Problem 85 - Chemistry Central Science

Question

(a) If the core electrons were totally effective at shielding the valence electrons and the valence electrons provided no shielding for each other, what would be the effective nuclear charge acting on the 3 s and \(3 p\) valence electrons in \(P\) ? (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 summary, if core electrons were totally effective at shielding the valence electrons and valence electrons provided no shielding for each other, the effective nuclear charge on the 3s and 3p valence electrons in P would be +5. However, using Slater's rules, the effective nuclear charge is approximately +3.95 for 3s and +3.4 for 3p, while detailed calculations provided values of +5.6 for 3s and +4.9 for 3p electrons. The difference in the values for 3s and 3p electrons arises due to the distinct orbital shapes and spatial distributions. Finally, when removing a single electron from a P atom, it will come from the highest energy orbital, which is the 3p orbital.

Step 1: Determine the electron configuration of phosphorus

To determine the effective nuclear charge on the valence electrons of Phosphorus (P), we need to first identify its electron configuration. The atomic number of P is 15, which means it has 15 electrons. The electron configuration can be written as: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3\]

Step 2: Calculate the effective nuclear charge without valence electron shielding

In this step, we calculate the effective nuclear charge (\(Z_{eff}\)) on the 3s and 3p valence electrons of Phosphorus, assuming core electrons shield valence electrons completely and valence electrons do not shield each other. The atomic number of P is 15, which means there are 15 protons in the nucleus. For the 3s and 3p valence electrons, there are 10 core electrons (1s, 2s, and 2p). Assuming complete shielding by the core electrons, the effective nuclear charge is: \(Z_{eff} = Z - S\) Where \(Z\) is the atomic number, and \(S\) is the shielding constant. For P, \(Z = 15\) and \(S = 10\), then we have: \(Z_{eff} = 15 - 10 = 5\) So, if the core electrons were fully effective at shielding the valence electrons and valence electrons provided no shielding for each other, the effective nuclear charge acting on the 3s and 3p valence electrons in P would be \(+5\).

Step 3: Calculate the effective nuclear charge using Slater's rules

Using Slater's rules, we can obtain more accurate values of the effective nuclear charge (\(Z_{eff}\)) for the 3s and 3p valence electrons. For the 3s electron in P (15 electrons): - Two 3s electrons contribute 0.35 each. - Three 3p electrons contribute 0.35 each. - Ten core electrons (1s, 2s, 2p) contribute 1 each. So the shielding constant (\(S\)) for the 3s electron: \(S_{3s} = (2-1)\times 0.35 + 3 \times 0.35 + 10 = 11.05\) The effective nuclear charge for the 3s electron: \(Z_{eff_{3s}} = Z - S_{3s} = 15 - 11.05 = 3.95\) For the 3p electron in P (15 electrons): - Two 3s electrons contribute 0.85 each. - Three 3p electrons contribute 0.35 each (excluding the one considered). - Ten core electrons (1s, 2s, 2p) contribute 1 each. So the shielding constant (\(S\)) for the 3p electron: \(S_{3p} = 2 \times 0.85 + (3-1) \times 0.35 + 10 = 11.6\) The effective nuclear charge for the 3p electron: \(Z_{eff_{3p}} = Z - S_{3p} = 15 - 11.6 = 3.4\) Using Slater's rules, the effective nuclear charge for the 3s valence electrons in P is approximately +3.95 and for the 3p valence electrons approximately +3.4.

Step 4: Discuss the difference in effective nuclear charges of 3s and 3p electrons

According to detailed calculations provided, the effective nuclear charge values are +5.6 for the 3s electrons and +4.9 for the 3p electrons. The difference in values for the 3s and 3p electrons arises due to the different shapes and spatial distributions of the s and p orbitals. The s orbitals are spherical and tend to be closer to the nucleus more often than the p orbitals, which have a dumbbell shape. As the 3s electrons spend more time closer to the nucleus, they experience a higher effective nuclear charge than the 3p electrons. The values from Slater's rules differ from detailed calculations, but the trend remains the same.

