In the Lewis structure for ICl2, how many lone pairs of electrons are around the central iodine atom?
The Correct Answer and Explanation is:
In the Lewis structure for ICl2, there are three lone pairs of electrons around the central iodine atom. Let’s break this down step by step.
Iodine (I) is in Group 17 of the periodic table, which means it has 7 valence electrons. Each chlorine (Cl) atom also has 7 valence electrons. ICl2 consists of one iodine atom and two chlorine atoms, which means the iodine atom is bonded to two chlorine atoms.
To determine the number of lone pairs on iodine, first count the total number of valence electrons:
- Iodine: 7 valence electrons
- Chlorine: 7 valence electrons × 2 = 14 valence electrons
So, in total, there are 7 + 14 = 21 valence electrons to distribute in the molecule.
Next, we place the iodine in the center and the chlorine atoms around it. Since iodine forms two single bonds with the chlorine atoms, this consumes 2 electrons for each bond, totaling 4 electrons (2 electrons per bond × 2 bonds).
After bonding, 21 total valence electrons − 4 electrons used in bonding = 17 electrons remain to be placed as lone pairs. These electrons are placed around the iodine and chlorine atoms.
Each chlorine atom will complete its octet by having three lone pairs, using up 6 electrons (3 lone pairs × 2 chlorine atoms). This leaves 11 electrons to be placed on iodine.
Now, iodine can use its remaining electrons to form lone pairs. Since iodine is in the third period or below, it can accommodate more than 8 electrons, and in this case, it will have three lone pairs. Thus, there are three lone pairs on iodine in ICl2.
The remaining 2 electrons (from the 11) form a single bond between iodine and chlorine, confirming the overall structure and bonding.
Therefore, iodine has three lone pairs of electrons in ICl2.
