Chapter 8: Trends in the Periodic Table

Section 8-2: Atomic and Ionic Radius

Section 8-3: Ionization Energy and Electron Affinity

Chapter 8 Practice Exercises and Review Quizzes

Section 8-1: Group Names

We will refer to particular groups on the periodic table by their common names, including:

Group	Name
1	alkali metals (H not included)
2	alkaline earth metals
17	halogens
18	noble gases

Metals in the central region of the periodic table are known as transition metals.

Section 8-2: Atomic and Ionic Radius

When comparing neutral atoms in the same group of the periodic table, an atom having valence electrons with the larger value of n will generally have the larger atomic radius. For example, let us compare the two alkali metals lithium and sodium, which have the following electron configurations:

Li: [He] 2s¹ Na: [Ne] 3s¹

Since the valence 3s electron in sodium has the larger value of n than the valence 2s electron in lithium, we would expect sodium to have the larger atomic radius.

Neutral atoms in the same row of the periodic table do not have a difference in the value of n for the valence electrons. For example, the alkaline earth metal magnesium and the halogen chlorine in the third row both have valence electrons with n = 3:

Mg: [Ne] 3s² Cl: [Ne] 3s² 3p⁵

When two atoms have the same value of n for the valence electrons, the atom with the greater number of protons will generally have a greater effective nuclear charge to draw the valence electrons closer to the nucleus and, thus, decrease the atomic radius. Since chlorine's 17 protons are greater than magnesium's 12 protons, chlorine will have a greater effective nuclear charge to draw chlorine's valence electrons closer to the nucleus and, thus, chlorine is expected to have the smaller atomic radius, while magnesium with the lower effective nuclear charge is expected to have the larger atomic radius.

Based on the discussion above, we can draw the following conclusion regarding the trend on the periodic table for atomic radius:

An element located closer to the lower left corner of the periodic table will generally have the larger atomic radius, while an element located closer to the upper right corner of the periodic table will generally have the smaller atomic radius.

Note that there are numerous exceptions to the trend above, particularly among the transition metals.

Sample Exercise 8A:

Rank the elements bromine, calcium, and fluorine from smallest to largest atomic radius. Explain, including electron configurations in your answer.

Solution:

Based on the trend stated above, we would expect F, which is closest to the upper right corner of the periodic table, to have the smallest atomic radius and Ca, which is closest to the lower left corner of the periodic table, to have the largest atomic radius. If we compare the electron configurations of the two halogens Br and F, we see that Br should have the larger atomic radius because Br's valence electrons have n = 4, whereas F should have the smaller atomic radius because F's valence electrons have n = 2:

Br: [Ar] 4s² 3d¹⁰ 4p⁵ F: [He] 2s² 2p⁵

If we compare the electron configurations of Br and Ca, we see that both elements have valence electrons with n = 4:

Ca: [Ar] 4s²

However, Br's 35 protons should give Br a greater effective nuclear charge and smaller atomic radius, whereas Ca's 20 protons should give Ca a lower effective nuclear charge and larger atomic radius. Therefore, the three elements ranked from smallest to largest atomic radius would be F < Br < Ca.

When a neutral atom gains electrons to form a negatively-charged anion, the ionic radius of the anion will be larger than the atomic radius of the neutral atom.

The electrons added to make an anion do not typically change the value of n for the valence electrons, so we can attribute the increased radius of the anion to increased repulsion between the negatively-charged electrons that then spread out to increase the radius.

Sample Exercise 8B:

Which is larger, the atomic radius of I or the ionic radius of I^- ? Explain, including electron configurations in your answer.

Solution:

The neutral atom I and the anion I^- have the following electron configurations:

I: [Kr] 5s² 4d¹⁰ 5p⁵ I^- : [Kr] 5s² 4d¹⁰ 5p⁶

Both have valence electrons with n = 5, but the extra electron added to a 5p orbital in I^- leads to increased electron repulsion that causes the electrons to spread out more, thus increasing the ionic radius of I^- and making it larger than the atomic radius of I.

When a neutral atom loses electrons to form a positively-charged cation, the ionic radius of the cation will be smaller than the atomic radius of the neutral atom.

In some cases, all the valence electrons from the neutral atom will be removed to create a cation. As a result, the cation will have a smaller radius than the neutral atom because the value of n for the valence electrons in the neutral atom will be greater than the value of n for any remaining electrons in the cation.

In other cases, not all the valence electrons may be removed to create a cation, so the value of n will be the same for the valence electrons in the neutral atom and the remaining valence electrons in the cation. In these cases, we can attribute the decreased radius of the cation to decreased electron repulsion when electrons are removed, which allows the remaining electrons to move closer together and decrease the radius.

Sample Exercise 8C:

(a) Which is larger, the atomic radius of K or the ionic radius of K⁺ ? Explain, including electron configurations in your answer.

