Problem 85 - Chemistry The Central Science

Question

(a) If the core electrons were totally effective at screening the valence electrons and the valence electrons provided no screening for each other, what would be the effective nuclear charge acting on the \(3 s\) and \(3 p\) valence electrons in \(\mathrm{P}\) ? (b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nuclear charge is \(5.6+\) for the \(3 s\) electrons and \(4.9+\) for the \(3 p\) electrons. Why are the values for the \(3 s\) and \(3 p\) electrons different? (d) If you remove a single electron from a \(\mathrm{P}\) atom, which orbital will it come from? Explain.

Step-by-Step Solution

Answer

In a phosphorus atom, the effective nuclear charge acting on the valence electrons would be 5+ if core electrons provide total shielding and valence electrons provide no shielding. Using Slater's rules, we calculated the effective nuclear charges to be 5.3+ for the 3s electron and 6.0+ for the 3p electrons. The given detailed calculation values are 5.6+ for the 3s electrons and 4.9+ for the 3p electrons. These differences occur because the 3s electrons are closer to the nucleus than the 3p electrons, experiencing more attractive force from the nucleus. When removing a single electron from a phosphorus atom, it will come from the 3p orbital, as they are less tightly bound to the nucleus due to a lower effective nuclear charge.

Step 1: (a) Calculate the effective nuclear charge without valance-electron screening

The phosphorus atom has an atomic number of 15, meaning it has 15 protons in the nucleus. The electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^3.\] If the core electrons (1s, 2s, and 2p) create perfect screening and valence electrons (3s and 3p) don't create any screening, then the effective nuclear charge acting on 3s and 3p valence electrons will be the total nuclear charge minus the core electrons charge: \[Z_\text{eff} = Z - \text{core electrons} = 15 - (2+2+6) = 15 - 10 = 5 .\] So, in this case, the effective nuclear charge acting on the 3s and 3p valence electrons would be 5+.

Step 2: (b) Calculate the effective nuclear charge using Slater's rules

To calculate the effective nuclear charge using Slater's rules, we need to consider the electron shielding contributions from different orbital electrons. According to Slater's rules, the shielding constants are given for different orbitals: - For n=1, \(S = 0.30\) - For n=2, \(S = 0.85 \times \text{number of electrons}\) - For n=3, \(S = 1.00 \times \text{number of electrons}\) Let's first calculate the shielding for the 3s electrons: \[S(3s) = (2 \times 0.35) + (8 \times 0.85) + (1 \times 1.00) = 9.7\] Now, let's calculate the effective nuclear charge for the 3s electrons: \[Z_\text{eff}(3s) = Z - S(3s) = 15 - 9.7= 5.3+\] Now, let's calculate the shielding for the 3p electrons: \[S(3p) = (2 \times 0.35) + (8 \times 0.85) + (2 \times 1.00) = 9.0\] Now, let's calculate the effective nuclear charge for the 3p electrons: \[Z_\text{eff}(3p) = Z - S(3p) = 15 - 9.0 = 6.0+\] Using Slater's rules, we calculated that the effective nuclear charges are 5.3+ for the 3s electron and 6.0+ for the 3p electrons.

Step 3: (c) Comparing the values and explaining the difference

The 3s and 3p effective nuclear charges differ because the 3s electrons are closer to the nucleus than the 3p electrons, experiencing more attractive force from the nucleus. The given detailed calculation values are 5.6+ for the 3s electrons and 4.9+ for the 3p electrons. These values differ from our findings in part (b) due to Slater's rules being an approximation. However, the order is the same, with the 3s effective nuclear charge being higher than the 3p effective nuclear charge.

Step 4: (d) Determine which orbital electron will be removed from

When you remove a single electron from a phosphorus atom, it will come from the orbital with the highest energy and least binding by the nucleus. In this case, the electron will come from the 3p orbital, as the effective nuclear charge for the 3p electrons is lower than that for the 3s electrons, meaning that they are less tightly bound to the nucleus.

Key Concepts

Electron Configuration

Electron configuration is a way of representing the arrangement of electrons around the nucleus of an atom. Each electron in an atom occupies a specific energy level or shell, which is designated by quantum numbers. It is useful because it gives us insight into how atoms bond and react with one another.

In the case of phosphorus (P), its atomic number is 15. Therefore, a phosphorus atom has 15 electrons. The electron configuration for phosphorus can be written as:

• \[1s^2 2s^2 2p^6 3s^2 3p^3\]

In this configuration:

• "1s^2" means that there are two electrons in the first energy level (1s orbital).
• "2s^2 2p^6" represents the second energy level fully occupied by eight electrons (2s and 2p orbitals).
• "3s^2 3p^3" indicates that in the third energy level, there are two electrons in the 3s orbital and three in the 3p orbital.

Understanding electron configuration is key to predicting chemical properties and behaviors of atoms.

Slater's Rules

Slater's rules provide a way to estimate the shielding effect experienced by an electron in an atom. The shielding effect is when inner electrons repel outer ones, reducing the effective nuclear charge (\(Z_{eff}\)) felt by these outer electrons.

To apply Slater's rules, you assign different shielding constants depending on the group of electrons (like 1s, 2s, 2p, etc.). Here's a simplified approach:

• Electrons within the same group: each electron contributes 0.35 to the shielding constant.
• Electrons in the \preceding group: these each typically contribute 0.85.
• Electrons in even lower groups contribute 1.00 each.

Using these rules helps us to calculate a more refined effective nuclear charge:\[ Z_{eff} = Z - S \]where \(Z\) is the atomic number and \(S\) is the shielding constant.

In the case of phosphorus:- The 3s electrons shield less because they are closer to the nucleus and experience more nuclear attraction.- Since shielding reduces the nuclear attraction to valence electrons, calculating it accurately helps predict physical and chemical properties.

Shielding Effect

The shielding effect describes how core electrons block some of the nuclear charge from reaching outer valence electrons. This makes outer electrons less strongly attracted to the nucleus.

This effect occurs due to the

• electronic repulsion between electrons in different energy levels or shells.

It means that, as you go to outer shells, the effective nuclear charge experienced by the electrons decreases.

In our example with the phosphorus atom:

• The core electrons (e.g., 1s, 2s, and 2p) provide significant shielding for the 3s and 3p electrons.
• As a result, the effective nuclear charge felt by 3s is slightly more than 3p because the 3s electrons are closer to the nucleus and less shielded.

Understanding the shielding effect is crucial for predicting which electrons are most likely to be removed during chemical reactions or when atoms form ions.