Question

5 using coulombs law, approximate the force of attraction between the oxygen nucleus and one of its valence electrons. use slaters rules to estimate the effective nuclear charge of oxygen.

Answer:

Part 1: Estimate Effective Nuclear Charge (\(Z_{eff}\)) using Slater’s Rules

Step 1: Electron Configuration of Oxygen

Oxygen has an atomic number \(Z = 8\), so its electron configuration is \(1s^2 2s^2 2p^4\).

Step 2: Apply Slater’s Rules for Valence Electrons (2p electrons)

• Core electrons (1s): \(n = 1\) electrons. For a 2p electron, electrons with \(n = 1\) contribute \(0.85\) per electron? No, wait—Slater’s rules:
• Electrons in the same \(n\) shell: \(0.35\) (except \(n=1\), which is \(0.30\) for \(n=1\), but actually, for \(n \geq 2\), same \(n\): \(0.35\); inner shells (\(n - 1\)): \(0.85\); inner shells (\(n - 2\) or lower): \(1.00\).

For a 2p electron in O (\(1s^2 2s^2 2p^4\)):

• Electrons in \(n = 2\) (same shell): \(2s^2 + 2p^3\) (since we exclude the target electron) → \(2 + 3 = 5\) electrons. Each contributes \(0.35\).
• Electrons in \(n = 1\) (inner shell): \(1s^2\) → \(2\) electrons. Each contributes \(0.85\).

So shielding constant \(S = (5 \times 0.35) + (2 \times 0.85)\).

Calculate \(S\):
\(5 \times 0.35 = 1.75\); \(2 \times 0.85 = 1.7\); total \(S = 1.75 + 1.7 = 3.45\).

Effective nuclear charge \(Z_{eff} = Z - S = 8 - 3.45 = 4.55\) (approximate value for O’s 2p electrons).

Part 2: Coulomb’s Law for Force of Attraction

Coulomb’s Law: \(F = \frac{k |q_1 q_2|}{r^2}\), where \(k = 8.988 \times 10^9 \, \text{N·m}^2/\text{C}^2\), \(q_1\) = charge of nucleus (\(Z_{eff} e\), since \(Z_{eff}\) accounts for shielding), \(q_2\) = charge of electron (\(-e\), but magnitude is \(e\)), and \(r\) = distance between nucleus and valence electron (approximate Bohr radius for \(n=2\), \(r \approx 0.529 \times 10^{-10} \, \text{m} \times n^2\); for \(n=2\), \(r \approx 0.529 \times 10^{-10} \times 4 \approx 2.116 \times 10^{-10} \, \text{m}\)).

Step 1: Define Charges

• Nucleus charge: \(q_1 = Z_{eff} e\) (where \(e = 1.602 \times 10^{-19} \, \text{C}\)).
• Electron charge: \(q_2 = e\) (magnitude).

Step 2: Substitute into Coulomb’s Law

\(F = \frac{k (Z_{eff} e) (e)}{r^2} = \frac{k Z_{eff} e^2}{r^2}\).

Step 3: Plug in Values

• \(k = 8.988 \times 10^9 \, \text{N·m}^2/\text{C}^2\)
• \(Z_{eff} \approx 4.55\) (from Slater’s rules)
• \(e = 1.602 \times 10^{-19} \, \text{C}\)
• \(r \approx 2.116 \times 10^{-10} \, \text{m}\) (Bohr radius for \(n=2\))

Calculate numerator: \(k Z_{eff} e^2 = 8.988 \times 10^9 \times 4.55 \times (1.602 \times 10^{-19})^2\).

First, \(e^2 = (1.602 \times 10^{-19})^2 \approx 2.566 \times 10^{-38} \, \text{C}^2\).

Then, \(8.988 \times 10^9 \times 4.55 \approx 40.9 \times 10^9 = 4.09 \times 10^{10}\).

Multiply by \(e^2\): \(4.09 \times 10^{10} \times 2.566 \times 10^{-38} \approx 1.05 \times 10^{-27} \, \text{N·m}^2\).

Denominator: \(r^2 = (2.116 \times 10^{-10})^2 \approx 4.48 \times 10^{-20} \, \text{m}^2\).

Now, \(F = \frac{1.05 \times 10^{-27}}{4.48 \times 10^{-20}} \approx 2.34 \times 10^{-8} \, \text{N}\) (attractive force, so negative sign indicates attraction, but magnitude is ~\(2.3 \times 10^{-8} \, \text{N}\)).

Final Answers

• Effective nuclear charge (\(Z_{eff}\)) of oxygen (for 2p valence electrons): \(\boldsymbol{\approx 4.55}\) (or more precisely, using exact Slater’s rules: sometimes O’s \(Z_{eff}\) for 2p is ~4.5, but our calculation gave 4.55).
• Force of attraction (approximate): \(\boldsymbol{\approx 2.3 \times 10^{-8} \, \text{N}}\) (attractive, so direction is towards the nucleus).

(Note: The force calculation is approximate due to simplified \(r\) and \(Z_{eff}\); more accurate methods use quantum mechanical models, but this is a classical approximation with Slater’s \(Z_{eff}\).)