Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Transition metals despite having higher values of standard reduction potentials, are poor reducing agents. This is due to (a) low heat of hydration (b) high ionization energies (c) low ionization energies (d) high enthalpy of vapourization

Short Answer

Expert verified
(b) high ionization energies.

Step by step solution

01

Understanding the Role of Standard Reduction Potentials

Transition metals generally have higher standard reduction potentials, which means they are more likely to gain electrons and be reduced rather than lose electrons and oxidize another substance. Therefore, high reduction potential typically indicates weaker reducing ability, as reducing agents donate electrons.
02

Evaluating Reducing Agent Properties

A strong reducing agent tends to have a low ionization energy because it can easily lose electrons. Transition metals, although having a high standard reduction potential, could still act as poor reducing agents due to having high ionization energies. This high ionization energy makes it difficult for them to lose electrons.
03

Analyzing the Given Options

Analyzing the given options, "(a) low heat of hydration," "(b) high ionization energies," "(c) low ionization energies," and "(d) high enthalpy of vaporization," the one that relates to the poor ability to lose electrons and thus to be a reducing agent is (b) high ionization energies.
04

Drawing the Conclusion

Since high ionization energies inhibit the metal's ability to lose electrons, leading to less effective reducing agents, the correct explanation for why transition metals are poor reducing agents in spite of having high standard reduction potentials is (b) high ionization energies.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Standard Reduction Potential
When discussing transition metals, it is crucial to understand the concept of standard reduction potentials. This value measures the tendency of a chemical species to acquire electrons and be reduced. Metals with high standard reduction potentials, like transition metals, are more inclined to gain electrons. As a result, they are more likely to undergo reduction themselves rather than donating electrons to other substances and causing their oxidation.
This behavior might initially seem confusing. Typically, you would expect a material with a high reduction potential, indicating readiness to gain electrons, to effectively act as a reducing agent. However, the opposite is true. A high standard reduction potential signifies a strong likelihood of reduction, which typically makes them less effective at being reducing agents.
Understanding this balance helps clarify why transition metals, despite their high standard reduction potentials, are not strong reducing agents.
Reducing Agents
Reducing agents are substances that promote the reduction of another species by donating electrons to it. To do so effectively, a reducing agent must easily part with its electrons. This property is often associated with low ionization energies. When a reducing agent has a low ionization energy, it requires little energy to remove electrons, facilitating this electron donation process.
Transition metals generally struggle in this capacity due to their inclination to hold onto their electrons tightly. Despite their potential to reduce other substances, their high ionization energies make it difficult for them to give away electrons freely.
This electron retention behavior is the reason why transition metals don't perform as well as other metals with lower ionization energies. Therefore, their reducing power is limited compared to metals that find it easy to lose electrons due to lower ionization energy.
Ionization Energy
Ionization energy is a fundamental concept when understanding the properties of elements, especially transition metals. It refers to the amount of energy necessary to remove an electron from a gaseous atom or ion. Higher ionization energy means that more energy is required to remove an electron from an atom, indicating a strong hold on its electrons.
Transition metals often exhibit high ionization energies. This makes them less willing to lose electrons, hence poorer at acting as reducing agents. While they may have desirable properties like a high melting point or tensile strength, their electricity retention capability makes electron donation challenging.
In essence, their high ionization energies, while contributing to some notable physical properties, inhibit their ability as reducing agents since they retain their electrons more strongly than other groups of metals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The lanthanide contraction is responsible for the fact that (a) \(\mathrm{Zr}\) and Y have about the same radius (b) \(\mathrm{Zr}\) and \(\mathrm{Zn}\) have the same oxidation state (c) \(\mathrm{Zr}\) and Hf have about the same atomic radius (d) \(\mathrm{Zr}\) and \(\mathrm{Nb}\) have similar oxidation state

Which of the following compounds are coloured due to charge transfer spectra? (a) \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{CuSO}_{4}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{KMnO}_{4}\)

The d-orbitals participating in hybridization of central metal atom may be from the outermost shell or the penultimate shell. This depends on the nature of metal and the nature of ligand. The complexes involving the inner \(\mathrm{d}\) level (inner orbital complexes) result when the ligand is a powerful or strong ligand resulting in diamagnetic or low spin complexes. A weak ligand usually results in the formation of outer orbital complex or high spin complex. The number of unpaired electrons present in \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) and \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) are, respectively (a) 0,0 (b) 0,4 (c) 1,2 (d) 2,4

Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being (a) 4 f-orbitals more diffused than the 5 f-orbitals (b) Lesser energy difference between \(5 \mathrm{f}\) and 6 d than between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\) orbitals (c) More energy difference between \(5 \mathrm{f}\) and \(6 \mathrm{~d}\) than between \(4 \mathrm{f}\) and \(5 \mathrm{~d}\) orbitals (d) More reactive nature of the actinoids than the lanthanoids.

In which one of the following transition metal complexes, does the metal exhibit zero oxidation state? (a) \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{X}_{3}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) (d) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{SO}_{4}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free