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?

To determine which ion is smaller, we need to consider the effective nuclear charge experienced by the outermost electrons. Since isoelectronic ions have the same number of electrons, the one with a higher effective nuclear charge will have its electrons pulled closer to the nucleus and thereby have a smaller ionic radius. F- and Na+ have 10 electrons each. F- has 9 protons in its nucleus, while Na+ has 11 protons. Therefore, Na+ will have a higher effective nuclear charge than F- and will be smaller in size.

Equation 7.1 is given as: \(Z_{\text {eff}} = Z - S\), where Z is the atomic number and S is the screening constant. Here, we are told to assume that core electrons contribute 1.00, and valence electrons contribute 0.00 to the screening constant, S. For F-, Z = 9, and S = 2 (two core electrons in 1s). For the Na+ ion, Z = 11, and S = 10 (2 electrons in 1s and 8 electrons in 2s and 2p). Now we can calculate \(Z_{\text {eff}}\) for each ion: \(Z_{\text {eff,F-}} = 9 - 2 = 7\) \(Z_{\text {eff,Na+}} = 11 - 10 = 1\)

According to Slater's rules, for ns and np electrons, the total screening constant S can be calculated as follows: 1. 0.35 for each electron to the right in the same group (ns, np) 2. 0.85 for each electron in one shell lower than the electron being considered (n-1) 3. 1.0 for each electron in all shells lower than n-1 For F-, there are 2 electrons in the 2s shell and 7 electrons in the 2p shell including the electron being considered. According to Slater's rules: \(S_{\mathrm{F}^{-}} = (1 \times 2) + (0.85 \times 2) + (0.35 \times 6) = 7.00\) Now, we can calculate the \(Z_{\text {eff,F-}}\): \(Z_{\text {eff,F-}} = 9 - 7.00 = 2.00\) For Na+, there are 2 electrons in the 2s shell and 5 electrons in the 2p shell. According to Slater's rules: \(S_{\mathrm{Na}^{+}} = (1 \times 10) + (0.85 \times 2) + (0.35 \times 5) = 10.15\) Now, we can calculate the \(Z_{\text {eff,Na+}}\): \(Z_{\text {eff,Na+}} = 11 - 10.15 = 0.85\)

For isoelectronic ions, their ionic radii are inversely proportional to their effective nuclear charges. As the effective nuclear charge increases, the electrons are pulled closer to the nucleus, resulting in a smaller ionic radius. In this case, Na+ has a higher effective nuclear charge than F-, so it has a smaller ionic radius.