Draw Lewis structures for each compound.
a. Cl2O7 (no Cl-Cl bond)
b. H3PO3 (two OH bonds)
Solution 107E Step 1 Let’s explain the molecule Cl 2 7 Since it’s specified dthat Cl - Cl bond doesn’t exists in this molecule. We have to assume that an O act like a bridge between the two atoms of Cl identified in the formula. The other six atoms of O have to be written around the atom of Cl (three O for each Cl), like in figure Step 2 Now we can calculate the number of electrons involved in this molecule by applying the formula Total electrons= 2(Valence electrons in Cl) + 7(Valence electrons in O)= 2(7) + 7(6)= 56 Step 3 Now we have to assign the bonding electrons for each couple of atoms We used 16 electrons Step 4 Now we have to assign the non bonding electrons, first to the terminal atom of Oxygen (three lone pairs each), then to the bridging atom of Oxygen (two lone pairs) We used 56 of 56 electrons Step 5 In order to lower the molecule’s Energy, both atoms of Cl expand their octet by forming three double bond each with the atoms of Oxygen (thus using one of the Oxygens’ lone pairs) We obtain this final Lewis structure Step 7 Let’s explain the molecule H P3 . 3 We start writing P in the middle and then the three atoms of Oxygen at Phosphorus’ left, right and upper side. Then, next to two atoms of Oxygen ,we have to write an atom of H. The last atom of H is in the lower side respect to P, thus we obtain this disposition Step 8 Now we can calculate the number of electrons involved in this molecule by applying the formula Total electrons= (Valence electrons in P) + 3(Valence electrons in O) + 3(Valence electrons in H)= (5) + 3(6) +3(1)= 5 +18 +3= 26