Discussing this chapter, a classmate says, "Since elements that form cations are metals and elements that form anions are nonmetals, elements that do not form ions are metalloids." Do you agree or disagree?

Short Answer

Expert verified
I disagree with the statement, "Since elements that form cations are metals and elements that form anions are nonmetals, elements that do not form ions are metalloids." This is because metalloids have intermediate properties between metals and nonmetals and can sometimes lose electrons to form cations or gain electrons to form anions, depending on the specific element and the conditions. Their ion-forming abilities depend on their specific properties and electron structures.

Step by step solution

01

Understand the properties of metals, nonmetals, and metalloids

Metals are elements that are generally shiny, conductive, and malleable. They tend to lose electrons and form positive ions (cations). Nonmetals are elements that do not share the characteristics of metals. They are generally poor conductors of heat and electricity and tend to gain electrons, forming negative ions (anions). Metalloids, also called semi-metals, have properties that are intermediate between metals and nonmetals. They can sometimes behave like metals and other times like nonmetals, depending on the conditions.
02

Analyze the ion-forming abilities of metals and nonmetals

Metals lose electrons to form cations, and nonmetals gain electrons to form anions. This ion-forming ability is mainly due to their respective properties and electron structures. Metals, having low ionization energy, tend to lose electrons easily, while nonmetals, with high electron affinities, have a tendency to gain electrons.
03

Assess the ion-forming abilities of metalloids

Metalloids have intermediate properties between metals and nonmetals, which means they can sometimes lose electrons to form cations or gain electrons to form anions, depending on the specific element and the conditions in which they are present. Therefore, saying that metalloids do not form ions is not entirely accurate.
04

Conclusion

Based on the above analysis, we can disagree with the statement, "Since elements that form cations are metals and elements that form anions are nonmetals, elements that do not form ions are metalloids." Metalloids can form ions under certain conditions, and their ion-forming abilities depend on their specific properties and electron structures.

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!

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

We will see in Chapter 12 that semiconductors are materials that conduct electricity better than nonmetals but not as well as metals. The only two elements in the periodic table that are technologically useful semiconductors are silicon and germanium. Integrated circuits in computer chips today are based on silicon. Compound semiconductors are also used in the electronics industry. Examples are gallium arsenide, GaAs; gallium phosphide, GaP; cadmium sulfide, CdS; and cadmium selenide, CdSe. (a) What is the relationship between the compound semiconductors’ compositions and the positions of their elements on the periodic table relative to Si and Ge? (b) Workers in the semiconductor industry refer to "II–VI" and "III–V" materials, using Roman numerals. Can you identify which compound semiconductors are II–VI and which are III–V? (c) Suggest other compositions of compound semiconductors based on the positions of their elements in the periodic table.

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 \(\mathrm{Xe} .\) Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2} ?\) If so, suggest a product or products of this reaction.

Write the electron configurations for the following ions, and determine which have noble-gas configurations: \((\mathbf{a})\mathrm{Co}^{2+}\) \((\mathbf{b})\mathrm{Sn}^{2+},(\mathbf{c}) \mathrm{Zr}^{4+},(\mathbf{d}) \mathrm{Ag}^{+},(\mathbf{e}) \mathrm{S}^{2-}.\)

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Chlorine reacts with water. (b) Barium metal is heated in an atmosphere of hydrogen gas. (c) Lithium reacts with sulfur. (d) Fluorine reacts with magnesium metal.

(a) As described in Section 7.7 , the alkali metals react with hydrogen to form hydrides and react with halogens to form halides. Compare the roles of hydrogen and halogens in these reactions. Write balanced equations for the reaction of fluorine with calcium and for the reaction of hydrogen with calcium. (b) What is the oxidation number and electron configuration of calcium in each product?

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