Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 3

Explain why atomic radii decrease very gradually from scandium to copper.

Short Answer

Expert verified
Atomic radii decrease gradually from Scandium to Copper primarily due to electron-electron repulsion in the d-orbital. This repulsion partially offsets the increased attraction between the increasing number of protons in the nucleus and the electrons, causing the atomic radius to decrease less significantly than expected.

Step by step solution

01

Understand Atomic Structure

The atomic radius is dependent on the number of energy levels (also known as shells) in an atom and the number of electrons in these shells. As you go across a period on the periodic table, the number of protons in the nucleus increases and consequently the positive charge of the nucleus increases. This leads to a greater attraction between the nucleus and the electrons which can pull the electrons closer to the nucleus, resulting in a smaller atomic radius. Scandium (Sc) and Copper (Cu) are in the same period of the periodic table indicating that they have the same number of energy levels.
02

Understand Specific Transition Metal Properties

Scandium (Sc) to Copper (Cu) are transition metals. These elements have an interesting property where additional electrons are added to an inner d-orbital. This forms a 'shield' of electrons that increases electron-electron repulsion. This repulsion counters the increased charge of the nucleus, not allowing the atomic radius to decrease as significantly as it does with other elements across a period.
03

Relate Atomic Structure and D-Orbital Shielding to Atomic Radii

The atomic radii decrease from Scandium to Copper but the decrease is smaller than expected. This is because with each successive element, the increasing number of protons in the nucleus is partially offset by the increasing number of electrons in the d-orbital. This d-orbital 'shield' results in less contraction of the electron cloud than expected, and the atomic radius decreases only gradually.

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.

Start your free trial

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

Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fluoride is added, a green precipitate is formed. When aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what is happening in these two cases.

Explain the following facts: (a) Copper and iron have several oxidation states, whereas zinc has only one. (b) Copper and iron form colored ions, whereas zinc does not.

A solution made by dissolving \(0.875 \mathrm{~g}\) of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{3}\) in \(25.0 \mathrm{~g}\) of water freezes at \(-0.56^{\circ} \mathrm{C}\) Calculate the number of moles of ions produced when 1 mole of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{3}\) is dissolved in water and suggest a structure for the complex ion present in this compound.

A student in 1895 prepared three coordination compounds containing chromium, with the following properties: $$ \begin{array}{llc} \hline & & \begin{array}{l} \mathrm{Cl}^{-} \text {Ions in } \\ \text { Solution per } \\ \text { Formula Unit } \end{array} \\ \text { Formula } & \text { Color } & 3 \\ \hline \text { (a) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Violet } & 3 \\ \text { (b) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Light green } & 2 \\ \text { (c) } \mathrm{CrCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O} & \text { Dark green } & 1 \\ \hline \end{array} $$ Write modern formulas for these compounds and suggest a method for confirming the number of \(\mathrm{Cl}^{-}\) ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates and \(\mathrm{Cr}\) has a coordination number of 6 in all the compounds.

Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)dichlorochromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammin eaquachlorocobalt(III) chloride.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation
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