Electron Gain Enthalpy

What Is Electron Gain Enthalpy?

So far we've looked at how hard it is to take an electron away from an atom (that was ionization enthalpy). Now we flip the question: what happens when we try to add an extra electron to a neutral gaseous atom?

Definition

The electron gain enthalpy ($\Delta_{\text{eg}}H$) of an element is the enthalpy change when an isolated gaseous atom accepts one electron to form a gaseous negative ion (anion).

The process is:

$$\text{X}(g) + e^{-} \;\longrightarrow\; \text{X}^{-}(g) \qquad \Delta_{\text{eg}}H$$

Think of it as the mirror image of ionization — except the sign can go either way. Sometimes the atom welcomes the electron (energy released, $\Delta_{\text{eg}}H < 0$) and sometimes it resists (energy absorbed, $\Delta_{\text{eg}}H > 0$).

Sign convention — this is where many students trip up

• $\Delta_{\text{eg}}H < 0$ (negative) → energy is released, the process is exothermic, atom happily accepts the electron. Example: halogens.
• $\Delta_{\text{eg}}H > 0$ (positive) → energy must be supplied, the process is endothermic, atom resists the extra electron. Examples: noble gases, alkaline-earth metals.

Common units: kJ/mol (preferred in NCERT) or eV/atom (1 eV/atom ≈ 96.485 kJ/mol).

[Board + JEE/NEET tip] — The most negative $\Delta_{\text{eg}}H$ in the periodic table belongs to chlorine: $-349$ kJ/mol. Memorise this single number — it's a recurring MCQ trap.

Why negative means "favourable"

The thermodynamic convention is universal: when a system releases energy, it becomes more stable, and we write enthalpy change as negative. So a large negative $\Delta_{\text{eg}}H$ means the anion is much more stable than the neutral atom + free electron, i.e., the atom readily accepts that extra electron.

Three Qualitative Cases — Who Wants an Electron, Who Doesn't

Let's look at the three most common situations you'll meet in problems. The energy-level diagram below shows why the sign of $\Delta_{\text{eg}}H$ depends on where the new electron has to go.

Case A — Halogens: the electron-loving champions

A halogen has the configuration $ns^{2}\,np^{5}$. Adding one more electron completes the $p^{6}$ octet — the same electron count as the next noble gas. Huge energy release. That's why halogens have the largest negative electron gain enthalpies:

Element	$\Delta_{\text{eg}}H$ (kJ/mol)
F	$-328$
Cl	$-349$
Br	$-325$
I	$-295$

Case B — Alkaline earths: polite refusal

Beryllium ($[He]\,2s^{2}$) and magnesium ($[Ne]\,3s^{2}$) already have a full $s$ sub-shell. The incoming electron has no choice but to go into the higher-energy $p$ sub-shell. That's energetically uphill, so $\Delta_{\text{eg}}H$ is positive:

• Be: $+66$ kJ/mol (approximate)
• Mg: $+67$ kJ/mol

Case C — Noble gases: outright rejection

A noble gas has a full $ns^{2}\,np^{6}$ outer shell. An extra electron would have to start an entirely new shell — expensive:

• Ne: $+116$ kJ/mol
• Ar: $+96$ kJ/mol

These are positive values, which is why noble gases do not form stable uni-negative anions under normal conditions. This is a favourite JEE/NEET statement.

[NEET important] — Group order of magnitude: halogens most exothermic; oxygen family next; then carbon, boron families; then the "reluctant" groups (N, alkaline earths, noble gases).

Variation Across a Period

Going from left to right within a period, the nuclear charge $Z$ increases while the added electrons go into the same shell. This means:

• the atom becomes smaller,
• effective nuclear charge on the outer shell rises,
• the atom "holds" new electrons more effectively.

Net effect: $\Delta_{\text{eg}}H$ becomes more negative (more exothermic, more favourable) as we move rightward — except at the two obvious speed-bumps (alkaline earths and noble gases, as we saw above).

The plot below shows this pattern for Periods 2 and 3.

Two characteristic anomalies across a period

Just like ionization enthalpy has Be>B and N>O dips, electron gain enthalpy has its own characteristic bumps. Watch for these:

1. N is LESS negative (actually positive) compared to C and O

Nitrogen has $[He]\,2s^{2}\,2p^{3}$ — a half-filled $p$ sub-shell, which is an extra-stable configuration. Forcing a 4th $p$ electron in means pairing it up with an existing one, which creates repulsion. Nitrogen therefore has $\Delta_{\text{eg}}H \approx +34$ kJ/mol — it actually resists an extra electron.

