The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are

Q: Why is the electron gain enthalpy of halogens negative?

Because halogens are just one electron short of a stable noble gas configuration. When they gain an electron, they achieve stability and release a large amount of energy, which is represented by a negative sign.

Q: Shouldn't Fluorine release the most energy since it's the most electronegative?

Electronegativity is the tendency to attract shared electrons in a bond, while electron gain enthalpy is about adding an electron to an isolated atom. Fluorine's extremely small size causes high inter-electronic repulsion, making it harder to add a new electron compared to Chlorine.

Q: Why does the value become less negative from Chlorine to Iodine?

As you move down the group, the atomic size increases. The incoming electron is added to a shell that is further away from the nucleus, so the electrostatic attraction is weaker, and less energy is released.

Q: What does a 'less negative' value mean physically?

A less negative value simply means that less energy is released during the process. For example, $-325 \text{ kJ/mol}$ releases less energy than $-349 \text{ kJ/mol}$.

Q: Is this anomaly only present in halogens?

No! A similar anomaly exists in Group 16, where Oxygen has a less negative electron gain enthalpy than Sulfur for the exact same reason: small size and high inter-electronic repulsion.

Question

The electron gain enthalpy (in $	ext{kJ/mol}$) of fluorine, chlorine, bromine and iodine, respectively, are

Options:

(A) $-333, -325, -349$ and $-296$

(B) $-296, -325, -333$ and $-349$

(C) $-333, -349, -325$ and $-296$

(D) $-349, -333, -325$ and $-296$

Sigma Percentile · Accepted Answer

### The Concept of Electron Gain Enthalpy

To truly master the periodic table, we must understand the energetic transactions that occur when atoms interact with electrons. **Electron gain enthalpy** ($\Delta_{eg}H$) is defined as the enthalpy change that takes place when an isolated gaseous atom accepts an extra electron to form a negative ion.

For halogens, this process is highly favorable. They are just one electron short of achieving a stable, noble gas electron configuration. Because they 'want' this electron so badly, a large amount of energy is released when they finally get it. In thermodynamics, energy released is denoted by a negative sign, meaning the process is highly **exothermic**.

### The Expected Trend

If we look at the general trends of the periodic table, as we move down a group, the atomic size increases. The outermost shell gets further and further away from the positively charged nucleus.

Because the incoming electron is added to a shell that is farther from the nucleus, the electrostatic force of attraction it experiences is weaker. Consequently, less energy is released. Therefore, we expect the electron gain enthalpy to become **less negative** as we go from Fluorine down to Iodine.

### The Fluorine Anomaly: A Crowded Room

But nature loves exceptions, and Fluorine is a classic one! Fluorine is exceptionally small. Imagine a tiny room already packed with seven people (its valence electrons). When an eighth person (the incoming electron) tries to enter, the people inside push back.

This is exactly what happens at the atomic level. The incoming electron faces intense **inter-electronic repulsion** from the dense electron cloud in Fluorine's compact $2p$ subshell. Because of this repulsion, the net attraction is reduced, and the energy released is actually *less* than expected.

Chlorine, on the other hand, is larger. Its $3p$ subshell is more spacious, so the incoming electron doesn't face as much repulsion. Thus, Chlorine releases more energy than Fluorine when it gains an electron.

### The Final Verdict

Because of this anomaly, Chlorine has the most negative electron gain enthalpy in the entire periodic table. The correct order of energy released is:

$$	ext{Cl} > 	ext{F} > 	ext{Br} > 	ext{I}$$

When we look at the given values, we must find the sequence where the second value (Chlorine) is the most negative, followed by the first (Fluorine), then the third (Bromine), and finally the fourth (Iodine).

The values $-333 	ext{ kJ/mol}$ for F, $-349 	ext{ kJ/mol}$ for Cl, $-325 	ext{ kJ/mol}$ for Br, and $-296 	ext{ kJ/mol}$ for I perfectly match this physical reality.

Animated Solution for Chemistry - Periodicity in Properties: The electron gain enthalpy (in $kJ/mol$ ) of fluorine, chlorine, bromine and iodine, respectively, are

Select Answer:

Visualized Solution

$Δ_{e g} H$

$Trend: Cl \to I$

$Anomaly: F vs Cl$

$Order of ∣ Δ_{e g} H ∣$

$Final Values$

The Sigma Insight: Periodic Table and Periodic Properties

The Concept of Electron Gain Enthalpy

The Expected Trend

The Fluorine Anomaly

The Final Verdict

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Animated Solution for Chemistry - Periodicity in Properties: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are