To draw the Lewis structure for ICl (Iodine monochloride), we first need to determine the total number of valence electrons available for bonding. Iodine (I) has 7 valence electrons, and chlorine (Cl) also has 7 valence electrons. Therefore, the total number of valence electrons is:
7 (from I) + 7 (from Cl) = 14 valence electrons.
Next, we start by placing the two atoms next to each other and create a single bond between them. Each bond between two atoms represents 2 electrons. After forming this bond, we will have used:
2 electrons (1 bond) from the 14 total, leaving us with 12 remaining electrons.
We will place the remaining electrons around the chlorine atom first, giving it a total of 3 lone pairs (6 electrons) as follows:
5 electrons remain after this, which will be allocated to the iodine. Iodine will have 1 lone pair (2 electrons) and will also be bonded to chlorine.
The final structure looks like this:
..
:Cl:
/
I
..
Where the two dots on the chlorine represent the lone pairs of electrons, and the line connecting I and Cl represents the bonding pair of electrons.