2. Noble gas at the end is always positive

Neon, argon etc., as already noted — the extra electron has to start a new shell.

What to remember for exams

Across a period (left → right), $\Delta_{\text{eg}}H$ generally becomes more negative, but with interruptions at Group 2 (alkaline earths), Group 15 (half-filled $p^{3}$), and a sharp jump to positive at Group 18 (noble gases).

[JEE Main tip] — Problems often give values for four consecutive period-3 elements and ask you to identify them. Memorise the rough Period-3 trend: Na($-53$), Mg($+67$), Al($-43$), Si($-134$), P($-72$), S($-200$), Cl($-349$), Ar($+96$). The two "dips toward zero" are Mg and P.

Variation Down a Group — and the Famous Cl > F Anomaly

Intuitive expectation

Going down a group, the outermost shell moves farther from the nucleus. The incoming electron feels a weaker nuclear pull, so we'd expect $\Delta_{\text{eg}}H$ to become less negative down a group. That's the general rule, and for most groups it roughly holds.

The famous exception: Cl is MORE negative than F

For halogens, the measured values are:

Halogen	$\Delta_{\text{eg}}H$ (kJ/mol)
F	$-328$
Cl	$-349$ ← largest
Br	$-325$
I	$-295$

Fluorine, despite being the smallest halogen (and having the highest $Z_{\text{eff}}$), does not have the most negative $\Delta_{\text{eg}}H$. Chlorine does. Why?

The explanation

Fluorine's outermost shell is the tiny $2p$ shell, which is unusually compact. Five electrons are already crammed into a very small volume. When the sixth electron tries to squeeze in, it experiences intense electron-electron repulsion from the existing electrons, and this repulsion partially cancels the nuclear attraction.

Chlorine's $3p$ shell is larger. The existing five $3p$ electrons are more spread out. The incoming sixth electron finds more room, so the repulsion is much smaller. Result: Cl releases more energy than F does.

The same pattern repeats

This "small-size repulsion" effect is not unique to fluorine. The same pattern shows up in other second-period elements:

• S has a more negative $\Delta_{\text{eg}}H$ than O for the same reason: oxygen's compact $2p$ shell causes stronger electron-electron repulsion for the incoming electron.
• Similarly, nitrogen is less favourable than phosphorus because the second-period shell is unusually compact and nitrogen also has the extra stability of a half-filled $2p^{3}$ configuration.

[JEE high-yield] — Rule of thumb: in the second period, an extra electron entering a small, densely-populated sub-shell faces extra repulsion. So the 3rd-row element of a p-block group often has a more negative $\Delta_{\text{eg}}H$ than the 2nd-row element.

Summary rule

Down a group, $\Delta_{\text{eg}}H$ generally becomes less negative — except that the very first member of a group (F, O, N in the second period) is often anomalous because its shell is abnormally compact.

Electron Affinity vs Electron Gain Enthalpy — Don't Confuse Them

This is a subtle but exam-favourite distinction.

Electron Affinity (EA)

Electron affinity is the energy released when a gaseous atom gains an electron. It is usually reported as a positive number for favourable cases.

Electron Gain Enthalpy ($\Delta_{\text{eg}}H$)

Electron gain enthalpy is the enthalpy change for the process:

$$\text{X}(g) + e^{-} \rightarrow \text{X}^{-}(g)$$

So when the process is favourable, $\Delta_{\text{eg}}H$ is negative.

Key NCERT-level relation

At the Class 11 / NCERT level, the two are treated as equal in magnitude and opposite in sign:

$$\Delta_{\text{eg}}H \approx -\text{EA}$$

So when we say "chlorine has an electron affinity of $+349$ kJ/mol" or "the electron gain enthalpy of chlorine is $-349$ kJ/mol," we are describing the same physical tendency.

Five factors that govern the magnitude of $\Delta_{\text{eg}}H$

1. Atomic size — smaller atom pulls the extra electron closer, usually giving a more negative value.
2. Nuclear charge — higher $Z_{\text{eff}}$ means stronger attraction, so usually more negative.
3. Electronic configuration — incomplete sub-shells (e.g. $p^{5}$) eagerly accept; full or half-filled sub-shells resist.
4. Shell compactness — very small shells (like F's $2p$) suffer extra electron-electron repulsion, pushing $\Delta_{\text{eg}}H$ toward less negative values.
5. Shielding by inner electrons — more shielding weakens the pull on the incoming electron.

