Problem 87
Question
Consider the change in effective nuclear charge experienced by a \(2 p\) electron as we proceed from \(C\) to \(N\). (a) Based on a simple model in which core electrons screen the valence electrons completely and valence electrons do not screen other valence electrons, what do you predict for the change in \(Z_{\text {eff }}\) from \(C\) to \(\mathrm{N}\) ? (b) What change do you predict using Slater's rules? (c) The actual change in \(Z_{\text {eff }}\) from \(C\) to \(N\) is \(0.70+.\) Which approach to estimating \(Z_{\text {eff }}\) is more accurate? (d) The change in \(\mathrm{Z}_{\text {eff }}\) from \(\mathrm{N}\) to \(\mathrm{O}\) is smaller than that from \(\mathrm{C}\) to \(\mathrm{N}\). Can you provide an explanation for this observation?
Step-by-Step Solution
VerifiedKey Concepts
Slater's rules
To apply Slater's rules, follow these steps:
- Write down the electron configuration of the element.
- Identify which electron's \(Z_{\text{eff}}\) you need to calculate.
- Assign shielding coefficients based on electron types (e.g., 1s, 2s).
- Calculate \(Z_{\text{eff}} = Z - S\), where \(Z\) is the atomic number and \(S\) is the shielding constant.
shielding effect
Core electrons are closer to the nucleus and repel valence electrons, reducing the attractive force these outer electrons experience. The more core electrons, the stronger the shielding effect.
As you move across a period (C to N), the number of core electrons stays the same, but the number of protons increases. This leads to a stronger pull on valence electrons, increasing \(Z_{\text{eff}}\). However, how we account for this effect changes depending on the model used.
atomic structure
Electrons are organized in shells and subshells: for example, the configuration for carbon and nitrogen involves the 1s, 2s, and 2p orbitals. Electrons closer to the nucleus have lower energy levels and feel a higher effective nuclear charge compared to those further away.
This distribution helps explain why elements behave differently in chemical reactions, and why calculating \(Z_{\text{eff}}\) accurately is important for predicting these behaviors.
periodic trends
As you move from left to right across a period, \(Z_{\text{eff}}\) generally increases. This increase is due to additional protons in the nucleus, which pull valence electrons closer, and also due to minimal changes in inner electron shielding.
However, the change in \(Z_{\text{eff}}\) can vary due to factors like electron-electron repulsions in the same orbital, reflected more accurately by Slater's rules. Recognizing these trends is essential for understanding chemical reactivity and bonding. By understanding these nuances, students can better predict how elements might interact in different scenarios.