Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which atom or ion in the following pairs would you expect to be larger? (a) S or \(\mathrm{S}^{2-}\) (b) \(\mathrm{Ca}\) or \(\mathrm{Ca}^{2+}\) (c) \(\mathrm{O}^{-}\) or \(\mathrm{O}^{2}\)

Short Answer

Expert verified
(a) $\mathrm{S}^{2-}$ is larger; (b) $\mathrm{Ca}$ is larger; (c) Assuming $\mathrm{O}^{2}$ is $\mathrm{O}^{2-}$, $\mathrm{O}^{2-}$ is larger.

Step by step solution

01

Understanding the Concept of Ion Size

Atoms gain or lose electrons to form ions. When an atom gains electrons to become an anion (like \(\mathrm{S}^{2-}\)), the additional electron(s) repel each other, often increasing the ion's size. Conversely, when an atom loses electrons to form a cation (like \(\mathrm{Ca}^{2+}\)), it releases electron shells or reduces electron-electron repulsion, often decreasing in size.
02

Analyzing Pair (a): $\mathrm{S}$ vs. $\mathrm{S}^{2-}$

Sulfur (S) gains two electrons to become \(\mathrm{S}^{2-}\). With additional electrons, the electron-electron repulsion increases, causing the electron cloud to expand. Therefore, the sulfide ion \(\mathrm{S}^{2-}\) is larger than the neutral sulfur atom (S).
03

Analyzing Pair (b): $\mathrm{Ca}$ vs. $\mathrm{Ca}^{2+}$

Calcium (Ca) loses two electrons to form \(\mathrm{Ca}^{2+}\). Losing electrons results in a decreased electron-electron repulsion and the remaining electrons are drawn closer to the nucleus. As a result, the calcium ion \(\mathrm{Ca}^{2+}\) is smaller than the neutral calcium atom (Ca).
04

Analyzing Pair (c): $\mathrm{O}^{-}$ vs. $\mathrm{O}^{2}$

The problem seems to contain an error as \(\mathrm{O}^{2}\) is not an ion but rather an atom or molecule. Assuming it meant \(\mathrm{O}^{2-}\), single electron gain in \(\mathrm{O}^{-}\) leads to increased size compared to oxygen by itself; however \(\mathrm{O}^{2-}\) gains second electrons, which increases size further due to more electron repulsion compared to \(\mathrm{O}^{-}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Cation
Cations are positively charged ions formed when an atom loses one or more electrons. The process of forming a cation is known as ionization. When an atom becomes a cation, it loses part of its electron cloud. Losing electrons diminishes the electron repulsion within the atom, allowing remaining electrons to be pulled closer to the nucleus. This contraction leads to cations being smaller in size compared to their parent atoms.
For example, consider calcium ( Ca ) and its cation, calcium ion ( Ca^{2+} ). Calcium loses two electrons to form Ca^{2+} . As electrons are lost, the electron cloud contracts, and the atom becomes smaller.
Anion
Anions are negatively charged ions formed when an atom gains one or more electrons. This electron gain increases electron-electron repulsion within the atom, causing the electron cloud to expand. As a result, anions are generally larger than their parent atoms since the additional electrons repel each other and occupy more space.
For instance, sulfur (S) becomes sulfide ion (S^{2-} ) by gaining two electrons. With more electrons, the repulsion among them increases, pushing the electron cloud to expand and making S^{2-} larger than the neutral sulfur atom.
Electron Repulsion
Electron repulsion plays a crucial role in determining the size of ions. Electrons, which are negatively charged, naturally repel each other. When additional electrons are added to an atom to form an anion, this repulsion force becomes stronger, which results in a larger electron cloud. This is why anions tend to be larger than the neutral atoms they originate from.
On the other hand, when electrons are removed to create a cation, the reduction in electron repulsion allows the remaining electrons to be drawn in closer to the nucleus, resulting in a smaller size than the original atom. Understanding electron repulsion helps explain the behavior and size change of both anions and cations.
Ion Formation
Ion formation is a fundamental chemical process that occurs when atoms gain or lose electrons to achieve a stable electronic configuration, often similar to that of noble gases. This process leads to the formation of either positively charged ions called cations or negatively charged ions termed anions.
  • Cations form by the loss of electrons, reducing the atom's electron count and usually resulting in a decrease in size.
  • Anions form by the gain of electrons, increasing the electron count and typically causing an increase in size due to heightened electron repulsion.
Understanding ion formation is essential for noticing the changes in physical properties like atomic size, which directly influences how atoms interact with each other in chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free