Question

Slater's rules are a simple way to estimate the effective nuclear charge experienced by an electron. In this approach, the "shielding constant," \(\mathrm{S}\), is calculated. The effective nuclear charge is then the difference between S and the atomic number, \(Z\). (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) $$ Z^{*}=Z-\mathrm{S} $$ The shielding constant, \(\mathrm{S}\), is calculated using the following rules: 1\. The electrons of an atom are grouped as follows: (1s) \((2 s, 2 p)(3 s, 3 p)\) (3d) \((4 s, 4 p)\) (4d), and so on. 2\. Electrons in higher groups (to the right) do not shield those in the lower groups. 3\. For \(n s\) and \(n p\) valence electrons (a) Electrons in the same \(n s, n p\) group contribute 0.35 (for \(1 s 0.30\) works better). (b) Electrons in the \(n-1\) group contribute 0.85 (c) Electrons in the \(n-2\) group (and lower) contribute 1.00 4\. For \(n d\) and \(n f\) electrons, electrons in the same \(n d\) or \(n f\) group contribute \(0.35,\) and those in groups to the left contribute 1.00 As an example, let us calculate \(Z^{*}\) for the outermost electron of oxygen: $$ \begin{array}{c} \mathrm{S}=(2 \times 0.85)+(5 \times 0.35)=3.45 \\ Z^{*}=8-3.45=4.55 \end{array} $$ Here is a calculation for a \(d\) electron in \(\mathrm{Ni}\) : $$ Z^{*}=28-[18 \times 1.00]-[7 \times 0.35]=7.55 $$ and for an s electron in \(\mathrm{Ni}\) : $$ \begin{aligned} Z^{*}=28-[10 \times 1.00]-[16 \times 0.85] & \\ -[1 \times 0.35] &=4.05 \end{aligned} $$ (Here \(3 s, 3 p,\) and \(3 d\) electrons are in the \((n-1)\) groups.) (a) Calculate \(Z^{*}\) for \(F\) and \(N e\). Relate the \(Z^{*}\) values for O, F, and Ne to their relative atomic radii and ionization energies. (b) Calculate \(Z^{* *}\) for one of the \(3 d\) electrons of \(\mathrm{Mn}\), and compare this with \(Z^{*}\) for one of the \(4 s\) electrons of the element. Do the \(Z^{*}\) values give us some insight into the ionization of Mn to give the cation?

Short answer

Z* for F is 5.25, Ne is 5.55. Mn's 3d Z* is 5.6, and 4s Z* is 2.75. Lower 4s Z* indicates easier removal in ionization.

Step-by-step solution

Determine Electron Configuration for F and Ne

Fluorine (F) has an atomic number of 9, so its electron configuration is 1s² 2s² 2p⁵. Neon (Ne) has an atomic number of 10, with an electron configuration of 1s² 2s² 2p⁶.

Calculate S for F's Outermost Electron

For the 2p electron in F, the shielding constant S is calculated as follows: - The 1s electrons (n-2 group) contribute 2.0 (2 x 1.0). - The electrons in the same 2s, 2p group contribute 5 x 0.35. Thus, S = 2 + 5 x 0.35 = 3.75.

Calculate Z* for Fluorine

Using Z = 9 and S = 3.75 from the previous step, the effective nuclear charge is calculated as:\( Z^{*} = Z - S = 9 - 3.75 = 5.25 \).

Calculate S for Ne's Outermost Electron

For the outer 2p electron in Ne: - 1s electrons (n-2 group) contribute as 2 x 1.00 = 2.00. - Same 2s, 2p group electrons (excluding the outermost) contribute 7 x 0.35. Thus, S = 2 + 7 x 0.35 = 4.45.

Calculate Z* for Neon

Using Z = 10 and S = 4.45, the effective nuclear charge is:\( Z^{*} = Z - S = 10 - 4.45 = 5.55 \).

Compare Z* Values

The calculated Z* values show Z* for F (5.25) is slightly less than for Ne (5.55). These values align with the trend that larger Z* values correspond to smaller atomic radii and higher ionization energies.

Determine Electron Configuration for Mn

Manganese (Mn) has an atomic number of 25, with an electron configuration of [Ar] 3d⁵ 4s².

