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

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 "=O)+("for"toO)+2("for"-O-)=0`
`:.2x-10=0orx=+5`
It, however, does not form a pentabalide. 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 pentahalied, we need five half filled orbitals. Since, nitrogen with n=2 cna have olny 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 to 3s orbital, we can still have five half-filled orbital. to form `NCl_(5)`. 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 thermodynamaically not feasible. Thus, N cannot have a covalency of 5. i.e., nitrogen does not form a pentahcalide.