Explain the periodicity of electron gain enthalpy of elements in the periodic table.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Electron gain enthalpy $(\Delta_{eg}H)$ shows less regularity compared to ionization enthalpy.
Across a period (left to right),electron gain enthalpy generally becomes more negative (increases in magnitude) with an increase in atomic number. This is because the effective nuclear charge increases,making it easier for an incoming electron to be added to the smaller atom,as the added electron is closer to the positively charged nucleus.
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,the value of electron gain enthalpy generally becomes less negative (e.g.,$[S(-200), Se(-195), Te(-190), Po(-174)]$ and $[Cl(-349), Br(-325), I(-295), At(-270)]$). This occurs because the atomic volume 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 when an electron is added to oxygen or fluorine ($n=2$ shell),it experiences significant inter-electronic repulsion due to the small size of the shell. In the $n=3$ shell (sulfur or chlorine),the added electron occupies a larger region of space,resulting in significantly less electron-electron repulsion.

Explore More

Similar Questions

Electron affinity is the

Assertion $(A)$: Fluorine has a smaller negative electron gain enthalpy than chlorine.
Reason $(R)$: The electron-electron repulsion is higher in chlorine than in fluorine.

For electron affinity of halogens,which of the following is correct?

Knowing the electron gain enthalpy values for $O \rightarrow O^{-}$ and $O \rightarrow O^{2-}$ as $-141 \ kJ \ mol^{-1}$ and $702 \ kJ \ mol^{-1}$ respectively,how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^{-}$?
(Hint: Consider lattice energy factor in the formation of compounds).

The correct order of electron gain enthalpy is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo