Question

effective nuclear charge and atomic radius

1. which electrons experience a greater effective nuclear charge, the valence electrons in beryllium, or the valence electrons in nitrogen? why?

Explanation:

To determine which valence electrons experience a greater effective nuclear charge (Zₑff), we use the concept that Zₑff = Z - S, where Z is the atomic number and S is the shielding constant. Beryllium (Be) has Z = 4, and nitrogen (N) has Z = 7. Both are in the same period (period 2), so the shielding from inner electrons is similar. Since N has a higher Z, its valence electrons experience a greater Zₑff.

Answer:

The valence electrons in nitrogen experience a greater effective nuclear charge. This is because nitrogen (atomic number 7) has a greater atomic number than beryllium (atomic number 4), and both elements have similar electron shielding from inner - shell electrons (as they are in the same period). Using the formula for effective nuclear charge \( Z_{\text{eff}}=Z - S \) (where \( Z \) is the atomic number and \( S \) is the shielding constant), with a similar \( S \) for both, the larger \( Z \) of nitrogen leads to a larger \( Z_{\text{eff}} \) for its valence electrons.