But first,
How does \(Cl\) gain an electron to become \(Cl^{-}\)? Obviously,
But how? And more importantly why? And not so obviously,
Yes! It is making fun of you.
As stated before diffotoms, where the inner orbital structure of the element is different, are responsible for varied oxidation states.
In the case of \(ClO_{4}^{-}\) with \(Cl^{+7}\), the outer \(n=3\) swapped with inner \(n=4\), each of the four outer paired orbits bonds with a Oxygen leaving the unpaired orbit outside. This outer unpaired orbit gains an electron and the ions gains a negative charge.
In the case of \(Cl_2O_6\) with \(Cl^{+6}\), the unpaired orbits of two Chlorine pairs up and bond an Oxygen. Each of the three paired orbits of one of the Chlorine bonds an Oxygen. The other Chlorine bonds with two other Oxygen.
In the case of \(ClO^{-}_3\) with \(Cl^{+5}\), the outer \(n=3\) shell bonds with three Oxygen. The unpaired orbit gains an extra electron and the ion gains a negative charge.
In the case of \(ClO_2\) with \(Cl^{+4}\), the inner \(n=4\) shell moves to the outer layer and bonds with two Oxygen. The filled unpaired orbit rest in the lower layer.
In the case of \(ClO^{-}_{2}\) with \(Cl^{+3}\), the outer \(n=3\) shell bonds with two Oxygen. The unpaired orbit gains an extra electron and the ion gains a negative charge.
Quantum number \(n=1,2,3,4..\) are all valid solutions. This means, around a nucleus there can be one, two, three, four...paired orbits. An unpaired orbit is also a solution, but has a single spin number.
Still, how does \(Cl\) gain an electron to become \(Cl^{-}\)?
