Question

Write the expected electron configuration for (a) \(\mathrm{Cr}^{3+}\), (b) \(\mathrm{Au}^{3+}, (\mathrm{c}) \mathrm{Ru}^{2+}\) (d) \(\mathrm{Cu}^{+}\), (e) \(\mathrm{Mn}^{4+},(\mathrm{f}) \mathrm{Ir}^{+}\)

Short answer

The expected electron configurations for the given ions are as follows: (a) Cr³⁺: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\] (b) Au³⁺: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\] (c) Ru²⁺: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^6\] (d) Cu⁺: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\] (e) Mn⁴⁺: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\] (f) Ir⁺: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^7\]

Step-by-step solution

(a) Cr³⁺ electron configuration

The atomic number of Cr (Chromium) is 24, which means it has 24 electrons in its neutral state. The neutral electron configuration for Cr is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5\] Now, we need to remove 3 electrons to form Cr³⁺. We will start by removing the 1 electron from 4s orbital, and then remove the remaining 2 electrons from the 3d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\]

(b) Au³⁺ electron configuration

The atomic number of Au (Gold) is 79, meaning it has 79 electrons in its neutral state. The neutral electron configuration for Au is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1 4f^{14} 5d^{10}\] Now, we need to remove 3 electrons to form Au³⁺. We start by removing the one 6s electron, and then remove two electrons from the 5d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\]

(c) Ru²⁺ electron configuration

The atomic number of Ru (Ruthenium) is 44, so it has 44 electrons in its neutral state. The neutral electron configuration for Ru is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 4d^7\] We need to remove 2 electrons to form Ru²⁺, so we start by removing the 1 electron from 5s orbital and then remove one electron from the 4d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^6\]

(d) Cu⁺ electron configuration

The atomic number of Cu (Copper) is 29, meaning it has 29 electrons in its neutral state. The neutral electron configuration for Cu is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\] We need to remove 1 electron to form Cu⁺ ion. We remove the one 4s electron: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}\]

(e) Mn⁴⁺ electron configuration

The atomic number of Mn (Manganese) is 25, so it has 25 electrons in its neutral state. The neutral electron configuration for Mn is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\] We need to remove 4 electrons to form Mn⁴⁺. We start by removing the two 4s electrons, and then remove two electrons from 3d orbital: \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\]

(f) Ir⁺ electron configuration

The atomic number of Ir (Iridium) is 77, meaning it has 77 electrons in its neutral state. The neutral electron configuration for Ir is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 6s^2 5d^7\] We need to remove 1 electron to form Ir⁺. We start by removing the one 6s electron: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^7\]

Key concepts

Transition Metals

Transition metals are elements found in the central block of the periodic table, specifically in groups 3 through 12. They are characterized by their ability to form various stable ionic states, often possessing incomplete d subshells.

These metals have unique properties such as colored compounds, variable oxidation states, and the ability to act as catalysts. Electron configurations for transition metals can be more complex due to the energy closeness of their s and d orbitals, leading to exceptions in their electron filling order. For instance, chromium and copper have unusual electron configurations (( Cr: [1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5) and (Cu: [1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}) respectively) due to increased stability when half-filled or fully filled d orbitals are present.

Ionic States

Ionic states, also known as oxidation states, refer to the charge that atoms exhibit when they lose or gain electrons to form ions. This is common in transition metals, which can lose different numbers of electrons, resulting in cations with positive charges.

For example, Chromium can lose three electrons to form Cr^3+, or Manganese can lose four to form Mn^4+. These varying ionic states contribute to the versatility of transition metals in forming compounds with different chemical and physical properties. It is essential to remember that when creating ions, electrons are typically removed from the orbitals starting from the highest principal quantum number first.

Atomic Number

The atomic number of an element, indicated by the symbol Z, is the number of protons found in the nucleus of an atom. It also determines the number of electrons in a neutral atom due to the charge balance between protons and electrons. For instance, Ruthenium has an atomic number of 44, which means both the neutral atom of Ru and its ion Ru^2+ have 44 protons, with the neutral atom having 44 electrons and the Ru^2+ ion having 42 electrons.

The atomic number is fundamental in determining the position of an element in the periodic table, its chemical behavior, and electron configuration pattern.

Electron Removal

Electron removal is the process of losing electrons from an atom or ion, which usually happens during the formation of cations. For transition metals, this process can affect the distribution of electrons among the different orbitals.

Electrons are generally removed from orbitals starting with the highest principal quantum number and then within each principal number, starting from orbitals with the highest energy. For example, when forming Cu^+ from copper, the electron removal will occur from the 4s orbital leaving the 3d orbital fully occupied ([1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10}). Understanding how electrons are removed can help predict ionic states and the reactivity of the resulting ions.