Why the valency of Nitrogen not 5 in N2O5?

Answer 1

The simplest answer is that none of the oxygens in N2O5 have a -2 formal charge, so giving nitrogen a +5 formal charge would lead to a charge imbalance. Since the molecule must be charge neutral, we know that the nitrogen must have a different formal charge. (Proof by Contradiction)

The more complicated answer requires a discussion of the bonding in N2O5. As explained in the video in the Related Link, each of the nitrogens sits touching three oxygens, with one of these three being shared with the other nitrogen. The "middle" oxygen is single-bonded to each nitrogen, meaning that the oxygen in question has a 0 formal charge. On the extremities, there are two oxygens that are single-bonded to a nitrogen and two oxygens that are double-bonded. The single-bonded ones have a -1 formal charge and the double bonded ones have a 0 formal charge. This makes the total formal charge coming from the five oxygens to be 0+0-1-1+0 = -2. Therefore the nitrogens must each be +1 since the structure is parallel and the charge must be neutral.

Oh dear what a strange question. In simple classical pre- GN Lewis octet rule following the definition of valency the valency of N is 5. two double bonds one single surrounding each N. (this old theory would give 10 valence electrons around the N) From an oxidation number point of view (sometimes termed valency these days) the N atoms oxidation #'s are +5. However if you apply the octet rule you get a different answer , involving charged structures which in valence bond theory resonate. This is I think what the answer above is getting at.