(b) Which is larger, the atomic radius of Sn or the ionic radius of Sn²⁺ ? Explain, including electron configurations in your answer.

Solution:

(a) When the complete electron configurations of K and K⁺ are compared, we see that the value of n = 4 for the valence 4s electron in K is greater than the value of n for any remaining electrons in K⁺ :

K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ K⁺: 1s² 2s² 2p⁶ 3s² 3p⁶

Therefore, the atomic radius of K will be larger than the ionic radius of K⁺.

(b) When the electron configurations of Sn and Sn²⁺ are compared, we see that both have valence electrons with n = 5:

Sn: [Kr] 5s² 4d¹⁰ 5p² Sn^{2+ :} [Kr] 5s² 4d¹⁰

However, electron repulsion is decreased when electrons are removed to create Sn²⁺, so the remaining electrons in Sn²⁺ can move closer together to decrease the radius of Sn²⁺. Therefore, the atomic radius of Sn will be larger than the ionic radius of Sn^{2+ .}

Isoelectronic ions have the same number of electrons and identical electron configurations. In a group of isoelectronic ions, the ion with the greatest number of protons will have the greatest effective nuclear charge to draw the valence electrons closer to the nucleus, thus making its ionic radius the smallest in the group.

Sample Exercise 8D:

Rank Al³⁺, F^-, Na⁺, and O^2-from smallest to largest ionic radius. Explain, including electron configurations in your answer.

Solution:

All are isoelectronic with 10 electrons and a complete electron configuration of 1s² 2s² 2p⁶. More protons = greater effective nuclear charge = smaller ionic radius, so from smallest to largest we have:

Al³⁺ (13 p) < Na⁺ (11 p) < F^- (9 p) < O^2- (8 p)

Section 8-3: Ionization Energy and Electron Affinity

First ionization energy (I₁) is the amount of energy that must be absorbed to remove one valence electron from a neutral gaseous atom to create a cation with a 1+ charge. An element with a larger atomic radius will generally have a lower first ionization energy. As such, we can draw the following conclusion regarding the trend on the periodic table for first ionization energy:

An element located closer to the lower left corner of the periodic table will generally have a lower first ionization energy, while an element located closer to the upper right corner of the periodic table will generally have a higher first ionization energy.

Sample Exercise 8E:

Rank the elements fluorine, magnesium, and potassium from smallest to largest atomic radius and lowest to highest first ionization energy.

Solution:

Closest to the upper right corner of the periodic table = smallest atomic radius = highest first ionization energy. Therefore, from smallest to largest atomic radius we have F < Mg < K, and from lowest to highest first ionization energy we have the opposite order, K < Mg < F.

Note that there are numerous exceptions to the general first ionization energy trend stated above. For example, sulfur is to the right of phosphorus in the same row of the periodic table, yet sulfur has a lower first ionization energy than phosphorus. We can compare orbital diagrams showing the valence 3p electrons in phosphorus and sulfur to explain this exception:

phosphorus:

↑

3p³

sulfur:

↑↓

↑

3p⁴

In phosphorus, the electron removed is one of the unpaired 3p electrons. In sulfur, the electron removed is one of the paired 3p electrons. It is believed that electron repulsion between the paired 3p electrons in sulfur makes removal of one electron in the pair easier, which results in a lower first ionization energy for sulfur.

Second ionization energy (I₂) is the energy that must be absorbed to remove one electron from the 1+ cation that remains after the first valence electron is removed from a neutral atom. Third ionization energy (I₃) is the energy that must be absorbed to remove one electron from the 2+ cation that remains after an electron is removed from a 1+ cation. Subsequent ionization energies are labeled I₄, I₅, etc. Ionization energy increases each time an electron is removed from an atom or ion:

I₁ < I₂ < I₃ . . .

Ionization energy increases gradually until all the valence electrons with the largest value of n are removed, but then there will be an enormous jump in the ionization energy required to remove the next, non-valence, electron from an (n – 1) orbital. For example, since silicon has the complete electron configuration 1s² 2s² 2p⁶ 3s² 3p², the ionization energies will increase gradually from I₁ through I₄ as the two valence 3p and then the two valence 3s electrons are removed. However, I₅ will then be enormous compared to I₄ as the 5^th electron removed from silicon comes from a non-valence 2p orbital and, thus, requires a tremendous amount of energy for removal.

Sample Exercise 8F:

Explain the huge increase in ionization energy between I₃ and I₄ for boron.

Solution:

Boron has the complete electron configuration 1s² 2s² 2p¹. The ionization energies will increase gradually from I₁ through I₃ as the valence 2p and then two valence 2s electrons are removed, but then I₄ will be enormous compared to I₃ as the 4^th electron removed from boron comes from the non-valence 1s orbital and, thus, requires a huge amount of energy for removal.

