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

What is the lanthanide contraction? How does the lanthanide contraction affect the properties of the \(4 d\) and \(5 d\) transition metals?

Short Answer

Expert verified
The lanthanide contraction is a phenomenon where the atomic radii of lanthanide elements decrease steadily as the atomic number increases, due to the poor shielding of the nuclear charge by the 4f orbitals. This affects the properties of the 4d and 5d transition metals, making their atomic radii and chemical properties more similar than expected, such as electronegativities, ionization energies, and melting points. The lanthanide contraction also leads to the formation of complex and stable metal ions and influences the catalytic properties of transition metals, with 5d transition metals generally having more favorable catalytic properties than their 4d counterparts.

Step by step solution

01

Define Lanthanide Contraction

The lanthanide contraction is a phenomenon where the atomic radii of lanthanide elements (elements in the f-block of the periodic table) decrease steadily as the atomic number increases. This occurs due to the poor shielding of the nuclear charge by the 4f orbitals, causing the outer electrons to be attracted more strongly to the nucleus, resulting in a smaller atomic radius.
02

How Lanthanide Contraction Affects 4d and 5d Transition Metals

The lanthanide contraction has considerable effects on the properties of the 4d and 5d transition metal elements found in the d-block of the periodic table: 1. Due to the lanthanide contraction, the atomic radii of the 5d transition metals are closer to the atomic radii of their 4d counterparts than would be expected from their positions in the periodic table. This results in similar chemical and physical properties, such as similar electronegativities, ionization energies, and melting points. 2. The similar properties of 4d and 5d transition metal elements due to the lanthanide contraction lead to the formation of complex and stable metal ions. This is because the smaller size of the 5d metal ions compared to the expected size makes them more dense and capable of forming stronger bonds with ligands. 3. The lanthanide contraction also affects the catalytic properties of transition metals. Both 4d and 5d transition metals are used as catalysts in a variety of chemical reactions. However, the 5d transition metals tend to have more favorable catalytic properties compared to their 4d counterparts due to their higher density and ability to form stronger bonds with reactants. In summary, the lanthanide contraction significantly impacts the properties of the 4d and 5d transition metal elements, leading to similar atomic radii, chemical properties, and catalytic behaviors.

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!

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

Use the data in Appendix 4 for the following. a. Calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction $$3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)$$ that occurs in a blast furnace. b. Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature. Calculate \(\Delta G^{\circ}\) at \(800 .{ }^{\circ} \mathrm{C}\) for this reaction.

You isolate a compound with the formula \(\mathrm{PtCl}_{4} \cdot 2 \mathrm{KCl}\). From electrical conductance tests of an aqueous solution of the compound, you find that three ions per formula unit are present, and you also notice that addition of \(\mathrm{AgNO}_{3}\) does not cause a precipitate. Give the formula for this compound that shows the complex ion present. Explain your findings. Name this compound.

A certain first-row transition metal ion forms many different colored solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in wa ter, the colors of the solutions are red, yellow, green, and blue Further experiments reveal that two of the complex ions are para magnetic with four unpaired electrons and the other two are dia magnetic. What can be deduced from this information about th four coordination compounds?

Acetylacetone, abbreviated acacH, is a bidentate ligand. It loses a proton and coordinates as acac \(^{-}\), as shown below, where \(\mathrm{M}\) is a transition metal: Which of the following complexes are optically active: cis\(\mathrm{Cr}(\mathrm{acac})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\), trans \(-\mathrm{Cr}(\mathrm{acac})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\), and \(\mathrm{Cr}(\mathrm{acac})_{3} ?\)

The compound cisplatin, \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: $$\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)$$ Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

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