The property/properties that show irregularity in first four elements of group-17 is/are:
(A) Covalent radius
(B) Electron affinity
(C) Ionic radius
(D) First ionization energy
Choose the correct answer from the options given below:

Group-17 elements in their normal sequence are $$F, Cl, Br, I$$. For each periodic property we compare the numerical values of these four elements and see whether the trend is strictly monotonic (regular) or has an exception (irregular).

General rule along a group: covalent radius increases because each successive element has an extra electron shell.

Experimental values (pm): $$F = 71,\; Cl = 99,\; Br = 114,\; I = 133$$. The values rise smoothly with no reversal. Hence the covalent radius shows a regular trend, not an irregularity.

General rule down a group: electron affinity becomes less exothermic (numerical value decreases) as atomic size increases.

Experimental values (kJ mol$$^{-1}$$): $$F = -328,\; Cl = -349,\; Br = -325,\; I = -295$$. Notice the first step $$F \rightarrow Cl$$ is opposite to the expected trend because $$|EA_{Cl}| \gt |EA_{F}|$$. This reversal is due to very small size of $$F$$ causing greater inter-electronic repulsion in the compact $$2p$$ subshell.

Since the trend is not monotonic, electron affinity shows an irregularity in the first four halogens.

For isoelectronic $$X^-$$ ions, radius increases smoothly with increasing nuclear charge because shielding outweighs attraction. Measured radii (pm): $$F^- = 133,\; Cl^- = 181,\; Br^- = 196,\; I^- = 220$$, a regular rise. Thus ionic radius does not show any irregularity.

General rule down a group: first ionization energy decreases because outer electrons are farther from the nucleus and better shielded.

Values (kJ mol$$^{-1}$$): $$F = 1681,\; Cl = 1251,\; Br = 1140,\; I = 1008$$, a smooth decrease with no reversal. Therefore the trend is regular.

Only electron affinity (property B) displays an irregular pattern for the first four Group-17 elements.

Correct choice: Option C  -  B only.