First electron affinity (EA₁) is the energy associated with the process of adding one electron to a neutral gaseous atom to create an anion with a 1- charge. In general, it is more favorable to add an electron to an element with a smaller atomic radius, although there are numerous exceptions to this. For example, the addition of an electron to a noble gas is unfavorable.

Chapter 8 Practice Exercises and Review Quizzes:

8-1) Rank the elements beryllium, nitrogen, and strontium from smallest to largest atomic radius. Explain, including electron configurations in your answer.

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8-1) Be: [He] 2s² N: [He] 2s² 2p³ Sr: [Kr] 5s²

N is closest to the upper right corner of the periodic table, so N should have the smallest atomic radius. Sr is closest to the lower left corner of the periodic table, so Sr should have the largest atomic radius. The valence 5s electrons in Sr have n = 5 as compared to the valence 2s electrons in Be having n = 2, so Sr should have a larger atomic radius than Be. Although both N and Be have valence electrons with n = 2, N has 7 protons as compared to Be's 4 protons, so N will have a greater effective nuclear charge and smaller atomic radius than Be. Therefore, the three elements ranked from smallest to largest atomic radius would be N < Be < Sr.

8-2) Which is larger, the atomic radius of S or the ionic radius of S^2-? Explain, including electron configurations in your answer.

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8-2) S: [Ne] 3s² 3p⁴ S^{2- :} [Ne] 3s² 3p⁶

Both have valence electrons with n = 3, but the two extra electrons added to 3p orbitals in S^2- lead to increased electron repulsion that causes the electrons to spread out more, thus increasing the ionic radius of S^2- and making it larger than the atomic radius of S.

8-3) (a) Which is larger, the atomic radius of Sb or the ionic radius of Sb³⁺ ? Explain, including electron configurations in your answer

(b) Which is larger, the atomic radius of Al or the ionic radius of Al³⁺? Explain, including electron configurations in your answer.

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8-3) (a) Sb: [Kr] 5s² 4d¹⁰ 5p³ Sb³⁺ : [Kr] 5s² 4d¹⁰

Both have valence electrons with n = 5. However, electron repulsion is decreased when electrons are removed to create Sb³⁺, so the remaining electrons in Sb³⁺ can move closer together to decrease the radius of Sb³⁺. Therefore, the atomic radius of Sb will be larger than the ionic radius of Sb³⁺.

(b) When the complete electron configurations of Al and Al³⁺ are compared, we see that the value of n = 3 for the valence electrons in Al is greater than the value of n for any remaining electrons in Al^{3+ :}

Al: 1s² 2s² 2p⁶ 3s² 3p¹ Al^{3+ :} 1s² 2s² 2p⁶

Therefore, the atomic radius of Al will be larger than the ionic radius of Al³⁺.

8-4) Rank Ca²⁺, Cl^-, P^3-, and Sc³⁺from smallest to largest ionic radius. Explain, including electron configurations in your answer.

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8-4) All are isoelectronic with 18 electrons and a complete electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶. More protons = greater effective nuclear charge = smaller ionic radius, so from smallest to largest we have:

Sc³⁺ (21 p) < Ca²⁺(20 p) < Cl^- (17 p) < P^3- (15 p)

8-5) Rank the elements calcium, oxygen, and silicon from smallest to largest atomic radius and lowest to highest first ionization energy.

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8-5) Closest to the upper right corner of the periodic table = smallest atomic radius = highest first ionization energy. Therefore, from smallest to largest atomic radius we have O < Si < Ca, and from lowest to highest first ionization energy we have the opposite order, Ca < Si < O.

8-6) Explain the huge increase in ionization energy between I₅ and I₆ for phosphorus.

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8-6) Phosphorus has the complete electron configuration 1s² 2s² 2p⁶ 3s² 3p³. The ionization energies will increase gradually from I₁ through I₅ as the three valence 3p electrons and then two valence 3s electrons are removed, but then I₆ will be enormous compared to I₅ as the 6^th electron removed from phosphorus comes from the non-valence 2p orbital and, thus, requires a huge amount of energy for removal.

8-7) Explain the exception to the general first ionization energy trend where oxygen has a lower first ionization energy than nitrogen even though oxygen is further to the right in the same row of the periodic table.

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8-7) We can compare orbital diagrams showing the valence 2p electrons in nitrogen and oxygen to explain this exception:

nitrogen:

↑

2p³

oxygen:

↑↓

↑

2p⁴

In nitrogen, the electron removed is one of the unpaired 2p electrons. In oxygen, the electron removed is one of the paired 2p electrons. It is believed that electron repulsion between the paired 2p electrons in oxygen makes removal of one electron in the pair easier, which results in a lower first ionization energy for oxygen.

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