To draw the Lewis structure for iodine pentafluoride (IF5), we start by determining the total number of valence electrons available. Iodine (I) has 7 valence electrons, and each fluorine (F) atom has 7 valence electrons. Since there are five fluorine atoms, we calculate the total number of valence electrons as follows:
- Valence electrons from I: 7
- Valence electrons from 5 F: 5 imes 7 = 35
- Total = 7 + 35 = 42 valence electrons
Next, we place the iodine atom in the center because it is less electronegative than fluorine. We then surround it with the five fluorine atoms. Each F atom will form a single bond with the iodine atom, using 2 electrons per bond (1 electron from I and 1 electron from F). So, we use:
- 5 bonds imes 2 electrons/bond = 10 electrons
After forming the bonds, we deduct these 10 electrons from the total:
- 42 – 10 = 32 electrons remaining
Next, we distribute the remaining electrons to the fluorine atoms to ensure they all have a complete octet. Each of the five fluorine atoms needs 6 more electrons to achieve the octet (since they already share 1 with I via the bond):
- 5 F atoms imes 6 electrons each = 30 electrons
This leaves us with:
- 32 – 30 = 2 electrons left
Lastly, we place the remaining 2 electrons on the iodine atom, giving it an expanded octet. The final Lewis structure shows the iodine atom in the center with five single bonds to each of the five fluorine atoms and two lone pairs of electrons on the iodine.
Electrons Count Summary:
- Iodine (I): 7 valence electrons + 2 lone pair electrons = 9 electrons
- Each Fluorine (F): 7 valence electrons + 1 bond electron = 8 electrons (5 F atoms total)
- Total for Fluorine = 5 imes 8 = 40 electrons
So, in total, the IF5 molecule comprises 42 electrons, with each species contributing accordingly.