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, \(\mathrm{ZrO}_{2}\), reacts 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 \(1: 2\). 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 \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

Short answer

Bohr's prediction is justified using electron configuration arguments, as both hafnium and zirconium are in Group 4 and have similar electron configurations with two electrons in their outermost d orbitals, giving them comparable chemical properties. The balanced chemical equation for the reaction between ZrCl4 and sodium metal is: \[ZrCl_4 + 4Na \rightarrow Zr + 4NaCl\] This is an oxidation-reduction reaction where sodium is oxidized and zirconium tetrachloride is reduced. The balanced chemical equation for the reaction between solid zirconium dioxide and chlorine gas is: \[ZrO_2 + 2Cl_2 + 3C \rightarrow ZrCl_4 + CO_2 + 2CO\] Using stoichiometry, we determine that 104.87 grams of ZrCl4 would be formed from a 55.4-g sample of ZrO2, assuming 100% yield. Zr and Hf both lose four electrons from their outer energy levels to form +4 oxidation states, leading to the formation of tetrachlorides, MCl4, and dioxides, MO2.

Step-by-step solution

Use Electron Configuration to Justify Bohr's Prediction

Bohr predicted that hafnium (element number 72) would be found along with zirconium (element number 40) rather than with the rare earth elements. Let's compare the electron configurations of hafnium and zirconium: Zirconium (Zr): \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^2\] Hafnium (Hf): \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 5d^2 4f^{14}\] Both hafnium and zirconium are in the same group (Group 4), and they have similar electron configurations. They each have two electrons in their outermost d orbitals. This similarity in electron configurations gives them comparable chemical properties, suggesting that hafnium would more likely be found with zirconium rather than the rare earth elements.

Write Balanced Chemical Equation of ZrCl4 and Sodium Metal Reaction, Identify Redox Reaction

First, let's write the balanced chemical equation for the reaction of zirconium tetrachloride (ZrCl4) with sodium metal (Na). \[ZrCl_4 + 4Na \rightarrow Zr + 4NaCl\] This is an oxidation-reduction (redox) reaction because the oxidation states of the elements change during the reaction. Sodium's oxidation state increases from 0 to +1 while zirconium's oxidation state decreases from +4 to 0. Therefore: - Sodium is oxidized (loses electrons). - Zirconium tetrachloride is reduced (gains electrons).

Write Balanced Chemical Equation of ZrO2 and Chlorine Gas Reaction

To find the balanced chemical equation for the reaction of solid zirconium dioxide (ZrO2) with chlorine gas (Cl2) in the presence of carbon, which produces ZrCl4, CO2, and CO, follow these steps: 1. Write the unbalanced chemical equation: \[ZrO_2 + Cl_2 + C \rightarrow ZrCl_4 + CO_2 + CO\] 2. Balance the equation: \[ZrO_2 + 2Cl_2 + 3C \rightarrow ZrCl_4 + CO_2 + 2CO\] The balanced chemical equation is: \[ZrO_2 + 2Cl_2 + 3C \rightarrow ZrCl_4 + CO_2 + 2CO\]

Calculate the Mass of ZrCl4 Formed in the Reaction

Given a 55.4-g sample of ZrO2 and assuming ZrO2 is the limiting reagent with a 100% yield, let's calculate the mass of ZrCl4 formed: 1. Calculate the moles of ZrO2 in the sample: \(moles_{ZrO_{2}} = \frac{mass_{ZrO_{2}}}{molar\,mass_{ZrO_{2}}}\), where the molar mass of ZrO2 = 123.22 g/mol. \[moles_{ZrO_2} = \frac{55.4}{123.22} = 0.450\,mol\] 2. Determine the stoichiometry between ZrO2 and ZrCl4 using the balanced chemical equation. There is a 1:1 ratio, so for each mole of ZrO2, we can produce an equal amount of ZrCl4. 3. Calculate the mass of ZrCl4 formed: \(mass_{ZrCl_{4}} = moles_{ZrCl_{4}} \times molar\,mass_{ZrCl_{4}}\), where the molar mass of ZrCl4 = 233.04 g/mol. \[mass_{ZrCl_{4}} = 0.450\,mol \times 233.04 \,g/mol = 104.87\,g\] Therefore, 104.87 grams of ZrCl4 would be formed.

Explain Why Zr and Hf Form Specific Chlorides and Oxides Using Electron Configurations

As mentioned before, Zr and Hf have similar electron configurations, which result in similar chemical properties. Zr and Hf can both lose four electrons from their outer energy levels in order to form +4 oxidation states: - Zirconium: \[Zr \rightarrow Zr^{4+} + 4e^-\] - Hafnium: \[Hf \rightarrow Hf^{4+} + 4e^-\] This +4 oxidation state leads to the formation of tetrachlorides, MCl4, and dioxides, MO2, for both Zr and Hf.

Key concepts

Understanding Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly known as redox reactions, are fundamental chemical processes where electrons are transferred between substances. In a redox reaction, one species (the oxidizing agent) gains electrons while another (the reducing agent) loses electrons.

In the context of our zirconium example when reacting zirconium tetrachloride (ZrCl₄) with sodium (Na), it's essential to identify who's losing and gaining electrons. Sodium goes from an elemental state (oxidation number 0) to a +1 oxidation state when it forms sodium chloride (NaCl), indicating it donates electrons and is oxidized. Conversely, Zr goes from a +4 to a 0 oxidation state, showcasing that it accepts electrons and is reduced. Recognizing these changes is crucial for correctly describing the chemistry behind material transformations.

Chemical Properties of Elements and Electron Configuration

Every chemical element has distinct properties based on its electron configuration, which refers to the distribution of electrons in atomic or molecular orbitals. In our exercise, we saw that hafnium (Hf) and zirconium (Zr) have similar electron configurations, particularly in the outermost d subshell where they each host two electrons. This similarity means they often exhibit alike chemical behaviors, relevant for predicting where Hf would likely be located in nature. For instance, both Zr and Hf form chlorides (ZrCl₄ and HfCl₄) due to their ability to achieve a +4 oxidation state readily, underscoring the importance of electron configuration in determining the reactivity and chemistry of elements.

Understanding the underlying principles of electron configurations can significantly enhance comprehension of various chemical reactions, including predicting possible compounds and their properties. It also helps in foreseeing the behavior of elements during chemical reactions, making it imperative for solving stoichiometry problems.

Deciphering Stoichiometry Calculations

Stoichiometry calculations are the quantitative backbone of chemistry, enabling scientists and students to calculate relative quantities of reactants and products in chemical reactions. It requires a balanced chemical equation, as an unbalanced equation could lead to inaccurate predictions of the product amounts.

The exercise provided involves a solid understanding of stoichiometry to calculate the amount of zirconium tetrachloride (ZrCl₄) formed from zirconium dioxide (ZrO₂). Starting with the mass of ZrO₂, we use molar masses to convert to moles, considering the 1:1 stoichiometric ratio from the balanced equation. This conversion is the heart of stoichiometry, allowing us to translate mass to moles and back to determine the mass of products created from given reactants. The concept is pivotal in predicting the outcomes of reactions and in planning the amounts needed for various chemical processes.