Though nitrogen exhibits `+5`oxidation state,it does not form pentahalide ,because

Explanation

Nitrogen shows an oxidation state of +5 in `N_(2)O_(5)` as calculated below.
`2x + 2 (-2) + 2(-2) + 2 (-1) = 0`
(for = 0) + (for `rarr` O) + 2 (for -O-) = 0
`therefore 2x - 10 = 0 or x = +5`

It, howeverm does not form a pentahalide. This may be explained as follows :
The electronic configuration of nitrogen is `1s^(2) 2s^(2) 2p_(x)^(1) 2p_(y)^(1)2p_(z)^(1)`. 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 and no d-orbitals, it cannot expand its valence shell to show a covalency of 5. However, N has a vacant 3s orbital. If one of the 2s-electrons gets promoted 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.