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Chapter 12: Problem 150

Which of the following ion has the highest value of ionic radius? (a) \(\mathrm{F}^{-}\) (b) \(\mathrm{O}^{2-}\) (c) \(\mathrm{B}^{3+}\) (d) \(\mathrm{Li}^{+}\)

Short Answer

Expert verified
The ion 0^{2-}") has the highest ionic radius.

Step by step solution

01

Understanding Ionic Radius

Ionic radius refers to the size of an ion in a crystal lattice. Generally, more negative charge means a larger radius due to increased electron-to-electron repulsion.
02

Examining Anions and Cations

Anions (negatively charged ions) are usually larger than their parent atoms, while cations (positively charged ions) are usually smaller. Among the options, 0^{-2}") and 0^{-}") are anions, and 3^{+}") and 7^{+}") are cations.
03

Comparing Radii

Between the anions, 0^{-2}") has an additional electron compared to 0^{-}"). This makes 0^{-2}") larger due to higher electron repulsion. Meanwhile, 2^{+}") and 7^{+}") as cations, both have fewer electrons than their neutral atoms, making them smaller than their neutral counterparts. 2^{+}") and 0^{-2}") both involve fewer electrons locally pulling away from the nucleus.
04

Choosing the Largest Ion

Since 0^{-2}") possesses higher negative charge and thus greater repulsion among electrons than the singly charged anions and cations, 0^{-2}") tends to be the largest in terms of ionic radius.

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Key Concepts

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

Anions and Cations
Understanding anions and cations is crucial to grasping the concept of ionic radius. Anions are negatively charged ions that have gained extra electrons. When an atom gains electrons to become an anion, it generally increases in size. This is because the additional electrons cause an increase in electron-electron repulsion.
  • Anions are typically larger than their original atoms.
  • Greater electron repulsion pushes the electrons further apart, leading to a larger radius.
In contrast, cations are positively charged ions that have lost electrons. The loss of electrons results in less electron repulsion and a greater effective nuclear charge on the remaining electrons. This pulls them closer to the nucleus, reducing the size of the ion.
  • Cations are usually smaller than their parent atoms.
  • There is less shielding and repulsion when electrons are removed.
Electron Repulsion
Electron repulsion is a key factor in determining the size of ions. It occurs because electrons, having the same negative charge, repel each other. This repulsion plays a pivotal role in expanding or constricting the space the electrons occupy around the nucleus.
For anions, which carry a more negative charge, more electrons contribute to increased repulsion. This pushes the electrons further apart, resulting in an increase in ionic radius. It's like having more people in a room where everyone wants their own space鈥攕pace becomes larger.
  • The more electrons in an anion, the more significant the repulsion.
  • This is why O2- can have a larger radius compared to F-.
On the flip side, for cations, electron loss means decreased repulsion and a stronger attractive force from the nucleus on the remaining electrons. This reduces the ionic radius.
  • Less electron repulsion results in a more compact ionic structure.
  • Higher positive charge increases nuclear pull.
Crystal Lattice
The arrangement of ions in a crystal lattice influences the ionic radius. A crystal lattice is the ordered structure that ions form in solid-state materials. It can impact how closely ions pack together and therefore their effective size.
In a crystal lattice, ions with larger radii, like those of anions, dictate spacing because they occupy more space. Hence, ions with larger radii fit less efficiently into the lattice. They can create more "breathing room" within the lattice as compared to compact cations.
  • Anions take up more space, affecting lattice structure and spacing.
  • Cations fit efficiently due to their smaller size, resulting in tighter packing.
The interactions between ions in a lattice also affect stability and ionic mobility. By understanding the size and interactions of ions like the O2-, we can predict their behavior in a crystal structure effectively. This knowledge allows us to predict properties like conductivity and durability of ionic compounds.

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Most popular questions from this chapter

In the following questions two statements (Assertion) (A) and Reason (R) are given. Mark (a) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct but \(\mathrm{R}\) is not the correct expalnation of \(\mathrm{A}\). (c) A is true but \(\mathrm{R}\) is false. (d) A is false but \(R\) is true. (e) \(\mathrm{A}\) and \(\mathrm{R}\) both are false. Assertion: The electron gain enthalpies have large negative values toward the upper right of the periodic table preceding the noble gases. Reason: The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the positively charged nucleus.

In the following questions two statements (Assertion) (A) and Reason (R) are given. Mark (a) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct but \(\mathrm{R}\) is not the correct expalnation of \(\mathrm{A}\). (c) A is true but \(\mathrm{R}\) is false. (d) A is false but \(R\) is true. (e) \(\mathrm{A}\) and \(\mathrm{R}\) both are false. Assertion: The 5 th period of periodic table contains 18 elements not 32 . Reason: \(\mathrm{n}=5,1=0,1,2,3\). The order in which the energy of available orbitals \(4 \mathrm{~d}, 5 \mathrm{~s}\) and \(5 \mathrm{p}\)

In the following questions two statements (Assertion) (A) and Reason (R) are given. Mark (a) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct but \(\mathrm{R}\) is not the correct expalnation of \(\mathrm{A}\). (c) A is true but \(\mathrm{R}\) is false. (d) A is false but \(R\) is true. (e) \(\mathrm{A}\) and \(\mathrm{R}\) both are false. Assertion: The first ionization energy of Be is greater than that of \(\mathrm{B}\). Reason: \(2 \mathrm{p}\) orbital is lower in energy than \(2 \mathrm{~s}\). 131 .

Arrange \(\mathrm{S}, \mathrm{O}\) and Se in the ascending order of electron affinity. (a) \(\mathrm{S}<\mathrm{Se}<\mathrm{O}\) (b) \(\mathrm{Se}<\mathrm{O}<\mathrm{S}\) (c) \(\mathrm{Se}<\mathrm{S}<\mathrm{O}\) (d) \(\mathrm{S}<\mathrm{O}<\mathrm{Se}\)

Which ionization potential in the following equations involves the greatest amount of energy? (a) \(\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+\mathrm{e}\) (b) \(\mathrm{K} \longrightarrow \mathrm{K}^{+} \mathrm{e}^{-}\) (c) \(\mathrm{Cs}^{2+} \longrightarrow \mathrm{Cs}^{3+}+\mathrm{e}^{-}\) (d) \(\mathrm{Ca}^{+} \longrightarrow \mathrm{Ca}^{2+}+\mathrm{e}^{-}\)
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