Q.Which of the following pairs of elements would have a more negative electron gain enthalpy?

Concept and Intuition

Electron gain enthalpy ($\Delta_{eg}H$) measures the energy change when an atom gains an electron. It is usually negative (energy released, favourable) because the added electron experiences net attraction to the nucleus and the atom moves toward a more stable configuration. It becomes positive (energy must be supplied) in special cases:

• Adding a second electron to an already-negative ion (extra electron-electron repulsion), or
• Adding an electron to an atom that already has an extra-stable, fully-filled configuration (noble gases, or atoms with exactly half-filled/filled subshells to a lesser extent).

Step-by-Step Solution

1. Li(g) → Li⁻(g): Li has configuration $1s^2 2s^1$; adding an electron completes the 2s subshell — favourable, $\Delta_{eg}H$ is negative.
2. O(g) → O⁻(g): first electron gain by oxygen (moving toward a more stable configuration) is exothermic; $\Delta_{eg}H$ is negative.
3. Xe(g) → Xe⁻(g): Xenon already has a complete octet ($5s^25p^6$, or the outer shell fully filled); forcing an extra electron in means placing it in a higher-energy orbital of the next shell against a very poor net attraction — this costs energy, so $\Delta_{eg}H$ is positive. …

Concept & Intuition

Electron gain enthalpy is the energy change when an atom gains an electron. A more negative value means a stronger tendency to gain an electron. Key factors:

• Nuclear charge: Higher charge pulls in an extra electron more strongly.
• Atomic size: Smaller atoms have a stronger hold on an added electron.
• Stability of configuration: Half-filled or fully-filled subshells resist gaining an electron (less negative or even positive values). Here, we compare four elements based on their configurations to predict their relative electron-gaining tendencies.

Step-by-step reasoning

1. Identify the elements

   • A: $1s^2 2s^2 2p^4$ → Oxygen (O), group 16, period 2.
   • B: $1s^2 2s^2 2p^6 3s^1$ → Sodium (Na), group 1, period 3.
   • C: $1s^2 2s^2 2p^6$ → Neon (Ne), noble gas, period 2.
   • D: $1s^2 2s^2 2p^5$ → Fluorine (F), group 17, period 2.

2. Recall general trends

   • Electron gain enthalpy becomes more negative across a period (left to right) due to increasing nuclear charge and decreasing size.
   • Down a group, it becomes less negative (or more positive) because the added electron goes into a larger orbital, farther from the nucleus.
   • Noble gases have highly positive electron gain enthalpy (they strongly resist gaining an electron).

3. Analyze each element

   • C (Neon): Has a completely filled $2p^6$ subshell. Adding an electron would disrupt this stable octet, requiring energy input. So its electron gain enthalpy is positive (endothermic). Tendency to gain electron: lowest.
   • B (Sodium): Has a single $3s$ electron. Losing that electron is easy, but gaining an electron? Sodium has a large atomic radius (period 3) and low nuclear charge relative to its size. Its electron gain enthalpy is actually slightly negative (about -53 kJ/mol) because adding an electron to the $3s$ orbital is not very favorable. Tendency: low, but higher than neon.
   • A (Oxygen): Has $2p^4$ — two short of a full octet. It has a high nuclear charge (8) and small size. Its electron gain enthalpy is quite negative (about -141 kJ/mol). However, adding an electron to oxygen is less exothermic than to fluorine because oxygen’s smaller size leads to greater electron-electron repulsion in the $2p$ subshell. …

Concept and Intuition

Electron gain enthalpy generally becomes more negative down a group due to increasing atomic size reducing electron-electron repulsion in the smaller atom of the same family — this is why chlorine (not fluorine) has the most negative electron gain enthalpy among halogens, since fluorine's very small size causes significant repulsion among its compact electron cloud when adding another electron. The same pattern (S more negative than O) holds in the oxygen family for the same reason.

Step-by-Step Solution

1. Known standard values (kJ/mol): F = −328, Cl = −349, O = −141, S = −200. …

Concept and Intuition

Electron gain enthalpy (the energy change when an atom accepts an electron) generally becomes more negative (more exothermic/favourable) as you move across a period toward the halogens, because nuclear charge increases and atomic size decreases, pulling the incoming electron in more strongly. Metals like Al have very little affinity for an extra electron, so their electron gain enthalpy is only weakly negative.

