/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 Note from the following table th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 88

Note from the following table that there is a significant increase in atomic radius upon moving from \(\mathrm{Y}\) to La, whereas the radii of Zr to Hf are the same. Suggest an explanation for this effect. $$ \begin{aligned} &\text { Atomic Radii (pm) }\\\ &\begin{array}{cccc} \hline \text { Sc } & 170 & \text { Ti } & 160 \\ \text { Y } & 190 & \text { Zr } & 175 \\ \text { La } & 207 & \text { Hf } & 175 \\ \hline \end{array} \end{aligned} $$

Short Answer

Expert verified
The increase from Y to La is due to added electron shells; the constant radii from Zr to Hf result from lanthanide contraction.

Step by step solution

01

Understand Periodic Trends

Atomic radius generally decreases across a period from left to right in the periodic table due to the increasing positive charge of the nucleus pulling the electron cloud closer.
02

Examine Y to La Transition

When moving from Y (Yttrium) to La (Lanthanum), we move into the lanthanide series. Lanthanides have an additional electron shell compared to their preceding elements, leading to a larger atomic radius due to the additional layer of electrons.
03

Examine Zr to Hf Transition

Moving from Zr (Zirconium) to Hf (Hafnium), both elements hold the same electron configuration beyond filled shells and have similar shielding effects. Hafnium's additional protons increase the nuclear charge, but the f-electron subshell does not shield these effectively, and thus the radius remains similar.
04

Explain the Lanthanide Contraction

The observation that the radii of Hf remain similar compared with Zr is a result of the lanthanide contraction. This occurs because of the poor shielding provided by 4f electrons in lanthanides, allowing added protons in subsequent elements to pull the electrons closer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Atomic Radius
The atomic radius refers to the size of an atom, typically measured from the nucleus to the outer boundary of the surrounding cloud of electrons. It is an essential indicator of the chemical properties of an element. As you move across a period from left to right in the periodic table, the atomic radius generally decreases. This reduction is due to the increased positive charge of the nucleus, which attracts the electron cloud more strongly and pulls it closer to the core. Conversely, as you move down a group, the atomic radius increases because additional electron shells are added, outweighing the nuclear attraction due to added protons.
  • The nucleus gets more positively charged as you move right.
  • Electrons are added to the same shell in a period, increasing nuclear pull.
  • While moving down, new shells are added, increasing the radius.
Understanding how atomic radius varies is crucial when predicting how elements react and form bonds, impacting their involvement in forming various compounds.
Lanthanide Contraction
Lanthanide contraction is a unique phenomenon observed in the elements of the period following the lanthanides. When analyzing elements such as Zirconium (Zr) and Hafnium (Hf), despite being in different groups, their atomic radii are quite similar. This surprise is due to the poor shielding effect of the 4f orbitals present in lanthanides preceding Hafnium. Even though these orbitals are deeply buried, they don't shield the nuclear charge effectively. This means that protons added as you move across the lanthanide series draw electrons closer, reducing atomic size despite the increase in shell number.
  • 4f orbitals have poor shielding ability.
  • Protons draw electrons closer as they add up.
  • Leads to small size variation in elements after lanthanides.
This factor plays a significant role in explaining why Hafnium and Zirconium have nearly identical atomic sizes even though Hafnium has more protons. This subtlety impacts chemistry significantly, influencing element reactivity and physical characteristics.
Electron Configuration
Electron configuration describes the distribution of electrons in an atom's orbitals. It follows the Aufbau principle, filling shells and subshells starting from the lowest energy level ascending to higher levels. This order is crucial in understanding the chemical properties and reactivity of elements. As you transition from Y (Yttrium) to La (Lanthanum), a new electron shell begins to fill, causing an increase in atomic radius. This is because the added electrons occupy a new, more distant shell, increasing the atom's size.
  • Predicts how electrons are arranged in atoms.
  • Determines chemical reactivity and bonding patterns.
  • Aufbau principle guides filling of electrons in orbitals.
On the other hand, when moving from Zr to Hf, despite an additional electron in a new f subshell, due to lanthanide contraction, the radius does not increase as expected. This is a perfect example of how intricate and impactful electron configuration can be in understanding elemental properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Arrange the following atoms in order of increasing effective nuclear charge experienced by the electrons in the \(n=2\) shell: Be, Br, Na, P, Se.

Which of the following chemical equations is connected to the definitions of (a) the first ionization energy of oxygen, (b) the second ionization energy of ox ' ygen, and \((\mathbf{c})\) the electron affinity of oxygen? (i) \(\mathrm{O}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{O}^{-}(g)\) (ii) \(\mathrm{O}(g) \longrightarrow \mathrm{O}^{+}(g)+\mathrm{e}^{-}\) (iii) \(\mathrm{O}(g)+2 \mathrm{e}^{-} \longrightarrow \mathrm{O}^{2-}(g)\) (iv) \(\mathrm{O}(g) \longrightarrow \mathrm{O}^{2+}(g)+2 \mathrm{e}^{-}\) \((\mathbf{v}) \mathrm{O}^{+}(g) \longrightarrow \mathrm{O}^{2+}(g)+\mathrm{e}^{-}\)

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in Na and \(\mathrm{K}\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Na}\) and \(\mathrm{K}\) by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }}\) ? (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

When magnesium metal is burned in air (Figure 3.6), 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-g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) 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.

(a) What is the trend in first ionization energies as one proceeds down the group 17 elements? Explain how this trend relates to the variation in atomic radii. (b) What is the trend in first ionization energies as one moves across the fourth period from \(\mathrm{K}\) to \(\mathrm{Kr}\) ? How does this trend compare with the trend in atomic radii?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.