Problem 13 - Chemistry The Central Science In Si Units

Question

For each of the following compounds, determine the electron configuration of the transition-metal ion. (a) \(\mathrm{CuO}\), (b) \(\mathrm{Cu}_{2} \mathrm{O}\) (c) \(\mathrm{V}_{2} \mathrm{O}_{5},\) (d) \(\mathrm{MnO}\).

Step-by-Step Solution

Answer

CuO: [Ar] 3d^9; Cu2O: [Ar] 3d^{10}; V2O5: [Ar]; MnO: [Ar] 3d^5.

Step 1: Identify Oxidation States

To determine the electron configuration of the transition metal ions in each compound, first identify the oxidation state of the metal in each compound using the charge of the other elements. Oxygen usually has an oxidation state of -2.

Step 2: Calculate Cu Oxidation States

For (a) CuO, Oxygen's -2 oxidation state means Cu has an oxidation state of +2. In (b) Cu2O, each O has a -2 charge so since there are two Cu atoms, each Cu has a +1 oxidation state.

Step 3: Calculate V Oxidation State

In (c) V2O5, each O has a -2 charge, leading to a total of -10 from Oxygen; therefore, V must provide +10, with each V having a +5 oxidation state.

Step 4: Calculate Mn Oxidation State

In (d) MnO, O has a -2 charge, so Mn must be in the +2 oxidation state to balance the charge.

Step 5: Determine Electron Configuration

Now, determine the electron configuration for each metal in its ionic form by first finding the neutral atom configuration, then removing the appropriate number of electrons from the outer orbitals: (a) Cu^{2+}: Remove 2 electrons from Cu [Ar] 3d^{10} 4s^1 to get [Ar] 3d^9. (b) Cu^{+}: Remove 1 electron from Cu [Ar] 3d^{10} 4s^1 to get [Ar] 3d^{10}. (c) V^{5+}: Remove 5 electrons from V [Ar] 3d^3 4s^2 to get [Ar]. (d) Mn^{2+}: Remove 2 electrons from Mn [Ar] 3d^5 4s^2 to get [Ar] 3d^5.

Key Concepts

Oxidation State

The oxidation state refers to the number of electrons that an atom gains, loses, or shares when it forms chemical bonds. In ionic compounds, it indicates the effective charge of the atom. It acts as a book-keeping tool to determine how electrons are distributed among atoms in a molecule.

• For oxygen, the common oxidation state is -2. This is because oxygen typically needs two electrons to fill its outer shell, so it often gains two electrons.
• In the compounds provided, the metal's oxidation state balances the overall charge dictated by oxygen.

To find the oxidation state, consider:

• Each element's preferred states and typical behavior in compounds.
• The sum of oxidation states in a neutral compound is zero.

This concept helps us to determine how many electrons need to be removed from a transition metal in forming its ionic state.

Transition Metals

Transition metals are elements found in the d-block of the periodic table and have partially filled d orbitals. This category includes elements like Cu, V, and Mn. Transition metals are known for their ability to form various oxidation states.

• These metals can lose varying numbers of d and s electrons, leading to different ionic forms.
• Transition metals like copper (Cu) and vanadium (V) can exhibit different electron configurations depending on their oxidation states. For example, copper can be Cu⁺ or Cu²⁺.

The diversity in oxidation states arises due to:

• The d subshell's ability to be filled variably.
• Complex electronic structures that allow for multiple stable ions.

These multiple oxidation states contribute to the colorful characteristics and catalytic properties of transition metals.

Ionic Compounds

Ionic compounds consist of positive and negative ions held together by ionic bonds. These strong electrostatic forces form between atoms when electrons are transferred from one atom to another.

• In the case of metal oxides, metals like Cu, V, and Mn donate electrons to oxygen, forming these ionic bonds.
• The metal atoms lose electrons to become cations, whereas oxygen gains electrons, becoming anion.

Essential properties of ionic compounds include:

• High electronegativity differences between atoms, typically between metals and non-metals.
• Ionic crystals, known for their high melting and boiling points due to strong bonds.

Understanding ionic compounds aids in predicting the composition of metal oxides and the electron arrangements leading to ionic forms.

Electron Removal

Electron removal is the process of losing electrons to form positive ions or cations. For transition metals, determining which electrons are removed first to form a specific oxidation state is crucial. This step follows once the oxidation state is identified.

• Electrons are generally removed first from the s orbital before d orbitals, due to energy considerations.
• For example, in Cu, the electron configuration of the neutral atom is [Ar] 3d¹⁰ 4s¹, but when forming Cu²⁺, electrons are removed from the 4s orbital first.

Factors influencing electron removal include:

• Electron affinity and energy levels of the orbitals.
• Stability of d orbitals in transition metals, which are typically lower in energy compared to the s-orbital electrons in their outer shell.

Appreciating electron removal guides how we understand the formation of metal ions and their electron configurations in ionic compounds.