Successive electron gain enthalpies

Adding a second electron to an already-negative anion is ALWAYS endothermic:

$$\text{O}^{-}(g) + e^{-} \;\longrightarrow\; \text{O}^{2-}(g) \qquad \Delta_{\text{eg}}H_{2} = +744\ \text{kJ/mol}$$

You're now forcing a negatively charged electron against an already-negative ion — huge repulsion. Yet oxides ($O^{2-}$) exist in solids because the lattice energy released upon crystal formation compensates. Energetics must always be evaluated across the complete process, not just the electron-addition step.

[NEET important] — A statement like "the second electron gain enthalpy of oxygen is negative because $O^{2-}$ is more stable than $O^{-}$" is FALSE as an atomic property. $O^{2-}$ stability comes from lattice energy in ionic solids, not from an intrinsic negative $\Delta_{\text{eg}}H_{2}$.

Exam Trap Check-list & Cheat Sheet

Six traps that catch students every year

1. Sign confusion — "Energy released = negative $\Delta_{\text{eg}}H$." The atom with the MOST negative value is the one that gains an electron MOST easily. More negative = more favourable.

2. Cl, not F, is the champion. The most negative electron gain enthalpy in the periodic table belongs to chlorine ($-349$ kJ/mol), not fluorine. If a question says "F has the most negative $\Delta_{\text{eg}}H$ among halogens" — it's wrong.

3. Noble gases do NOT accept electrons readily. Any option claiming a negative $\Delta_{\text{eg}}H$ for He, Ne, Ar, Kr, Xe, or Rn is a distractor.

4. EA and $\Delta_{\text{eg}}H$ have opposite signs. A high electron affinity means a very negative electron gain enthalpy.

5. O$^{2-}$ is not formed by a single-step exothermic gain. The second electron gain is endothermic; oxide ion stability comes from lattice energy in solids.

6. Comparison problems: when ranking $\Delta_{\text{eg}}H$ values, remember that "largest" in magnitude is not the same as "most negative". Always check sign first.

One-line quick rules

• Halogens: most negative (favourable). Order: Cl > F > Br > I (most negative to least).
• Alkaline earths: positive ($s^{2}$ fully filled).
• Noble gases: positive (full shell).
• Group 15 (N, P, As…): much smaller magnitude, and nitrogen is positive because of half-filled stability.
• Down a group: generally less negative, except anomalous second-row members (F, O, N) due to compact-shell repulsion.

Compact periodic-table heuristic

Region of PT	Typical $\Delta_{\text{eg}}H$
Group 17 (halogens)	Most negative, about $-295$ to $-349$ kJ/mol
Group 16 (chalcogens)	Negative
Group 14	Moderately negative
Group 15	Small negative or slightly positive
Group 13	Small negative
Group 2 (Be, Mg)	Positive ($s^{2}$ full)
Group 1 (Li, Na)	Small negative
Group 18 (noble gases)	Positive

Five-second reasoning template

When asked to compare $\Delta_{\text{eg}}H$ between two elements, run through this checklist:

1. Are either of them noble gases? → that one is positive.
2. Are either of them alkaline earths or Group 15? → those resist relative to neighbours.
3. Is one a halogen? → that one is usually the strongest candidate for the most negative value.
4. If same group, is one from the 2nd period (F, O, N)? → compact-shell repulsion can reverse the expected trend.
5. Otherwise, apply the general "rightward means more negative" rule across a period.

Armed with these five tests, you can handle most $\Delta_{\text{eg}}H$ comparison problems quickly.

Solved Examples

Example 1: Identify the most negative electron gain enthalpy

Among H, F, Cl, Br, which element has the most negative electron gain enthalpy?

Solution:

The electron gain enthalpies (kJ/mol) are:

• H: $-73$
• F: $-328$
• Cl: $-349$
• Br: $-325$

The most negative belongs to Cl ($-349$ kJ/mol).

Takeaway: Cl, not F, is the champion. F is anomalous because its 2p shell is very compact, causing extra electron-electron repulsion for the incoming electron.

Example 2: Sign of $\Delta_{\text{eg}}H$ for Mg

Is $\Delta_{\text{eg}}H$ of magnesium positive or negative? Explain.

Solution:

Electronic configuration of Mg is $[Ne]\,3s^{2}$, a fully-filled $s$ sub-shell.

