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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

01

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): \([1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^25d^24f^{14}]\)
02

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—La 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

(a) What is an isoelectronic series? (b) Which neutral atom is isoelectronic with each of the following ions: \(\mathrm{Al}^{3+}, \mathrm{Ti}^{4+}, \mathrm{Br}^{-}, \mathrm{Sn}^{2+}\)

Write balanced equations for the following reactions: (a) barium oxide with water, (b) iron(II) oxide with perchloric acid, (c) sulfur trioxide with water, (d) carbon dioxide with aqueous sodium hydroxide.

(a) Write the electron configuration for \(\mathrm{Li}\), and estimate the effective nuclear charge experienced by the valence electron. (b) The energy of an electron in a one-electron atom or ion equals \(\left(-2.18 \times 10^{-18} \mathrm{~J}\right)\left(\frac{\mathrm{Z}^{2}}{n^{2}}\right)\) where \(\mathrm{Z}\) is the nuclear charge and \(n\) is the principal quantum number of the electron. Estimate the first ionization energy of Li. (c) Compare the result of your calculation with the value reported in table \(7.4\), and explain the difference. (d) What value of the effective nuclear charge gives the proper value for the ionization energy? Does this agree with your explanation in (c)?

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 \(\mathrm{B}\) ?

For each of the following sets of atoms and ions, arrange the members in order of increasing size: (a) \(\mathrm{Se}^{2-}, \mathrm{Te}^{2-}, \mathrm{Se}\); (b) \(\mathrm{Co}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}\), (c) \(\mathrm{Ca}, \mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}\) (d) \(\mathrm{Be}^{2+}, \mathrm{Na}^{+}, \mathrm{Ne}\)

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