Questions: The shielding of electrons gives rise to an effective nuclear charge, Zeff, which explains why boron is larger than oxygen. Estimate the approximate Zeff felt by a valence electron of boron and oxygen, respectively?

Solution Steps

The effective nuclear charge, \( Z_{\text{eff}} \), is the net positive charge experienced by an electron in a multi-electron atom. It accounts for the actual nuclear charge \( Z \) minus the shielding effect caused by other electrons.

Boron has an atomic number \( Z = 5 \). The electron configuration is \( 1s^2 2s^2 2p^1 \). The valence electron in the 2p orbital is shielded by the 2 core electrons in the 1s orbital and the 2 electrons in the 2s orbital.

Using Slater's rules:

• Electrons in the same group (2p) contribute 0.35 each.
• Electrons in the n-1 shell (1s) contribute 0.85 each.

Thus, the shielding constant \( S \) is: \[ S = 2 \times 0.85 + 2 \times 0.35 = 1.7 + 0.7 = 2.4 \]

Therefore, the effective nuclear charge for boron is: \[ Z_{\text{eff}} = Z - S = 5 - 2.4 = 2.6 \]

Oxygen has an atomic number \( Z = 8 \). The electron configuration is \( 1s^2 2s^2 2p^4 \). The valence electrons in the 2p orbital are shielded by the 2 core electrons in the 1s orbital and the 2 electrons in the 2s orbital.

Using Slater's rules:

• Electrons in the same group (2p) contribute 0.35 each.
• Electrons in the n-1 shell (1s) contribute 0.85 each.

Thus, the shielding constant \( S \) is: \[ S = 2 \times 0.85 + 2 \times 0.35 + 3 \times 0.35 = 1.7 + 0.7 + 1.05 = 3.45 \]

Therefore, the effective nuclear charge for oxygen is: \[ Z_{\text{eff}} = Z - S = 8 - 3.45 = 4.55 \]

Final Answer