To draw Lewis structures for the specified species, we need to follow a systematic approach. Below are the Lewis structures for each molecule, detailing the arrangement of electrons, lone pairs, bond pairs, and formal charges where applicable.
A. Lewis Structure for AsF6
1. **Count the valence electrons:** Arsenic (As) has 5 valence electrons and each fluorine (F) has 7, giving us a total of: 5 + (6 x 7) = 47 valence electrons.
2. **Arrange the atoms:** Place As in the center with 6 F atoms surrounding it.
3. **Form bonds:** Connect each F to As with a single bond, using up 12 electrons (6 pairs) for the bonds.
4. **Distribute remaining electrons:** Distribute the remaining electrons to each F atom, ensuring they each have 3 lone pairs (6 electrons). This consumes all 47 electrons without any formal charges, as As has 0 formal charge (5 from As, 6 bonds: 5-0-0), and all F atoms achieve -1 formal charge (7 from F, 1 bond: 7-1-6).
Formal Charges: As: 0, F: -1 each.
B. Lewis Structure for Cl3PO
1. **Count the valence electrons:** Chlorine (Cl) has 7 valence electrons and oxygen (O) has 6, so the total is: (3 x 7) + 6 = 27 valence electrons.
2. **Arrange the atoms:** Cl atoms are placed around the central P atom.
3. **Form bonds:** Connect Cl atoms with single bonds to P. This uses 6 electrons, leaving 21 electrons.
4. **Add remaining electrons:** Distribute to satisfy octets; place 3 lone pairs on Cl (3 x 6 = 18), and place a double bond between P and O, using 4 electrons (since double bond counts as 4). This ends with 2 lone pairs on O. The probable arrangement leads to Cl having a lone pair and P having a +1 charge, while O has a -1 formal charge.
Formal Charges: Cl: 0, P: +1, O: -1.
C. Lewis Structure for IF5
1. **Count the valence electrons:** Iodine (I) has 7 valence electrons and each fluorine (F) has 7, resulting in: 7 + (5 x 7) = 42 valence electrons.
2. **Arrange the atoms:** I is in the center surrounded by 5 F atoms.
3. **Form bonds:** Connect each F to I with a single bond (5 x 2 = 10 electrons, 32 left).
4. **Lone pairs on F:** Each F gets 3 lone pairs (3 x 6 = 18), totaling 38 electrons, leaving 4 electrons. Place a lone pair on I which will result in 1 lone pair on I.
Formal Charges: I: 0, F: -1 each.