Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+,\) respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Si and Cl 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) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in P, phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Step by step solution

01

Understand the effective nuclear charge concept

The effective nuclear charge (\(Z_{\text{eff}}\)) is the net positive charge experienced by an electron in an atom. It is the difference between the positive charge of the nucleus and the negative charge of the screening electrons. The shielding effect of these screening electrons reduces the electromagnetic force experienced by the outermost electron, leading to a lower effective nuclear charge.

02

Calculate \(Z_{\text{eff}}\) with the simplified method

According to the simplified method, core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant. For Silicon (Si): Atomic number Z = 14 (14 protons in nucleus) Core electrons: 1s, 2s, 2p (total 10) Valence electrons: 3s, 3p (total 4) Screening constant S = 10 x 1.00 + 4 x 0 = 10 \(Z_{\text{eff, Si}} = Z - S = 14 - 10 = 4\) For Chlorine (Cl): Atomic number Z = 17 (17 protons in nucleus) Core electrons: 1s, 2s, 2p, 3s (total 10) Valence electrons: 3p (total 7) Screening constant S = 10 x 1.00 + 7 x 0 = 10 \(Z_{\text{eff, Cl}} = Z - S = 17 - 10 = 7\)

03

Calculate \(Z_{\text{eff}}\) using Slater's rules

To calculate \(Z_{\text{eff}}\) using Slater's rules, we need to consider the contributions from all surrounding electrons to the screening constant. Slater's rules state that electrons in the same group shield the electron more effectively than those further away. For Silicon (Si): Following Slater's rules: S = 1s (2 * 0.3) + 2s, 2p (8 * 0.85) + 3s, 3p (4 * 1) = 0.6 + 6.8 + 4 = 11.4 \(Z_{\text{eff, Si}} = Z - S = 14 - 11.4 = 2.6\) For Chlorine (Cl): Following Slater's rules: S = 1s (2 * 0.35) + 2s, 2p (8 * 0.85) + 3s (2 * 1) + 3p (5 * 1) = 0.7 + 6.8 + 2 + 5 = 14.5 \(Z_{\text{eff, Cl}} = Z - S = 17 - 14.5 = 2.5\)

04

Compare the methods and determine the accuracy

Now we have calculated the \(Z_{\text{eff}}\) values using both methods, let's compare them with the given values: Simplified method: \(Z_{\text{eff, Si}}\) = 4 (compared to 4.29+) \(Z_{\text{eff, Cl}}\) = 7 (compared to 6.12+) Slater's rules: \(Z_{\text{eff, Si}}\) = 2.6 (compared to 4.29+) \(Z_{\text{eff, Cl}}\) = 2.5 (compared to 6.12+) The simplified method gives more accurate results, as its values are closer to the given values. Also, it can account for the steady increase in \(Z_{\text{eff}}\) that occurs upon moving left to right across a period.

05

Predict \(Z_{\text{eff}}\) for Phosphorus (P)

To predict the \(Z_{\text{eff}}\) for a valence electron in Phosphorus (P), we use the values calculated for Si and Cl: For Phosphorus (P), which lies between Si and Cl in the periodic table, we can make an approximation based on the simplified method (since it was more accurate): \(Z_{\text{eff, P}} \approx \frac{Z_{\text{eff, Si}} + Z_{\text{eff, Cl}}}{2} = \frac{4 + 7}{2} = 5.5\) Therefore, we predict the \(Z_{\text{eff}}\) value for a valence electron in Phosphorus (P) to be around 5.5.

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