Why is the first electron gain enthalpy of Oxygen negative, but its second electron gain enthalpy strongly positive?

When a neutral Oxygen atom gains an electron to form $O^-_{(g)}$, energy is released ($\Delta_{eg} H_1 = -141 \text{ kJ mol}^{-1}$) because of nuclear attraction. However, adding a second electron to the already negatively charged $O^-_{(g)}$ ion experiences strong electrostatic repulsion. Overcoming this repulsion requires significant energy input, making $\Delta_{eg} H_2 = +780 \text{ kJ mol}^{-1}$ strongly endothermic.

How do I quickly determine if an oxide is acidic, basic, amphoteric, or neutral in NEET problems?

Remember these heuristics: Metallic oxides are generally basic (e.g., $Na_2 O, CaO$), non-metallic oxides are acidic (e.g., $SO_3, N_2 O_5$), and neutral oxides include $CO, NO, N_2 O, H_2 O$. Amphoteric oxides react with both acids and bases—memorize the key amphoteric metals using the Kota mnemonic: "Zn, Al, Sn, Pb, Be, Ga" (e.g., $ZnO, Al_2 O_3, BeO, SnO_2, PbO_2$).

Why does Ionization Energy decrease down a group despite an increase in nuclear charge $Z$?

Down a group, two opposing factors occur: nuclear charge $Z$ increases (tending to increase attraction), but the principal quantum number $n$ increases, adding new electron shells. The increased distance of valence electrons from the nucleus, combined with effective shielding ($\sigma$) by inner shells, outweighs the increase in $Z$. Consequently, $Z_{\text{eff}}$ felt by outer electrons remains nearly constant while atomic size increases, causing $\Delta_i H$ to decrease.

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Solved High-Yield Sample Questions

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Question #1NEET 2024

Which of the following orders is correct for the first ionization enthalpy ( $\Delta_i H_1$ ) of the second-period elements $Be, B, C, N, O$ ?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write electronic configurations for each element: $Be: 1s^2 2s^2$ , $B: 1s^2 2s^2 2p^1$ , $C: 1s^2 2s^2 2p^2$ , $N: 1s^2 2s^2 2p^3$ , $O: 1s^2 2s^2 2p^4$ .

Step 2

Identify structural stability anomalies: $Be$ has a stable fully-filled $2s^2$ subshell, making removing an electron harder than from $B$ ( $2p^1$ ). Thus, $\Delta_i H_1(B) < \Delta_i H_1(Be)$ . Nitrogen has a stable half-filled $2p^3$ configuration, making its $\Delta_i H_1$ higher than Oxygen ( $2p^4$ ). Thus, $\Delta_i H_1(O) < \Delta_i H_1(N)$ .

Step 3

Arrange the overall trend across Period 2: general increase with anomalies yields $B < Be < C < O < N$ . Thus, Option A is the correct answer.

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Question #2JEE Main 2023

Consider the isoelectronic species $O^{2-}$ , $F^-$ , $Na^+$ , and $Mg^{2+}$ . The correct order of their ionic radii is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Identify that all given species ( $O^{2-}, F^-, Na^+, Mg^{2+}$ ) have $10$ electrons each, making them an isoelectronic series.

Step 2

Apply the rule for isoelectronic species: ionic radius decreases as nuclear charge $Z$ increases because a higher positive charge pulls the electron cloud closer. The nuclear charges are $Z(Mg^{2+}) = 12$ , $Z(Na^+) = 11$ , $Z(F^-) = 9$ , $Z(O^{2-}) = 8$ .

Step 3

Arrange in order of increasing radius: as $Z$ decreases, radius increases. Hence, $r(Mg^{2+}) < r(Na^+) < r(F^-) < r(O^{2-})$ . Therefore, Option A is correct.

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Mastering Classification of Elements and Periodicity in Properties: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Effective Nuclear Charge ($Z_{\text{eff}}$)

Pauling Scale of Electronegativity

Mulliken Electronegativity Scale

Ionization Enthalpy Variation

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

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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