Adding one more electron would force it into the higher-energy $3p$ sub-shell:

$$\text{Mg}(g) + e^{-} \longrightarrow \text{Mg}^{-}(g) \quad \Delta_{\text{eg}}H = +67\ \text{kJ/mol}$$

So $\Delta_{\text{eg}}H$ is positive — energy must be supplied. Mg resists gaining an electron.

Takeaway: atoms with a fully-filled $s$ sub-shell often have positive electron gain enthalpy.

Example 3: Unit conversion

The electron affinity of chlorine is 3.62 eV/atom. Convert this to kJ/mol and state the corresponding $\Delta_{\text{eg}}H$.

Solution:

Use 1 eV/atom = 96.485 kJ/mol.

$$\text{EA}(\text{Cl}) = 3.62 \times 96.485 = 349.3\ \text{kJ/mol}$$

Since $\Delta_{\text{eg}}H \approx -\text{EA}$:

$$\Delta_{\text{eg}}H(\text{Cl}) = -349\ \text{kJ/mol}$$

Takeaway: EA is reported as a positive value for energy released, but electron gain enthalpy carries the opposite sign for the same favourable process.

Example 4: Why is $\Delta_{\text{eg}}H$ of N positive?

Explain why nitrogen has a positive electron gain enthalpy whereas oxygen has a negative one, even though O is to the right of N.

Solution:

Configuration of N: $[He]\,2s^{2}\,2p^{3}$ — exactly half-filled $2p$ (one electron in each p orbital). Half-filled sub-shells are unusually stable due to maximum exchange energy.

Adding a 4th p electron means pairing it with an existing one, creating repulsion and destroying the half-filled stability:

$$\text{N}(g) + e^{-} \longrightarrow \text{N}^{-}(g) \quad \Delta_{\text{eg}}H \approx +34\ \text{kJ/mol}$$

Oxygen ($[He]\,2s^{2}\,2p^{4}$) already has one paired p electron; gaining one more moves it toward the stable p$^{5}$ arrangement, so the first electron gain enthalpy is negative.

Takeaway: half-filled and fully-filled sub-shells resist additional electrons.

Example 5: Successive electron gain — the O$^{2-}$ puzzle

If $\Delta_{\text{eg}}H_{1}(\text{O}) = -141$ kJ/mol and $\Delta_{\text{eg}}H_{2}(\text{O}) = +744$ kJ/mol, find the total enthalpy for forming $O^{2-}(g)$ from $O(g)$.

Solution:

The two-step process is:

$$\text{O}(g) + e^{-} \longrightarrow \text{O}^{-}(g) \quad \Delta_{\text{eg}}H_{1} = -141\ \text{kJ/mol}$$ $$\text{O}^{-}(g) + e^{-} \longrightarrow \text{O}^{2-}(g) \quad \Delta_{\text{eg}}H_{2} = +744\ \text{kJ/mol}$$

Overall:

$$\Delta_{\text{eg}}H_{\text{total}} = -141 + 744 = \mathbf{+603\ \text{kJ/mol}}$$

So forming isolated gaseous $O^{2-}$ from $O$ is strongly endothermic.

Takeaway: $O^{2-}$ does not form spontaneously in the gas phase. Oxide ions exist in solids because the lattice energy released upon crystal formation more than compensates.

Example 6: Rank the elements by $\Delta_{\text{eg}}H$

Arrange F, Cl, Br, I in order of decreasing electron gain enthalpy (more negative to less negative).

Solution:

Values (kJ/mol):

• Cl: $-349$ (most negative)
• F: $-328$
• Br: $-325$
• I: $-295$ (least negative)

So the order is:

$$\boxed{\text{Cl} > \text{F} > \text{Br} > \text{I}}$$

(where "greater" means more negative).

Takeaway: F's anomalously small 2p shell produces extra repulsion for the incoming electron — that's why Cl beats F. After Cl, the normal down-a-group decrease takes over.

Example 7: Identifying an element from its $\Delta_{\text{eg}}H$

An element X in Period 3 has $\Delta_{\text{eg}}H = -200$ kJ/mol. Identify X.

Solution:

Period-3 $\Delta_{\text{eg}}H$ values (kJ/mol):

• Na: $-53$
• Mg: $+67$
• Al: $-43$
• Si: $-134$
• P: $-72$
• S: $-200$
• Cl: $-349$
• Ar: $+96$

The value $-200$ matches sulfur (S).