Step-by-Step Solution

1. Al is a metal (group 13); metals have low tendency to accept an additional electron, so its electron gain enthalpy is the least negative of the four.
2. N (group 15) has a stable half-filled $2p^3$ configuration, which resists accepting another electron, keeping its enthalpy close to zero — but still slightly more negative than Al in this comparison.
3. O (group 16) is a strong electron acceptor (forms O$^{2-}$ in many compounds), giving a substantially more negative value. …

Concept and Intuition

Electron gain enthalpy generally becomes more negative across a period, but small second-period atoms like O and F are exceptions to the "more negative going up a group" pattern: their compact valence shells already hold significant electron density, so adding one more electron faces higher electron-electron repulsion than in the larger third-period elements (S, Cl) with more diffuse valence orbitals.

Step-by-Step Solution

1. List approximate values: O ≈ −141, F ≈ −328, S ≈ −200, Cl ≈ −349 kJ/mol.
2. Compare magnitudes: |O| is the smallest of the four. …

Concept and Intuition

Electron gain enthalpy becomes more negative going down a group in general, but fluorine is an exception: due to its very small atomic size, the added electron experiences unusually strong electron-electron repulsion in the compact 2p subshell, making its $\Delta_{eg}H$ less negative than chlorine's. So the order of increasingly negative electron gain enthalpy among the halogens is F > I > Br > Cl (i.e. Cl is most negative), with standard values approximately F = -328, Cl = -349, Br = -325, I = -295 kJ/mol.

Step-by-Step Solution

1. Assign values from memory: Cl = -349 kJ/mol, Br = -325 kJ/mol, F = -328 kJ/mol, I = -295 kJ/mol.
2. Match against List II labels: a = -295 (I), b = -328 (F), c = -349 (Cl), d = -325 (Br). …

Concept and Intuition

Electron gain enthalpy generally becomes more negative (more energy released) across a period and less negative down a group — but there are well-known anomalies: small, very electronegative atoms like O and F have compact valence shells causing extra electron-electron repulsion in the new electron's orbital, restraining their electron affinity somewhat below expectations. Also, N has a stable half-filled $2p^3$ configuration and is unusually reluctant to accept an extra electron (its electron gain enthalpy is close to zero or slightly positive).

Step-by-Step Solution

1. F, being small and highly electronegative, has the most negative electron gain enthalpy of these four ($\approx -328$ kJ/mol) — dominant term. …

Concept and Intuition

Electron gain enthalpy ($\Delta_{eg}H$) is the energy change when a gaseous atom accepts an electron. Atoms with an already extra-stable electronic configuration — a fully-filled subshell, a half-filled subshell, or a full noble-gas shell — resist accepting another electron, because doing so would force it into a higher-energy orbital and disturb that stability. $Be$ ($1s^2 2s^2$) and $Mg$ ($1s^2 2s^2 2p^6 3s^2$) both have completely filled outermost $s$-subshells, so — much like the noble gases — they buck the usual "more negative across a period" trend and instead show anomalously unfavourable, positive electron gain enthalpies.

Step-by-Step Solution

1. General periodic trend: electron gain enthalpy becomes progressively more negative across a period (electron addition becomes more favourable) due to rising effective nuclear charge and shrinking atomic size.
2. Known exceptions to this trend: Group 2 elements ($ns^2$, e.g. $Be$, $Mg$), Group 15 elements ($ns^2np^3$, half-filled $p$, e.g. $N$), and the noble gases ($ns^2np^6$) — all have extra-stable configurations that resist electron addition.
3. Standard reference values: $\Delta_{eg}H(Be) \approx +100\ kJ/mol$ and $\Delta_{eg}H(Mg) \approx +230\ kJ/mol$ — both clearly positive, meaning energy must be supplied to force an electron onto the atom, not a "small negative" value as the Assertion states. …

Concept and Intuition

Electron gain enthalpy magnitude (with negative sign, i.e., how exothermic electron addition is) depends on effective nuclear charge, atomic size, and electron-electron repulsion in the orbital receiving the extra electron. Nitrogen's extra stability from a half-filled $2p^3$ configuration makes adding an electron unfavourable. Within halogens, fluorine's very small size causes higher electron-electron repulsion than expected, making chlorine (larger, less repulsion) have the highest magnitude electron gain enthalpy of all elements.

Step-by-Step Solution

1. Nitrogen ($2p^3$, half-filled, extra stability) resists gaining an electron — near-zero/least favourable value.
2. Phosphorus ($3p^3$, also half-filled but larger atom, less repulsion than N) — mildly favourable, more negative than N. …

Concept and Intuition

Electron gain enthalpy generally becomes more negative (more exothermic) across a period and less negative down a group, but there are two classic anomalies tested here: (i) N (half-filled $2p^3$, extra stable, resists adding an electron — its $\Delta_{eg}H$ is near zero or slightly positive) and (ii) O vs S (O's very small atomic size causes strong inter-electron repulsion in its compact 2p subshell, making $O$'s electron gain enthalpy less negative than the larger S, even though O is more electronegative).

Step-by-Step Solution …