To draw the Lewis structure for IF3, we start by determining the total number of valence electrons from all atoms involved. Iodine (I) is in Group 17 and has 7 valence electrons, while each of the three fluorine (F) atoms also has 7 valence electrons. Therefore, the total number of valence electrons is:
- 1 Iodine atom: 7 electrons
- 3 Fluorine atoms: 3 x 7 = 21 electrons
Total: 7 + 21 = 28 valence electrons
Next, we identify the central atom. In this case, iodine is the central atom surrounded by three fluorine atoms. Each fluorine atom will form a single bond with the iodine atom, using up 6 electrons (2 electrons for each bond), leaving us with:
- 28 – 6 = 22 electrons remaining
Now, we place the remaining 22 electrons around the fluorine atoms to complete their octets. Each fluorine needs 6 more electrons to complete its octet:
- 3 Fluorine atoms x 6 electrons = 18 electrons
- Electrons remaining after fulfilling F’s octets: 22 – 18 = 4 electrons
These 4 electrons will be placed as lone pairs on the iodine atom. Thus, iodine will have:
- 3 shared pairs (one with each fluorine) and
- 2 lone pairs of electrons.
Summarizing:
- A. Total number of valence electrons: 28
- B. Number of shared electron pairs around the central atom: 3
- C. Number of unshared electron pairs around the central atom: 2
- D. This part of the question seems to be incomplete.
The Lewis structure of IF3 can be represented as follows:
