Problem 15 - Chemistry The Central Science

Question

Write out the ground-state electron configurations of (a) \(\mathrm{Ti}^{3+},(\mathbf{b}) \mathrm{Ru}^{2+},(\mathbf{c}) \mathrm{Au}^{3+},(\mathbf{d}) \mathrm{Mn}^{4+} .\)

Step-by-Step Solution

Answer

The ground-state electron configurations of the ions are: a) \(\mathrm{Ti}^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^1\) b) \(\mathrm{Ru}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^6\) c) \(\mathrm{Au}^{3+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\) d) \(\mathrm{Mn}^{4+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\)

Step 1: (a) Titanium ion (\(\mathrm{Ti}^{3+}\)) electron configuration

1. Find the atomic number of Titanium (Ti). In the periodic table, it is 22. 2. Determine the electron configuration of the neutral atom: \[\mathrm{Ti}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2\] 3. Remove three electrons from the atom to form the ion \(\mathrm{Ti}^{3+}\), starting from the outermost shell: \[\mathrm{Ti}^{3+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^1\]

Step 2: (b) Ruthenium ion (\(\mathrm{Ru}^{2+}\)) electron configuration

1. Find the atomic number of Ruthenium (Ru). In the periodic table, it is 44. 2. Determine the electron configuration of the neutral atom: \[\mathrm{Ru}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^6\] 3. Remove two electrons from the atom to form the ion \(\mathrm{Ru}^{2+}\), starting from the outermost shell: \[\mathrm{Ru}^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^6\]

Step 3: (c) Gold ion (\(\mathrm{Au}^{3+}\)) electron configuration

1. Find the atomic number of Gold (Au). In the periodic table, it is 79. 2. Determine the electron configuration of the neutral atom: \[\mathrm{Au}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^{10} 6s^1\] 3. Remove three electrons from the atom to form the ion \(\mathrm{Au}^{3+}\), starting from the outermost shell: \[\mathrm{Au}^{3+}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^8\]

Step 4: (d) Manganese ion (\(\mathrm{Mn}^{4+}\)) electron configuration

1. Find the atomic number of Manganese (Mn). In the periodic table, it is 25. 2. Determine the electron configuration of the neutral atom: \[\mathrm{Mn}: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5\] 3. Remove four electrons from the atom to form the ion \(\mathrm{Mn}^{4+}\), starting from the outermost shell: \[\mathrm{Mn}^{4+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^3\]

Key Concepts

Oxidation States

Oxidation states, also known as oxidation numbers, are a way to keep track of the electrons in atoms as they form compounds or ions. In essence, they describe how many electrons an atom gains or loses when it forms an ion or compound. For example, Titanium (Ti), with an oxidation state of +3, indicates the loss of three electrons to become \(\mathrm{Ti}^{3+}\). This is essential because understanding oxidation states helps you predict how different elements will react with each other in chemical reactions.

• Positive oxidation states, like \(\mathrm{Fe}^{3+}\), suggest lost electrons.
• Negative oxidation states indicate gained electrons.

Grasping oxidation states is critical in electron configurations because the oxidation state tells us how many electrons are removed from or added to the element's neutral atom configuration. Remember, electrons are removed first from the outermost shell, which is why Titanium as \(\mathrm{Ti}^{3+}\) loses electrons from the 4s orbital in its electron configuration, leading to the configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^1\). This step-by-step removal is vital in predicting the chemical behavior of ions in compounds.

Transition Metals

Transition metals are elements found in the d-block of the periodic table. They are characterized by having d orbitals that are being filled with electrons. These metals, such as Titanium (Ti), Ruthenium (Ru), and Gold (Au), often have multiple oxidation states. This flexibility arises because electrons in the d orbital can be lost or shared in chemical reactions, allowing for a variety of cations. Some key features of transition metals:

• They often form colored compounds.
• They can have multiple stable oxidation states.
• They are typically good conductors of electricity.

The diversity in possible oxidation states makes the chemical behavior of transition metals complex but fascinating. For example, Mn (Manganese) can exist in oxidation states from +2 to +7. This versatility is crucial in many biological and industrial processes, such as the role of iron in hemoglobin or the use of noble metals in catalysts. Understanding this concept will help you not only with writing electron configurations but also with predicting how these metals interact in various chemical environments.

Atomic Number

The atomic number is fundamental in understanding elements. It represents the number of protons in an atom's nucleus and determines the chemical identity of the element. For instance, Titanium has an atomic number of 22, meaning every titanium atom contains 22 protons. This number also equals the number of electrons in a neutral atom. Knowing the atomic number allows you to:

• Identify an element on the periodic table.
• Determine the standard electron configuration for neutral atoms.
• Understand the charge and behavior of an element in ions and compounds.

In electron configuration notations, the atomic number guides how electrons are disposed across various orbitals. When writing electron configurations for ions, like Gold which has the atomic number 79, one removes electrons starting from the highest energy level based on the atomic arrangement. Thus, understanding the atomic number is crucial for predicting how elements transition between different states and form compounds or ions.