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Nitrogen has no available d-orbitals. Yet, it forms compounds where its oxidation state is +5, like in nitric acid. How?

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    $\begingroup$ What makes you think that available d-orbitals are essential for achieving high oxidation states? $\endgroup$ – bon Jul 16 '15 at 16:27
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    $\begingroup$ @AgnishomChattopadhyay No, they don't. $\endgroup$ – permeakra Jul 16 '15 at 16:54
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    $\begingroup$ Generally, the simpliest intuition for this is to imagine that nitrogen looses one electron, becoming isoelectronic to carbon. After forming four bonds with, say, fluorine, we gain $\ce{NF4^+}$ cation where nitrogen is connected to four more electronegative atoms and have positive charge to begin with, thus +5 oxidation state. In some compounds, say, nitric acid, the electron is not moving too far, instead migrating to nearest oxygen, so nitrate anion is actually isoelectronic to carbonate dianion. $\endgroup$ – permeakra Jul 16 '15 at 16:57
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    $\begingroup$ d-orbital hybridisation in expanded octet compounds is a subject of much debate. It has been discussed extensively on this site and the general consensus is that hypervalency provides a better description of the bonding in expanded octet molecules. See here, here and here for some initial reading. $\endgroup$ – bon Jul 16 '15 at 17:06
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    $\begingroup$ By the way, nitrogen (and any atom type) has d-orbitals. They just aren't typically occupied by electrons. $\endgroup$ – jerepierre Jul 16 '15 at 17:19
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The concept of oxidation states is just that: A theoretical concept. You assign electronegativity values based on one of the numerous scales available, split the bonds heterolytically according to those numbers, and then count electrons.

As such, it is very easy to make a cation like @permeakra stated in the comments that satisfies the octet rule and still gets to state +5 (the molecular cation in question is $\ce{NF4+}$).

There are no $d$ orbitals necessary (which are also theoretical, but slightly closer to reality) to explain how a +5 state is possible.

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