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Chapter 7: Problem 37

For each of the following sets of atoms and ions, arrange the members in order of increasing size: (a) \(\mathrm{Se}^{2-}, \mathrm{Te}^{2-}, \mathrm{Se}\) (b) \(\mathrm{Co}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\) (d) \(\mathrm{Be}^{2+}, \mathrm{Na}^{+}, \mathrm{Ne}\). (c) \(\mathrm{Ca}, \mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}\)

Short Answer

Expert verified
The short version of the answer is: (a) Se < Se虏鈦� < Te虏鈦� (b) Fe鲁鈦� < Co鲁鈦� < Fe虏鈦� (d) Be虏鈦� < Na鈦� < Ne (c) Ti鈦粹伜 < Sc鲁鈦� < Ca

Step by step solution

01

(a) Arrange Se虏鈦�, Te虏鈦�, Se in increasing size

Firstly, we need to locate Se (Selenium) and Te (Tellurium) on the periodic table. Se has an atomic number of 34, while Te has an atomic number of 52. Both Se虏鈦� and Te虏鈦� have gained two extra electrons, increasing electron-electron repulsion, and thus these ions are larger than their neutral counterparts. Since Se and Te belong to the same group in the periodic table, they have the same number of valence electrons. However, as we move down the group, the atomic size increases due to the increase in the number of electron shells. In conclusion, the order is Se < Se虏鈦� < Te虏鈦�.
02

(b) Arrange Co鲁鈦�, Fe虏鈦�, Fe鲁鈦� in increasing size

Co (Cobalt) and Fe (Iron) are next to each other in the periodic table, with Fe being ahead of Co. Co has an atomic number of 27, and Fe has an atomic number of 26. Higher positive charges pull the electrons closer to the nucleus, resulting in a smaller size for the ion. As Fe鲁鈦� has a higher positive charge than Fe虏鈦�, its size is smaller. Since Co and Fe have a relatively similar number of protons, their size difference is not significant. However, Co鲁鈦� will be slightly smaller than Fe虏鈦�. The order is Fe鲁鈦� < Co鲁鈦� < Fe虏鈦�.
03

(d) Arrange Be虏鈦�, Na鈦�, Ne in increasing size

For this set, we have Be虏鈦� (Beryllium ion), Na鈦� (Sodium ion), and Ne (Neon atom). Beryllium has an atomic number of 4, Sodium has an atomic number of 11, and Neon has an atomic number of 10. Be虏鈦� loses two electrons, which results in a smaller size due to less electron-electron repulsion and a higher nuclear charge pulling the electrons closer. Na鈦� loses one electron, making it smaller than neutral Na but larger than Be虏鈦�. Meanwhile, Ne is a noble gas with a full electron shell, which results in a relatively large size compared to these ions. Thus, the order is Be虏鈦� < Na鈦� < Ne.
04

(c) Arrange Ca, Ti鈦粹伜, Sc鲁鈦� in increasing size

Lastly, we have Ca (Calcium), Ti鈦粹伜 (Titanium ion), and Sc鲁鈦� (Scandium ion). Calcium has an atomic number of 20, Titanium has an atomic number of 22, and Scandium has an atomic number of 21. Ti鈦粹伜 loses four electrons, resulting in a smaller size as it has a higher nuclear charge that pulls the remaining electrons closer. Sc鲁鈦� loses three electrons, making it smaller than Ca but larger than Ti鈦粹伜. Calcium, as a neutral atom, is larger than both Ti鈦粹伜 and Sc鲁鈦� due to more electron-electron repulsion and a lower nuclear charge. The order for this set is Ti鈦粹伜 < Sc鲁鈦� < Ca.

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Key Concepts

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

Periodic Table
The periodic table is a fundamental tool in chemistry, structuring elements based on atomic number and recurring chemical properties. Each element is placed in a specific position, which tells us a lot about its characteristics and behavior.
  • Groups: The columns in the periodic table are called groups. Elements in the same group have similar chemical and physical properties because they have the same number of valence electrons.

  • Periods: The rows are called periods. Moving from left to right in a period, the elements become less metallic. As you move down a group, elements have more electron shells, increasing atomic size.

When comparing atomic or ionic sizes, remember that size often increases as you move down a group because additional electron shells are added. Conversely, within the same period, atomic size generally decreases from left to right due to increasing nuclear charge, which pulls electrons closer to the nucleus.
Electron Configuration
Electron configuration is the arrangement of electrons in an atom or ion. It determines much about an element's chemical behavior. The electrons are distributed across various orbitals, and their arrangement follows the "aufbau principle," which fills orbitals in order of increasing energy.
  • Orbitals: Electrons are positioned in orbitals (s, p, d, f), which are defined by the principal quantum number (\(n\)) and angular momentum quantum number (\(l\)).

  • Valence Electrons: These are the outermost electrons in an atom and play a significant role in chemical reactions and bonding.

When atoms form ions, they either lose or gain electrons to achieve a stable electron configuration. Typically, this means obtaining a noble gas configuration, which is particularly stable due to the full outer electron shell. For example, when a neutral atom becomes a cation by losing electrons, its electron configuration changes, sometimes dramatically altering its chemical properties.
Ionization
Ionization refers to the process by which neutral atoms or molecules gain or lose electrons, forming charged particles called ions. This process is crucial in determining the chemical and physical properties of an element in its ionic form.
  • Cations and Anions: Cations are positively charged ions that result from an atom losing electrons, whereas anions are negatively charged ions formed by gaining electrons.

  • Size Changes: The size of an ion compared to its neutral atom can vary significantly. Cations tend to be smaller than their parent atoms because the loss of electrons results in less electron-electron repulsion and a stronger pull on the remaining electrons from the nucleus. Conversely, anions are larger due to added electron-electron repulsion.

Ionization energy is also a concept related to ionization; it's the energy required to remove an electron from an atom or ion. Elements with low ionization energy tend to form cations easily, while those with high ionization energy are more likely to form anions or remain neutral.

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Most popular questions from this chapter

Consider the \(\mathrm{A}_{2} \mathrm{X}_{4}\) molecule depicted here, where \(\mathrm{A}\) and \(\mathrm{X}\) are elements. The \(\mathrm{A}-\mathrm{A}\) bond length in this molecule is \(d_{1},\) and the four \(\mathrm{A}-\mathrm{X}\) bond lengths are each \(d_{2}\). (a) In terms of \(d_{1}\) and \(d_{2}\), how could you define the bonding atomic radii of atoms \(A\) and \(X ?(b)\) In terms of \(d_{1}\) and \(d_{2}\), what would you predict for the \(\mathrm{X}-\mathrm{X}\) bond length of an \(\mathrm{X}_{2}\) molecule? \([\) Section 7.3\(]\)

Find three examples of ions in the periodic table that have an electron configuration of \(n d^{8}(n=3,4,5 \ldots) .\)

Consider the isoelectronic ions \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, \(S,\) calculate \(Z_{\text {eff }}\) for the \(2 p\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S\). (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Write electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cr}^{3+}\), (b) \(\mathrm{N}^{3-},(\mathrm{c}) \mathrm{Sc}^{3+},(\mathrm{d}) \mathrm{Cu}^{2+}\) (e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{Au}^{+}\).

It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alternatively, one could measure how well electricity is conducted by each of the elements to determine how "metallic" the elements are. On the basis of conductivity, there is not much of a trend in the periodic table: Silver is the most conductive metal, and manganese the least. Look up the first ionization energies of silver and manganese; which of these two elements would you call more metallic based on the way we define it in this book?
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