the Lewis structure for ClO₃⁻, the goal is to arrange the valence electrons around the atoms such that the octet rule is satisfied with an emphasis on minimal formal charges. The central chlorine (Cl) atom can expand its octet to accommodate extra electrons if necessary. 1. **Count Total Valence Electrons:** - Chlorine (Cl): 7 valence electrons - Oxygen (O): 6 valence electrons × 3 = 18 - Additional electron for the negative charge: 1 - Total: 7 + 18 + 1 = 26 valence electrons 2. **Determine Structure:** - Place chlorine as the central atom, surrounded by three oxygen atoms. - Form single bonds between Cl and each O atom. - Distribute remaining electrons to satisfy the octet rule, prioritizing terminal oxygen atoms first. - If needed, form double bonds with an oxygen atom to reduce formal charges or accommodate remaining electrons on chlorine. 3. **Calculate Formal Charges:** - Adjust the electron placement such that the formal charge on chlorine is minimized using expanded octets if possible. 4. **Check and Count Lone Pairs:** - Calculate and verify the number of lone pairs on each atom, considering any double bonds or expanded electrons on Cl. **Outcome:** Ensure no atom exceeds or is deficient in the traditional octet (other than the expanded octet on Cl), and calculate final lone pairs for clarity in non-bonding electron arrangements. **Conclusion:** Final structural evaluation should confirm that the formal charges are minimized, ideally resulting in the chlorine having the lowest possible formal charge, with precise counting of non-bonding electron pairs on each atom.
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