Answer
See explanation
Work Step by Step
a. To determine an approximate value for \( Z_{\text{eff}} \) for the \( 1s \) electron in a silver atom, we need to understand the concept of effective nuclear charge (\( Z_{\text{eff}} \)). In the Bohr model, the effective nuclear charge experienced by an electron in a multi-electron atom is given by the equation:
\[ Z_{\text{eff}} = Z - S \]
where \( Z \) is the atomic number (number of protons in the nucleus) and \( S \) is the screening constant or shielding constant.
In the case of the \( 1s \) electron in a silver atom, we can assume that it is shielded by the \( 2s \) and \( 2p \) electrons. Since the \( 1s \) electron is closest to the nucleus, it experiences the full nuclear charge (\( Z \)) but is partially shielded by the other electrons.
The ionization energy (\( IE \)) for the \( 1s \) electron is a measure of the energy required to remove that electron from the atom. It is directly related to the effective nuclear charge experienced by the electron. The higher the ionization energy, the stronger the effective nuclear charge.
Given that the ionization energy for the \( 1s \) electron in a silver atom is \( 2.462 \times 10^6 \, \text{kJ/mol} \), we can use this information to determine an approximate value for \( Z_{\text{eff}} \).
$2.462\times 10^6\times \frac{1}{6.0221\times 10^{23}}\times 1000=2.178\times 10^{-18}(Z_{eff})^2$
$Z_{eff}\approx 43.33$
b. Comparing \( Z_{\text{eff}} \) from part a to \( Z \) for silver (\( \text{Ag} \)), we can rationalize the relative numbers.
In the periodic table, the atomic number (\( Z \)) represents the number of protons in the nucleus of an atom. For silver (\( \text{Ag} \)), the atomic number is 47.
The calculated value of $Z_{eff}$ is $43.33$, therefore less than $47$. Electrons in other orbitals can penetrate the 1s orbital. So a 1s electron can be slightly
shielded from the nucleus by these penetrating electrons, giving a value of $Z_{eff}$ close to but less than Z.
