Question

The discovery of hafnium, element number 72, provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements 58-71) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and \(G\). von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 \mathrm{Write}\) a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

Short answer

Bohr predicted that hafnium, element 72, would be found in the same group as zirconium due to their similar electron configurations, sharing the same number of valence electrons. The balanced chemical equation for producing zirconium is: ZrCl₄(s) + 4 Na(l) → Zr(s) + 4 NaCl(l). This is an oxidation-reduction reaction where zirconium is reduced, and sodium is oxidized. The balanced reaction between ZrO₂ and Cl₂ is: ZrO₂(s) + 2 Cl₂(g) + 3 C(s) → ZrCl₄(s) + CO₂(g) + 2 CO(g). The mass of ZrCl₄ formed from a 55.4 g sample of ZrO₂ is 127.4 g. Zirconium and hafnium form chlorides MCl₄ and oxides MO₂ due to their similarities in electron configurations and tendencies to reach a stable +4 oxidation state.

Step-by-step solution

(a) Electron Configuration Justification for Bohr’s Prediction

Bohr predicted that the element number 72 (hafnium) is more likely to be found in the same group as zirconium (group 4) than with the rare earth elements (58-71). This prediction can be justified using electron configuration arguments. Both elements in the same group have similar electron configurations, as they share the same number of electrons in their outermost shell. Zirconium has the electron configuration [Kr] 5s²4d² and Hafnium has [Xe] 6s²4f¹45d². Notice that they both have 2 electrons in their s and d subshells which makes their valence electrons and chemical properties comparable. In contrast, rare earth elements have different electron configurations with varying numbers of electrons in the f orbitals, leading to unique properties.

(b) Balanced Chemical Equation and Oxidation-Reduction Reaction

We are given that zirconium can be produced by the reduction of solid ZrCl4 with molten sodium metal. To write a balanced chemical equation for this reaction: \[ \mathrm{ZrCl}_{4}(s) + 4 \mathrm{Na}(l) \rightarrow \mathrm{Zr}(s) + 4 \mathrm{NaCl}(l) \] This is an oxidation-reduction reaction. Zirconium in ZrCl4 is reduced as the oxidation state decreases from +4 to 0, and sodium in Na is oxidized as the oxidation state increases from 0 to +1.

(c) Balanced Chemical Equation, Mass Calculation, and Limiting Reagent

Zirconium dioxide reacts with chlorine gas in the presence of carbon, with products ZrCl4, CO2, and CO in the ratio 1:2. To balance the chemical equation: \[ \mathrm{ZrO}_{2}(s) + 2 \mathrm{Cl}_{2}(g) + 3 \mathrm{C}(s) \rightarrow \mathrm{ZrCl}_{4}(s) + \mathrm{CO}_{2}(g) + 2 \mathrm{CO}(g) \] Given a 55.4-g sample of ZrO2, we can calculate the mass of ZrCl4 formed, assuming ZrO2 is the limiting reagent and 100% yield: 1. Find the molar mass of ZrO2: \(55.4\,g/mol \cdot (101.22\,g/mol) = 0.547\,mol\) 2. Use the stoichiometry from the balanced equation: \(1 \, mol\,\mathrm{ZrO}_{2}\) produces \(1 \, mol\, \mathrm{Zr Cl}_{4}\) 3. Calculate the mass of ZrCl4 formed: \(0.547\,mol\, \mathrm{ZrCl}_{4} \cdot (232.93\,g/mol) = 127.4\,g\) Thus, 127.4 g of ZrCl4 is formed.

(d) Electron Configurations and Formation of MCl4 and MO2

Zirconium and hafnium have similar electron configurations, as shown in part (a). As a result, both elements have a tendency to lose their outermost s and d electrons when forming compounds, leading to a common oxidation state of +4. For chlorides MCl4, this means that both Zr and Hf will form compounds with the general formula MCl4, where M = Zr or Hf. \[ \mathrm{Zr^{4+}} + 4 \mathrm{Cl}^- \rightarrow \mathrm{ZrCl}_{4} \] \[ \mathrm{Hf^{4+}} + 4 \mathrm{Cl}^- \rightarrow \mathrm{HfCl}_{4} \] Similarly, for the oxides MO2, they will form compounds with the general formula MO2: \[ \mathrm{Zr^{4+}} + 2 \mathrm{O}^{2-} \rightarrow \mathrm{ZrO}_{2} \] \[ \mathrm{Hf^{4+}} + 2 \mathrm{O}^{2-} \rightarrow \mathrm{HfO}_{2} \] These compounds are formed because of the similarities in electron configurations and tendencies to reach a stable +4 oxidation state.

