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Arrange each of the following sets of atoms and ions, in order of increasing size: (a) \(\mathrm{Pb}, \mathrm{Pb}^{2+}, \mathrm{Pb}^{4+}\) (b) \(\mathrm{V}^{3+}, \mathrm{Co}^{2+}, \mathrm{Co}^{3+}\) (c) \(\mathrm{Se}^{2-}, \mathrm{S}^{2-}, \mathrm{Sn}^{2+}\) (d) \(\mathrm{K}^{+}, \mathrm{Rb}^{+}, \mathrm{Br}^{-}\)

Short Answer

Expert verified
(a) \(\mathrm{Pb}^{4+} < \mathrm{Pb}^{2+} < \mathrm{Pb}\) (b) \(\mathrm{Co}^{3+} < \mathrm{V}^{3+} < \mathrm{Co}^{2+}\) (c) \(\mathrm{Sn}^{2+} < \mathrm{Se}^{2-} < \mathrm{S}^{2-}\) (d) \(\mathrm{Rb}^{+} < \mathrm{K}^{+} < \mathrm{Br}^{-}\)

Step by step solution

01

Ionic Radius Comparison

Comparing the atomic and ionic radii of the same element, we simply need to consider the different number of electrons that an ion has. As the charge of an ion increases, there will be fewer electrons, and as a result, the electrons are pulled closer to the nucleus. This leads to a decrease in size. So, \(\mathrm{Pb}^{4+}\) is smaller than \(\mathrm{Pb}^{2+}\), which is smaller than \(\mathrm{Pb}\).
02

Arrange the Sizes

Based on the above analysis, we can arrange the atoms and ions in order of increasing size: \[\mathrm{Pb}^{4+} < \mathrm{Pb}^{2+} < \mathrm{Pb}\] (b) For \(\mathrm{V}^{3+}, \mathrm{Co}^{2+}, \mathrm{Co}^{3+}\):
03

Effect of Charge on Size

Considering the different charges, we can determine that \(\mathrm{Co}^{3+}\) will be smaller than \(\mathrm{Co}^{2+}\) because it has one less electron.
04

Effect of Atomic Number on Size

Given that \(\mathrm{V}^{3+}\) and \(\mathrm{Co}^{3+}\) have the same charge, we can compare their radii based on position in the periodic table. Vanadium (Z=23) has fewer protons than Cobalt (Z=27), therefore \(\mathrm{V}^{3+}\) will have a larger size compared to \(\mathrm{Co}^{3+}\) as the electrons are pulled less towards the nucleus.
05

Arrange the Sizes

Based on the above analysis, we can arrange the atoms and ions in order of increasing size: \[\mathrm{Co}^{3+} < \mathrm{V}^{3+} < \mathrm{Co}^{2+}\] (c) For \(\mathrm{Se}^{2-}, \mathrm{S}^{2-}, \mathrm{Sn}^{2+}\):
06

Effect of Charge on Size

Since the negative ions have equal charges, we must compare their sizes based on their positions in the periodic table.
07

Effect of Atomic Number on Size

Selenium (Z=34) has more protons than Sulfur (Z=16), so \(\mathrm{Se}^{2-}\) will be smaller than \(\mathrm{S}^{2-}\) due to the greater pull of its nucleus. Tin (Z=50) has more protons than Selenium, so \(\mathrm{Sn}^{2+}\) will be smaller than \(\mathrm{Se}^{2-}\).
08

Arrange the Sizes

Based on the above analysis, we can arrange the atoms and ions in order of increasing size: \[\mathrm{Sn}^{2+} < \mathrm{Se}^{2-} < \mathrm{S}^{2-}\] (d) For \(\mathrm{K}^{+}, \mathrm{Rb}^{+}, \mathrm{Br}^{-}\):
09

Effect of Charge on Size

Since we have both positive and negative ions, we should first consider the negative ion to be larger than the positive ions due to their extra electrons.
10

Effect of Atomic Number on Size

Potassium (Z=19) has fewer protons than Rubidium (Z=37), so \(\mathrm{K}^{+}\) will have a larger size compared to \(\mathrm{Rb}^{+}\).
11

Arrange the Sizes

Based on the above analysis, we can arrange the atoms and ions in order of increasing size: \[\mathrm{Rb}^{+} < \mathrm{K}^{+} < \mathrm{Br}^{-}\]

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Key Concepts

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

Atomic Number
The atomic number is a crucial concept that represents the number of protons in the nucleus of an atom. It determines the element's position in the periodic table and significantly influences its atomic structure and properties. Every element has a unique atomic number which distinguishes it from others. For instance, vanadium has an atomic number of 23, while cobalt's is 27. This means cobalt has more protons in its nucleus, contributing to stronger nuclear charges and influencing atomic and ionic sizes.

