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Chapter 7: Problem 91

Note from the following table that the increase in atomic radius in moving from \(Z r\) to \(H f\) is smaller than in moving from \(Y\) to La. Suggest an explanation for this effect. \begin{tabular}{llll} \hline \multicolumn{3}{l} { Atomic Radii \((\AA)\)} \\ \hline \(\mathrm{Sc}\) & \(1.44\) & \(\mathrm{Ti}\) & \(1.36\) \\ \(\mathrm{Y}\) & \(1.62\) & \(\mathrm{Zr}\) & \(1.48\) \\ \(\mathrm{La}\) & \(1.69\) & \(\mathrm{Hf}\) & \(1.50\) \\ \hline \end{tabular}

Short Answer

Expert verified
The smaller increase in atomic radius from Zr to Hf compared to Y to La can be explained by the presence of the 4f orbital electrons in Hf, which are poor at shielding due to their compact shape. Although the 4f orbital electrons increase the electron shielding effect to some extent, the increase in nuclear charge (protons) from Zr to Hf is more significant. The increased nuclear charge causes a larger attraction between the nucleus and electrons, resulting in a smaller atomic radius.

Step by step solution

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1. Determining the electron configuration of the given elements

Start by finding the electron configurations of the given elements. It is important to remember that the electronic configuration is organized in terms of subshells or orbitals, and these are filled according to the periodic table order. Here are the electron configurations for the elements mentioned in the table: - Sc (atomic number 21): \([1s^22s^22p^63s^23p^64s^23d^1]\) - Ti (atomic number 22): \([1s^22s^22p^63s^23p^64s^23d^2]\) - Y (atomic number 39): \([1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^1]\) - Zr (atomic number 40): \([1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^2]\) - La (atomic number 57): \([1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^25d^14f^0 ]\) - Hf (atomic number 72): \([1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^25d^24f^{14}]\)
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2. Understanding the changes in atomic radius

Keep in mind the following factors that impact the atomic radius: 1. Nuclear charge: The number of protons in the nucleus increases as we move across the periodic table. This results in a stronger attraction between the nucleus and electrons, causing electron cloud to contract, thus reducing atomic radius. 2. Electron shielding: The inner shells shield the outer electrons from the nuclear charge. As we move across the table, additional electrons are added in the inner shells, which result in an increased shielding effect, leading to a slight increase in atomic radius. 3. Energy levels: When we move to a new period in the periodic table, an additional energy level is added, causing a significant increase in atomic radius.
03

3. Comparing the change in atomic radius from Y/La and Zr/Hf

In moving from Y to La, we observe an increase in atomic radius. This is due to the fact that La has an additional energy level (5d) compared to Y (4d), which increases electron shielding and results in a larger atomic radius. However, in moving from Zr to Hf, the increase in atomic radius is smaller. This can be explained by the presence of the 4f orbital electrons in Hf, which are poor at shielding due to their compact shape. Although the 4f orbital electrons increase the electron shielding effect to some extent, the increase in nuclear charge (protons) from Zr to Hf is more significant. The increased nuclear charge causes a larger attraction between the nucleus and electrons, resulting in a smaller atomic radius. In conclusion, the smaller increase in atomic radius from Zr to Hf is due to poor electron shielding by 4f orbital electrons relative to the increase in nuclear charge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Understanding electron configuration is crucial when discussing atomic radius and properties of elements. Electron configuration refers to the arrangement of electrons in the orbitals of an atom. Electrons are filled into orbitals based on their energy levels, following the Aufbau principle and Hund's rule. Each orbital can hold a specific number of electrons, and the configuration influences chemical behavior and periodic trends.

For example, Scandium (Sc) has its outermost electron in the 3d subshell, while Yttrium (Y) and Lanthanum (La) have their outermost electrons in the 4d and 5d subshells, respectively. These configurations impact the shielding effect and the effective nuclear charge experienced by the outer electrons, affecting the atomic radius in turn. With a proper grasp of electron configurations, one can predict changes in properties like atomic size across different elements in the periodic table.
Periodic Table Trends
The periodic table showcases trends in atomic radius that can be predicted based on an element's position. As you move from left to right across a period, atomic radii generally decrease due to the increase in nuclear charge without an increase in shielding. However, when moving down a group, atomic radii increase because electrons are added to new energy levels which are further from the nucleus, outweighing the increase in nuclear charge.

Understanding these trends helps explain why Y to La experiences a significant increase in atomic radius鈥擫a lies in a lower period and introduces a new electron shell. But from Zr to Hf, located in the same period, the additional protons have a more pronounced effect due to poor shielding by 4f electrons, causing a relatively smaller increase in atomic radius.
Electron Shielding
Electron shielding describes the phenomenon where inner electron shells block the outer electrons from the full effect of the nuclear charge. This weakens the hold the nucleus has on outermost electrons, potentially increasing atomic radius. As the number of inner electrons rises, they absorb some of the force exerted by the nucleus.

However, not all electron orbitals shield equally. For Hafnium (Hf), the 4f electrons do not shield as effectively as other orbitals due to their shape and proximity to the nucleus. Consequently, even though an additional inner shell might suggest greater shielding and therefore a larger atomic size, the type of orbital matters significantly. This peculiarity in shielding by 4f electrons in Hf versus the absence of such orbitals in Zr contributes to the unexpected trend in atomic radii.
Nuclear Charge
Nuclear charge is the total charge of the nucleus, determined by the number of protons. It plays a fundamental role in determining the size of an atom. Greater nuclear charge exerts a stronger pull on the electrons, reducing the atomic radius. This concept explains why, despite La having more electrons and a greater degree of electron shielding than Y, it also has more protons, which limits the increase in its atomic radius.

Moreover, in comparing Zr and Hf, the greater number of protons in Hf results in a stronger nuclear charge, thus pulling the electrons closer to the nucleus despite the increase in electron shielding. The interplay between nuclear charge and electron shielding directly impacts atomic radius and is a key factor when explaining the differences in atomic sizes between different elements. Understanding nuclear charge helps students predict how the atomic radius can change with the addition of protons in the nucleus.

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Most popular questions from this chapter

When magnesium metal is burned in air (Figure 3.5), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a \(6.3-\mathrm{g} \mathrm{Mg}\) ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(\mathrm{~g})\) and the reaction goes to \(\mathrm{com}\) pletion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(\mathrm{~g})\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

It is possible to produce compounds of the form \(\mathrm{GeClH}_{3}, \mathrm{GeCl}_{2} \mathrm{H}_{2}\), and \(\mathrm{GeCl}_{3} \mathrm{H}\). What values do you predict for the Ge \(-\mathrm{H}\) and \(\mathrm{Ge}-\mathrm{Cl}\) bond lengths in these compounds?

Use orbital diagrams to illustrate what happens when an oxygen atom gains two electrons. Why is it extremely difficult to add a third electron to the atom?

For each of the following pairs, which element will have the greater metallic character: (a) Li or Be, (b) Li or Na, (c) Sn or \(\mathrm{P}_{\text {, }}\) (d) \(\mathrm{Al}\) or \(\mathrm{B}\) ?

An element \(X\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(\left.1000^{\circ} \mathrm{C}\right)\). Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of \(58^{\circ} \mathrm{C}\). (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(X\) is a metal, nonmetal, or metalloid? Explain. (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element \(\mathrm{X}\).
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