Q.Why is N2 less reactive at room temperature?

Step 1 — Bonding in $N_2$.

Each nitrogen atom (2s²2p³) contributes 3 unpaired p electrons; between two N atoms these form one σ bond and two π bonds, i.e., a triple bond: $:N \equiv N:$.

Step 2 — Bond strength.

Because nitrogen is a small atom, its 2p orbitals overlap very effectively side-on and head-on, giving an unusually short (110 pm) and strong N≡N bond with bond dissociation enthalpy ≈ 941 kJ mol⁻¹ — among the highest of any diatomic molecule.

Step 3 — Kinetic consequence.

Breaking this bond (a prerequisite for N2 to react, e.g. with H2 or O2) requires a very large activation energy that is simply not available at room temperature; hence $N_2$ behaves as essentially inert under ordinary conditions, reacting only at high temperature/pressure or with catalysts (as in the Haber process) or with highly reactive metals (e.g. Li).

Identify the bond order/type in N₂ from nitrogen's valence electron configuration, quote the resulting exceptionally high bond dissociation enthalpy, and connect that directly to kinetic inertness at ambient temperature.

• Don't say $N_2$ is "unreactive" because it is thermodynamically stable in an absolute sense for all reactions — many reactions of $N_2$ are thermodynamically favourable (e.g., Haber process); the real barrier at room temperature is kinetic (high activation energy to break the triple bond).
• Don't forget the two contributing factors: it's both the small size (good orbital overlap) AND the bond order (triple, not single) that give the very high bond enthalpy.