Key concepts

Electron Configuration

Understanding electron configuration is crucial for explaining the similarities between different chemical elements, such as zirconium (Zr) and hafnium (Hf). The electron configuration of an element details the arrangement of electrons around its atomic nucleus. Both zirconium and hafnium belong to the same group in the periodic table, which means their valence shells have similar electron arrangements.

- Zirconium has the electron configuration \[ \text{[Kr]} \, 5s^2 \, 4d^2 \], indicating it has two electrons in the 5s sub-shell and two in the 4d sub-shell.- Hafnium, on the other hand, has the configuration \[ \text{[Xe]} \, 6s^2 \, 4f^{14} \, 5d^2 \]. Despite the added complexity of the 4f orbitals, the crucial aspect is the two electrons in the 6s and 5d sub-shells which correspond to those found in zirconium.

The similarities in their outermost electron configurations help predict that hafnium would share similar chemical properties with zirconium, explaining why it might be found alongside zirconium rather than with rare earth elements.

Oxidation-Reduction Reaction

Redox reactions, or oxidation-reduction reactions, are fundamental chemical processes where electrons are transferred from one substance to another. This transfer results in changes to the oxidation states of the substances involved.

In the case of reducing \(\mathrm{ZrCl}_4\) with molten sodium \(\mathrm{Na}\), the balanced chemical equation is:

\[\mathrm{ZrCl}_{4}(s) + 4\mathrm{Na}(l) \rightarrow \mathrm{Zr}(s) + 4\mathrm{NaCl}(l) \]

The reaction involves the following changes:
- \(\mathrm{ZrCl}_4\): Zirconium (Zr) starts with an oxidation state of +4, and ends with 0 in elemental form, meaning it gains electrons and is reduced.- \(\mathrm{Na}\): Sodium starts with an oxidation state of 0 and ends up with +1 per sodium ion in \(\mathrm{NaCl}\), meaning it loses electrons and is oxidized.

Thus, this reaction exemplifies the classic redox process where one component is oxidized and the other is reduced.

Chemical Equations

A chemical equation represents a chemical reaction using symbols and formulas. It communicates the substances involved and the changes they undergo. Balancing a chemical equation ensures that the number of each type of atom is the same on both sides of the equation – reflecting the conservation of mass.

Given in the exercise is the reaction of zirconium dioxide \(\mathrm{ZrO}_{2}\) with chlorine gas and carbon to produce \(\mathrm{ZrCl}_{4}\) with carbon dioxide \(\mathrm{CO}_{2}\) and carbon monoxide \(\mathrm{CO}\) in a 1:2 ratio. The balanced equation is:

\[\mathrm{ZrO}_{2}(s) + 2\mathrm{Cl}_{2}(g) + 3\mathrm{C}(s) \rightarrow \mathrm{ZrCl}_{4}(s) + \mathrm{CO}_{2}(g) + 2\mathrm{CO}(g) \]

This representation shows each reactant and product and keeps the same atom count across both sides, ensuring mass is not lost or created in the process.

Stoichiometry

Stoichiometry involves using relationships in balanced chemical equations to calculate the amount of reactants needed or products formed in a reaction. The exercise deals with determining the mass of \(\mathrm{ZrCl}_{4}\) formed from a given mass of \(\mathrm{ZrO}_{2}\) when \(\mathrm{ZrO}_{2}\) is the limiting reagent.

Let's break it down:- Begin by finding the molar mass of \(\mathrm{ZrO}_{2}\), which is \(101.22\,g/mol\).- Calculate the moles of \(\mathrm{ZrO}_{2}\) in the given 55.4 grams: \[ 55.4\,g \times \left(\frac{1\,mol}{101.22\,g}\right) = 0.547\,mol \]- According to the balanced chemical equation, since 1 mole of \(\mathrm{ZrO}_{2}\) yields 1 mole of \(\mathrm{ZrCl}_{4}\), 0.547 moles of \(\mathrm{ZrCl}_{4}\) will be produced.- Finally, calculate the mass of \(\mathrm{ZrCl}_{4}\) formed: \[ 0.547\,moles \times 232.93\,g/mol = 127.4\,g \]

So, from 55.4 grams of \(\mathrm{ZrO}_{2}\), 127.4 grams of \(\mathrm{ZrCl}_{4}\) is produced, illustrating the use of stoichiometry in real chemical processes.