Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $$p\pi - p\pi$$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
In the light of given statements, choose the correct answer from the options given below.

Nitrogen belongs to the second period, therefore its valence shell contains only the 2s and three 2p orbitals; no vacant 2d orbitals are available.

Case 1 : Verification of Statement I
• Nitrogen gives a whole series of oxides: $$N_2O$$ (+1), $$NO$$ (+2), $$N_2O_3$$ (+3), $$NO_2$$ / $$N_2O_4$$ (+4) and $$N_2O_5$$ (+5).
• In these compounds nitrogen and oxygen are of comparable, small atomic size. Their 2p orbitals can overlap side-by-side, producing strong $$p\pi - p\pi$$ multiple bonds (for example $$N=O$$, $$N\equiv O$$).
• The stability imparted by these multiple bonds allows nitrogen to exist comfortably in every oxidation state from +1 up to +5.
Hence Statement I is true.

Case 2 : Verification of Statement II
• To obtain a +5 oxidation state in a halide, nitrogen would need to accommodate ten electrons from five halogen atoms (example: $$NX_5$$).
• That requires an expanded octet, possible only when empty d orbitals are available for valence-shell expansion (as in $$PCl_5$$, $$AsF_5$$ etc.).
• Nitrogen lacks d orbitals in its valence shell (only 2s and 2p are present), so it cannot expand its octet beyond eight electrons. Therefore $$NX_5$$ type halides do not exist.
Thus Statement II is also true.

Both statements are true, so the correct choice is Option D.