   # Slater’s rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the “shielding constant,” 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 ‒ S The shielding constant, S , is calculated using the following rules: (1) The electrons of an atom are grouped as follows: (1 s ) (2 s , 2 p ) (3 s , 3 p ) (3 d ) (4 s , 4 p ) (4 d ), and so on. (2) Electrons in higher groups (to the right) do not shield those in the lower groups. (3) For ns and np valence electrons a) Electrons in the same ns, np group contribute 0.35 (for 1 s 0.30 works better). b) Electrons in n ‒ 1 groups contribute 0.85. c) Electrons in n ‒ 2 groups (and lower) contribute 1.00. (4) For nd and nf electrons, electrons in the same nd or nf 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: S = (2 × 0.85) + (5 × 0.35) = 3.45 Z * = 8 ‒ 3.45 = 4.55 Here is a calculation for a d electron in Ni: Z * = 28 ‒ [18 × 1.00] ‒ [7 × 0.35] = 7.55 and for an s electron in Ni: Z * = 28 ‒ [10 × 1.00] ‒ [16 × 0.85] ‒ [1 × 0.35] = 4.05 (Here 3 s , 3 p , and 3 d electrons are in the ( n ‒ 1) groups.) a) Calculate Z * for F and Ne. 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 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? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 7, Problem 83SCQ
Textbook Problem
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## Slater’s rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the “shielding constant,” 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 ‒ SThe shielding constant, S, is calculated using the following rules: (1) The electrons of an atom are grouped as follows: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d), and so on. (2) Electrons in higher groups (to the right) do not shield those in the lower groups. (3) For ns and np valence electrons a) Electrons in the same ns, np group contribute 0.35 (for 1s 0.30 works better). b) Electrons in n ‒ 1 groups contribute 0.85. c) Electrons in n ‒ 2 groups (and lower) contribute 1.00. (4) For nd and nf electrons, electrons in the same nd or nf 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: S = (2 × 0.85) + (5 × 0.35) = 3.45 Z* = 8 ‒ 3.45 = 4.55 Here is a calculation for a d electron in Ni: Z* = 28 ‒ [18 × 1.00] ‒ [7 × 0.35] = 7.55 and for an s electron in Ni: Z* = 28 ‒ [10 × 1.00] ‒ [16 × 0.85] ‒ [1 × 0.35] = 4.05 (Here 3s, 3p, and 3d electrons are in the (n ‒ 1) groups.) a) Calculate Z* for F and Ne. Relate the Z* values for O, F, and Ne to their relative atomic radii and ionization energies. b) Calculate Z* for one of the 3d electrons of Mn, and compare this with Z* for one of the 4s electrons of the element. Do the Z* values give us some insight into the ionization of Mn to give the cation?

a)

Interpretation Introduction

Interpretation:

Effective nuclear charge Z* for F and Ne and the Z* values for o, F, and Ne to their relative atomic radii and ionization energies has to be related.

Concept Introduction:

Nuclear charge (Z*): The effective nuclear charge generally denoted by (Zeff or Z*) it is the net positive charge experienced by an electron in a multi-electron atom. This word “effective” is used because the shielding effect of negatively charged electron prevents higher orbital electrons experience the full nuclear charge.

Increase and decrease electro negativity: The less vacancy electrons an atoms has the least it will gain of electrons. Moreover the electron affinity decrease down the groups and from right to left across the periods on the periodic table, the reason is electrons are placed in a higher energy level far from the nucleus thus a decrease from its pull.

Ionization energy (IE): The ionization energy is the required to remove an electron from an atom in the gas phase.

General formula of ionization energy= Atom in the ground state(g)Atom+(g)+eΔU=Ionizationenergy(IE)

As predicted by coulomb’s law, energy must be supplied to overcome the attraction between an electron and the nucleus and to separate the electron from the atom. Thus ionization energies always have positive values. An electron father from the nucleus generally has smaller ionization energy and electron closer to the nucleus has larger ionization energy.

### Explanation of Solution

The shielding constant (S) calculation methods points are given the problems

The effective nuclear charge (Z*) for the fluorine (F) atoms is 5.20, than nuclear charge (Z*) of Neon is 5.85.

LetusconsidertheZ*fortheoutermostelectronsforflurine(F)F=(2×0.85)+(6×0.35)=3.8Z*=93.8=5

b)

Interpretation Introduction

Interpretation:

Effective nuclear charge Z* for one of the 3d electrons of Mn has to be calculated and the Z* values for one of the 4s electrons  has to be compared  and Z* values make some insight into the ionization of Mn to give the cation has to be related.

Concept Introduction:

Nuclear charge (Z*): The effective nuclear charge generally denoted by (Zeff or Z*) it is the net positive charge experienced by an electron in a multi-electron atom. This word “effective” is used because the shielding effect of negatively charged electron prevents higher orbital electrons experience the full nuclear charge.

Increase and decrease electro negativity: The less vacancy electrons an atoms has the least it will gain of electrons. Moreover the electron affinity decrease down the groups and from right to left across the periods on the periodic table, the reason is electrons are placed in a higher energy level far from the nucleus thus a decrease from its pull.

Ionization energy (IE): The ionization energy is the required to remove an electron from an atom in the gas phase.

General formula of ionization energy= Atom in the ground state(g)Atom+(g)+eΔU=Ionizationenergy(IE)

As predicted by coulomb’s law, energy must be supplied to overcome the attraction between an electron and the nucleus and to separate the electron from the atom. Thus ionization energies always have positive values. An electron father from the nucleus generally has smaller ionization energy and electron closer to the nucleus has larger ionization energy.

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