Chapter 23, Problem 1SAQ

What is the electron configuration of the Cu+ ion?

[Ar]4s03d10

[Ar]4s23d9

[Ar]4s13d9

[Ar]4s23d8

Interpretation Introduction

Interpretation:

To find out the electronic configuration of Cu+ ions.

Concept introduction:

The distribution of electrons among the various orbitals in the atom is said to be the electronic configuration which is done by following three rules:

- Aufbau Principle

- Pauli exclusion principle

- Hund's rule

For example: The configuration is written as ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{4}}{\text{4s}}^{\text{2}}$ which is further represented as ${\text{[Ar]3d}}^{\text{4}}{\text{4s}}^{\text{2}}$

Correct answer: Option (a) ${\text{[Ar]4s}}^0{\text{3d}}^{10}$ is the correct answer.

Explanation of Solution

Reason for correct answer: Argon is a noble gas that falls before copper in periodic table and it is represented in square bracket [ $\text{Ar}$ ].

$\text{Cu}$ is placed in the fourth period so orbitals are written as $\text{[Ar]4s3d}$

$\text{Cu}$ has 11 more electrons than argon. So, it is filled like ${\text{[Ar]4s}}^2{\text{3d}}^9$

Due to half-filled and completely filled orbitals the compound is stable, so the correct configuration of copper would be ${\text{[Ar]4s}}^1{\text{3d}}^{10}$

${\text{Cu}}^{\text{+}}$ has lost one electron which is relative to $\text{Cu}$ so a single electron is removed from s orbital. Thus the final electronic configuration will be ${\text{[Ar]4s}}^0{\text{3d}}^{10}$

Thus, option (a) ${\text{[Ar]4s}}^0{\text{3d}}^{10}$ is the correct answer.