How is the type of hybridization determined?

Step 1. Determine the central atom. The smallest electronegative atom will be the center atom. For example, iodine.

Step 2. Create bonds associated with the central atom. Since there are 5 fluorine atoms, you need 5 bonds.

Step 3. Connect the remaining atoms to the bonds.

Step 4. Count the number of valence electrons. The group numbers show how many valence electrons an atom has. Both iodine and fluorine are in group 7A, giving each of the atoms 7 electrons.

Step 5. Add the total number of electrons. Iodine has 7 and every fluorine has 7. Since there are five fluorides, you need to multiply the seven electrons of one fluorine atom by five. 1 (7) + 5 (7) = 42 total electrons.

Step 6. Fill the outer atoms with electrons. Remember to follow the octet rule. Single bonds are calculated for 2 electrons. So give each fluorine atom 6 electrons.

Step 7. Count the remaining electrons.
Each fluorine atom has 8 electrons around it, including bound electrons. This gives you 40 out of 42 total electrons.

Step 8. Use the remaining electrons. You must use the remaining two electrons; since all five fluorine atoms have eight electrons, place the remaining electrons on iodine. Iodine is an exception to the octet rule because it can expand its orbitals since it has a “d” subshell.

Step 9. Use the Lewis structure to draw the shape. Remember that single electron pairs are not involved in molecular geometry. Use bonds to draw the sides of the shape.

Step 10. Connect each atom to a line. In the case of iodine pentafluoride, iodine and four fluorides create a square for the base.

Step 11. Define the shape. The shape of iodine pentafluoride looks like a pyramid with a square base. Thus, the molecular form name for iodine pentafluoride is square pyramidal.

Step 12. Count the bonds and single electron pairs associated with the atom. Iodine has 5 bonds and 1 single electron pair. Each fluorine has 1 bond and 3 single electron pairs.

Step 13. Determine hybridization. Since iodine has a total of 5 bonds and 1 single pair, the hybridization is sp3d2. Indicators on subshells should add up to the number of bonds and single pairs. Fluorine has 1 bond and 3 single pairs, giving a total of 4, making hybridization: sp3. By adding exhibitors, you get 4.