Calculate S for Mn's 3d Electron

For a 3d electron: - Electrons in the same group (3d) contribute 4(0.35). - Electrons to the left (1s² 2s² 2p⁶ 3s² 3p⁶) contribute 18 x 1.00. Thus, S = 18 + 4(0.35) = 19.4.

Calculate Z* for Mn's 3d Electron

Using Z = 25 and S = 19.4, the effective nuclear charge is:\( Z^{*} = 25 - 19.4 = 5.6 \).

Calculate S for Mn's 4s Electron

For a 4s electron: - Electrons in the (n-1) groups (3d electrons) contribute 5 x 0.85. - Inner electrons (1s² 2s² 2p⁶ 3s² 3p⁶) contribute 18 x 1.00. Thus, S = 18 + 5 x 0.85 = 22.25.

Calculate Z* for Mn's 4s Electron

Using Z = 25 and S = 22.25, the effective nuclear charge is:\( Z^{*} = 25 - 22.25 = 2.75 \).

Insight into Mn Ionization

The Z* for Mn's 4s electron (2.75) is lower than that for the 3d electron (5.6), indicating that 4s electrons are easier to remove, which explains the initial ionization of Mn to form Mn²⁺.

Key concepts

Slater's Rules

Slater's Rules are a set of guidelines used to estimate the effective nuclear charge ( Z^* ) felt by an electron in an atom. This concept is crucial because it helps explain how tightly an electron is held by the nucleus. The effective nuclear charge affects properties like atomic radii and ionization energy.

To compute this charge, Slater's Rules first require dividing the atom's electrons into specific groups based on their shells and subshells:

• (1s)
• (2s, 2p)(3s, 3p)
• (3d)(4s, 4p)

Electrons in higher groups (to the right) don't shield those in the lower groups. For the valence electrons ( np ), the contributions are:

• Electrons in the same shell: 0.35
• Electrons in n-1 : 0.85
• Electrons in n-2 or lower: 1.00

These rules help compute the shielding constant ( S ), which is subtracted from the atomic number ( Z ) to find Z^{**} . This method offers a simplified calculation that provides insight into the behavior of electrons.

Atomic Radii

Atomic radius is a measure of the size of an atom, typically the distance from the nucleus to the outer boundary of the surrounding cloud of electrons. The concept is fundamental to understanding how an atom's size varies across the periodic table.

The effective nuclear charge ( Z^* ) heavily influences atomic radii. A higher Z^* indicates that electrons are pulled closer to the nucleus, resulting in a smaller atomic radius. Conversely, a lower Z^* means the electrons are less tightly bound and can reside farther from the nucleus, resulting in a larger atomic size.

As you move across a period in the periodic table, Z^* generally increases, causing atomic radii to decrease. This is because electrons are added to the same principal energy level while additional protons are added to the nucleus, enhancing the pull on the electrons. As you move down a group, the increased shielding from additional inner electron shells usually results in larger atomic radii.

Ionization Energy

Ionization energy is the energy required to remove the outermost electron from an atom in its gaseous state. This key property is closely related to both atomic radii and effective nuclear charge.

The relationship with Z^* is quite direct: as Z^* increases, so does the ionization energy. A strong nuclear attraction means that more energy is needed to remove an electron. Therefore, elements with higher Z^* , typically found at the end of periods, have high ionization energies.

This relation also explains the trend of ionization energies across the periodic table. Across a period, ionization energy tends to increase because Z^* is higher with each additional proton, leading to a greater pull on the electrons. Down a group, ionization energy decreases because the outer electrons are further away from the nucleus, despite an increase in Z^* , as added electron shells increase shielding.

Electron Configuration

Electron configuration is the distribution of electrons of an atom or molecule in atomic or molecular orbitals. This arrangement is fundamental to understanding chemical properties and reactivity.

For an element, knowing its electron configuration helps you determine how electrons are organized around the nucleus. This informs predictions about bonding behavior and reactivity.

• An example is Fluorine (F): 1s² 2s² 2p⁵, indicating how specific electrons fill orbitals in a manner governed by principles like the Pauli exclusion principle and Hund's rule.
• For Neon (Ne), with a configuration of 1s² 2s² 2p⁶, it's clear that this full outer p orbital confirms the stability and inertness characteristic of noble gases.

Understanding electron configurations enables predictions about the position of elements in the periodic table, their chemical properties, and how they will interact in chemical reactions.