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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." Metals have a tendency to lose their outer electrons to achieve stability by replicating noble gas electron configurations, which results in the formation of cations. Examples include Sodium (Na) forming Na⁺, Magnesium (Mg) forming Mg²⁺, and Iron (Fe) forming Fe²⁺ and Fe³⁺ ions.

Step by step solution

01

Understanding cations and metals

Cations are positively charged ions, which are formed when an atom loses one or more of its electrons. Metals are elements that are characterized by their ability to conduct electricity, malleability, ductility, and the tendency to lose electrons from their outermost shells to form positive ions. Metals are located on the left side of the periodic table, including alkali metals, alkaline earth metals, and transition metals.
02

Metal's ability to form cations

The outermost electron shell of a metal atom generally contains one, two, or three electrons, which are relatively loosely held by the nucleus. As a result, metals are more likely to lose their outermost electrons in order to achieve a stable electron configuration, often resembling the electron configuration of a noble gas. This loss of electrons results in a net positive charge, leading to the formation of a cation.
03

Agreeing or disagreeing with the statement

With the understanding of cations and the tendency of metals to form them, we can now evaluate the statement made by the classmate. The statement, "An element that commonly forms a cation is a metal," is accurate. Metals have the propensity to lose their outer electrons to become cations because of their desire to achieve stability by replicating noble gas electron configurations.
04

Provide examples and explanation for agreement

Some examples of metals that commonly form cations are: 1. Sodium (Na): Sodium readily loses one electron to become a positively charged ion, Na⁺. 2. Magnesium (Mg): Magnesium loses two electrons to become a positively charged ion, Mg²⁺. 3. Iron (Fe): Iron can lose two or three electrons to form Fe²⁺ and Fe³⁺ ions. In each case, the metal element loses electrons to achieve a stable electron configuration, and in doing so, it forms a cation. Therefore, we agree with the statement made by the classmate.

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Most popular questions from this chapter

Using only the periodic table, arrange each set of atoms in order from largest to smallest: (a) K, Li, Cs; (b) Pb, Sn, Si; (c) F, O, N.

Identify two ions that have the following ground-state electron configurations: (a) \([\mathrm{Ar}]\), (b) \([\mathrm{Ar}] 3 d^{5}\), (c) \([\mathrm{Kr}] 5 s^{2} 4 d^{10}\).

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Si}\) and \(\mathrm{Cl}\) 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) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text {eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text {eff }}\) for a valence electron in \(\mathrm{P}\), phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Elemental cesium reacts more violently with water than does elemental sodium. Which of the following best explains this difference in reactivity? (i) Sodium has greater metallic character than does cesium. (ii) The first ionization energy of cesium is less than that of sodium. (iii) The electron affinity of sodium is smaller than that of cesium. (iv) The effective nuclear charge for cesium is less than that of sodium. (v) The atomic radius of cesium is smaller than that of sodium.

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7}\). (a) What is the name of this product (see Table \(2.6\) )? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Under usual conditions, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a colorless liquid with a boiling point of \(81^{\circ} \mathrm{C}\). Is this boiling point expected or surprising? (d) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q)\) ? (e) If the oxygen in \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is considered to have the \(-2\) oxidation state, what is the oxidation state of the \(\mathrm{Cl}\) ? What is the electron configuration of \(\mathrm{Cl}\) in this oxidation state?

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