Problem 30 - Chemistry: Principles and Reactions

Question

Follow the directions of Question 29 for (a) \(\mathrm{Mo}^{3+}\) (b) \(\mathrm{Pd}^{4+}\)

Step-by-Step Solution

Answer

Answer: The electron configurations for the Mo³⁺ and Pd⁴⁺ ions are as follows: - Mo³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹ - Pd⁴⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d⁴

Step 1: Identify the atomic numbers of Mo and Pd

Molybdenum (Mo) is atomic number 42, while Palladium (Pd) is atomic number 46. This information can be found in the periodic table.

Step 2: Write the electron configuration for neutral atoms

Using the periodic table or electron configuration rules, write the electron configurations for the neutral atoms of Mo and Pd. For Mo, with atomic number 42, the electron configuration is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^4\]For Pd, with atomic number 46, the electron configuration is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8\]

Step 3: Write the electron configuration for the ions

Now, we will remove the required number of electrons from the neutral atom configurations to get the ion configurations. Keep in mind that electrons are removed from the most recently added quantum level shells. (a) For Mo³⁺, we will remove three electrons from the neutral configuration:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^4 \Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^1\]So, the electron configuration for Mo³⁺ is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^1\] (b) For Pd⁴⁺, we will remove four electrons from the neutral configuration:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^8 \Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^4\] So, the electron configuration for Pd⁴⁺ is:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 4d^4\]

Key Concepts

Transition Metals

Transition metals are a group of elements found in the middle of the periodic table. They are known for their versatile chemistry and unique properties. These metals, which include Molybdenum (Mo) and Palladium (Pd), are characterized by d-orbitals, which play a crucial role in their behavior.

• Transition metals can exhibit multiple oxidation states. This is because they have a large number of orbitals available for electrons to occupy.
• They often form colored compounds due to the d-d electron transitions that occur within their d-orbitals.
• These elements can also act as catalysts in numerous chemical reactions, making them essential in industrial processes.

In the case of molybdenum and palladium, you can see the flexibility in their electron configurations. Molybdenum has its electrons in 4d orbitals, whereas palladium goes up to 4d as well. These elements, because of their d-orbital configuration, can easily lose or share electrons.

Oxidation States

Oxidation states are an important concept in chemistry, particularly when dealing with transition metals. The oxidation state of an element is a measure of the degree of oxidation, which essentially tells us how many electrons have been lost or gained compared to the neutral atom.

• It's crucial for understanding the reactivity and bonding properties of elements.
• Transition metals can have a wide variety of oxidation states, unlike main-group elements.

For example, in this exercise, we have molybdenum as a Mo³⁺ ion, denoting it has lost three electrons. Similarly, palladium as a Pd⁴⁺ ion means it has lost four electrons. This property of having different oxidation states allows transition metals to participate in a wide range of chemical reactions, acting as important agents in processes like catalysis or electron transport.

Molecular Ions

Molecular ions are formed when electrons are removed from or added to molecules, producing charged particles. In the context of transition metals, molecular ions often consist of complexes where the metal is surrounded by ligands - molecules or ions that donate pairs of electrons.

• Molecular ions are crucial in numerous chemical reactions and can affect the stability of compounds.
• Transition metals tend to form positive molecular ions because they can lose electrons easily from their outer s and d orbitals.

For instance, when forming Mo³⁺ and Pd⁴⁺ ions, the respective atoms lose electrons, leading to a cationic form. Such changes in electronic structure are key to the reactivity and function of transition metal ions in chemical systems.