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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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Most popular questions from this chapter

Give the symbol of the element (a) in group 14 that has the smallest atoms; (b) in period 5 that has the largest atoms; (c) in group 17 that has the lowest first ionization energy.

Which of the following ions are likely to be found in chemical compounds: \(\mathrm{Na}^{2+}, \mathrm{Li}^{+}, \mathrm{Al}^{4+}, \mathrm{F}^{2-},\) or \(\mathrm{Te}^{2-} ?\) Explain briefly.

Two elements, \(A\) and \(B\), have the electron configurations shown. $$ \mathrm{A}=[\mathrm{Kr}] 4 s^{2} \quad \mathrm{B}=[\mathrm{Ar}] 3 d^{10} 4 s^{2} 4 p^{5} $$ (a) Which element is a metal? (b) Which element has the greater ionization energy? (c) Which element has the larger atomic radius? (d) Which element has the greater electron affinity?

Use ionization energies and electron affinities listed in the text to determine whether the following reaction is endothermic or exothermic. $$ \mathrm{Mg}(\mathrm{g})+2 \mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{g})+2 \mathrm{F}^{-}(\mathrm{g}) $$

Neither \(\mathrm{Co}^{2+}\) nor \(\mathrm{Co}^{3+}\) has \(4 \mathrm{s}\) electrons in its electron configuration. How many unpaired electrons would you expect to find in each of these ions? Explain.
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