Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\mathrm{eff}}\) for these two 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?

For isoelectronic ions, the number of protons in the nucleus (atomic number, Z) determines their size. The ion with fewer protons will have a smaller effective nuclear charge and, as a consequence, a larger size. Cl^- has an atomic number of 17 and K^+ has an atomic number of 19. Therefore, Cl^- is larger than K^+.

According to Equation 7.1, the effective nuclear charge can be calculated as: \[Z_{\mathrm{eff}} = Z - S \] where \(Z\) is the atomic number and \(S\) is the screening constant. Given that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, we can calculate the effective nuclear charges for Cl^- and K^+ ions. For Cl^- ion: Atomic number, \(Z = 17\) Core electrons = 10 (1s², 2s², 2p^6) Valence electrons = 7 (3s², 3p^5) Screening constant, \(S = 10\) \[Z_{\mathrm{eff}} = 17 - 10 = 7\] For K^+ ion: Atomic number, \(Z = 19\) Core electrons = 18 (1s², 2s², 2p^6, 3s², 3p^6) Valence electrons = 1 (4s^1) Screening constant, \(S = 18\) \[Z_{\mathrm{eff}} = 19 - 18 = 1\]

Using Slater's rules to estimate the screening constant for both ions, the following results are obtained: For Cl^- ion: \(S = 10 \times 0.85 (1s², 2s², 2p^6) + 7 \times 0.12 (3s², 3p^5) = 8.5 + 0.84 = 9.34\) For K^+ ion: \(S = 18 \times 0.85 (1s², 2s², 2p^6, 3s², 3p^6) + 1 \times 0.12 (4s^1) = 15.3 + 0.12 = 15.42\) Now, we can calculate the effective nuclear charges using these estimated screening constants: For Cl^- ion: \[Z_{\mathrm{eff}} = 17 - 9.34 = 7.66\] For K^+ ion: \[Z_{\mathrm{eff}} = 19 - 15.42 = 3.58\]

For isoelectronic ions, the number of electrons is the same, but the number of protons in the nucleus (atomic number, Z) differs. As the effective nuclear charge increases, the attraction between the nucleus and the electrons also increases, causing the ionic radius to decrease. In other words, there is an inverse relationship between the effective nuclear charge and the ionic radius for isoelectronic ions. The ion with a higher effective nuclear charge will have a smaller ionic radius, and vice versa.