Question

Which will experience the greater effective nuclear charge, the electrons in the \(n=3\) shell in Ar or the \(n=3\) shell in Kr? Which will be closer to the nucleus? Explain.

Short answer

The n=3 electrons in Krypton (Kr) will experience a greater effective nuclear charge, with \(Z_{eff}(Kr) = 26\), compared to Argon (Ar), where \(Z_{eff}(Ar) = 8\). This stronger attraction between the electrons and nucleus in Kr results in the n=3 electrons being closer to the nucleus than those in Ar.

Step-by-step solution

Understand the concept of effective nuclear charge

Effective nuclear charge refers to the net positive charge experienced by electrons in an atom. Because electrons in inner shells shield the electrons in outer shells from the full positive charge of the nucleus, the effective nuclear charge experienced by outer shell electrons is less than the total positive charge of the protons in the nucleus.

Calculate the effective nuclear charge for the n=3 shell electrons in Ar and Kr

The formula for calculating effective nuclear charge (Zeff) is given as follows: \[Z_{eff} = Z - S\] Where \(Z\) is the atomic number (number of protons) and \(S\) is the shielding constant, which represents the shielding effect of inner shell electrons. Argon (Ar) has an atomic number of 18, meaning it has 18 protons. Since the electrons in question are in the n=3 shell, they will experience significant shielding from the 10 electrons in the n=1 and n=2 shells. For simplicity, we'll assume that the shielding constant, S, is equal to the number of inner shell electrons. Thus, for Ar: \[Z_{eff}(Ar) = 18 - 10 = 8\] Krypton (Kr) has an atomic number of 36, meaning it has 36 protons. It also has 10 electrons in the n=1 and n=2 shells, which cause a shielding effect on the electrons in the n=3 shell. Therefore, for Kr: \[Z_{eff}(Kr) = 36 - 10 = 26\]

Compare the effective nuclear charges and determine which electrons will be closer to the nucleus

Comparing the calculated effective nuclear charges, we can see that: \[Z_{eff}(Kr) > Z_{eff}(Ar)\] This means electrons in the n=3 shell in Krypton experience a greater effective nuclear charge than those in Argon. A greater effective nuclear charge results in a stronger attraction between the electrons and the nucleus. As a result, the n=3 electrons in Kr will be closer to the nucleus than those in Ar.