Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Discussing this chapter, a classmate says, "An element that commonly forms a cation is a metal." Do you agree or disagree? Explain your answer.

Short Answer

Expert verified
I agree with the statement "An element that commonly forms a cation is a metal." This is because metals have a tendency to lose their valence electrons due to their electron configuration, which results in the formation of positive ions or cations. For example, sodium (Na) loses one electron to form a sodium cation (Na+), and magnesium (Mg) loses two electrons to form a magnesium cation (Mg2+). Therefore, it is true that metals commonly form cations.

Step by step solution

01

Understanding cations

A cation is a positively charged ion that has lost one or more electrons. This process is called ionization. Elements can form cations by losing electrons, which increases their overall positive charge.
02

Understanding metals

Metals are elements that have some characteristic properties such as being shiny, malleable, ductile, and good conductors of heat and electricity. In general, metals have one to three electrons in their outermost energy levels (valence electrons). These electrons have a relatively weak attachment to their nucleus, which makes it easier for metals to lose electrons and form cations.
03

Determining if metals commonly form cations

Metals, due to their properties and electron configuration, have a tendency to lose their valence electrons, which results in the formation of positive ions or cations. For example, sodium (Na) loses one electron to form a sodium cation (Na+), and magnesium (Mg) loses two electrons to form a magnesium cation (Mg2+). Therefore, it is true that an element that commonly forms a cation is a metal.
04

Conclusion

We agree with the statement "An element that commonly forms a cation is a metal," as metals tend to lose their valence electrons due to their electron configuration, forming cations in the process.

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 Vaia!

Key Concepts

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

Metals
Metals are fascinating elements with distinct properties that make them unique. They are known for their shiny appearance and are typically malleable and ductile, which means they can be shaped or stretched into wires without breaking.
Metals are excellent conductors of heat and electricity, which is why they are commonly used in many electrical applications.
Additionally, metals are generally found on the left side and towards the center of the periodic table.
Due to their structure, metals have a tendency to lose electrons easily, primarily to achieve a stable electron configuration. This property makes them more likely to form cations, which are positively charged ions usually created when a metal atom loses one or more electrons.
Electron Configuration
Electron configuration is a way to describe how electrons are distributed around an atom's nucleus. Each atom has distinct energy levels that can hold a certain number of electrons.
The electrons closest to the nucleus occupy the lowest energy levels, while those further away reside in higher energy levels.
For metals, the electrons in their outermost energy level, or shell, are called valence electrons.
These valence electrons are key players in chemical reactions because they are the most loosely bound to the nucleus.
When metals lose their valence electrons, they achieve a more stable electron configuration, often resembling the nearest noble gas, which has a full outer shell.
This stability is a driving factor behind the formation of cations.
Ionization
Ionization is a process where an atom loses or gains electrons, resulting in the formation of ions.
For metals, ionization involves losing electrons to form cations. In a typical ionization process, especially for metals, energy called ionization energy is needed to remove electrons from the atom.
The tendency of metals to lose electrons during ionization is largely because their valence electrons are held less tightly due to the larger atomic radius and the shielding effect from inner electrons.
Therefore, metals have relatively low ionization energies, making it easier for them to lose valence electrons, leading to cation formation.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a critical role in chemical bonding and reactions.
In metals, these valence electrons are usually one to three in number and are less tightly bound to the nucleus.
  • This weak hold allows the electrons to be easily removed during reactions.
  • The loss of valence electrons helps the atom achieve a stable electron configuration.
This vulnerability is what enables metals to readily form cations.
When a metal atom loses valence electrons, it often achieves the electron configuration of the nearest noble gas, stabilizing the ion through energy minimization.

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

(a) Use orbital diagrams to illustrate what happens when an oxygen atom gains two electrons. (b) Why does \(\mathrm{O}^{3-}\) not exist?

We can draw an analogy between the attraction of an electron to a nucleus and seeing a lightbulb -in essence, the more \(n u=\) clear charge the electron "sees," the greater the attraction. (a) Within this analogy, discuss how the screening by core electrons is analogous to putting a frosted-glass lampshade between the lightbulb and your eyes, as shown in the illustration. (b) Explain how we could mimic moving to the right in a row of the periodic table by changing the wattage of the lightbulb. (c) How would you change the wattage of the bulb and/or the frosted glass to mimic the effect of moving down a column of the periodic table? [Section 7.2]

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

Compare the elements bromine and chlorine with respect to the following properties: (a) electron configuration, (b) most common ionic charge, (c) first ionization energy, (d) reactivity toward water, (e) electron affinity, (f) atomic radius. Account for the differences between the two elements.

In the chemical process called electron transfer, an electron is transferred from one atom or molecule to another. (We will talk about electron transfer extensively in Chapter 20.) A simple electron transfer reaction is $$ \mathrm{A}(g)+\mathrm{A}(g) \longrightarrow \mathrm{A}^{+}(g)+\mathrm{A}^{-}(g) $$ In terms of the ionization energy and electron affinity of atom A, what is the energy change for this reaction? For a representative nonmetal such as chlorine, is this process exothermic? For a representative metal such as sodium, is this process exothermic? [Sections 7.4 and 7.5\(]\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.

Sign-up for free