Question

Write electron configurations forthe following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Cr}^{3+}\), (b) \(\mathrm{N}^{3-}\), (c) \(\mathrm{Sc}^{3+}\) (d) \(\mathrm{Cu}^{2+}\), (e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{Au}^{+}\).

Short answer

The electron configurations for the given ions are: (a) \(\mathrm{Cr}^{3+}\) - [Ar]3d\(^3\) (b) \(\mathrm{N}^{3-}\) - [Ne] (c) \(\mathrm{Sc}^{3+}\) - [Ar] (d) \(\mathrm{Cu}^{2+}\) - [Ar]3d\(^9\) (e) \(\mathrm{Tl}^{+}\) - [Xe]4f\(^{14}\)5d\(^{10}\)6s\(^2\)6p\(^5\) (f) \(\mathrm{Au}^{+}\) - [Xe]4f\(^{14}\)5d\(^{10}\)6s\(^1\) The ions with noble-gas configurations are: \(\mathrm{N}^{3-}\) and \(\mathrm{Sc}^{3+}\).

Step-by-step solution

Identify atomic numbers and electron counts

Based on the given ions, look at the periodic table to find each ion's atomic number and the number of electrons in the ion. (a) \(\mathrm{Cr}^{3+}\) - Chromium has an atomic number of 24, so it originally has 24 electrons. Since the ion is Cr\(^{3+}\), it has lost 3 electrons. Therefore, the ion has 21 electrons. (b) \(\mathrm{N}^{3-}\) - Nitrogen has an atomic number of 7, so it originally has 7 electrons. Since the ion is N\(^{3-}\), it has gained 3 electrons. Therefore, the ion has 10 electrons. (c) \(\mathrm{Sc}^{3+}\) - Scandium has an atomic number of 21, and the Sc\(^{3+}\) ion has lost 3 electrons. Therefore, the ion has 18 electrons. (d) \(\mathrm{Cu}^{2+}\) - Copper has an atomic number of 29, and the Cu\(^{2+}\) ion has lost 2 electrons. Therefore, the ion has 27 electrons. (e) \(\mathrm{Tl}^{+}\) - Thallium has an atomic number of 81, and the Tl\(^{+}\) ion has lost 1 electron. Therefore, the ion has 80 electrons. (f) \(\mathrm{Au}^{+}\) - Gold has an atomic number of 79, and the Au\(^{+}\) ion has lost 1 electron. Therefore, the ion has 78 electrons. ##Step 2: Write the electron configurations for each ion##

Write electron configurations

Using the number of electrons for each ion, write the electron configurations: (a) \(\mathrm{Cr}^{3+}\) - [Ar]3d\(^3\) (b) \(\mathrm{N}^{3-}\) - [Ne] (c) \(\mathrm{Sc}^{3+}\) - [Ar] (d) \(\mathrm{Cu}^{2+}\) - [Ar]3d\(^9\) (e) \(\mathrm{Tl}^{+}\) - [Xe]4f\(^{14}\)5d\(^{10}\)6s\(^2\)6p\(^5\) (f) \(\mathrm{Au}^{+}\) - [Xe]4f\(^{14}\)5d\(^{10}\)6s\(^1\) ##Step 3: Identify noble-gas configurations##

Determine noble-gas configurations

Observe the electron configurations we obtained in step 2 and identify ions with noble-gas configurations: (a) \(\mathrm{Cr}^{3+}\) - Not a noble-gas configuration (b) \(\mathrm{N}^{3-}\) - Noble-gas configuration ([Ne]) (c) \(\mathrm{Sc}^{3+}\) - Noble-gas configuration ([Ar]) (d) \(\mathrm{Cu}^{2+}\) - Not a noble-gas configuration (e) \(\mathrm{Tl}^{+}\) - Not a noble-gas configuration (f) \(\mathrm{Au}^{+}\) - Not a noble-gas configuration The ions with noble-gas configurations are: \(\mathrm{N}^{3-}\) and \(\mathrm{Sc}^{3+}\).

Key concepts

Noble-Gas Configurations

Understanding noble-gas configurations is central to grasping electron distributions in atoms and ions. Noble gases are elements that are particularly stable due to having a full valence shell, which is why other elements aspire to mimic this configuration through gaining or losing electrons. This is often referred to as the 'octet rule' whereby atoms and ions strive to have eight electrons in their valence shell, emulating the electron configuration of a noble gas.

For example, when looking at the ion \(\mathrm{N}^{3-}\), it gains three electrons to reach a total of ten electrons. This aligns with the neon (Ne) noble-gas configuration, which is why we say \(\mathrm{N}^{3-}\) has a noble-gas configuration. Similarly, \(\mathrm{Sc}^{3+}\) loses three electrons to hold a total of eighteen, corresponding to the electron configuration of argon (Ar), another noble gas.

Ion Electron Count

To decipher the electron count of ions, one must understand the process of ionization. Ions are formed when atoms gain or lose electrons to achieve greater chemical stability; cations are positively charged ions that result from losing electrons, while anions are negatively charged due to gaining electrons.

The step-by-step solution illustrates this: \(\mathrm{Cr}^{3+}\) loses three electrons, leaving it with twenty-one instead of its original twenty-four. The practice involves balancing the charge indicated on the ion with the atomic number to accurately determine the electron count. As showcased, determining the ion's electron count is pivotal for writing electron configurations and discerning whether an ion has a noble-gas configuration.

Periodic Table Atomic Numbers

The periodic table is an invaluable tool for understanding electron configurations and the properties of elements. The atomic number of an element, located above the elemental symbol on the periodic table, indicates the number of protons in an element's nucleus when it's in its neutral state. It also tells us how many electrons the atom has before ionization occurs, as the number of electrons equals the number of protons in a neutral atom.

In the exercise, the atomic numbers give us the starting point to work out how the electron count changes as atoms form ions. For instance, copper (Cu) has the atomic number 29, but as a \(\mathrm{Cu}^{2+}\) ion, it loses two electrons, resulting in the ion having twenty-seven electrons, rather than twenty-nine. Understanding atomic numbers is crucial to solving electron configuration problems and to a general understanding of chemical behavior.