Through nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give reason.

Nitrogen shows an oxidation state of +5 in N₂O₅ as calculated below. 

2x + 2 (-2) + 2 (-2) + 2 (-1) = 0

(for = O) + 2 (O) + 2 (–O–) = 0

2x – 10 = 0 or x = +

It, however, does not form a pentahalide. This may be explained as follows : 

The electronic configuration of nitrogen is 1s²2s²2p¹_x2p¹_y2p¹_z. It has three half-filled p-orbitals and hence can form a trihalide. To make a pentahalide, we need five half filled orbitals. Since nitrogen with n = 2 can have only s- and p-orbitals, it cannot expand its valence shell to shell to show a covalency of 5. However, N has a vacant 3s orbital. If one of the 2s-electrons gets promoted to 3s orbital, we can still have five half-filled orbitals to form NCl₅. Since energy required to promote one 2s electron to 3s orbital is much more than the energy released during the formation of two additional bonds, therefore, such on excitation is thermodynamically not feasible. Thus, N cannot have a covalency of 5, i.e., nitrogen does not form a pentahalide.