Oxidation state of Halogens So the question is which is the most common oxidation state of halogens and I have to justify my answer. Is my following thought correct? Do I have to add something? 

The most common oxidation state of halogens is $-1$. Halogens are the most electronegative elements of the periodic table. Their outer electron configuration is $n\mathrm{s^2}n\mathrm{p^5}$. If chlorine, for example, gains one more electron, the outer p orbitals are completely filled (resulting in a full octet). Since halogens are electronegative they can easily remove an electron from the nearby atom. So then the halogen becomes isoelectronic with a noble gas and it is more stable in this state.
There is not one single most common oxidation state across all halogens. While fluorine is the most electronegative element of which compounds are known — meaning that it can only exist in the $\mathrm{-I}$ and $\pm0$ oxidation states — the electronegativity continually reduces down the group to iodine. Iodide in the oxidation state $\mathrm{-I}$ is easily oxidised to $\ce{I2}$ or $\ce{IO3-}$ with the oxidation states $\pm0$ and $\mathrm{+V}$, respectively.

What is true, however, is that all halogens except fluorine show oxidation states from $\mathrm{-I}$ to $\mathrm{+VII}$ as can be predicted from their electronic structure $n\mathrm{s^2}\,n\mathrm{p^5}$.