Step 5: Determine which orbital electron will be removed from for a positively-charged phosphorus ion.

When a single electron is removed from a P atom, the electron is most likely to come from the orbital with the highest energy (and lowest ionization energy). From the electron configuration, we know that the 3p orbitals have the highest energy compared to previous orbitals. So, when P loses one electron, it will come from a 3p orbital.

Key Concepts

Slater's Rules

Slater's rules provide a method to estimate the shielding effect or screening constant for valence electrons, thereby allowing the calculation of the effective nuclear charge exerted by the nucleus. These rules provide correction factors that take into account the reduced attractive force due to the electron-electron repulsion within the atom.
In essence, to apply Slater's rules, the electrons in an atom are grouped into particular sets or "groups" based on their energy levels and types of orbitals (s, p, d, f). The shielding contributions from these sets to a specific electron in question are then calculated.

• Electrons in the same group (like other 3p electrons when determining the charge on a 3p electron) contribute a reduced amount to the shielding.
• Electrons in lower energy levels contribute more to shielding.

This method gives a more accurate description of the forces acting on an electron compared to assuming all inner electrons completely shield outer electrons, as Slater's rules account for the partial shielding often observed in reality.

Electron Configuration

Electron configuration describes the arrangement of electrons in the orbitals of an atom. The phosphorus atom, for instance, has an atomic number of 15, which means it houses 15 electrons. For phosphorus, these electrons are arranged as follows: \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3\]This configuration is key to understanding its chemical properties, such as reactivity and bonding behavior.
Each number and letter combination represents the filling of electron orbitals, starting from the lowest energy level. The superscript represents the number of electrons in each orbital.

• The numbers (1, 2, 3) denote the energy levels or shells.
• The letters (s, p) indicate the shape of the subshells, with 's' being spherical and 'p' being dumbbell-shaped.
• The superscripts (e.g., \(^2\) in \(1s^2\)) specify the number of electrons in that subshell.

Understanding electron configuration helps predict the chemical interactions that an atom can undergo, crucial in fields like chemistry and material science.

Shielding Effect

The shielding effect, or screening effect, refers to the reduction in attraction between an electron and the atomic nucleus due to the presence of other electrons. Electrons located between the nucleus and the electron in question can diminish the full nuclear charge that the outer electron would otherwise experience.
This effect plays a pivotal role in multi-electron atoms, where inner electrons effectively "shield" the outer electrons from experiencing the full positive charge of the nucleus. As a result, the outer (or valence) electrons encounter an "effective nuclear charge" that is less than the actual nuclear charge.

• Inner or core electrons are most effective at shielding.
• Electrons in the same shell provide less shielding than those in inner shells.

The shielding effect influences many atomic properties, such as atomic size, ionization energy, and the chemistry of the atom. Due to this effect, as you move across a period in the periodic table, even though additional electrons are being added, the effective nuclear charge typically increases, and atoms tend to become smaller.

Phosphorus Atom

The phosphorus atom is of particular interest in this context due to its position on the periodic table and its electron configuration. With an atomic number of 15, phosphorus has five valence electrons, making it a key element in biological molecules like DNA and RNA, where it contributes to forming the phosphate backbone.
Phosphorus resides in the third period and the 15th group (or 5th main group), sharing common chemical traits with other members of this group, such as nitrogen and arsenic.

• Its electron configuration is \[1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3\], highlighting its trio of unpaired p electrons.
• The presence of unpaired electrons makes phosphorus relatively reactive, seeking to complete its octet by forming covalent bonds.

Phosphorus' valence electrons (\(3s^2\) \, \(3p^3\)) are crucial for its chemical behavior. The difference in effective nuclear charges on these orbitals influences how readily phosphorus can lose or share electrons in chemical reactions, leading to its versatility in forming compounds like phosphates, phosphides, and in organic chemistry.