Question

The correct order of radii is (a) \(\mathrm{N}<\mathrm{Be}<\mathrm{B}\) (b) \(\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}\) (c) \(\mathrm{Na}<\mathrm{Li}<\mathrm{K}\) (d) \(\mathrm{Fe}^{+3}<\mathrm{Fe}^{+2}<\mathrm{Fe}^{+4}\)

Short answer

The correct order is (b) \\(F^{-} < O^{2-} < N^{3-}\\).

Step-by-step solution

Analyzing Atomic and Ionic Radii

To determine the correct order of radii, we need to consider atomic and ionic size trends. In general, within a period, atomic radii decrease from left to right, and within a group, atomic radii increase from top to bottom in the periodic table. For ions, more negative charge typically means larger size due to increased electron-electron repulsion and less positive charge on the nucleus to attract the electrons.

Breaking Down Options

Let's examine each option based on these trends: (a) For neutral atoms across a period (N, Be, B), radii generally decrease from left to right. Hence, Be > B > N is expected. (b) Among anions (F⁻, O²⁻, N³⁻), radii increase with increasing negative charge due to additional electron-electron repulsion. Thus, N³⁻ > O²⁻ > F⁻ is expected. (c) For neutral atoms down a group (Na, Li, K), radii increase as we move down from Li to K. So, Li < Na < K is expected. (d) Comparing cations of the same element (Fe³⁺, Fe²⁺, Fe⁴⁺), less positive charge typically leads to larger radii, thus Fe³⁺ < Fe²⁺ < Fe⁴⁺ generally does not hold unless predicting oxidation state differences beyond atomic effects.

Evaluating Correctness of Each Option

(a) is incorrect because Be should be larger than N and is incorrect in the context of a period. (b) correctly follows the trend as explained for anions: the more negatively charged, the larger the radius. (c) incorrectly identifies the order; radii should increase down a group, so K > Na > Li is expected instead. (d) incorrectly suggests that Fe³⁺ has the smallest radius; typically, the least positive state should have the largest radius, so Fe²⁺ > Fe³⁺ > Fe⁴⁺ reflects common oxidation state perspectives.

Conclusion

Among the options, (b) \(F^{-} < O^{2-} < N^{3-}\) is the correct order of radii as it aligns with the trend of increasing ionic size with increasing negative charge.

Key concepts

Periodic Table Trends

The periodic table is an essential tool in understanding the properties of elements, such as atomic and ionic radii.
One key trend is that atomic radii generally decrease from left to right across a period. This happens because the number of protons in the nucleus increases, enhancing the nucleus' ability to pull electrons closer.
Consequently, the atomic size reduces. However, moving down a group, the atomic radii increase as more electron shells are added. Each additional shell places the outermost electrons farther away from the nucleus, leading to a larger atomic size.
These trends help predict the behavior of elements and their ions in chemical reactions. By understanding these visual patterns, students can make educated guesses about the relative sizes of different elements and their ions, which are critical in solving exercises like the one presented.

Ionic Size

Ionic size differs from atomic size due to the loss or gain of electrons that form ions.
Cations, formed by losing electrons, are typically smaller than their parent atoms because the loss allows the remaining electrons to be pulled closer by the unchanged nuclear charge.
Anions, on the other hand, become larger as the addition of electrons increases electron-electron repulsion and expands the electron cloud.
Understanding ionic sizes is crucial, especially when dealing with isoelectronic ions, which have the same number of electrons but different nuclear charges. In these cases, the ion with the higher positive charge will generally be smaller due to a stronger attraction by the protons. For example, in the sequence of F⁻, O²⁻, N³⁻, each step adds more negative charge, which increases the radius of these anions.

Electron-Electron Repulsion

When considering ionic and atomic size, electron-electron repulsion plays a pivotal role. Electrons are negatively charged particles that repel each other.
As more electrons are added, usually in forming anions, this repulsion causes the electron cloud to expand even if the nuclear charge remains constant.
Moreover, when comparing ions such as F⁻, O²⁻, and N³⁻, the increase in number of electrons with each additional charge increases repulsion significantly, causing an increment in size.
This phenomenon explains why anions are generally larger than their neutral atoms and also clarifies the order of size among comparable anions.
Recognizing the effect of electron-electron repulsion helps in understanding not only the sizes of different ions but also provides insights into their chemical behaviors and reactions.

Oxidation State Effects

The oxidation state of an ion significantly affects its size.
Typically, the less positive the charge of the ion, the larger its radius. This is because there are fewer protons to pull the electron cloud closer.
For example, in cations of a single element, such as the Fe series (Fe²⁺, Fe³⁺, Fe⁴⁺), higher oxidation states correspond to fewer electrons and thus smaller sizes as the remaining electrons feel a stronger attraction to the nucleus.
In practical settings, understanding oxidation state effects can help predict the reactivity and stability of compounds.
A change in oxidation state affects not only the size but also properties like hardness, solubility, and color of compounds.
Thus, by grasping the concept of oxidation states, students can predict and rationalize not only the size but also the broader chemical and physical behaviors of ions.