Question

\(\mathrm{Cu}^{+}\) has an ionic radius of \(77 \mathrm{pm}\), but \(\mathrm{Cu}^{2+}\) has an ionic radius of \(73 \mathrm{pm}\). Explain.

Short answer

Cu²⁺ is smaller than Cu⁺ due to greater effective nuclear charge on fewer electrons.

Step-by-step solution

Understanding Ionic Radius

The ionic radius is a measure of the size of an ion in a crystal lattice. Generally, when an ion loses electrons (oxidized), it becomes smaller, and when it gains electrons (reduced), it becomes larger due to increased electron-electron repulsion in the shell.

Analyzing Copper Ions

The copper ions given are \(\mathrm{Cu}^{+}\) and \(\mathrm{Cu}^{2+}\). The \(\mathrm{Cu}^{+}\) ion has lost one electron compared to its neutral atom, and \(\mathrm{Cu}^{2+}\) has lost two electrons.

Increased Effective Nuclear Charge

When an electron is lost, the effective nuclear charge experienced by the remaining electrons increases. This is because the same number of protons is acting on fewer electrons, pulling them closer to the nucleus.

Comparison of Radii

In \(\mathrm{Cu}^{2+}\), the effective nuclear charge is greater than in \(\mathrm{Cu}^{+}\), because it has lost more electrons. This leads to \(\mathrm{Cu}^{2+}\) having a smaller ionic radius (73 pm) compared to \(\mathrm{Cu}^{+}\) (77 pm) due to the stronger pull of the protons on the electrons.

Key concepts

Copper Ions

Copper is a fascinating element because it can exist in different ionic forms. When we talk about copper ions, we usually refer to two common types:

• \( ext{Cu}^+\) or copper(I) ion
• \( ext{Cu}^{2+}\) or copper(II) ion

These ions form when the neutral copper atom, which has the chemical symbol Cu, loses electrons. The neutral copper atom normally has 29 electrons.
The copper(I) ion forms by losing one electron, leaving it with 28 electrons. On the other hand, the copper(II) ion forms by losing two electrons, thus it has 27 electrons.
This loss of electrons affects the size of the ion, which we term as the ionic radius.

Effective Nuclear Charge

The concept of effective nuclear charge is crucial for understanding changes in ionic size. In an atom, electrons are attracted to the positively charged nucleus, where protons reside.
When electrons are removed from an atom to form positive ions, the number of protons stays the same, while the electron count decreases. This means fewer electrons are sharing the attractive force from the same number of protons.
The term 'effective nuclear charge' refers to the net positive charge experienced by electrons in an ion. With electron removal, the effective nuclear charge on the remaining electrons increases.
This results from a stronger pull on the electrons by the protons, making the ion smaller. In the case of copper ions,

• \( ext{Cu}^+\) has a certain effective nuclear charge
• \( ext{Cu}^{2+}\) experiences a higher effective nuclear charge

This is why \( ext{Cu}^{2+}\) is smaller than \( ext{Cu}^+\), despite both ions being positive.

Electron Configuration

Electron configuration helps us understand the arrangement of electrons around the nucleus of an atom or ion. Copper's electron configuration in its neutral state is \([ ext{Ar} ] 3d^{10} 4s^1\).
When copper forms ions, it loses electrons starting from the outermost shell. For \( ext{Cu}^+\), one electron is lost from the 4s orbital, resulting in the electron configuration \([ ext{Ar} ] 3d^{10}\).

• This configuration is relatively stable due to the filled 3d subshell.

For \( ext{Cu}^{2+}\), another electron is removed, but this time from the 3d orbital, giving it the electron configuration \([ ext{Ar} ] 3d^9\).

• The removal of an additional electron increases the effective nuclear charge.

Understanding electron configuration is essential because it influences how atoms interact, bond, and change size as they lose electrons to form ions.