Question

Write the electron configurations for the following ions: (a) \(\ln ^{3+}\) (b) \(\mathrm{Sb}^{3+}\) (c) \(\mathrm{Te}^{2-}\), (d) \(\mathrm{Te}^{6+}\) (e) \(\mathrm{Hg}^{2+}\) (f) \(\mathrm{Rh}^{3+}\),

Short answer

The electron configurations for the given ions are: (a) \(\ln ^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^1\) (b) \(\mathrm{Sb}^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^3\) (c) \(\mathrm{Te}^{2-}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (d) \(\mathrm{Te}^{6+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) (e) \(\mathrm{Hg}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^9\) (f) \(\mathrm{Rh}^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^7\)

Step-by-step solution

(a) Finding electron configuration for \(\ln ^{3+}\)

First, we need to identify the atomic number for this element (\(\ln\)) which represents Lanthanum (La). The atomic number for Lanthanum (La) is 57. In a neutral atom, the number of electrons is equal to the atomic number. Therefore, La has 57 electrons. Now, we need to account for the 3+ charge of the ion, which means it loses 3 electrons. So, we have 57 - 3 = 54 electrons. The electron configuration for this ion is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^1\]

(b) Finding electron configuration for \(\mathrm{Sb}^{3+}\)

First, we need to identify the atomic number for this element (\(\mathrm{Sb}\)) which represents Antimony. The atomic number for Antimony (Sb) is 51. In a neutral atom, the number of electrons is equal to the atomic number. Therefore, Sb has 51 electrons. Now, we need to account for the 3+ charge of the ion, which means it loses 3 electrons. So, we have 51 - 3 = 48 electrons. The electron configuration for this ion is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^3\]

(c) Finding electron configuration for \(\mathrm{Te}^{2-}\)

First, we need to identify the atomic number for this element (\(\mathrm{Te}\)) which represents Tellurium. The atomic number for Tellurium (Te) is 52. In a neutral atom, the number of electrons is equal to the atomic number. Therefore, Te has 52 electrons. Now, we need to account for the 2- charge of the ion, which means it gains 2 electrons. So, we have 52 + 2 = 54 electrons. The electron configuration for this ion is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\]

(d) Finding electron configuration for \(\mathrm{Te}^{6+}\)

We already have the atomic number and number of electrons for neutral Tellurium (Te) from the previous step (c). Now, we need to account for the 6+ charge of the ion, which means it loses 6 electrons. So, we have 52 - 6 = 46 electrons. The electron configuration for this ion is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\]

(e) Finding electron configuration for \(\mathrm{Hg}^{2+}\)

First, we need to identify the atomic number for this element (\(\mathrm{Hg}\)) which represents Mercury. The atomic number for Mercury (Hg) is 80. In a neutral atom, the number of electrons is equal to the atomic number. Therefore, Hg has 80 electrons. Now, we need to account for the 2+ charge of the ion, which means it loses 2 electrons. So, we have 80 - 2 = 78 electrons. The electron configuration for this ion is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^9\]

(f) Finding electron configuration for \(\mathrm{Rh}^{3+}\)

First, we need to identify the atomic number for this element (\(\mathrm{Rh}\)) which represents Rhodium. The atomic number for Rhodium (Rh) is 45. In a neutral atom, the number of electrons is equal to the atomic number. Therefore, Rh has 45 electrons. Now, we need to account for the 3+ charge of the ion, which means it loses 3 electrons. So, we have 45 - 3 = 42 electrons. The electron configuration for this ion is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^7\]

Key concepts

Atomic Number

The atomic number of an element is a fundamental concept that describes the number of protons in the nucleus of an atom. It is denoted by the letter "Z." The atomic number determines the identity of an element and its position on the periodic table. Each element has a unique atomic number which corresponds to the number of electrons in a neutral atom, allowing us to predict the electron configuration.
For example, the atomic number for Lanthanum (La) is 57, indicating it has 57 protons and, in a neutral state, 57 electrons. When calculating electron configurations for ions, the atomic number helps us determine how many electrons are gained or lost. Understanding atomic numbers is key to grasping chemical properties and behaviors of elements.

Lanthanides

Lanthanides, also known as rare earth elements, encompass elements from Lanthanum (La) to Lutetium (Lu) on the periodic table. These elements are in block f of the periodic table, characterized by filling the 4f orbitals. Lanthanides are known for their similar chemical properties which often makes them challenging to separate in their natural form.
A defining feature of lanthanides is their large atomic and ionic sizes as they progress across the series. This is due to the "lanthanide contraction," which refers to the phenomenon where the 4f orbitals do not shield the nuclear charge effectively, leading to smaller atomic radii.
Lanthanum (La), as seen in the original exercise, is the first element in the lanthanide series. Lanthanides often form +3 cations due to the stability provided by losing three electrons, aligning with the "Ln\(^{3+}\)" ion used in solving the electron configuration problem.

Transition Metals

Transition metals are elements found in the center of the periodic table, specifically in groups 3 to 12. These metals are characterized by their filling d orbitals. Transition metals are unique because they have partially filled d orbitals which contribute to their ability to form colorful compounds, act as catalysts, and possess multiple oxidation states.
Key properties of transition metals include:

• High melting and boiling points
• Good electrical conductors
• Complex ion formation capability
• Variable oxidation states, important for electron configuration

In the original exercise, elements like Rhodium (Rh) and Mercury (Hg) fall into the category of transition metals. Their ability to lose electrons and form ions, such as Rh\(^{3+}\) or Hg\(^{2+}\), reflects their complex chemistry and the importance of understanding electron configurations.

Periodic Table

The periodic table is an organized chart of all known chemical elements, arranged by increasing atomic number. Modern periodic tables are structured in rows called periods and columns called groups or families. Each element is positioned based on its atomic number which corresponds to the number of protons in its nucleus.
The table is designed to highlight periodic trends in element properties such as electronegativity, atomic radius, and ionization energy. For instance, moving from left to right across a period, atomic size decreases while electronegativity increases. Conversely, elements in the same group tend to exhibit similar chemical properties due to having the same number of valence electrons.
Understanding the layout of the periodic table is crucial when determining electron configurations, as it allows easy prediction of an element's behavior and reactivity, helping students apply these concepts in exercises like the one provided.

Ion Charge

Ion charge plays a fundamental role in determining the electron configuration of ions, which is the focus of many chemistry problems like the one provided in the exercise. An ion is an atom or molecule that has gained or lost one or more electrons, resulting in a net electric charge.
- Positive ions, or cations, form when an atom loses electrons.- Negative ions, or anions, form when an atom gains electrons.
The charge of an ion is indicated by a superscript next to the element symbol. For example, "\(^{3+}\)" indicates a loss of three electrons, while "\(^{2-}\)" indicates a gain of two electrons. Understanding ion charge helps in determining how many electrons to add or remove from the neutral atomic electron configuration. Calculating these changes is vital for balancing chemical reactions and predicting molecular structures.