The Electron Configuration: Ions: Videos & Practice Problems

Cation Electron Configuration Example

To determine the condensed electron configuration for the titanium 3 ion (Ti³⁺), we start by identifying the electron configuration of the neutral titanium atom. Titanium has an atomic number of 22, which means it has 22 electrons. The condensed electron configuration for neutral titanium can be expressed as follows:

Ti: [Ar] 4s² 3d²

In this configuration, [Ar] represents the electron configuration of argon, the nearest noble gas preceding titanium. The next step involves adjusting this configuration to reflect the loss of electrons corresponding to the +3 charge of the titanium ion.

According to the Aufbau principle, when removing electrons, we start with the highest energy level. For titanium, the highest energy level is 4s, followed by 3d. To form the Ti³⁺ ion, we need to remove a total of three electrons:

1. Remove the two electrons from the 4s subshell:

4s² becomes empty.

2. Remove one electron from the 3d subshell:

3d² becomes 3d¹.

Thus, the final condensed electron configuration for the titanium 3 ion is:

Ti³⁺: [Ar] 3d¹

This configuration indicates that the titanium 3 ion has lost three electrons, resulting in a stable electron arrangement.

Anion Electron Configuration

Anions are negatively charged ions formed when an atom accepts one or more electrons. This process involves adding electrons to the available orbitals of the atom. For nonmetals, the naming convention for anions involves retaining the base name of the element while changing the ending to "ide." For example, when oxygen gains electrons, it becomes oxide (O^2-), and when fluorine gains an electron, it becomes fluoride (F^-). Understanding this naming convention is crucial when determining the electron configurations of anions, as it helps identify the specific anion based on its elemental origin.

Anion Electron Configuration Example

The nitride ion, represented as N^3-, is derived from nitrogen, which is located in group 3A of the periodic table. The negative charge of -3 indicates that the nitride ion has gained three additional electrons. To determine its full electron configuration, we first need to establish the configuration for neutral nitrogen, which has an atomic number of 7. The electron configuration for neutral nitrogen is:

1s² 2s² 2p³

Next, we account for the three extra electrons gained by the nitride ion. The 1s and 2s orbitals are already filled to their maximum capacity of 2 electrons each, so they remain unchanged:

1s² 2s²

The 2p orbital can hold up to 6 electrons, and since it currently has 3, we can add the 3 additional electrons to fill it completely. Thus, the 2p configuration becomes:

2p⁶

Combining these, the full electron configuration for the nitride ion is:

1s² 2s² 2p⁶

This configuration indicates that the nitride ion now has the same electron configuration as the noble gas neon, which is a stable configuration. Therefore, while the full electron configuration is the most complete answer, it is also valid to note that the nitride ion shares its electron configuration with neon, represented in shorthand as:

[Ne]

What is the full electron configuration of the selenide ion?

• A. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴
• B. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
• C. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁴ 5s²
• D. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁸ 4p⁴ 5s² 5p²

Answer: (B)

Determine the electron configuration for the Cl⁺ ion.

• A. [Ne] 3s¹ 3p⁵
• B. [Ne] 3s² 3p⁶
• C. [Ne] 3s² 3p⁵
• D. [Ne] 3s² 3p⁴

Answer: (D)

Determine the electron configuration and electron orbital diagram for the silver ion.

• A. [Kr] 4d¹⁰
• B. [Kr] 4d⁸ 5p²
• C. [Kr] 4d⁸
• D. [Kr] 4d⁸ 5s²

Answer: (A)

Here's what students ask on this topic:

How do you determine from which orbital electrons are removed when forming cations?

When forming cations, electrons are removed from the orbitals with the highest principal quantum number, $n$, which corresponds to the shell number or energy level. For example, electrons are removed from the 3p orbital before the 3s orbital because both have $n=3$, but 3p has higher energy. If orbitals share the same $n$ value, electrons are removed first from the orbital listed last in the electron configuration sequence, as it has higher energy. This means for orbitals like 2s and 2p, electrons are removed from 2p first. This approach ensures that electrons are removed from the highest energy orbitals, consistent with the Aufbau principle in reverse.

What is the difference between cations and anions in terms of electron configuration changes?

Cations are positively charged ions formed when an atom loses electrons, while anions are negatively charged ions formed when an atom gains electrons. For cations, electrons are removed from the orbitals with the highest principal quantum number $n$, starting with the highest energy orbital within that shell. For anions, electrons are added to orbitals that have available space, typically filling the next available orbital in the same shell or subshell. This difference reflects the loss of electrons in cations and gain of electrons in anions, which affects their electron configurations and chemical properties.

How are anions named based on their electron configurations?

Anions are named by taking the base name of the nonmetal element and changing its ending to "-ide." For example, oxygen becomes oxide, fluorine becomes fluoride, and chlorine becomes chloride. This naming convention indicates that the element has gained electrons to form a negatively charged ion. The electron configuration of an anion reflects the addition of electrons to orbitals with available space, often completing or partially filling the outermost shell, which stabilizes the ion.

Why are electrons removed from 3p orbitals before 3s orbitals when forming cations?

Electrons are removed from 3p orbitals before 3s orbitals because, although both have the same principal quantum number $n=3$, the 3p orbitals have higher energy than 3s orbitals. The order of energy within the same shell is determined by the subshell type, with $p$ orbitals having higher energy than $s$ orbitals. Therefore, when electrons are lost to form cations, they are removed first from the higher energy 3p orbitals to minimize the energy of the resulting ion.

What role does the principal quantum number play in electron removal for ion formation?

The principal quantum number, $n$, indicates the shell or energy level of an electron. When forming cations, electrons are removed first from orbitals with the highest $n$ value because these electrons are in the outermost shell and are less tightly bound to the nucleus. This makes them easier to remove. For example, electrons in the 3rd shell ($n=3$) are removed before those in the 2nd shell ($n=2$). This principle helps predict the electron configuration of ions accurately.