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Which periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 and \(8 ?(\mathbf{a})\) The number of valence electrons reaches a maximum at group 8. (b) The effective nuclear charge increases on moving left across each period. (c) The radii of the transition-metal elements reach a minimum for group \(8,\) and as the size of the atoms decreases it becomes easier to remove electrons.

Short Answer

Expert verified
The correct answer is (c): The radii of the transition-metal elements reach a minimum for group 8, and as the size of the atoms decreases, it becomes easier to remove electrons. This periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 and 8.

Step by step solution

01

Option (a): The number of valence electrons reaches a maximum at group 8.

For the transition-metal elements, the number of valence electrons increases from left to right across a period. However, the maximum oxidation state does not necessarily correlate directly with the number of valence electrons. Additionally, the maximum valence electrons for transition-metal elements are not reached in groups 7 and 8, so this statement does not explain the observed trend. #Step 2: Analyzing Option (b)#
02

Option (b): The effective nuclear charge increases on moving left across each period.

The effective nuclear charge is the net positive charge experienced by an electron in the outermost shell of an atom. While it is true that the effective nuclear charge increases from left to right across a period, this statement says it increases moving left, which is incorrect. Therefore, this option does not explain the observed trend in the maximum oxidation state. #Step 3: Analyzing Option (c)#
03

Option (c): The radii of the transition-metal elements reach a minimum for group 8, and as the size of the atoms decreases, it becomes easier to remove electrons.

As we move from left to right across a period, atomic radii generally decrease due to increases in the effective nuclear charge. Smaller atomic radii mean that outer electrons are more tightly bound to the nucleus, making it easier to remove them and form higher oxidation states. This statement aligns with the observed trend that the maximum oxidation states peak near groups 7 and 8. Thus, it is the most likely explanation for the observed trend. #Conclusion# The correct answer to the exercise is: (c) The radii of the transition-metal elements reach a minimum for group 8, and as the size of the atoms decreases, it becomes easier to remove electrons. This periodic trend is partially responsible for the observation that the maximum oxidation state of the transition-metal elements peaks near groups 7 and 8.

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Key Concepts

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

Transition Metals
Transition metals are elements found in the d-block of the periodic table, specifically in groups 3 through 12. They include well-known metals such as iron, copper, and gold. These metals are characterized by their ability to form multiple oxidation states, which are the charge states an atom can adopt when electrons are lost, gained, or shared in chemical reactions.

One key feature of transition metals is their partially filled d-orbitals. This allows them to form a wide variety of complex ions and compounds, contributing to their diverse chemistry and utility in industrial applications. This partially filled d-subshell also explains why they exhibit various oxidation states, leading to colorful compounds and reversible redox reactions.

Transition metals are pivotal in catalysis, including enzymatic processes. Such properties arise from their capacity to switch between different oxidation states easily, facilitating chemical transformations.
Oxidation States
Oxidation states are a crucial concept when discussing the chemistry of transition metals. An oxidation state reflects the effective charge of an atom after electrons have been transferred in a chemical reaction. For transition metals, exhibiting a range of oxidation states is common due to their electron configurations, specifically the involvement of d and s electrons.

The high variety of oxidation states in transition metals is due to the small energy difference between their s and d orbitals. This allows them to lose different numbers of d electrons to create various stable oxidation states. For example, manganese can have oxidation states from +2 to +7. This variability is significant because it underpins many chemical processes, including catalysis and biological functions.

Learning about oxidation states helps in predicting the types of compounds that transition metals can form and their reactivity during chemical reactions. It is also crucial in understanding the electron transfer processes involved in redox reactions.
Effective Nuclear Charge
The effective nuclear charge, often abbreviated as Z*, is the net positive charge experienced by an electron in an atom. It accounts for not only the positive charge of the protons in the nucleus but also the shielding effect caused by inner-shell electrons.

In transition metals, as you move from left to right across a period, the effective nuclear charge increases. This occurs because more protons are added to the nucleus without a significant increase in inner-shell electron shielding. Consequently, electrons in the outer shells are pulled closer to the nucleus more strongly. This impacts several atomic properties, such as atomic radii and electronegativity.

Understanding effective nuclear charge is crucial for explaining the contraction of atomic radii and the increase in ionization energies across the d-block, which in turn influences the oxidation states that transition metals can exhibit.
Atomic Radii
Atomic radii refer to the size of an atom, typically measured from the nucleus to the boundary of surrounding electron clouds. In transition metals, the trend in atomic radii is influenced heavily by effective nuclear charge.

As one moves across a period in the periodic table, the atomic radii of transition metals decrease. This is because the effective nuclear charge increases, pulling the outer electrons closer to the nucleus and thus reducing the overall size of the atom.

The decrease in atomic radii across transition metals has significant implications. It leads to stronger attraction of electrons by the nucleus, making it easier to remove them and achieve higher oxidation states. This trend explains why the oxidation states of transition metals peak around groups 7 and 8, such as manganese and iron, where the radii are minimized, lowering the energy required for electron removal and formation of stable high oxidation states.

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Most popular questions from this chapter

If the lobes of a given \(d\) -orbital point directly at the ligands, will an electron in that orbital have a higher or lower energy than an electron in a \(d\) -orbital whose lobes do not point directly at the ligands?

The lobes of which \(d\) orbitals point directly between the ligands in (a) octahedral geometry, (b) tetrahedral geometry?

Indicate the coordination number and the oxidation number of the metal for each of the following complexes: (a) \(\mathrm{K}_{2} \mathrm{PtCl}_{4}\) (b) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \mathrm{Br}_{2}\) (c) \(\mathrm{OsO}_{4}\) (d) \(\left[\mathrm{Mn}(\mathrm{en})_{3}\right]\left(\mathrm{NO}_{3}\right)_{2}\) (e) \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{3}\) (f) \(\left[\mathrm{Zn}(\mathrm{bipy})_{2}\right]\left(\mathrm{ClO}_{4}\right)_{2}\)

Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains \(\mathrm{Zn}^{2+}\) in its active site, is responsible for rapidly interconverting dissolved \(\mathrm{CO}_{2}\) and bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\). The zinc in carbonic anhydrase is tetrahedrally coordinated by three neutral nitrogencontaining groups and a water molecule. The coordinated water molecule has a \(\mathrm{p} K_{a}\) of \(7.5,\) which is crucial for the enzyme's activity. (a) Draw the active site geometry for the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase, just writing "N" for the three neutral nitrogen ligands from the protein. (b) Compare the \(\mathrm{p} K_{a}\) of carbonic anhydrase's active site with that of pure water; which species is more acidic? (c) When the coordinated water to the \(\mathrm{Zn}(\mathrm{II})\) center in carbonic anhydrase is deprotonated, what ligands are bound to the \(\mathrm{Zn}(\mathrm{II})\) center? Assume the three nitrogen ligands are unaffected. \((\mathbf{d})\) The \(\mathrm{p} K_{a}\) of \(\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is \(10 .\) Suggest an explanation for the difference between this \(\mathrm{p} K_{a}\) and that of carbonic anhydrase. (e) Would you expect carbonic anhydrase to have a deep color, like hemoglobin and other metal-ion-containing proteins do? Explain.

Draw the structure for \(\mathrm{Pt}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{NH}_{3}\right)_{2}\) and use it to answer the following questions: (a) What is the coordination number for platinum in this complex? (b) What is the coordination geometry? (c) What is the oxidation state of the platinum? (d) How many unpaired electrons are there? [Sections 23.2 and 23.6\(]\)

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