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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.

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Key Concepts

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

Periodic Table
The periodic table is a comprehensive chart organizing elements based on similar properties and increasing atomic numbers. This layout helps us understand trends such as atomic radii, ionization energy, and electronegativity. During the traversal from left to right across a period, the atomic number increases because more protons are added to the nucleus. This increasing positive nuclear charge pulls the electron cloud closer, generally causing a decrease in atomic size.

However, this sleek trend isn't linear due to additional effects from electron configurations, especially in transition metals. The periodic table is not just a mere collection of rows and columns; it speaks the language of chemistry by displaying the interplay of nuclear forces, electron configurations, and other atomic properties.
Transition Metals
Transition metals, recognized as the "d-block" of the periodic table, have unique properties owing to their electron structure. Typically recognizable by their ability to form various oxidation states, these metals have partially filled d-orbitals, leading to complex behaviors.

Transition metals like scandium to copper cover the stretch within one period where d-orbitals are gradually filled. Adding electrons to these d-orbitals in the atom doesn't drastically change atomic size each time, unlike in the main group elements where adding a proton and an electron per element typically shrinks the atom more noticeably.
  • Transition metals have high melting points and densities.
  • They often demonstrate magnetic and catalytic properties.
  • Their electron transitions within d-orbitals allow for unique coloration in compounds.
This makes transition metals fascinating subjects of study, expanding the story and utility of the periodic table into fields like materials science and catalysis.
D-Orbital Shielding
D-orbital shielding refers to how electrons in these orbitals affect the nuclear charge experienced by outer electrons. When transition metals add electrons to the d-orbitals, these electrons don't efficiently shield the increasing positive charge of the nucleus for subsequent electrons.

This means that as more protons and electrons are added (like moving from scandium to copper), the effective nuclear charge doesn't increase as much as it potentially could due to this d-electron "shielding" effect. This factor is key in understanding why the atomic radii of transition metals don't decrease as sharply across a period: the electron cloud shrinks less because d-orbital electrons partially counteract the pull of the increasingly positive nucleus.
  • The gradual filling of d-orbitals leads to additional electron-electron repulsions.
  • These inner d-electrons act like a buffer, reducing nucleus-electron attractions.
Understanding this concept clarifies various properties and behaviors of the transition metals in the periodic table.

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Most popular questions from this chapter

Draw qualitative diagrams for the crystal field splittings in (a) a linear complex ion \(\mathrm{ML}_{2}\), (b) a trigonalplanar complex ion \(\mathrm{ML}_{3}\), (c) a trigonal-bipyramidal complex ion ML_{5. }

Define the following terms: coordination compound, ligand, donor atom, coordination number, and chelating agent.

What are the systematic names for the following ion and compounds? (a) \(\left[c i s-\operatorname{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{3}\) (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}\)

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 student has prepared a cobalt complex that has one of the following three structures: \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]\) \(\mathrm{Cl}_{3},\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2},\) or \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl} .\) Explain how the student would distinguish between these possibilities by an electrical conductance experiment. At the student's disposal are three strong electrolytes \(-\mathrm{NaCl}, \mathrm{MgCl}_{2},\) and \(\mathrm{FeCl}_{3}-\) which may be used for comparison purposes.
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