Question

Why does nitrogen shows variable valency of +5...inspite of absence of d orbital?

Answer 1

Because nitrogen can be quaternized...i.e. the nitrogen lone pair can act in a dative interaction … it is also stereochemically active.

Explanation:

Consider the nitrogen oxides: `N_2O`; `NO`; `NO_2-=N_2O_4`; `N_2O_5`...

The lone pair on nitrogen is definitely stereochemically active...and this can be appreciated by the acid-base reaction that ammonia undergoes...

`NH_3(aq) + H_2O(l) rightleftharpoons ""^(+)NH_4+HO^-`

For `N_2O`, `"nitrous oxide"` aka `"laffing gas"` (as a grad student we had a large cylinder of this in the lab...since raves and concerts were often held at the university we often joked that we should wheel the cylinder down to the concert, and charge $5-00 per balloon...of course we would have ended up in gaol pdq). Here we got a linear molecule with charge separation...

`N-=stackrel+N-O^(-)`

`NO_2` is an interesting customer, in that the MONOMER features an UNPAIRED electron...and we conceive that the TWO `NO_2` molecules can dimerize and form a nitrogen-nitrogen bond

`2O=dotN^(+)-O^(-)rarr(""^(-)O)O=stackrel(+)N-stackrel(+)N=O(O^-)`

And of course we can use this simple electronic scheme to rationalize the experimental result...

And we go all the way up to `N(V+)` in `"dinitrogen pentoxide"` `N_2O_5`...again nitrogen is quaternized, with charge separation...

`(""^(-)O)O=stackrel(+)N-O-stackrel(+)N=O(O^-)`

And in this beast we got formal `N(V)` centres...

Please note that I framed this answer in terms of oxidation state rather than valency...