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The electron affinities of \(\mathrm{N}, \mathrm{O}, \mathrm{S}\) and \(\mathrm{Cl}\) are: (a) \(\mathrm{O} \approx \mathrm{Cl}<\mathrm{N} \approx \mathrm{S}\) (b) \(\mathrm{O}<\mathrm{S}<\mathrm{Cl}<\mathrm{N}\) (c) \(\mathrm{N}<\mathrm{O}<\mathrm{S}<\mathrm{Cl}\) (d) \(\mathrm{O}<\mathrm{N}<\mathrm{Cl}<\mathrm{S}\)

Short Answer

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(c) \(\mathrm{N}<\mathrm{O}<\mathrm{S}<\mathrm{Cl}\)

Step by step solution

01

Understanding Electron Affinity

Electron affinity is the energy change that occurs when an electron is added to a neutral atom in the gas phase to form a negative ion. Generally, as you move from left to right across a period, electron affinity tends to increase. However, there are exceptions among specific elements like the inert nature of Nitrogen due to its half-filled p orbital.
02

Reviewing Periodic Table Trends

Across the periodic table, the general trend is that electron affinity increases across a period and decreases down a group. Therefore, the elements in the same group should have a decreasing trend in electron affinity when moving down. Elements in the same period generally show increasing electron affinity moving from left to right, except for exceptions.
03

Analyzing Element Groups and Periods

Nitrogen and Oxygen are in the second period, whereas Sulfur and Chlorine are in the third period. Based on periodic trends, Oxygen should have less electron affinity compared to Chlorine. Nitrogen, with a half-filled stable p orbital, will have a lower affinity than Oxygen despite being to its left in the same period.
04

Arranging the Electron Affinity

Considering periodic trends and exceptions: Nitrogen less than Oxygen due to stability, Oxygen less than Sulfur because Sulfur is lower in the same group, and both are less than Chlorine as it is a highly electronegative atom ready to accept an electron. Thus, arrange them as: \( ext{N} < ext{O} < ext{S} < ext{Cl} \).
05

Matching With Options

Identify the option sequence matching the determined order of electron affinity: With the order \( ext{N} < ext{O} < ext{S} < ext{Cl} \), the correct choice in the options provided is option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Table Trends
Periodic table trends guide us in predicting how various properties of elements change as we move across periods and groups. Generally, in terms of electron affinity, these trends suggest some clear patterns. As you progress from left to right across a period, electron affinity typically increases. This is because elements are more eager to complete their outer electron shells by gaining electrons. However, there are exceptions to this trend, such as nitrogen, which due to its half-filled p orbital configuration, exhibits peculiar behavior. Down a group, electron affinity tends to decrease. Why? The added electron enters an orbital further from the nucleus, experiencing less attraction due to increased shielding effect. These periodic trends in electron affinity play an essential role in understanding the reactivity and stability of elements in different chemical reactions.
Group and Period Analysis
To understand electron affinity variations among elements like \(\mathrm{N}, \\mathrm{O}, \\mathrm{S}, \\mathrm{and} \\mathrm{Cl}\), it's crucial to examine their positions in the periodic table. \\mathrm{N} \\mathrm{and} \\mathrm{O} ature lies in the second period, where electron affinity increases as you move right. Interestingly, nitrogen has a stable half-filled p orbital, contributing to its relatively low electron affinity compared to oxygen, despite oxygen being to its right.In the third period lies \\mathrm{S} \and \\mathrm{Cl}. Within the same period, \\mathrm{Cl} \typically has a higher electron affinity than \\mathrm{S}, due to its higher electronegativity and desire to complete its valence shell. Comparing periods, \\mathrm{O} and \\mathrm{S}, share group traits, yet \\mathrm{S} \exerts slightly stronger electron affinity due to its lower position in the periodic table which incorporates additional electron shells.
Electron Addition Energy
Electron addition energy, colloquially known as electron affinity, represents the energy change when an electron is added to an atom in the gas phase. The numerical value of electron affinity helps to quantify this process. The general rule is that if adding an electron to an atom releases energy, it means that the atom has a high electron affinity. The ability of an atom to accept an electron isn't just about energy values; it's also influenced by its electronic configuration. Nitrogen, for example, resists gaining an additional electron due to its semi-stable electron configuration with three half-filled p orbitals. On the other hand, elements like chlorine, with high electron affinity, release more energy as they reach a more stable electronic arrangement by gaining an electron, filling their valence shell. Grasping electron addition energy concepts allows us to rationalize why some elements readily attract electrons, reflecting in their chemical behavior.

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Most popular questions from this chapter

The formation of \(\mathrm{O}^{2-}\) (g) starting from \(\mathrm{O}\) (g) is endothermic by \(603 \mathrm{~kJ} \mathrm{~mol}^{-1} .\) If electron affinity of \(\mathrm{O}(\mathrm{g})\) is \(-141 \mathrm{~kJ} \mathrm{~mol}^{-1}\), the second electron affinity of oxygen would be: (a) \(+744 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-744 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+462 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-462 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Which of the following statements are correct? (1) HF is a stronger acid than \(\mathrm{HCl}\). (2) Among halide ions, iodide is the most powerful reducing agent. (3) Fluorine is the only halogen that does not show a variable oxidation state. (4) HOCl is a stronger acid than HOBr. (a) 2 and 4 (b) 2 and 3 (c) 1,2 and 3 (d) 2,3 and 4

Which of the following orders are correct? (1) Thermal stability \(\mathrm{BeCO}_{3}<\mathrm{MgCO}_{3}<\mathrm{CaCO}_{3}<\mathrm{SrCO}_{3}<\mathrm{BaCO}_{3}\) (2) Basic nature \(\mathrm{LiOH}>\mathrm{NaOH}>\mathrm{KOH}>\mathrm{RbOH}>\mathrm{CsOH}\) (3) Solubility in water \(\mathrm{BeSO}_{4}<\mathrm{MgSO}_{4}<\mathrm{CaSO}_{4}<\mathrm{SrSO}_{4}<\mathrm{BaSO}_{4}\) (4) Melting point \(\mathrm{NaCl}>\mathrm{KCl}>\mathrm{RbCI}>\mathrm{CsCl}>\mathrm{LiCl}\) (a) (1), (4) (b) \((1),(2),(4)\) (c) \((2),(3)\) (d) (1), (2) (3), (4)

Which of the following is the atomic number of a metal? (a) 32 (b) 34 (c) 36 (d) 38

Which of the electronic configuration represents a noble gas? (a) \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}\) (b) \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{3}\) (c) \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{6}, 3 \mathrm{~s}^{2}, 3 \mathrm{p}^{1}\) (d) \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{4}\)

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