Draw The Best Lewis Structure For ICI, Ion On Your Scratch Paper. Use It To Answer The Following Questions: How Many Bonds Does The Central Atom Have? How Many Lone Pairs (Pairs, Not Individual Dots) Does The Entire Structure Have? What Is The Formal Charge On The I? Single Bond Counts As 1, Double Bonds Count As 2, Triple Bonds Count As 3
The Correct Answer and Explanation is:
Correct Answers:
- Number of bonds on the central atom (Iodine, I): 2
- Number of lone pairs in the entire structure: 6
- Formal charge on Iodine (I): 0
Explanation
To determine the correct Lewis structure for the ICl₂⁻ ion (iodine dichloride anion), we follow these steps:
1. Count the Total Valence Electrons
- Iodine (I) has 7 valence electrons.
- Each Chlorine (Cl) also has 7 valence electrons.
- The ion has a -1 charge, so we add 1 electron.
Total = 7 (I) + 7×2 (Cl) + 1 = 22 electrons
2. Determine the Central Atom
Iodine is less electronegative than chlorine, so it becomes the central atom. The Cl atoms are terminal.
3. Form Bonds
We form single bonds between I and each Cl. That uses up 4 electrons (2 per bond), leaving 18 electrons.
4. Distribute Remaining Electrons
Distribute lone pairs (in pairs of 2 electrons) to complete octets:
- Each Cl gets 6 electrons (3 lone pairs), using 12 electrons total.
- The remaining 6 electrons go to iodine as 3 lone pairs.
5. Check Octet Rule and Geometry
Iodine has 2 bond pairs + 3 lone pairs = 10 electrons (which is okay, since iodine can expand its octet).
This gives an AX₂E₃ geometry (linear molecular shape) according to VSEPR theory.
6. Count Bonds and Lone Pairs
- Bonds on central atom (I): 2 (both single bonds to Cl)
- Lone pairs in entire structure:
- 3 on I
- 3 on each Cl = 6 from Cl
- Total = 3 (I) + 6 (Cl) = 9 pairs, but only 6 are lone pairs (excluding bonding electrons).
7. Calculate Formal Charge on Iodine
Formal Charge=Valence−Non-bonding electrons−12×Bonding electrons\text{Formal Charge} = \text{Valence} – \text{Non-bonding electrons} – \frac{1}{2} \times \text{Bonding electrons} =7−6−12(4)=7−6−2=−1= 7 – 6 – \frac{1}{2}(4) = 7 – 6 – 2 = -1
Wait — that implies Iodine has -1, but that’s wrong. Let’s check more carefully.
- I: 7 valence electrons
- 6 non-bonding electrons (3 lone pairs)
- 2 single bonds → 4 bonding electrons → contributes 2
So: FC=7−6−2=−1\text{FC} = 7 – 6 – 2 = -1
Ah! Actually, this confirms the formal charge on Iodine is -1.
Thus:
- Number of bonds on I: 2
- Number of lone pairs in total: 9
- Formal charge on I: -1
Revised Final Answers (corrected):
- Number of bonds on I (central atom): 2
- Number of lone pairs in the entire structure: 9
- Formal charge on I: -1
