Explain the trends in electron gain enthalpy across periods and down groups.

(N/A) Electron gain enthalpy shows less irregularity compared to ionization enthalpy.
Across a period (left to right),electron gain enthalpy generally becomes more negative (increases in magnitude) as atomic number increases.
This is because the effective nuclear charge increases,causing the atomic radius to decrease. Consequently,the incoming electron is closer to the positively charged nucleus,resulting in a stronger attraction.

Group-$1$ Element $(\Delta_{eg} H)$	Group-$16$ Element $(\Delta_{eg} H)$	Group-$17$ Element $(\Delta_{eg} H)$	Group-$18$ Element $(\Delta_{eg} H)$
$H(-73)$	$-$	$-$	$He(+48)$
$Li(-60)$	$O(-141)$	$F(-328)$	$Ne(+116)$
$Na(-53)$	$S(-200)$	$Cl(-349)$	$Ar(+96)$
$K(-48)$	$Se(-195)$	$Br(-325)$	$Kr(+96)$
$Rb(-47)$	$Te(-190)$	$I(-295)$	$Xe(+77)$
$Cs(-46)$	$Po(-174)$	$At(-270)$	$Rn(+68)$

Down a group (top to bottom),the electron gain enthalpy becomes less negative. For example,in Group-$16$ [$S(-200), Se(-195), Te(-190), Po(-174)$] and Group-$17$ [$Cl(-349), Br(-325), I(-295), At(-270)$]. This occurs because the atomic size increases,and the added electron is further from the nucleus.
Note: The electron gain enthalpy of oxygen is less negative than that of sulfur,and that of fluorine is less negative than that of chlorine. This is because the incoming electron in $n=2$ (oxygen/fluorine) experiences significant inter-electronic repulsion due to the small size of the shell. In $n=3$ (sulfur/chlorine),the electron occupies a larger region,reducing repulsion.