Question

The electron configuration of \(\mathrm{C}\) is \(1 s^{2} 2 s^{2} 2 p^{2}\). (a) If each core electron (i.e., the \(1 s\) electrons) were totally effective in shielding the valence electrons (i.e., the \(2 s\) and \(2 p\) electrons) from the nucleus and the valence electrons did not shield one another, what would be the shielding constant \((\sigma)\) and the effective nuclear charge \(\left(Z_{\text {eff }}\right)\) for the \(2 s\) and \(2 p\) electrons? (b) In reality, the shielding constants for the \(2 s\) and \(2 p\) electrons in \(\mathrm{C}\) are slightly different. They are 2.78 and \(2.86,\) respectively. Calculate \(Z_{\text {eff }}\) for these electrons, and explain the differences from the values you determined in part (a).

Short answer

(a) \(\sigma = 2\), \(Z_{\text{eff}} = 4\). (b) \(Z_{\text{eff}}(2s) = 3.22\), \(Z_{\text{eff}}(2p) = 3.14\).

Step-by-step solution

Identify the Atomic Structure of Carbon

Carbon (\(C\)) has an atomic number of 6, meaning it has 6 protons and typically 6 electrons. Its electron configuration is: \(1s^2 2s^2 2p^2\). Here, the core electrons are \(1s\) electrons, and the valence electrons are \(2s\) and \(2p\) electrons.

Determine the Shielding Constant \(\sigma\) for Part (a)

If core electrons fully shield the valence electrons, each \(1s\) electron contributes a shielding effect of +1. Thus, the shielding constant \(\sigma = 2\) because there are two \(1s\) electrons.

Calculate Effective Nuclear Charge \(Z_{\text{eff}}\) for Part (a)

\(Z_{\text{eff}}\) is calculated using the formula: \(Z_{\text{eff}} = Z - \sigma\). For carbon, \(Z = 6\), so \(Z_{\text{eff}} = 6 - 2 = 4\). Hence the effective nuclear charge experienced by \(2s\) and \(2p\) electrons in this ideal scenario is 4.

Calculate Realistic \(Z_{\text{eff}}\) for 2s Electrons in Part (b)

Given the realistic shielding constant for \(2s\) electrons is \(2.78\). Use \(Z_{\text{eff}} = Z - \sigma\) again, so \(Z_{\text{eff}} = 6 - 2.78 = 3.22\) for \(2s\) electrons.

Calculate Realistic \(Z_{\text{eff}}\) for 2p Electrons in Part (b)

The shielding constant for \(2p\) electrons is \(2.86\). Thus, \(Z_{\text{eff}} = 6 - 2.86 = 3.14\) for \(2p\) electrons.

Explain the Differences in \(Z_{\text{eff}}\)

The differences in \(Z_{\text{eff}}\) are due to the fact that \(2s\) and \(2p\) electrons shield each other slightly, and their penetration closer to the nucleus differs. This causes the \(2s\) electron to experience a slightly higher effective nuclear charge than the \(2p\) electron.

Key concepts

Electron Configuration

The electron configuration of an atom tells us how its electrons are distributed among its atomic orbitals. For carbon, this is denoted as \(1s^2 2s^2 2p^2\). Let's break this down:

• "\(1s^2\)" indicates two electrons in the first shell (s-orbital).
• "\(2s^2\)" and "\(2p^2\)" show four electrons in the second shell with both s and p orbitals occupied.

Each number before the letter tells us which energy level or shell the electron is in. The letter (s, p, d, f) represents the type of atomic orbital. Lastly, the superscript tells us the number of electrons in those orbitals. This makes studying electron configuration crucial, as it helps in understanding other concepts like the shielding effect.

Shielding Effect

The shielding effect describes how core electrons can reduce the attraction between the nucleus and the valence electrons. In simpler terms, it is the ability of inner-shell electrons to partially block the nuclear charge from being felt by outer-shell (valence) electrons.

To think of it practically, imagine a spotlight (nucleus) shining on an outer audience member (valence electrons). The core electrons act as a stage curtain partially blocking some of that light. This will make the audience member perceive the light as being less intense than it originally was.

In our example, the \(1s\) electrons are core electrons that provide the same amount of shielding within their shell, each contributing substantially, and leaving valence electrons with a lower effective nuclear charge.

Atomic Structure

Atomic structure serves as the backbone for understanding the behavior of electrons around the nucleus. An atom consists of a nucleus that contains protons and neutrons, surrounded by electrons that occupy different energy levels or shells.

For carbon:

• The atomic number is 6, which means there are 6 protons in the nucleus.
• The electron configuration \(1s^2 2s^2 2p^2\) reveals how electrons fill these shells.

Understanding atomic structure is fundamentally important as it allows predicting an element's chemical behavior, bonding characteristics, and its role in different chemicals processes.

Valence Electrons

Valence electrons are the electrons that reside in the outermost energy level of an atom. They are crucial as they mostly determine the chemical properties and reactivity of an element. For carbon with the electron configuration \(1s^2 2s^2 2p^2\), the valence electrons are \(2s\) and \(2p\), making a total of 4 valence electrons.

This group of electrons participates in forming chemical bonds – such as covalent bonds – which are essential for constructing molecules. They are also the electrons that gain, lose, or share when atoms interact with each other, leading to various chemical reactions and compounds.

Understanding valence electrons can help explain why elements bond the way they do, and predict the compounds they can form.