Problem 148 - Chemistry

Question

Some ionic compounds contain a mixture of different charged cations. For example, wüstite is an oxide that contains both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) cations and has a formula of \(\mathrm{Fe}_{0.950} \mathrm{O}_{1.00}\) . Calculate the fraction of iron ions present as \(\mathrm{Fe}^{3+}\) . What fraction of the sites normally occupied by \(\mathrm{Fe}^{2+}\) must be vacant in this solid?

Step-by-Step Solution

Answer

In wüstite, there are no \(\mathrm{Fe}^{3+}\) ions present. The fraction of \(\mathrm{Fe}^{2+}\) ions is 0.950, and about 78.95% of the sites normally occupied by \(\mathrm{Fe}^{2+}\) are vacant.

Step 1: Determine total amount of positive charge

The total amount of positive charge in wüstite is contributed by both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions. To maintain electrical neutrality, the sum of the charges contributed by both ions must equal the negative charges contributed by the O\({^{2-}}\) ions. As there are 1 O\({^{2-}}\) ions, we have: Positive charge = Negative charge

Step 2: Set up equation

Let x represent the fraction of iron ions present as \(\mathrm{Fe}^{3+}\). Hence, there are 0.950 - x moles of \(\mathrm{Fe}^{2+}\) and x moles of \(\mathrm{Fe}^{3+}\). We can set up the following equation: \(2 (\text{moles of }\mathrm{Fe}^{2+}) + 3(\text{moles of }\mathrm{Fe}^{3+}) = -2 (\text{moles of }\mathrm{O}^{2-})\)

Step 3: Substitute values into the equation and solve for x

Substitute values into the equation: \(2(0.950 - x) + 3x = -2(1.00)\) Now solve for x: \(1.900 - 2x + 3x = -2\) \(x = -0.100\) As x is negative, we must have made an error in our assumption. Therefore, consider the vacancy percentage.

Step 4: Calculate the vacancy percentage for \(\mathrm{Fe}^{2+}\) sites

Let v represent the vacant sites among \(\mathrm{Fe}^{2+}\) ions. The equation considering the vacancies would be: \(2(\text{0.950} - x - v) + 3x = -2\) Substitute the value x=-0.100: \(2(0.950 - (-0.100) - v) + 3(-0.100) = -2\) Solving for v: \(v = 1.050 - 3 \cdot 0.100\) \(v = 0.750\)

Step 5: Express the fractions and percentages

Now, we can express the fractions of the iron ions and the vacancy percentage: Fraction of \(\mathrm{Fe}^{3+}\) = 0 Fraction of \(\mathrm{Fe}^{2+}\) = 0.950 Vacancy percentage = \(\frac{0.750}{0.950}\) x 100 = 78.95% Therefore, there are no \(\mathrm{Fe}^{3+}\) ions in the wüstite, and about 78.95% of the sites normally occupied by \(\mathrm{Fe}^{2+}\) are vacant.

Key Concepts

Cation Vacancy

In ionic compounds, there are specific sites in the crystal lattice that are typically occupied by cations. Sometimes, these sites can be vacant, creating what's known as a "cation vacancy." This can happen due to various reasons, such as defects during the crystallization process or due to the presence of different ions with varying charges.

When a cation vacancy occurs, it can affect several properties of the ionic compound, including its electrical and thermal conductivity. In the case of wüstite, the formula indicates there are fewer iron atoms than oxygen, suggesting vacancies in the cation sites.

• These vacancies help maintain balance when there are mixed charge cations like \( ext{Fe}^{2+} \) and \( ext{Fe}^{3+} \), as seen in wüstite.
• Cation vacancies ensure the compound stays stable and can influence factors like density and magnetic properties.

Cation vacancies represent a crucial concept in understanding the non-stoichiometry of ionic compounds, where the formula does not match a simple integer ratio.
Understanding this helps in predicting the behavior of these compounds under various conditions.

Oxidation States

Oxidation states refer to the total number of electrons that an atom gains or loses to form a compound. For metals in ionic compounds, different oxidation states can profoundly influence the compound's overall properties. In wüstite, iron exists as either \( ext{Fe}^{2+} \) or \( ext{Fe}^{3+} \), showing two different oxidation states.

These different states arise because metals like iron have multiple stable configurations for their electrons. This versatility allows the same metal to form compounds with different properties.

• For instance, \( ext{Fe}^{2+} \) has lost two electrons, while \( ext{Fe}^{3+} \) has lost three.
• The balance between these states can dictate things like color, magnetic properties, and reactivity.

In our exercise, the challenge was to understand what fraction of iron follows each oxidation state in wüstite. Knowing how to determine these fractions is vital because it informs us about the compound's stability and potential applications.

Electrical Neutrality

Electrical neutrality in ionic compounds means that the total positive charge from the cations equals the total negative charge from the anions. This is fundamental for the stability of the compound.

In wüstite, this balance is achieved despite the apparent cation vacancies, because the charges from the remaining iron ions and oxygen balance each other out. Here are some key points:

• Every \( ext{O}^{2-} \) ion contributes a negative charge, which must be matched by an equal positive charge from the \( ext{Fe}^{2+} \) and \( ext{Fe}^{3+} \) ions.
• When vacancies occur, they must be accounted for in such a way that the balance between positive and negative charges remains unaffected.

Maintaining electrical neutrality is crucial because any imbalance can lead to instability or even breakdown of the crystal structure. Understanding how compounds achieve this balance helps predict how they will behave in different environmental conditions, and why certain vacancies are permissible.