Understanding atomic numbers helps us predict the behavior of elements, such as their reactivity and size. Elements with higher atomic numbers have more protons, which increases the nuclear charge. This means electrons are pulled closer to the nucleus, resulting in a generally smaller atomic radius. As a result, when comparing ions such as \([V^{3+}](https://www.periodni.com/v.html)\) and \([Co^{3+}](https://www.periodni.com/co.html)\), the one originating from the element with the higher atomic number usually has a smaller radius. Hence, \(V^{3+}\) is larger than \(Co^{3+}\) due to a lower atomic number.
Charge Effect
The charge effect refers to the impact of an atom’s charge on its size. Ions form when atoms gain or lose electrons, leading to a positive or negative charge. Positive ions, or cations, are smaller than their parent atoms because losing electrons results in a greater positive charge pulling the remaining electrons closer. Conversely, negative ions, or anions, are larger due to added electrons causing increased electron-electron repulsion that pushes them apart.

For example, consider \(Pb, Pb^{2+},\mathrm{ and } Pb^{4+}\). Here, as lead loses electrons to form \(Pb^{2+}\) and \(Pb^{4+}\), the ionic radius decreases with increasing positive charge. With fewer electrons, they are held more tightly by the nucleus. Meanwhile, negative ions like \([Br^{-}](https://www.periodni.com/br.html)\) expand due to the added electron(s) despite having the same number of protons to pull the electrons in. This charge effect is pivotal in predicting and understanding ionic sizes when examining ions.
Periodic Table Trends
Periodic table trends are essential for understanding the behavior and properties of elements. Key trends include atomic and ionic sizes, electronegativity, and ionization energy. These trends arise from the arrangement of the periodic table and provide insights into the chemical properties of elements.

One significant trend is that atomic size increases moving down a group due to added electron shells; for instance, moving from potassium (\([K^{+}](https://www.periodni.com/k.html)\)) to rubidium (\([Rb^{+}](https://www.periodni.com/rb.html)\)). Conversely, atomic size decreases as you move across a period from left to right as the increased nuclear charge pulls electrons closer, causing smaller radii.

Such trends are helpful in predicting the sizes of ions as well since the major difference between neutral atoms and ions involves changes due to changes in the electron configuration. For \(Br^{-}\), being on the rightmost part of its period, means more profound trends like size expansion due to negative charge are evident.
Atomic and Ionic Size
Atomic size refers to the dimensions of an atom, often measured as the distance from the nucleus to the outer edge of the electron cloud. Ionic size, on the other hand, refers to the dimensions of an ion. Both are influenced by factors like the atomic number, electron configuration, and charge effect.

Understanding these sizes is vital when comparing ions like \([K^{+}](https://www.periodni.com/k.html)\)) and \(\mathrm{Br^{-}}\), or elements like \(\mathrm{Sn^{2+}}\), \(\mathrm{Se^{2-}}\), and \(\mathrm{S^{2-}}\). Generally, as atoms become ions, their sizes can either increase or decrease as influenced by their gain or loss of electrons. Positive ions result in a smaller ionic radius and negative electrons expand it. Therefore, \(\mathrm{Sn^{2+}}\) is considerably smaller compared to the ions of elements beneath it in the same group like \(\mathrm{S^{2-}}\).

When examining such ions' sizes, it is also essential to consider the balance between nuclear charge and electron-electron repulsion to understand their ultimate size.

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Most popular questions from this chapter

Would you expect zirconium(II) oxide, \(\mathrm{ZrO},\) to react more readily with \(\mathrm{HCl}(a q)\) or \(\mathrm{NaOH}(a q) ?\)

Mercury in the environment can exist in oxidation states \(0,\) \(+1,\) and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111 ), but instead of using ultraviolet light to eject valence electrons, X rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy \(\left(1 \mathrm{ev}=1.602 \times 10^{-19} \mathrm{~J}\right)\) The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the \(1 s\) orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for O. (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case?

We will see in Chapter 12 that semiconductors are materials that conduct electricity better than nonmetals but not as well as metals. The only two elements in the periodic table that are technologically useful semiconductors are silicon and germanium. Integrated circuits in computer chips today are based on silicon. Compound semiconductors are also used in the electronics industry. Examples are gallium arsenide, GaAs; gallium phosphide, GaP; cadmium sulfide, CdS; and cadmium selenide, CdSe. (a) What is the relationship between the compound semiconductors' compositions and the positions of their elements on the periodic table relative to \(\mathrm{Si}\) and Ge? \((\mathbf{b})\) Workers in the semiconductor industry refer to "II-VI" and "III-V" materials, using Roman numerals. Can you identify which compound semiconductors are II-VI and which are III-V? (c) Suggest other compositions of compound semiconductors based on the positions of their elements in the periodic table.

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (PES), a technique based on the photoelectric effect. exo (Section 6.2 ) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{nm} .\) (a) What is the energy of a photon of this light, in joules? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(1.72 \times 10^{-18} \mathrm{~J}\). What is the first ionization energy of \(\mathrm{Hg}\), in \(\mathrm{kJ} / \mathrm{mol} ?\) (d) Using Figure 7.10 , determine which of the halogen elements has a first ionization energy closest to that of mercury.

Using only the periodic table, arrange each set of atoms in order from largest to smallest: \((\mathbf{a}) \mathrm{Ar},\) As, \(\mathrm{Kr} ;\) (b) \(\mathrm{Cd}, \mathrm{Rb}, \mathrm{Te} ;(\mathbf{c})\) \(\mathrm{C}, \mathrm{Cl}, \mathrm{Cu}\).

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