Match List-I with List-II.

List-I (Electronic configuration)	List-II ($$\Delta_i H$$ in kJ mol$$^{-1}$$)
(a) $$1s^2 2s^2$$	(p) 801
(b) $$1s^2 2s^2 2p^4$$	(q) 899
(c) $$1s^2 2s^2 2p^3$$	(r) 1314
(d) $$1s^2 2s^2 2p^1$$	(s) 1402

Choose the most appropriate answer from the options given below:

We need to identify the elements from their electronic configurations and match them with their first ionization enthalpies ($$\Delta_i H$$).

(a) $$1s^2 2s^2$$ is Beryllium (Be, Z = 4). (b) $$1s^2 2s^2 2p^4$$ is Oxygen (O, Z = 8). (c) $$1s^2 2s^2 2p^3$$ is Nitrogen (N, Z = 7). (d) $$1s^2 2s^2 2p^1$$ is Boron (B, Z = 5).

The first ionization enthalpies of these elements are: B = 801 kJ mol$$^{-1}$$, Be = 899 kJ mol$$^{-1}$$, O = 1314 kJ mol$$^{-1}$$, and N = 1402 kJ mol$$^{-1}$$. Note that nitrogen has a higher ionization enthalpy than oxygen due to the extra stability of its half-filled $$2p^3$$ configuration, and beryllium has a higher ionization enthalpy than boron due to the stability of its fully filled $$2s^2$$ subshell.

Therefore: (a) Be $$\to$$ (q) 899, (b) O $$\to$$ (r) 1314, (c) N $$\to$$ (s) 1402, (d) B $$\to$$ (p) 801.

The correct matching is **(a)→(q), (b)→(r), (c)→(s), (d)→(p)**.