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Elements in group 7A in the periodic table are called the halogens; elements in group \(6 \mathrm{~A}\) are called the chalcogens. (a) What is the most common oxidation state of the chalcogens compared to the halogens? (b) For each of the following periodic properties, state whether the halogens or the chalcogens have larger values: atomic radii, ionic radii of the most common oxidation state, first ionization energy, second ionization energy.

Short Answer

Expert verified
(a) The most common oxidation state of chalcogens is -2, while for halogens, it is -1. (b) Chalcogens have larger atomic radii and larger ionic radii of the most common oxidation state than halogens. Halogens have higher first and second ionization energies than chalcogens.

Step by step solution

01

Determine the most common oxidation states of chalcogens and halogens.

The most common oxidation states of an element can be generally derived from the group number in the periodic table. For chalcogens (group 6A), the most common oxidation state is -2, as they have 6 valence electrons and require 2 more electrons to achieve a full outer shell. For halogens (group 7A), the most common oxidation state is -1, as they have 7 valence electrons and require 1 more electron to achieve a full outer shell.
02

Compare atomic radii of chalcogens and halogens.

Atomic radii generally decrease across a period from left to right due to an increase in the effective nuclear charge, which means that as we move from left to right in a period, the atomic size decreases. Since chalcogens are to the left of halogens on the periodic table, chalcogens have a larger atomic radius than halogens.
03

Compare ionic radii of the most common oxidation state of chalcogens and halogens.

We know that chalcogens achieve a -2 oxidation state and halogens achieve a -1 oxidation state. When chalcogens gain 2 electrons, they will have a larger ionic radius than when halogens gain 1 electron. The ionic size increases as more electrons are added to the ion due to increased electron-electron repulsion. Thus, chalcogens have a larger ionic radius than halogens for their most common oxidation states.
04

Compare first ionization energies of chalcogens and halogens.

Ionization energy increases as we move from left to right across a period in the periodic table. This is because the atomic size decreases across a period, and it gets difficult to remove an electron from the compact electron cloud. Since chalcogens are to the left of halogens on the periodic table, they have lower first ionization energies compared to halogens.
05

Compare second ionization energies of chalcogens and halogens.

Considering the trend observed in the first ionization energies, we can apply the same concept to second ionization energies. As we move from left to right across a period in the periodic table, ionization energies increase due to a decrease in atomic size. Therefore, second ionization energies of chalcogens are lower than those of halogens, as chalcogens are to the left of halogens on the periodic table.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States of Chalcogens and Halogens
Understanding the oxidation states of chalcogens and halogens is fundamental when exploring chemical reactions involving these elements. The chalcogens, found in group 6A of the periodic table, most commonly exhibit an oxidation state of -2. This is because they have six valence electrons and generally gain two electrons to fill their outermost electron shell, achieving a stable noble gas configuration.

In contrast, halogens, which are positioned in group 7A, typically have an oxidation state of -1. This occurs as they have seven valence electrons, and only one more electron is needed to attain a full outer shell. These specific oxidation states are crucial when predicting the types of compounds that chalcogens and halogens will form. For instance, Oxygen (O), a chalcogen, often forms oxides with a -2 charge, while Chlorine (Cl), a halogen, typically forms chloride compounds with a -1 charge.
Atomic and Ionic Radii Comparison
The size of an atom or ion plays a critical role in its chemical properties and is compared in terms of atomic and ionic radii. Atomic radii refer to the distance from the nucleus to the outer perimeter of the electron cloud. Across a period in the periodic table, atomic radii decrease due to an increase in the effective nuclear charge, which pulls electrons closer to the nucleus.

The comparison between chalcogens and halogens reveals that chalcogens have larger atomic radii as they are situated to the left of halogens in a given period. When these elements achieve their most common oxidation states, chalcogens form -2 ions, and halogens form -1 ions. Consequently, chalcogens typically have larger ionic radii due to the addition of more electrons, which increases electron-electron repulsion, leading to a larger ionic size.
First and Second Ionization Energies
Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. The first ionization energy is specifically the energy needed to remove the first electron, revealing an element's reactivity and trend in electron affinity. As one moves from left to right across a period, first ionization energies increase because smaller atomic sizes make it harder to remove an electron from the denser electron cloud.

Chalcogens have lower first ionization energies than halogens due to their larger atomic size and lesser effective nuclear charge. Similarly, the second ionization energy, the energy needed to remove a second electron, also follows this trend. Chalcogens, being further to the left on the periodic table, exhibit lower second ionization energies compared to halogens.
Periodic Trends in the Periodic Table
The periodic table is not just a chart of elements; it showcases trends that help us predict various chemical behaviors and properties of elements. Periodic trends such as atomic radius, ionization energies, and electronegativity are intrinsically related to an element's position on the table.

Horizontally, from left to right, one notable trend is the decrease in atomic radii and the increase in ionization energies due to the stronger nuclear charge attracting electrons more efficiently. Vertically, elements show generally increasing atomic radii and decreasing ionization energies down a group, as additional electron shells increase the distance between the outer electrons and the nucleus. These trends provide a systematic and predictable framework for understanding the chemical properties and reactivities of different elements within the table.

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Most popular questions from this chapter

(a) Why does the quantum mechanical description of manyelectron atoms make it difficult to define a precise atomic radius? (b) When nonbonded atoms come up against one another, what determines how closely the nuclear centers can approach?

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of \(\mathrm{Xe}\). Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

Consider the isoelectronic ions \(\mathrm{F}^{-}\)and \(\mathrm{Na}^{+}\). (a) Which ion is smaller? (b) Using Equation \(7.1\) and assuming that core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant, \(S\), calculate \(Z_{\text {eff }}\) for the \(2 p\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S\). (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

If the electron affinity for an element is a negative number, does it mean that the anion of the element is more stable than the neutral atom? Explain.

Zinc in its 2+ oxidation state is an essential metal ion for life. \(\mathrm{Zn}^{2+}\) is found bound to many proteins that are involved in biological processes, but unfortunately \(\mathrm{Zn}^{2+}\) is hard to detect by common chemical methods. Therefore, scientists who are interested in studying \(\mathrm{Zn}^{2+}\)-containing proteins frequently substitute \(\mathrm{Cd}^{2+}\) for \(\mathrm{Zn}^{2+}\), since \(\mathrm{Cd}^{2+}\) is easier to detect. (a) On the basis of the properties of the elements and ions discussed in this chapter and their positions in the periodic table, describe the pros and cons of using \(\mathrm{Cd}^{2+}\) as a \(\mathrm{Zn}^{2+}\) substitute. (b) Proteins that speed up (catalyze) chemical reactions are called enzymes. Many enzymes are required for proper metabolic reactions in the body. One problem with using \(\mathrm{Cd}^{2+}\) to replace \(\mathrm{Zn}^{2+}\) in enzymes is that \(\mathrm{Cd}^{2+}\) substitution can decrease or even eliminate enzymatic activity. Can you suggest a different metal ion that might replace \(\mathrm{Zn}^{2+}\) in enzymes instead of \(\mathrm{Cd}^{2+}\) ? Justify your answer.

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