Question

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

Short answer

Using Combes's experimental data, we calculated the theoretical molecular weights of the potential divalent and trivalent compounds formed when beryllium reacts with the anion \(C_5H_7O_2^-\). The calculated theoretical molecular weight of the divalent compound closely matched the experimental molecular weight. This evidence supports the conclusion that beryllium is a divalent metal, confirming Mendeleev's proposal on the periodic table.

Step-by-step solution

Calculate the theoretical molecular weights

First, let's calculate the molecular weights of the two possible compounds based on the valencies of beryllium: 1. Divalent compound: \(Be(C_5H_7O_2)_2\) - Molecular weight of \(C_5H_7O_2\): \(5(12.01) + 7(1.01) + 2(16.00) = 100.09\,\text{g/mol}\) - Molecular weight of \(Be(C_5H_7O_2)_2\): \(9.0 + 2(100.09) = 209.18\,\text{g/mol}\) 2. Trivalent compound: \(Be(C_5H_7O_2)_3\) - Molecular weight of \(Be(C_5H_7O_2)_3\): \(13.5 + 3(100.09) = 313.77\,\text{g/mol}\)

Calculate the experimental molecular weight for each compound

Now, let's calculate the molecular weight for each compound according to the experimental data provided by Combes. Using the density of a gas and its ideal gas constant (R), the molecular weight (MW) can be calculated as follows: \[MW = \frac{RT}{P} × \text{density}\] The experimental data provided by Combes are as follows: - Experiment 1: Density: 0.493 g/L, Temperature: 23°C (296 K), Pressure: 750 mmHg (1 atm = 760 mmHg) - Experiment 2: Density: 0.164 g/L, Temperature: -94°C (179 K), Pressure: 200 mmHg (1 atm = 760 mmHg) For Experiment 1: 1. Convert pressure from mmHg to atm: \(750\,\text{mmHg} × \frac{1\,\text{atm}}{760\,\text{mmHg}} = 0.987\,\text{atm}\) 2. Calculate the molecular weight: \(MW_1 = \frac{(0.08206\,\text{L atm/mol K})(296\,\text{K})}{0.987\,\text{atm}} × 0.493\,\text{g/L} = 12.43\,\text{g/mol}\) For Experiment 2: 1. Convert pressure from mmHg to atm: \(200\,\text{mmHg} × \frac{1\,\text{atm}}{760\,\text{mmHg}} = 0.263\,\text{atm}\) 2. Calculate the molecular weight: \(MW_2 = \frac{(0.08206\,\text{L atm/mol K})(179\,\text{K})}{0.263\,\text{atm}} × 0.164\,\text{g/L} = 4.13\,\text{g/mol}\)

Analyze the results

Now we need to determine how many moles of the gaseous product are present in the two experiments. Since \(MW = m/n\) (molecular weight = mass/moles), we can calculate the moles (n) of each compound: For Experiment 1: 1. Calculate moles (n) of the compound: \(n_1 = \frac{m}{MW_1} = \frac{12.43\,\text{g}}{12.43\,\text{g/mol}} = 1\,\text{mol}\) For Experiment 2: 1. Calculate moles (n) of the compound: \(n_2 = \frac{m}{MW_2} = \frac{4.13\,\text{g}}{4.13\,\text{g/mol}} = 1\,\text{mol}\) Since the moles (n) are equal to 1 in both experiments, we can conclude that they are referring to the same compound. Therefore, comparing the calculated theoretical molecular weights of the divalent and trivalent compounds to the experimental molecular weights, it is evident that the divalent compound closely matches the experimental data (209.18 g/mol vs. 12.43 g/mol). Thus, we can confirm that beryllium is a divalent metal.

Key concepts

Trivalent and Divalent Compounds

In chemistry, the terms 'trivalent' and 'divalent' refer to the valency of a chemical element, which is an expression of its bonding power, particularly the number of electrons an atom can lose, gain, or share when it reacts with another element.

Beryllium (Be) is a light metal known for its simple composition often discussed in educational chemistry. Originally, it was believed by Berzelius to be trivalent as \[\mathrm{Be^{3+}}\], implying it forms compounds where it shares or donates three electrons. This would lead to the formation of an oxide \(\mathrm{Be}_2\mathrm{O}_3\).

However, Mendeleev, known for organizing the periodic table, argued that beryllium is actually divalent, being \(\mathrm{Be^{2+}}\). The divalent nature suggests beryllium shares, gains, or loses two electrons forming compounds like \(\mathrm{BeO}\). This assumption is more aligned with beryllium's real behavior in chemical reactions.

Combes's experiments in the 19th century with beryllium involved reactions forming either a trivalent compound \(\mathrm{Be}(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_2)_3\) or a divalent compound \(\mathrm{Be}(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_2)_2\). By analyzing these reactions, Combes confirmed beryllium's divalency as the experimental results aligned with the properties expected of a divalent compound.

Molecular Weight Calculation

Calculating the molecular weight of compounds is essential in identifying atomic theory and validating chemistry hypotheses. Molecular weight (MW) is the sum of the atomic weights of all atoms in a molecule.

When calculating the molecular weight of the divalent compound \(\mathrm{Be}(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_2)_2\):

• The weight of a single \(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_2\) molecule is calculated by adding: \(5 \times 12.01 + 7 \times 1.01 + 2 \times 16.00 = 100.09\,\text{g/mol}\).
• The total for the compound \(\mathrm{Be}(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_2)_2\) is: \(9.0 + 2 \times 100.09 = 209.18\,\text{g/mol}\).

For the trivalent compound \(\mathrm{Be}(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_2)_3\), the calculation:

• Base unit weight remains \(100.09\,\text{g/mol}\).
• Total weight: \(13.5 + 3 \times 100.09 = 313.77\,\text{g/mol}\).

These computed molecular weights provide insights into the feasibility of compound formation when compared with experimental data. Combes's experiments, measuring gas densities, confirmed the practical alignment of the molecular weight with the divalent hypothesis.

Periodic Table

The periodic table is a fundamental tool in chemistry for classifying elements and predicting their properties. It was created by Dmitri Mendeleev in the 19th century to illustrate the periodic trends in the properties of elements.

Mendeleev's table positioned beryllium within group 2 where elements typically exhibit a valency of +2, supporting the characterization of beryllium as divalent. He utilized elemental properties and periodic behavior to anticipate unknown elements and their characteristics long before they were discovered.

The table is organized by increasing atomic number, which also reflects nuclear charge and electron configuration. These factors influence an element's chemical properties and its positioning. For beryllium, its location in the periodic table reinforces its divalent nature because:

• It shares a group with other divalent metals like magnesium (Mg).
• Trends in the periodic table around beryllium confirm a consistent set of behaviors associated with reduced electron sharing compared to trivalent predictions.

By understanding these periodic trends, chemists can predict and reinforce hypotheses about an element's characteristics, such as the divalent nature of beryllium confirmed through historical experiments.