Takeaway: memorising the Period-3 row of approximate values lets you identify the element quickly.

Example 8: Comparing alkaline earths and alkali metals

Which has more negative $\Delta_{\text{eg}}H$: Na or Mg? Justify.

Solution:

Na: $[Ne]\,3s^{1}$ — adding one electron completes the $3s^{2}$ sub-shell. Modest exothermic release.

$$\Delta_{\text{eg}}H(\text{Na}) = -53\ \text{kJ/mol}$$

Mg: $[Ne]\,3s^{2}$ — fully filled $s$; new electron must enter the higher $3p$ sub-shell.

$$\Delta_{\text{eg}}H(\text{Mg}) = +67\ \text{kJ/mol}$$

So Na's $\Delta_{\text{eg}}H$ is more negative than Mg's.

Takeaway: filling up an already-existing sub-shell is favourable; entering a higher sub-shell is not.

Example 9: Why noble gases are "noble"

Explain, in terms of electron gain enthalpy, why the noble gases do not readily form uninegative anions like $Ar^{-}$ or $Ne^{-}$.

Solution:

Noble gases have fully-filled $ns^{2}\,np^{6}$ outer shells. Any extra electron must go into the next higher shell — a brand new level. This costs a large amount of energy:

$$\Delta_{\text{eg}}H(\text{Ar}) = +96\ \text{kJ/mol}$$ $$\Delta_{\text{eg}}H(\text{Ne}) = +116\ \text{kJ/mol}$$

Because these values are positive, the process is endothermic and anions like $Ar^{-}$ are not thermodynamically stable under normal conditions.

Takeaway: closed-shell noble-gas configurations are among the most stable arrangements in chemistry, which is why noble gases are chemically unreactive.

Example 10: Compact-shell repulsion — O vs S

Electron gain enthalpy values are O: $-141$ kJ/mol and S: $-200$ kJ/mol. Why is S's more negative, even though O is higher up the group and has a higher $Z_{\text{eff}}$?

Solution:

Expected trend down a group is usually toward less negative values, but O and S show an exception.

Reason: oxygen's $2p$ shell is unusually compact. The existing electrons are packed tightly, so when another electron is added, it experiences strong electron-electron repulsion, partly cancelling the nuclear attraction.

Sulfur's $3p$ shell is larger, so the incoming electron experiences less repulsion and the process is more exothermic.

Takeaway: small second-period shells can reverse the expected group trend.

Example 11: Percentage comparison

Calculate the percentage by which the electron gain enthalpy of Cl is more negative than that of F.

Solution:

Given:

• $\Delta_{\text{eg}}H(\text{F}) = -328$ kJ/mol
• $\Delta_{\text{eg}}H(\text{Cl}) = -349$ kJ/mol

Difference in magnitude = $349 - 328 = 21$ kJ/mol

Percentage (relative to F):

$$\text{\% more negative} = \frac{21}{328} \times 100 \;=\; 6.4\%$$

So Cl's electron gain enthalpy is about 6.4% more negative than F's.

Takeaway: although the effect is modest in percentage terms, it's conceptually very important — it is why Cl, not F, tops the list.

Example 12: Multi-step lattice energy reasoning

The formation of solid NaCl from $Na(g)$ and $Cl(g)$ in the gas phase involves electron transfer followed by lattice formation. Given $\Delta_{\text{i}}H_{1}(\text{Na}) = +496$ kJ/mol and $\Delta_{\text{eg}}H(\text{Cl}) = -349$ kJ/mol, how much net energy does the electron-transfer step alone cost or release?

Solution:

Electron-transfer step (ignoring lattice energy):

$$\text{Na}(g) \longrightarrow \text{Na}^{+}(g) + e^{-} \quad \Delta_{\text{i}}H = +496\ \text{kJ/mol}$$ $$\text{Cl}(g) + e^{-} \longrightarrow \text{Cl}^{-}(g) \quad \Delta_{\text{eg}}H = -349\ \text{kJ/mol}$$

Net:

$$\Delta H_{\text{e-transfer}} = +496 - 349 = \mathbf{+147\ \text{kJ/mol}}$$

So the electron transfer alone is endothermic by 147 kJ/mol.

Why then does NaCl form spontaneously? Because when $Na^{+}$ and $Cl^{-}$ come together to make the ionic lattice, the lattice energy released more than compensates.

Takeaway: ionic-bond formation cannot be understood from electron gain enthalpy alone — lattice energy is the crucial extra term.