# How does absence of nd orbital in Oxygen affect its valency?

The original question was 'Oxygen exhibits Oxidation state -2 to +2 but other elements of grp 16 exhibit only +2,+4,+6. Explain Why'. In the solution given i can't understand meaning of this line 'It doesn't have nd-orbitals due to which its valency doesn't increase'. Please explain how?

PS:I am a chemistry noob. So please provide as much details as possible :)

Whether or not oxygen can reach oxidation states higher than $\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\mathrm{-III}$ and $\mathrm{+V}$ but would also be restricted to <insert arbitrary number here>.

The actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\ce{XeF6, NF3}$ or $\ce{SF6}$ to $\ce{XeO4, NO3-}$ or $\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\mathrm{+II}$ on oxygen given our current knowledge.

Furthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. $\ce{Na2S}$, $\ce{Na2Se}$, $\ce{Na2Te}$ and $\ce{Na2Po}$ are all known and in all of these compounds the chalcogen has an oxidation state of $\mathrm{-II}$.

• While NF5 is still unknown, there was a report of possible NF6- anion and NF4+ is known for a long time. There's a theoretical study indicating possibility of analogous OF3+. – Mithoron Sep 13 '19 at 16:50

You need to review the concepts surrounding atomic shell structure and quantum numbers, and how they relate to the organization of the periodic table, I think. Briefly, since oxygen is in the second row of the periodic table, we know that its valence electrons occupy the second orbital shell, corresponding to orbitals with principal quantum number $n=2$. Recall also that for a given principal quantum number $n$, only sub-shells with orbital angular quantum number $l$ up to $n-1$ are allowed. Therefore, the valence shell of osygen only includes s ($l=0$) and p ($l=1$) orbitals. On the other hand, the other members of group 16 will by definition have higher principal quantum numbers, which means that their valence shells include d ($l=2$) orbitals. It is these orbitals that can be used to rationalize the existence of higher oxidation states for those elements, for example by the $sp^3d$ and $sp^3d^2$ hybridization schemes in VSEPR (valence shell electron pair repulsion) theory.

Now, that explains why heavier elements in group 16 can have higher (i.e. more positive) oxidation numbers than +2. Your question as stated appears to have an error however, since it says that those elements "exhibit only +2,+4,+6" ... that's not technically correct .. for example there are compounds where sulfur has an oxidation state of -2 (e.g. $H_2S$). However I don't think it matters for the answer to your question here.

• I know that Oxygen doesnt have d-orbital n other elements down the group have it but HOW exactly does that restricts oxygen OR say allows other elements to have more than +2 Oxidation State? – Maggi Iggam Jan 9 '15 at 16:45