Hello,
I have seen that chlorine can achieve 5 oxidation states
-1, +1, +3, +5, +7
The positive oxidation states are for the following chlorates
ClO- (+1), ClO2- (+3), ClO3- (+5), ClO4- (+7)
I'm trying to understand where the electrons “go” when these anions are formed
The electronic configuration of the chlorine atom is 2.8.7
Now if I think of promoting electrons in the chlorine atom from the 3p orbitals into 4s and 4p I can create the correct number of unpaired electrons to account for the formulae. But is the following correct?
ClO- (+1). Promote one 3s electron to 4s. Creates three unpaired electrons, two in 3p one in 4s. An oxygen atom bonds with two 3p electrons leaving an unpaired 4s electron. One further electron goes into the 4s orbital making the charge -1
Similarly
ClO2- (+3). two 3p electrons promoted to 4s and 4p (a hybridized 4sp1 ?). Two oxygen atoms bond with 4 unpaired electrons in three lots 3p and one 4sp1. Extra electron in 4sp1 to give negative charge
ClO3- (+5) One 3s electron and two 3p electrons promoted to 4s and 4p (a hybridized 4sp2 ?). Three oxygen atoms bond with 6 unpaired electrons in one 3s three lots 3p and two 4sp2. Extra electron in 4sp2 to give negative charge.
Thanks
Clive