Detailed calculations show that the value of \(Z_{\text { eff }}\) for the outermost electrons in Si and Cl atoms is \(4.29+\) and \(6.12+\) , respectively.(a) What value do you estimate for \(Z\) 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 \(Z_{\text { eff? }} ?(\mathbf{d})\) 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 \(\mathrm{P}\) , phosphorus, based on the calculations for Si and Cl.

Step by step solution

01

Estimation using core and valence electron contribution

First, let's estimate the Z_eff values by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant. The equation for Z_eff can be written as: \[Z_{eff} = Z - S\] where Z is the atomic number and S is the screening constant. For silicon (Si): Z = 14 (10 core electrons and 4 valence electrons) For chlorine (Cl): Z = 17 (10 core electrons and 7 valence electrons) Let's calculate the screening constant (S) using core and valence electron contributions for both Si and Cl.

02

Calculate the screening constant (S) values using core/valence electron contributions

(a) For Si: S(Si) = 10(core electrons) * 1.00 + 4(valence electrons) * 0.00 = 10 For Cl: S(Cl) = 10(core electrons) * 1.00 + 7(valence electrons) * 0.00 = 10 Now, we will calculate the estimated Z_eff values for Si and Cl.

03

Calculate the Z_eff values using core and valance electron contribution method

For Si: Z_eff(Si) = Z(Si) - S(Si) = 14 - 10 = 4 For Cl: Z_eff(Cl) = Z(Cl) - S(Cl) = 17 - 10 = 7 So, the estimated Z_eff using core and valance electron contributions are 4 for Si and 7 for Cl.

04

Estimation using Slater's rules

(b) Now, we will use Slater's rules to estimate the Z_eff values for Si and Cl. For silicon (Si), the electron configuration is: 1s² 2s² 2p⁶ 3s² 3p² For chlorine (Cl), the electron configuration is: 1s² 2s² 2p⁶ 3s² 3p⁵ Now, let's calculate the S values using Slater's rules for both Si and Cl.

05

Calculate screening constant (S) values using Slater's rules

For Si: S(Si) = (2 * 0.30) + (8 * 0.85) + (4 * 1) = 10.60 For Cl: S(Cl) = (2 * 0.30) + (8 * 0.85) + (7 * 1) = 11.80 Now, we will calculate the estimated Z_eff values for Si and Cl using Slater's rules.

06

Calculate the Z_eff values using Slater's rules

For Si: Z_eff(Si) = Z(Si) - S(Si) = 14 - 10.60 = 3.40 For Cl: Z_eff(Cl) = Z(Cl) - S(Cl) = 17 - 11.80 = 5.20 So, the estimated Z_eff using Slater's rules are 3.40 for Si and 5.20 for Cl.

07

Compare the two methods' accuracy

(c) The calculated Z_eff values are 4.29 for Si and 6.12 for Cl. Comparing these values, we can say that: Core and valence electron contribution method: Error for Si: |4.29 - 4| = 0.29 Error for Cl: |6.12 - 7| = 0.88 Slater's rules method: Error for Si: |4.29 - 3.40| = 0.89 Error for Cl: |6.12 - 5.20| = 0.92 We can see that the errors for both Si and Cl when using core and valence electron contributions are smaller than when using Slater's rules.

08

Determine which method accounts for Z_eff's increase across a period

(d) When we move from left to right across a period, Z_eff should increase. Let's see if this is the case for the two methods: Core and valence electron contribution method: Z_eff increases from 4 (Si) to 7 (Cl). Slater's rules method: Z_eff increases from 3.40 (Si) to 5.20 (Cl). Both methods show an increase in Z_eff when moving from Si to Cl. However, the core and valence electron contribution method had a more accurate estimation of Z_eff for Si and Cl.

09

Predict Z_eff for phosphorus (P)

(e) Now we will predict the Z_eff for phosphorus (P) based on the calculations for Si and Cl. We can try to find a trend using the core and valence electron contribution method since it was more accurate for Si and Cl. P has an atomic number (Z) of 15 (10 core electrons and 5 valence electrons). Using the core and valence electron contribution method, we can estimate Z_eff(P) to be: Z_eff(P) = Z(P) - S(P) = 15 - 10 = 5 So, based on the calculations for Si and Cl, we predict the effective nuclear charge Z_eff for a valence electron in phosphorus to be 5.

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