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

Step 1 — Electronic configuration of N.

Nitrogen is $[He]\,2s^2 2p^3$. Its valence shell is $n=2$, which has only one $s$ and three $p$ orbitals — there is no $2d$ subshell in the second energy level.

Step 2 — Consequence for bonding.

To form a pentahalide $NX_5$, nitrogen would need five hybrid orbitals ($sp^3d$) to bond with five halogen atoms, using one electron from each of 5 orbitals. Since a $2d$ orbital is not available at $n=2$, N cannot promote an electron into a $d$ orbital and cannot expand its octet beyond 4 bonds (covalency 4, as in $NH_4^+$).

Step 3 — Contrast with heavier congeners.

so $PCl_5, AsCl_5, SbCl_5$ do form via $sp^3d$ hybridisation, while nitrogen is restricted to trihalides ($NX_3$) and, with fluorine only, the tetrafluoride cation $NF_4^+$.

Compare the valence-shell orbitals available at nitrogen's principal quantum number (n = 2: only s, p) against the heavier Group 15 members (n ≥ 3: s, p, d available), and link orbital availability directly to maximum covalency.

• Do not say "nitrogen doesn't have enough electrons" — the issue is absence of d orbitals at n = 2, not electron count.
• Don't confuse this with oxidation state; N can show +5 (e.g., in $HNO_3$, $N_2O_5$) through multiple bonding/ionic character, it just cannot use that state to form a 5-covalent halide.