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(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?

Short Answer

Expert verified
In reactions involving alkali metals, hydrogen and halogens act as electron acceptors, forming metal hydrides and halides. The balanced equations for the reaction of fluorine and hydrogen with calcium are: \(Ca + F_2 \rightarrow CaF_2\) and \(Ca + H_2 \rightarrow CaH_2\), respectively. In both products, calcium has an oxidation number of +2 and an electron configuration of \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6\), which is similar to the noble gas Argon.

Step by step solution

01

Formation of hydrides and halides by alkali metals

Alkali metals react with hydrogen to form metal hydrides and with halogens to form metal halides. For example: - Formation of a metal hydride: \(2Na + H_2 \rightarrow 2NaH\) - Formation of a metal halide: \(2Na + F_2 \rightarrow 2NaF\) In these reactions, hydrogen and halogens act as electron acceptors, with each atom gaining an electron from the alkali metal to form a stable anion (H- or halide). Alkali metals lose one electron to form a stable cation (M+).
02

Write balanced equations for the given reactions

(a) Reaction of fluorine with calcium: The balanced equation for the reaction of fluorine with calcium is: \(Ca + F_2 \rightarrow CaF_2\) (b) Reaction of hydrogen with calcium: The balanced equation for the reaction of hydrogen with calcium is: \( Ca + H_2 \rightarrow CaH_2 \)
03

Determine the oxidation number of calcium

In both products, the oxidation number of calcium (Ca) is +2. In each reaction, calcium loses 2 electrons to form a stable cation (Ca2+). In the reaction with fluorine, each fluorine gains one electron to form a stable anion (F-), while in the reaction with hydrogen, each hydrogen gains one electron to form a stable anion (H-).
04

Determine the electron configuration of calcium

The electron configuration of a neutral calcium atom is: \[1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2\] In both products (CaF2 and CaH2), calcium loses 2 electrons, and its oxidation state is +2. The electron configuration of Ca2+ in these products becomes: \[1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6\] This is also the electron configuration of the noble gas Argon (Ar). In conclusion, calcium forms hydrides and halides by reacting with hydrogen and halogens, respectively, and attains a +2 oxidation state with an electron configuration similar to the noble gas Argon in both products.

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Key Concepts

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

Halides
Halides form when halogens react with metals. Halogens are a group of elements in the periodic table, including fluorine, chlorine, bromine, iodine, and astatine. These are very reactive non-metals and will readily accept electrons from metals to form halides.

In the reaction between a metal and a halogen, the metal donates its outermost electron to the halogen, forming a stable ionic compound. The metal becomes a cation (positively charged ion), while the halogen becomes an anion (negatively charged ion).
  • For example, sodium (Na) reacting with fluorine (F) forms sodium fluoride ( NaF ).
  • Calcium (Ca) reacting with fluorine forms calcium fluoride ( CaF_2 ).
Halides are important in various applications, including salts used in everyday life and materials in medical and industrial settings.
Oxidation Number
The oxidation number describes the degree of oxidation of an atom in a chemical compound. It reflects how many electrons an atom gains or loses when forming a compound.

For calcium in compounds like CaF_2 and CaH_2 , the oxidation number is +2. This means calcium has lost two electrons. It becomes positively charged and forms a Ca^{2+} ion.
  • In CaF_2 , each fluorine atom gains one electron from calcium, resulting in two F^- ions.
  • In CaH_2 , each hydrogen atom gains one electron from calcium, resulting in two H^- ions.
The concept of oxidation numbers is crucial for understanding redox reactions and predicting how elements interact.
Electron Configuration
Electron configuration arranges the electrons in an atom across different energy levels and subshells. It's like a map of where electrons reside around the nucleus. For a neutral calcium atom, the electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 .

When calcium forms a +2 ion, it loses two electrons, achieving the electron configuration of the noble gas argon, 1s^2 2s^2 2p^6 3s^2 3p^6 . Losing two electrons makes calcium stable as it mimics the electron configuration of a noble gas.
  • This loss of electrons is what gives Ca its +2 oxidation state.
  • Understanding electron configuration helps explain chemical stability and reactivity.
Alkali Metals
Alkali metals are highly reactive elements found in Group 1 of the periodic table. These include lithium, sodium, potassium, rubidium, cesium, and francium.

These metals have a single electron in their outermost shell, making them eager to lose that electron to achieve noble gas electron configuration.
  • When reacting with hydrogen, alkali metals form hydrides, such as 2Na + H_2 ightarrow 2NaH .
  • When reacting with halogens, they form halides, like 2Na + F_2 ightarrow 2NaF .
Due to their reactivity, alkali metals are never found in nature in their elemental form but in mineral compounds.
Balanced Equations
Balanced chemical equations represent the conservation of mass in reactions. They show that the number of atoms of each element remains constant before and after a reaction.

Each chemical equation must include equal numbers of each type of atom on both sides of the equation. It's like balancing a chemical checkbook, ensuring everything is accounted for.
  • For example, the reaction of calcium with fluorine: Ca + F_2 ightarrow CaF_2 .
  • Calcium with hydrogen: Ca + H_2 ightarrow CaH_2 .
Balancing equations helps understand the stoichiometry in reactions, predicting the amounts of substances consumed and produced.

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Most popular questions from this chapter

Tungsten has the highest melting point of any metal in the periodic table: \(3422^{\circ} \mathrm{C}\) . The distance between \(\mathrm{W}\) atoms in tungsten metal is 2.74 A. (a) What is the atomic radius of atungsten atom in this environment? (This radius is called the metallic radius.) If you put tungsten metal under high pressure, predict what would happen to the distance between \(\mathrm{W}\) atoms.

Little is known about the properties of astatine, At, because of its rarity and high radioactivity. Nevertheless, it is possible for us to make many predictions about its properties. (a) Do you expect the element to be a gas, liquid, or solid at room temperature? Explain. (b) Would you expect At to be a metal, nonmetal, or metalloid? Explain. (c) What is the chemical formula of the compound it forms with Na?

Potassium superoxide, \(\mathrm{KO}_{2},\) is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that 2 \(\mathrm{mol}\) of \(\mathrm{KO}_{2}(s)\) react with each mole of= \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(s)\) and \(\mathrm{CO}_{2}(g) .(\mathbf{b})\) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume 18.0 \(\mathrm{g} \mathrm{CO}_{2}(g) ?\) What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

Write equations that show the process for (a) the first two ionization energies of lead and (b) the fourth ionization energy of zirconium.

Identify each statement as true or false: (a) Ionization energies are always negative quantities. (b) Oxygen has a larger first ionization energy than fluorine. (c) The second ionization energy of an atom is always greater than its first ionization energy. (d) The third ionization energy is the energy needed to ionize three electrons from a neutral atom.

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