Question

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

Answer:

To solve this problem, we follow these steps:

Step 1: Determine the atomic number of selenium

Selenium (Se) has an atomic number \( Z = 34 \). This means the nucleus has 34 protons, so the charge of the nucleus \( q_{nucleus} = +34e \) (where \( e = 1.602 \times 10^{-19} \, \text{C} \) is the elementary charge).

Step 2: Apply Slater's rules to find the effective nuclear charge (\( Z_{\text{eff}} \)) for a valence electron of Se

Selenium is in period 4, group 16. Its electron configuration is \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^4 \). We consider a valence electron in the \( 4p \) subshell.

• Electrons in the same \( n \) (principal quantum number) and \( l \) (azimuthal quantum number) as the valence electron: For a \( 4p \) electron, the other \( 4p \) electrons (there are 3 more in \( 4p \)) and the \( 4s \) electrons (2). But for Slater's rules, electrons in the same group (same \( n \) and \( l \)) as the valence electron contribute 0.35 each, except for the first electron in a group, which contributes 0.30? Wait, no, the correct Slater's rules for a \( p \) electron in the outermost shell:

• Electrons with \( n > \) valence \( n \): 0 (since there are no electrons with \( n > 4 \) in Se)
• Electrons with \( n = \) valence \( n \) (same shell):
• For \( n = 4 \), electrons in \( 4s \) and \( 4p \): Each electron in \( 4s \) or \( 4p \) (excluding the valence electron itself) contributes 0.35. There are \( 2 + 3 = 5 \) electrons in \( n = 4 \) (excluding the valence electron), so contribution from \( n = 4 \): \( 5 \times 0.35 = 1.75 \)
• Electrons with \( n = \) valence \( n - 1 \) (one shell inner):
• For \( n = 3 \), electrons in \( 3s \), \( 3p \), \( 3d \): Each electron contributes 0.85. There are \( 2 + 6 + 10 = 18 \) electrons in \( n = 3 \)
• Electrons with \( n \leq \) valence \( n - 2 \) (two or more shells inner):
• For \( n = 1 \) and \( n = 2 \), electrons contribute 1.00 each. There are \( 2 + 8 = 10 \) electrons in \( n = 1 \) and \( n = 2 \)

So the shielding constant \( S \) is calculated as:

$$ S = (\text{contribution from } n = 1) + (\text{contribution from } n = 2) + (\text{contribution from } n = 3) + (\text{contribution from } n = 4) $$

$$ S = (2 \times 1.00) + (8 \times 1.00) + (18 \times 0.85) + (5 \times 0.35) $$

$$ S = 2 + 8 + 15.3 + 1.75 = 27.05 $$

Then the effective nuclear charge \( Z_{\text{eff}} = Z - S \):

$$ Z_{\text{eff}} = 34 - 27.05 = 6.95 $$

Wait, that seems low. Wait, maybe I made a mistake in Slater's rules for Se. Let's re - check. The correct electron configuration for Se is \( [Ar] 3d^{10} 4s^2 4p^4 \). When considering a \( 4p \) valence electron:

• Electrons with \( n < 3 \) (i.e., \( n = 1 \) and \( n = 2 \)): \( 2 + 8 = 10 \) electrons, each contributing 1.00, so \( 10\times1.00 = 10 \)
• Electrons with \( n = 3 \) ( \( 3s^2 3p^6 3d^{10} \)): There are \( 2 + 6+10 = 18 \) electrons. For electrons in \( n = 3 \) (which is \( n=\text{valence }n - 1\) when valence \( n = 4 \)), the contribution is 0.85 per electron, so \( 18\times0.85=15.3 \)
• Electrons with \( n = 4 \) but not in the same \( l \) as the valence electron: The \( 4s \) electrons (\( 2 \) electrons) are in \( n = 4 \), \( l = 0 \), while the valence electron is in \( n = 4 \), \( l = 1 \). Electrons in the same \( n \) but different \( l \) (here \( 4s \) and \( 4p \)): each \( 4s \) electron contributes 0.35, so \( 2\times0.35 = 0.7 \)
• Electrons in the same \( n \) and \( l \) (other \( 4p \) electrons): There are \( 3 \) other \( 4p \) electrons, each contributing 0.35, so \( 3\times0.35=1.05 \)

Now total shielding \( S=10 + 15.3+0.7 + 1.05=27.05 \)

Then \( Z_{\text{eff}}=Z - S=34 - 27.05 = 6.95\approx7 \) (a more accurate calculation or maybe a different interpretation of Slater's rules for Se might give a slightly different value, but for approximation, \( Z_{\text{eff}}\approx7 \))

So the charge of the nucleus experienced by the valence electron is \( q_{nucleus, \text{eff}}=Z_{\text{eff}}e\approx7e \)

The valence electron has a charge \( q_{electron}=-e \)

Step 3: Determine the distance between the nucleus and the valence electron

We can approximate the distance \( r \) between the nucleus and the valence electron. For a \( 4p \) electron in Se, we can use the Bohr radius formula for multi - electron atoms approximately. The average distance of a \( 4p \) electron from the nucleus can be approximated. The Bohr radius \( a_0 = 0.529\times 10^{-10}\, \text{m} \). For a \( n = 4 \) electron, the average distance \( r\approx n^2a_0=16\times0.529\times 10^{-10}\, \text{m}\approx8.464\times 10^{-10}\, \text{m} \) (this is a rough approximation, but we can use this for our calculation)

Step 4: Apply Coulomb's Law

Coulomb's Law is given by \( F = k\frac{|q_1q_2|}{r^2} \), where \( k = 8.988\times 10^{9}\, \text{N}\cdot\text{m}^2/\text{C}^2 \) is the Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.

Substituting \( q_1 = Z_{\text{eff}}e \), \( q_2=-e \) (we take the magnitude, so \( |q_1q_2| = Z_{\text{eff}}e^2 \)) and \( r\approx8.464\times 10^{-10}\, \text{m} \), \( Z_{\text{eff}}\approx7 \), \( e = 1.602\times 10^{-19}\, \text{C} \)

$$ F=k\frac{Z_{\text{eff}}e^2}{r^2} $$

$$ F = 8.988\times 10^{9}\times\frac{7\times(1.602\times 10^{-19})^2}{(8.464\times 10^{-10})^2} $$

First, calculate the numerator: \( 7\times(1.602\times 10^{-19})^2=7\times2.566\times 10^{-38}=1.796\times 10^{-37} \)

Denominator: \( (8.464\times 10^{-10})^2 = 7.164\times 10^{-19} \)

Then \( F=8.988\times 10^{9}\times\frac{1.796\times 10^{-37}}{7.164\times 10^{-19}} \)

\( \frac{1.796\times 10^{-37}}{7.164\times 10^{-19}}=2.507\times 10^{-19} \)

\( F = 8.988\times 10^{9}\times2.507\times 10^{-19}\approx2.25\times 10^{-9}\, \text{N} \)

Final Answer

The force of attraction between the selenium nucleus and one of its valence electrons is approximately \( \boldsymbol{2.3\times 10^{-9}\, \text{N}} \) (the value may vary slightly depending on the accuracy of the \( Z_{\text{eff}} \) calculation and the distance approximation).