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Chapter 9: Problem 18

The following species are isoelectronic with the noble gas krypton. Arrange them in order of increasing radius and comment on the principles involved in doing so: \(\mathrm{Rb}^{+}, \mathrm{Y}^{3+}, \mathrm{Br}^{-}, \mathrm{Sr}^{2+}, \mathrm{Se}^{2-}.\)

Short Answer

Expert verified
The order of increasing radius of the given isoelectronic species is \( \mathrm{Y}^{3+}< \mathrm{Sr}^{2+}< \mathrm{Rb}^{+}< \mathrm{Br}^{-}< \mathrm{Se}^{2-}\).

Step by step solution

01

Identify Isoelectronic Species

List the isoelectronic species. They have the same electron configuration as Krypton (\(Kr\)), which is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\). The species are \(\mathrm{Rb}^{+}, \mathrm{Y}^{3+}, \mathrm{Br}^{-},\mathrm{Sr}^{2+}, \mathrm{Se}^{2-}\).
02

Analyze Nuclear Charge

Analyze the number of protons for each atom/ion. The number of protons in the nuclear core (atomic number) is greater for \(\mathrm{Rb}^{+}\) (37), \(\mathrm{Sr}^{2+}\) (38) and \(\mathrm{Y}^{3+}\) (39) than for \(\mathrm{Br}^{-}\) (35), and \(\mathrm{Se}^{2-}\) (34). Higher charge implies more attraction on the electron cloud.
03

Arrange based on Atomic Radii

The higher the nuclear charge, the more effective it is at holding the electrons closer to the nucleus, thereby reducing the radius of the atom. Therefore, among the isoelectrons, the one with the highest nuclear charge will be smallest and the one with the lowest nuclear charge will be biggest. This gives us the order as: \(\mathrm{Y}^{3+}< \mathrm{Sr}^{2+}< \mathrm{Rb}^{+}< \mathrm{Br}^{-}< \mathrm{Se}^{2-}\).
04

Reason the Order

The above order is deduced by examining the nuclear charge of each ion. Higher nuclear charge pulls electrons closer to the nucleus, hence the ion size decreases. On the other hand, lower nuclear charge allows electrons to be further from the nucleus, hence the ion size increases.

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Key Concepts

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

Understanding Ionic Radius
When comparing ionic radius, it's important to look at the size of ions that have an equal number of electrons; such ions are called isoelectronic. The radius of an ion is a measure of the size of the ion, generally measured by the distance between the nucleus and the outermost electrons. In general, when an atom loses electrons and becomes a cation, its radius decreases. Conversely, when an atom gains electrons and becomes an anion, its radius increases.

Isoelectronic species, like extit{Rb} extsuperscript{+}, extit{Y} extsuperscript{3+}, extit{Br} extsuperscript{-}, extit{Sr} extsuperscript{2+}, and extit{Se} extsuperscript{2-}, all have the same electron configuration as krypton. Yet, their ionic radii differ. This difference arises from the varying nuclear charge of each species, with larger positive charges leading to smaller ionic radii. Thus, the ionic radius order accounts for the differences in their sizes:
  • Cations have smaller radii due to more protons pulling on the same number of electrons.
  • Anions have larger radii since fewer protons attract an increased electron cloud.


Understanding these principles will help predict the sizes of ions.
Influence of Nuclear Charge
Nuclear charge plays a key role in determining the size of ions within isoelectronic species. It refers to the total charge of the nucleus, which is determined by the number of protons present in the atom's nucleus. Normally, a higher nuclear charge implies a greater ability to attract and hold electrons, making them be pulled closer to the nucleus.

For isoelectronic species, the one with the highest nuclear charge will have the smallest radius because the electron cloud is more effectively pulled inward. Conversely, a lower nuclear charge means less pull on the electron cloud, allowing electrons to be farther from the nucleus, resulting in a larger radius. For example:
  • extit{Y} extsuperscript{3+} has the highest nuclear charge among the species in question, making its ionic radius the smallest.
  • extit{Se} extsuperscript{2-} has the lowest nuclear charge, thus the largest ionic radius due to weaker attraction of the electrons.


Recognizing how nuclear charge influences the attraction between the nucleus and the electron cloud is vital for determining ionic size.
Role of Electron Configuration
Electron configuration describes the arrangement of electrons in an atom or ion, indicating where the electrons reside around the nucleus within the shells and subshells. For these isoelectronic species, they all share an electron configuration identical to that of krypton. This is ( 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4p鈦� ).

Having the same electron configuration means these species have the same number of electrons distributed in identical subshells. However, differences arise from the nuclear charge.

Key points include:
  • Same electron configuration indicates the same number of energy levels and subshells filled.
  • The nuclear charge changes the arrangement by affecting how tightly the electrons are held.
This is why, despite having the same electron configuration, extit{Rb} extsuperscript{+}, extit{Y} extsuperscript{3+}, extit{Br} extsuperscript{-}, extit{Sr} extsuperscript{2+}, and extit{Se} extsuperscript{2-} differ in size. Understanding electron configuration helps us see how it provides a baseline for electron arrangement, while nuclear charge and resulting interactions create variation in ionic sizes.

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Most popular questions from this chapter

Refer only to the periodic table on the inside front cover, and arrange the following ionization energies in order of increasing value: \(I_{1}\) for \(\mathrm{F} ; I_{2}\) for \(\mathrm{Ba}\) \(I_{3}\) for \(\mathrm{Sc} ; I_{2}\) for \(\mathrm{Na} ; I_{3}\) for \(\mathrm{Mg}\). Explain the basis of any uncertainties.

Refer only to the periodic table on the inside front cover and indicate which of the atoms, \(\mathrm{Bi}, \mathrm{S}, \mathrm{Ba}, \mathrm{As}\) and \(\mathrm{Ca},\) (a) is most metallic; (b) is most nonmetallic; (c) has the intermediate value when the five are arranged in order of increasing first ionization energy.

Assuming that the seventh period is 32 members long, what should be the atomic number of the noble gas following radon (Rn)? Of the alkali metal following francium (Fr)? What would you expect their approximate atomic masses to be?

Indicate the smallest and the largest species (atom or ion) in the following group: Al atom, F atom, As atom, \(\mathrm{Cs}^{+}\) ion, \(\mathrm{I}^{-}\) ion, \(\mathrm{N}\) atom.

Answer each of the following questions: (a) Which of the elements \(P, A s,\) and \(S\) has the largest atomic radius? (b) Which of the following has the smallest radius: \(\mathrm{Xe}, \mathrm{O}^{2-}, \mathrm{N}^{3-},\) or \(\mathrm{F}^{-} ?\) (c) Which should have the largest difference between the first and second ionization energy: \(\mathrm{Al}, \mathrm{Si}, \mathrm{P},\) or \(\mathrm{Cl} ?\) (d) Which has the largest ionization energy: \(\mathrm{C}, \mathrm{Si}\), or \(\mathrm{Sn}\) ? (e) Which has the largest electron affinity: \(\mathrm{Na}, \mathrm{B}\) \(\mathrm{Al},\) or \(\mathrm{C} ?\)
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