Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Na}\) and \(\mathrm{K}\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Na}\) and \(\mathrm{K}\) by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }}\) ? (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

First, we need to calculate the effective nuclear charges for the outermost electrons in Na and K, considering that each core electron contributes 1.00 to the screening constant, and valence electrons contribute 0.00. We can estimate the Z_eff using the formula: \[Z_{eff} = Z - \sigma\] Where Z is the atomic number, and σ is the screening constant. In Na, Z = 11, and there are 10 core electrons, so σ = 10. For K, Z = 19, and there are 18 core electrons, so σ = 18. Applying the formula: For Na: \(Z_{eff} = 11 - 10 = 1\) For K: \(Z_{eff} = 19 - 18 = 1\)

Now, we will calculate the effective nuclear charges using Slater's rules. For Na and K, the outermost electrons are in the 3s and 4s orbitals, respectively. According to Slater's rules, the screening constant for the last s electron can be calculated using the following rules: - Electrons in the same group (n-1) and n (if the electron considered is ns) contribute 0.35 each - Electrons in the groups between 1 and n-1 contribute 0.85 each - Electrons in the 1s group contribute 1.00 each For Na: - 2 electrons in the 2s and 2p groups contribute 0.85 each: 2 * 0.85 - 1 electron in the 1s group contributes 1.00: 1 sigma_Na = 2(0.85) + 1 = 2.70 For K: - 2 electrons in the 3s and 3p groups contribute 0.85 each: 2 * 0.85 - 8 electrons in the 2s and 2p groups contribute 1.00 each: 8 - 1 electron in the 1s group contributes 1.00: 1 sigma_K = 2(0.85) + 8 + 1 = 10.70 Now plug these values into the formula for Z_eff: For Na: \(Z_{eff} = 11 - 2.70 = 8.30\) For K: \(Z_{eff} = 19 - 10.70 = 8.30\)

We have inferred that: - Using method (a) (core and valence electrons contributions): Z_eff is 1 for both Na and K. - Using Slater's rules: Z_eff in Na and K is 8.30. The actual values given: - For Na: Z_eff = 2.51+ - For K: Z_eff = 3.49+ Comparing these numbers, we can conclude that neither method estimates Z_eff accurately. Both methods overestimate the effective nuclear charge.

Both method (a) and Slater's rules fail to account for the gradual increase in Z_eff that occurs upon moving down a group. In the case of method (a), the Z_eff remains constant, and for Slater's rules, the Z_eff is the same for both Na and K.

Based on the calculations for Na and K, it is impossible to make an accurate prediction for Z_eff of the outermost electron in Rb. However, we can roughly estimate that it should be higher than Z_eff of K (3.49+). It would be advisable to use more accurate methods to estimate the Z_eff for the outermost electron in Rb.