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

Firstly, consider the atomic number: Se (34) and Te (52). Se^2- and Te^2- have the same charge, which means they have the same number of electrons. Since Te has a higher atomic number, its increased number of protons result in a stronger attraction to the electrons, making Te^2- smaller than Se^2-. Now, compare Se^2- and Se. Se^2- has gained two extra electrons, which causes increased electron-electron repulsion and a subsequent expansion of the electron cloud. Therefore, Se is smaller than Se^2-. The final order of increasing size is Se < Te^2- < Se^2-.

Co^3+ (27) and Fe^2+ (26) are in the same period and close in atomic number. However, Co^3+ has a higher charge, which results in a greater effective nuclear charge experienced by its electrons, making it smaller than Fe^2+. Now, compare Fe^2+ and Fe^3+. Both have the same atomic number (26), but Fe^3+ has one less electron, leading to a greater effective nuclear charge for the remaining electrons. Thus, Fe^3+ is smaller than Fe^2+. The final order of increasing size is Fe^3+ < Co^3+ < Fe^2+.

Ca (20), Ti^4+ (22), and Sc^3+ (21) are in the same period. Since they are close in atomic number, we will focus on the effect of the charges on their sizes. The higher the charge in an ion, the greater the effective nuclear charge experienced by its electrons and therefore, the smaller the ion's size. Thus, the order of increasing size is Ti^4+ < Sc^3+ < Ca.

Be^2+ (4), Na^+ (11), and Ne (10) are in different periods, so the atomic number is an important factor here. Ions with a higher atomic number have more protons, which exerts a stronger attractive force on the electron cloud. Additionally, we must consider the charges on the ions, as a greater charge corresponds to a smaller ion size. Be^2+ has a higher charge than Na^+ and Ne, making it the smallest of these three species. In this specific case, though Ne has one less proton and one less electron than Na^+, their electron configurations result in a similar effective nuclear charge experienced by the outer electrons. Ne's configuration is complete, whereas Na^+ has lost its outermost electron. This makes Na^+ larger than Ne since its electron cloud becomes less tightly bound. The final order of increasing size is Be^2+ < Ne < Na^+.