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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 < Se²⁻ < Te²⁻ (b) Fe³⁺ < Co³⁺ < Fe²⁺ (d) Be²⁺ < Na⁺ < Ne (c) Ti⁴⁺ < Sc³⁺ < Ca

Step by step solution

01

(a) Arrange Se²⁻, Te²⁻, 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 Se²⁻ and Te²⁻ 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 < Se²⁻ < Te²⁻.
02

(b) Arrange Co³⁺, Fe²⁺, Fe³⁺ 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 Fe³⁺ has a higher positive charge than Fe²⁺, its size is smaller. Since Co and Fe have a relatively similar number of protons, their size difference is not significant. However, Co³⁺ will be slightly smaller than Fe²⁺. The order is Fe³⁺ < Co³⁺ < Fe²⁺.
03

(d) Arrange Be²⁺, Na⁺, Ne in increasing size

For this set, we have Be²⁺ (Beryllium ion), Na⁺ (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. Be²⁺ 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⁺ loses one electron, making it smaller than neutral Na but larger than Be²⁺. 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 Be²⁺ < Na⁺ < Ne.
04

(c) Arrange Ca, Ti⁴⁺, Sc³⁺ in increasing size

Lastly, we have Ca (Calcium), Ti⁴⁺ (Titanium ion), and Sc³⁺ (Scandium ion). Calcium has an atomic number of 20, Titanium has an atomic number of 22, and Scandium has an atomic number of 21. Ti⁴⁺ loses four electrons, resulting in a smaller size as it has a higher nuclear charge that pulls the remaining electrons closer. Sc³⁺ loses three electrons, making it smaller than Ca but larger than Ti⁴⁺. Calcium, as a neutral atom, is larger than both Ti⁴⁺ and Sc³⁺ due to more electron-electron repulsion and a lower nuclear charge. The order for this set is Ti⁴⁺ < Sc³⁺ < Ca.

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Most popular questions from this chapter

(a) What is the general relationship between the size of an atom and its first ionization energy? (b) Which element in the periodic table has the largest ionization energy? Which has the smallest?

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?

Zinc in its \(2+\) oxidation state is an essential metal ion for life. \(\mathrm{Zn}^{2+}\) is found bound to many proteins that are involved in biological processes, but unfortunately \(\mathrm{Zn}^{2+}\) is hard to detect by common chemical methods. Therefore, scientists who are interested in studying \(\mathrm{Zn}^{2+}\) -containing proteins will frequently substitute \(\mathrm{Cd}^{2+}\) for \(\mathrm{Zn}^{2+}\), since \(\mathrm{Cd}^{2+}\) is easier to detect. (a) On the basis of the properties of the elements and ions discussed in this chapter and their positions in the periodic table, describe the pros and cons of using \(\mathrm{Cd}^{2+}\) as a \(\mathrm{Zn}^{2+}\) substitute. (b) Proteins that speed up (catalyze) chemical reactions are called enzymes. Many enzymes are required for proper metabolic reactions in the body. One problem with using \(\mathrm{Cd}^{2+}\) to replace \(\mathrm{Zn}^{2+}\) in enzymes is that \(\mathrm{Cd}^{2+}\) substitution can decrease or even eliminate enzymatic activity. Can you suggest a different metal ion that might replace \(\mathrm{Zn}^{2+}\) in enzymes instead of \(\mathrm{Cd}^{2+} ?\) Justify your answer.

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