Ch.6 - Electronic Structure of Atoms

Problem 107a

Chapter 6, Problem 107a

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?

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insert step 1: Begin by reviewing the periodic table and identifying the position of hafnium (Hf) and zirconium (Zr). Note that hafnium is element 72 and zirconium is element 40.

insert step 2: Understand the electron configuration of zirconium (Zr). Zirconium has an atomic number of 40, so its electron configuration is [Kr] 4d^2 5s^2.

insert step 3: Determine the electron configuration of hafnium (Hf). Hafnium, with an atomic number of 72, has the electron configuration [Xe] 4f^14 5d^2 6s^2.

insert step 4: Compare the electron configurations of hafnium and zirconium. Notice that both elements have similar outer electron configurations, with hafnium having a filled 4f subshell in addition to the 5d and 6s electrons.

insert step 5: Conclude that the similarity in electron configurations between hafnium and zirconium supports Bohr's prediction that hafnium would be found with zirconium rather than with the rare earth elements, which have different electron configurations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electron Configuration

Electron configuration describes the distribution of electrons in an atom's orbitals. For elements, this configuration follows the Aufbau principle, Hund's rule, and the Pauli exclusion principle, which dictate how electrons fill available energy levels. Understanding the electron configuration of hafnium and zirconium is crucial, as it reveals their chemical properties and helps explain why hafnium is found alongside zirconium in nature.

Periodic Trends

Periodic trends refer to the predictable patterns in elemental properties that arise from their positions in the periodic table. These trends include atomic radius, ionization energy, and electronegativity. Hafnium and zirconium are in the same group (Group 4), which suggests they share similar chemical behaviors, supporting Bohr's prediction that hafnium would be found with zirconium rather than with the rare earth elements.

Chemical Similarity

Chemical similarity is the concept that elements with similar electron configurations exhibit comparable chemical properties. This is particularly evident in transition metals, where elements in the same group often form similar compounds. Bohr's assertion that hafnium would be found with zirconium is based on their analogous electron configurations, leading to similar reactivity and bonding characteristics, which are essential for understanding their occurrence in nature.

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Textbook Question

In the experiment shown schematically below, a beam of neutral atoms is passed through a magnetic field. Atoms that have unpaired electrons are deflected in different directions in the magnetic field depending on the value of the electron spin quantum number. In the experiment illustrated, we envision that a beam of hydrogen atoms splits into two beams. (c) What do you think would happen if the beam of hydrogen atoms were replaced with a beam of helium atoms? Why?

Microwave ovens use microwave radiation to heat food. The energy of the microwaves is absorbed by water molecules in food and then transferred to other components of the food. (a) Suppose that the microwave radiation has a wavelength of 10 cm. How many photons are required to heat 200 mL of water from 25 to 75 °C?

Open Question

The stratospheric ozone (O3) layer helps to protect us from harmful ultraviolet radiation. It does so by absorbing ultraviolet light and falling apart into an O2 molecule and an oxygen atom, a process known as photodissociation. O3(g) → O2(g) + O(g). Use the data in Appendix C to calculate the enthalpy change for this reaction. What is the maximum wavelength a photon can have if it is to possess sufficient energy to cause this dissociation? In what portion of the spectrum does this wavelength occur?

Textbook 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). (d) Using their electron configurations, account for the fact that Zr and Hf form chlorides MCl₄ and oxides MO₂.

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Textbook 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). (c) Solid zirconium dioxide, ZrO₂, reacts with chlorine gas in the presence of carbon. Starting with a 55.4-g sample of ZrO₂, calculate the mass of ZrCl₄ formed, assuming that ZrO₂ is the limiting reagent and assuming 100% yield.

Open Question

(c) Consider the metal oxides whose enthalpies of formation (in kJ mol⁻¹) are listed here: Oxide K₂O₍s₂₎, CaO₍s₂₎, TiO₂₍s₂₎, V₂O₅₍s₂₎, ΔHf° -363.2, -635.1, -938.7, -1550.6. Calculate the enthalpy changes in the following general reaction for each case: MnOm₍s₂₎ + H₂(g) → nM₍s₂₎ + mH₂O(g). (You will need to write the balanced equation for each case and then compute ΔH°.)