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Chapter 3: Problem 123

Arrange the following in order of increasing radius and increasing ionization energy. a. \(\mathrm{N}^{+}, \mathrm{N}, \mathrm{N}^{-}\) b. \(\mathrm{Se}, \mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\) c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)

Short Answer

Expert verified
a. Increasing radius: \(\mathrm{N}^{+} < \mathrm{N} < \mathrm{N}^{-}\) and increasing ionization energy: \(\mathrm{N}^{-} < \mathrm{N} < \mathrm{N}^{+}\) b. Increasing radius: \(\mathrm{Cl}^{+} < \mathrm{Cl} < \mathrm{Se} < \mathrm{Se}^{-}\) and increasing ionization energy: \(\mathrm{Se}^{-} < \mathrm{Se} < \mathrm{Cl} < \mathrm{Cl}^{+}\) c. Increasing radius: \(\mathrm{Sr}^{2+} < \mathrm{Rb}^{+} < \mathrm{Br}^{-}\) and increasing ionization energy: \(\mathrm{Br}^{-} < \mathrm{Rb}^{+} < \mathrm{Sr}^{2+}\)

Step by step solution

01

Comparative atomic radius

All of these species have the same number of electron shells because they are different forms of nitrogen. However, \(\mathrm{N}^{+}\) has one less electron than \(\mathrm{N}\), and \(\mathrm{N}^{-}\) has one more electron than \(\mathrm{N}\). Therefore, the effective nuclear charge is highest for \(\mathrm{N}^{+}\) and lowest for \(\mathrm{N}^{-}\), so the order of increasing radius is \(\mathrm{N}^{+} < \mathrm{N} < \mathrm{N}^{-}\).
02

Comparative ionization energy

Ionization energy is the energy required to remove an electron from an atom or ion. In this case, it is directly related to the effective nuclear charge. Thus, \(\mathrm{N}^{+}\) has the highest ionization energy, and \(\mathrm{N}^{-}\) has the lowest ionization energy. The order of increasing ionization energy is \(\mathrm{N}^{-} < \mathrm{N} < \mathrm{N}^{+}\). b. \(\mathrm{Se}, \mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\)
03

Comparative atomic radius

Comparing \(\mathrm{Se}\) and \(\mathrm{Se}^{-}\), we see that \(\mathrm{Se}^{-}\) has an extra electron and thus a lower effective nuclear charge. Similarly, \(\mathrm{Cl}^{+}\) has one less electron than \(\mathrm{Cl}\), leading to a higher effective nuclear charge. Comparing \(\mathrm{Se}\) and \(\mathrm{Cl}\), we see that \(\mathrm{Se}\) is located below \(\mathrm{Cl}\) in the periodic table, indicating a larger atomic radius. Therefore, the order of increasing radius is \(\mathrm{Cl}^{+} < \mathrm{Cl} < \mathrm{Se} < \mathrm{Se}^{-}\).
04

Comparative ionization energy

Using the same logic as in Step 1, we can arrange the species in order of increasing ionization energy as: \(\mathrm{Se}^{-} < \mathrm{Se} < \mathrm{Cl} < \mathrm{Cl}^{+}\). c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)
05

Comparative atomic radius

Comparing \(\mathrm{Br}^{-}\) and \(\mathrm{Rb}^{+}\), the \(\mathrm{Br}^{-}\) has an extra electron and same number of shells as \(\mathrm{Rb}^{+}\). Also, \(\mathrm{Sr}^{2+}\) has two more protons and the same number of electron shells as \(\mathrm{Rb}^{+}\), which leads to a higher effective nuclear charge and a smaller atomic radius. The order of increasing radius is \(\mathrm{Sr}^{2+} < \mathrm{Rb}^{+} < \mathrm{Br}^{-}\).
06

Comparative ionization energy

Similarly, using the same logic as in Step 1, we can arrange the species in order of increasing ionization energy as: \(\mathrm{Br}^{-} < \mathrm{Rb}^{+} < \mathrm{Sr}^{2+}\).

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Most popular questions from this chapter

Name each of the following compounds: a. Cul b. \(\mathrm{CuI}_{2}\) c. \(\mathrm{CoI}_{2}\) d. \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) e. \(\mathrm{NaHCO}_{3}\) \(\mathbf{f .} \mathrm{S}_{4} \mathrm{N}_{4}\) g. \(\mathrm{SF}_{4}\) h. NaOCl i. \(\mathrm{BaCrO}_{4}\) J. \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)

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