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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 with alkali metals, hydrogen and halogens both act as electron acceptors. For the reaction between calcium (Ca) and fluorine (F鈧): \(Ca + F_2 \rightarrow CaF_2\). For the reaction between calcium (Ca) and hydrogen (H鈧) gas: \(Ca + H_2 \rightarrow CaH_2\). In both products, the oxidation number of calcium is +2, and its electron configuration is [Ar], as it loses two valence electrons from the 4s orbital.

Step by step solution

01

Comparing the Roles of Hydrogen and Halogens in Reactions with Alkali Metals

When alkali metals react with hydrogen, they form metal hydrides. And when they react with halogens, they form metal halides. In both these reactions, alkali metals lose electrons and get oxidized. Hydrogen and halogens act as electron acceptors; hydrogen gains an electron to form a hydride ion (H鈦), while halogens, like fluorine, gain an electron to form a halide ion (e.g., F鈦).
02

Writing Balanced Equations for the Reaction of Calcium with Fluorine and Hydrogen

For the reaction between calcium (Ca) and fluorine (F鈧): Ca + F鈧 鈫 CaF鈧 For the reaction between calcium (Ca) and hydrogen (H鈧) gas: Ca + H鈧 鈫 CaH鈧
03

Finding Oxidation Number and Electron Configuration of Calcium in Both Products

(a) In CaF鈧: The oxidation number of calcium is +2 because it loses two electrons to form a Ca虏鈦 ion, and two fluorine atoms each gain one electron to form F鈦 ions. Thus, the electron configuration of calcium in CaF鈧 is [Ar], where it has lost the two valence electrons from the 4s orbital. (b) In CaH鈧: The oxidation number of calcium is also +2 because it loses two electrons to form a Ca虏鈦 ion, and two hydrogen atoms each gain one electron to form H鈦 ions (hydride ions). The electron configuration of calcium in CaH鈧 is also [Ar], where it has lost its two valence electrons from the 4s orbital.

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Key Concepts

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

Alkali Metals
Alkali metals are a fascinating group of elements found in Group 1 of the periodic table. These metals include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). They share some unique characteristics that define their reactions with other elements.
  • Alkali metals are highly reactive, especially with water and air, due to their single electron in the outermost shell.
  • This reactivity increases as you move down the group, with francium being the most reactive.
  • Their low ionization energy makes it easy for them to lose their outer electron and form positive ions, or cations (e.g., Na鈦).

These metals react with hydrogen to form metal hydrides and with halogens to form metal halides. When alkali metals encounter hydrogen, they donate their single outer electron to form hydrides; similarly, they donate an electron to halogens, forming halides. Both reactions highlight the readiness of alkali metals to achieve a stable electronic configuration by shedding their lone valence electron.
Oxidation Number
Understanding oxidation numbers is crucial in identifying how atoms change during chemical reactions. An oxidation number signifies the total number of electrons that an atom either gains or loses to form a chemical bond.

In chemical reactions involving calcium:
  • When calcium reacts to form CaF鈧, it loses two electrons becoming a Ca虏鈦 ion. The oxidation number here is +2.
  • Similarly, in the formation of CaH鈧, calcium loses two electrons to become a Ca虏鈦 ion, maintaining the oxidation number of +2.
The oxidation number helps to balance equations and predict the outcome of chemical reactions, showcasing the electron transfer that occurs. In both CaF鈧 and CaH鈧, calcium's consistent oxidation number reflects its permanence in these reactions, making it easier to identify the products formed.
Balanced Equations
Balanced chemical equations ensure that the same number of each type of atom appears on both sides of the equation. This balance aligns with the law of conservation of mass, ensuring that mass is neither created nor destroyed during a chemical reaction.

For reactions involving calcium, we have:
  • Calcium reacts with fluorine (F鈧): \[\text{Ca} + \text{F}_2 \rightarrow \text{CaF}_2\]
  • Calcium reacts with hydrogen (H鈧): \[\text{Ca} + \text{H}_2 \rightarrow \text{CaH}_2\]
These equations show a 1:1 ratio for reactants and products, meaning calcium combines with a fixed amount of fluorine or hydrogen, resulting in even exchanges of electrons. Ensuring equations are balanced is key to accurately representing chemical changes and predicting how chemicals will interact and form new substances.
Electron Configuration
Electron configuration is a representation of the arrangement of electrons in an atom's orbitals, crucial for understanding chemical properties and reactivity.

Calcium, with an atomic number of 20, has the initial electron configuration of [\(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\)]. This indicates that its outermost shell is the 4s orbital with 2 electrons.
  • In CaF鈧 and CaH鈧, calcium loses these 4s electrons, resulting in the electron configuration of [Ar].
  • This shift reflects calcium's ability to form cations (Ca虏鈦) by releasing its valence electrons, achieving a more stable electron configuration like that of a noble gas.
Understanding electron configurations not only provides insight into why elements undergo certain reactions but also explains how they bond and what compounds they will form. By knowing an element's configuration, one can predict its behavior and interactions in various chemical contexts.

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Most popular questions from this chapter

Identify each statement as true or false: (a) lonization energies are always endothermic. (b) Potassium has a larger first ionization energy than lithium. (c) The second ionization energy of the sodium atom is larger than the second ionization energy of the magnesium atom. (d) The third ionization energy is three times the first ionization energy of an atom.

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Na}\) and \(\mathrm{K}\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Na}\) and \(\mathrm{K}\) 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) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

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

In Table 7.8 , the bonding atomic radius of neon is listed as \(58 \mathrm{pm},\) whereas that for xenon is listed as \(140 \mathrm{pm}\). A classmate of yours states that the value for Xe is more realistic than the one for Ne. Is she correct? If so, what is the basis for her statement?

Use electron configurations to explain the following observations: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

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