| Step 1: | Count the total number of valence electrons available on this group of atoms. For H2O: 2(2H) + 6(1O from Group 16) = 8 For PCl3: 5(P from Group 15) + 3 × 7 (3Cl from Group 17) = 26 Subtract electrons if overall positive; Add electrons if overall negative. |
| Step 2: | Join atoms by single bonds. Subtract the total number of electrons used from the Step 1 total. For H2O: Subtract 4 because each of two bonds use two electrons (this leaves 4). For PCl3: Subtract 6 because each of three bonds uses two electrons (20 remain) |
| Step 3: | Complete octets at outer atoms other than H. Subtract electrons used from the Step 2 total. An octet is complete if there are eight electrons around the atom. These may be bonding and non-bonding. Each single bond to an atom counts as two electrons. Cl | For: Cl – P – Cl Six additional electrons are required at each chlorine. These will be non-bonding. (Subtract 18 from 20 to give 2 remaining) For H-O-H: No change because the outer atoms are hydrogen (4 electrons remain). |
Step 4: | Place the remaining valence electrons at the central atom. For H2O Central atoms from the second row of the periodic table have a maximum of eight electrons.4 electrons remaining; place two nonbonding pairs at O as above. For PCl3 2 electrons remaining: place one non-bonding pair at P Central atoms from other rows have a minimum of eight electrons. |