/*! 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 40 Why don't alkaline earth metal c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 40

Why don't alkaline earth metal cations substitute for alkali metal cations in cases where the ionic radii are similar?

Short Answer

Expert verified
Answer: Alkaline earth metal cations do not typically substitute for alkali metal cations, even in cases where the ionic radii are similar, because of the differences in their electronic configurations, charges, and lattice energy. The higher charge on alkaline earth metal cations leads to stronger ionic bonds and higher lattice energy, which makes it less likely for them to be substituted by lower charged alkali metal cations.

Step by step solution

01

Recall the general electronic configuration of alkali and alkaline earth metals

Alkali metals belong to Group 1 in the periodic table and have an electronic configuration of [noble gas] ns^1, where 'n' is the period number. Alkaline earth metals belong to Group 2 and have an electronic configuration of [noble gas] ns^2.
02

Relate the electronic configuration to the formation of cations

When alkali metals and alkaline earth metals lose electrons to form cations, they achieve a stable noble gas configuration. Alkali metals lose one electron from the outermost shell, forming +1 cations (e.g., Na^+, K^+), while alkaline earth metals lose two electrons from the outermost shell, forming +2 cations (e.g., Ca^2+, Mg^2+).
03

Consider the charges of the cations in substitution

Since the alkaline earth metal cations have a +2 charge and alkali metal cations have a +1 charge, their electrostatic interactions with the surrounding anions are different. Alkaline earth metal cations form stronger ionic bonds with anions due to their higher charge, making it less likely for them to be substituted by alkali metal cations with a lower charge.
04

Compare the lattice energy of ionic compounds

Lattice energy is the energy required to break an ionic compound into its constituent ions, and it depends on the charges of the ions and the distance between them. According to Coulomb's law, lattice energy is proportional to the product of the charges of the ions and inversely proportional to the distance between their centers. Due to the higher charge on alkaline earth metal cations, ionic compounds containing them will have a higher lattice energy compared to those with alkali metal cations of similar ionic radii, making substitution less favorable.
05

Conclusion

Alkaline earth metal cations do not typically substitute for alkali metal cations, even in cases where the ionic radii are similar, because of the differences in their electronic configurations, charges, and lattice energy. The higher charge on alkaline earth metal cations leads to stronger ionic bonds and higher lattice energy, which makes it less likely for them to be substituted by lower charged alkali metal cations.

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!

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

Some medicines used in treating depression contain lithium carbonate. Draw the Lewis structure for \(\mathrm{Li}_{2} \mathrm{CO}_{3}\).

Which ion channel must accommodate the larger cation, a potassium or a sodium ion channel?

Several isotopes of arsenic are used in medical imaging. Which isotope, \(^{72}\) As or \(^{77}\) As, is more likely to be useful for PET imaging?

What are the products of radioactive decay of \(^{137} \mathrm{Cs} ?\) Write a balanced equation for the nuclear decay reaction.

Effects of Excess Fluoridation on Teeth Too much fluoride might lead to the formation of calcium fluoride according to the reaction $$\begin{aligned}\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s)+& 10 \mathrm{F}^{-}(a q) \rightleftharpoons \\\& 5 \mathrm{CaF}_{2}(s)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{OH}^{-}(a q)\end{aligned}$$ Write the equilibrium constant expression for the reaction. Given the \(K_{\mathrm{sp}}\) values for the following two reactions, calculate \(K\) for the reaction between \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})\) and fluoride ion that forms \(\mathrm{CaF}_{2}\) \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3}(\mathrm{OH})(s) \rightleftharpoons 5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{PO}_{4}^{3-}(a q)+\mathrm{OH}^{-}(a q)\) \(K_{\mathrm{sp}}=2.3 \times 10^{-59}\) \(\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathrm{sp}}=3.9 \times 10^{-11}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Nuclear 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.