Question

The lanthanide elements react with oxygen to give, generally, compounds of the type \(\mathrm{Ln}_{2} \mathrm{O}_{3}\) (where Ln stands for a lanthanide element). However, there are interesting exceptions, such as a common oxide of terbium, \(\mathrm{Tb}_{\mathrm{x}} \mathrm{O}_{y}\). Given that the compound is \(73.945 \%\) Tb, what is its formula? What is the oxidation number of terbium in this compound? Write a balanced equation for the reaction of terbium and oxygen to give this oxide.

Short answer

The formula is \( \mathrm{Tb}_{4} \mathrm{O}_{7} \), Tb has a +3.5 oxidation state. Balanced equation: \( 8 \mathrm{Tb} + 7 \mathrm{O}_2 \rightarrow 2 \mathrm{Tb}_{4} \mathrm{O}_{7} \).

Step-by-step solution

Calculate the masses of Tb and O

Given that the compound contains 73.945% terbium, we know that there are 100 - 73.945 = 26.055% oxygen in the compound. Assume 100g of the compound for ease of calculation. This means there are 73.945g of Tb and 26.055g of O in the compound. The atomic masses are approximately 158.93 g/mol for Tb and 16.00 g/mol for O.

Find the moles of Tb and O

Calculate the moles of terbium: \( \text{moles of Tb} = \frac{73.945}{158.93} \approx 0.465 \) moles. Calculate the moles of oxygen: \( \text{moles of O} = \frac{26.055}{16.00} \approx 1.628 \) moles.

Simplify mole ratio

To find the simplest whole number ratio, divide both mole values by the smallest number of moles. Thus, \( \frac{0.465}{0.465} = 1 \) for terbium and \( \frac{1.628}{0.465} \approx 3.5 \) for oxygen. This gives a simplified ratio of 2:7 when multiplying by 2 to avoid fractions. Hence, the formula of the compound is \( \mathrm{Tb}_{4} \mathrm{O}_{7} \).

Determine the oxidation number of Tb

Assume the oxidation state of oxygen is -2. For \( \mathrm{Tb}_{4} \mathrm{O}_{7} \), the total oxidation number from oxygen is \(7 \times -2 = -14\). Let the oxidation state of Tb be \( x \). Therefore, \( 4x - 14 = 0 \). Solving gives \( 4x = 14 \), hence \( x = 3.5 \). Thus, the oxidation state of Tb is +3.5.

Write the balanced equation

Writing the balanced chemical equation for the reaction of terbium with oxygen, consider \( 4 \mathrm{Tb} + 7/2 \mathrm{O}_2 \rightarrow \mathrm{Tb}_{4} \mathrm{O}_{7} \). Multiplying throughout by 2 to avoid fractions gives: \( 8 \mathrm{Tb} + 7 \mathrm{O}_2 \rightarrow 2 \mathrm{Tb}_{4} \mathrm{O}_{7} \).

Key concepts

Oxidation State

In lanthanide chemistry, understanding the oxidation state is crucial as it represents the charge of an atom within a molecule. The oxidation state helps to balance chemical equations and understand the electron transfer during reactions. In the compound \( \mathrm{Tb}_{4} \mathrm{O}_{7} \), oxygen typically holds an oxidation state of \(-2\). To deduce the oxidation state of terbium, one must ensure the entire compound is electrically neutral.
The calculation involves setting up an equation with the oxidation state of terbium \( x \):

• The total oxidation from oxygen is \( 7 \times -2 = -14 \).
• The entire oxidation sum has to be zero for neutrality: \( 4x - 14 = 0 \).
• Solving \( 4x = 14 \) gives \( x = 3.5 \).

This unusual oxidation state of \(+3.5\) in terbium highlights the unique chemistry of lanthanides, where fractional oxidation states can occur due to their partially filled \( f \)-orbitals.
Recognizing these states facilitates a deeper comprehension of lanthanide behavior and their compounds.

Chemical Reactions

Lanthanide elements like terbium are highly reactive, especially with oxygen. Understanding these reactions involves balancing equations to reflect the law of conservation of mass. When analysis reveals a compound such as \( \mathrm{Tb}_{4} \mathrm{O}_{7} \), it indicates a specific reaction pathway.
The balanced chemical reaction for terbium with oxygen is:

• Create a basic framework: \( 4 \mathrm{Tb} + \frac{7}{2} \mathrm{O}_2 \rightarrow \mathrm{Tb}_{4} \mathrm{O}_{7} \).
• To eradicate fractions, multiply the entire equation by 2: \( 8 \mathrm{Tb} + 7 \mathrm{O}_2 \rightarrow 2 \mathrm{Tb}_{4} \mathrm{O}_{7} \).

This equation ensures that the number of atoms becomes equal on both sides, signifying a complete and balanced reaction.
The steps in counting atoms and balancing them cater to understanding chemical processes deeply, emphasizing the precision required in chemical reactions.

Mole Ratios

Mole ratios are essential in chemistry for the determination of empirical formulas and the stoichiometry of reactions. They allow us to translate the microscopic world of atoms into measurable quantities.

• Understand that mole ratios derive from the stoichiometric coefficients in balanced chemical reactions.
• In the example of \( \mathrm{Tb}_{4} \mathrm{O}_{7} \), the calculation begins with determining moles from the given masses: \( 0.465 \) moles of Tb and \( 1.628 \) moles of O.
• Simplify the ratio: divide both values by the smallest number, \( 0.465 \), yielding \( 1:3.5 \).
• To eliminate fraction, multiply each by 2, producing \( 2:7 \), thus leading to the empirical formula \( \mathrm{Tb}_{4} \mathrm{O}_{7} \).

Mole ratios provide the bedrock for calculating reactant and product amounts, bridging the gap between the abstract atomic scale and tangible laboratory measurements.