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Chapter 4: Problem 66

Consider the following electron configuration: $$\left(\sigma_{3 s}\right)^{2}\left(\sigma_{3 s}^{* *}\right)^{2}\left(\sigma_{3 p}\right)^{2}\left(\pi_{3 p}\right)^{4}\left(\pi_{3 p}^{*}\right)^{4}$$ Give four species that, in theory, would have this electron configuration.

Short Answer

Expert verified
Four species that have the electron configuration \(\left(\sigma_{3 s}\right)^{2}\left(\sigma_{3 s}^{*}\right)^{2}\left(\sigma_{3 p}\right)^{2}\left(\pi_{3 p}\right)^{4}\left(\pi_{3 p}^{*}\right)^{4}\) are Silicon tetrachloride (SiCl₄), Sulfur hexafluoride (SF₆), Carbon tetraiodide (CI₄), and Germanium tetrabromide (GeBr₄).

Step by step solution

01

1. Count the electrons in the configuration

First, let's count the number of electrons in the given electron configuration. We will sum the superscripts of the molecular orbitals: \((\sigma_{3 s})^{2}+(\sigma_{3 s}^{*})^{2}+(\sigma_{3 p})^{2}+(\pi_{3 p})^{4}+(\pi_{3 p}^{*})^{4}\) \(= 2 + 2 + 2 + 4 + 4 = 14\) So, there is a total of 14 electrons in this configuration.
02

2. Identify species with 14 valence electrons

Now, we want to identify atomic or molecular species that have 14 valence electrons. Recall that elements in the same group of the periodic table have the same number of valence electrons. To help figure out which species might have 14 valence electrons, let's look at how many valence electrons are in some common groups: - Group 1 (alkali metals): 1 valence electron - Group 2 (alkaline earth metals): 2 valence electrons - Group 13 (boron group): 3 valence electrons - Group 14 (carbon group): 4 valence electrons - Group 15 (nitrogen group): 5 valence electrons - Group 16 (chalcogens): 6 valence electrons - Group 17 (halogens): 7 valence electrons - Group 18 (noble gases): 8 valence electrons Now, let's try to generate species with 14 valence electrons. A few examples are: 1. Silicon tetrachloride (SiCl₄): Si has 4 valence electrons, and each Cl has 7 valence electrons. So, SiCl₄ has \(4 + 4(7) = 4+28 = 14\) valence electrons. 2. Sulfur hexafluoride (SF₆): S has 6 valence electrons, and each F has 7 valence electrons. So, SF₆ has \(6 + 6(7) = 6+42 = 14\) valence electrons. 3. Carbon tetraiodide (CI₄): C has 4 valence electrons, and each I has 7 valence electrons. So, CI₄ has \(4 + 4(7) = 4+28 = 14\) valence electrons. 4. Germanium tetrabromide (GeBr₄): Ge has 4 valence electrons, and each Br has 7 valence electrons. So, GeBr₄ has \(4 + 4(7) = 4+28 = 14\) valence electrons. These are just four examples; there could be other molecules with 14 valence electrons. But for the purpose of this exercise, we have found four species that have the same electron configuration as the one given in the problem statement.

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Key Concepts

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

Molecular Orbitals
Molecular orbitals are at the heart of understanding how electrons are organized in molecules. They're formed when atomic orbitals overlap as atoms combine to form molecules. This overlap allows electrons to move between atoms, creating a molecular orbital that belongs to the molecule as a whole, rather than a single atom.

These orbitals can be bonding, anti-bonding, or non-bonding, influencing the stability and properties of the molecule. In the example of the electron configuration \(\left(\sigma_{3s}\right)^{2}\left(\sigma_{3s}^{*}\right)^{2}\left(\sigma_{3p}\right)^{2}\left(\pi_{3p}\right)^{4}\left(\pi_{3p}^{*}\right)^{4}\), the \(\sigma\) and \(\pi\) notations refer to the type of overlap - \(\sigma\) being head-to-head or tail-to-tail and \(\pi\) being side-to-side. The stars (*) indicate anti-bonding orbitals, which can house electrons but destabilize the bond.

Understanding how these molecular orbitals interact to accommodate electrons helps explain the behavior and properties of molecules, giving us insight into their chemical reactivity and physical properties.
Valence Electrons
The concept of valence electrons is crucial for explaining how elements connect and interact with each other to form compounds. Valence electrons are the electrons in an atom's outermost shell and are primarily responsible for an element’s chemical characteristics. They play a key role in the formation of chemical bonds, such as ionic or covalent bonds.

In the context of the electron configuration exercise, we counted the total number of electrons that would be considered valence electrons in a molecular context. It's these electrons that participate in chemical bonding and largely determine how a molecule interacts with other substances. The electron configuration gave us an insight into the stability of the molecule as well, showing us how the valence electrons are distributed among bonding and anti-bonding molecular orbitals.
Periodic Table Groups
Elements in the periodic table are arranged into columns called groups, which share similar chemical properties. This grouping is because they have the same number of valence electrons, hence showing similar behavior in chemical reactions. For example, all alkali metals in group 1 have one valence electron and are highly reactive.

When searching for species with a specific electron configuration, such as one with 14 valence electrons as given in the exercise, recognizing the periodic table groups is extremely useful. By knowing that elements from the same group have the same number of valence electrons, you can predict and construct molecules that fit a particular electron configuration. As seen in the exercise, molecules were constructed combining elements from different groups to achieve the desired number of valence electrons and hence, the underlying electron configuration.

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Most popular questions from this chapter

Draw the Lewis structures, predict the molecular structures, and describe the bonding (in terms of the hybrid orbitals for the central atom) for the following. a. \(\mathrm{XeO}_{3}\) b. \(\mathrm{XeO}_{4}\) c. \(\mathrm{XeOF}_{4}\) d. \(\mathrm{XeOF}_{2}\) e. \(\mathrm{XeO}_{3} \mathrm{F}_{2}\)

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. a. \(\mathrm{CN}^{+}\) b. CN c. \(\mathrm{CN}^{-}\)

Which of the following statements is/are true? Correct the false statements. a. The molecules \(\operatorname{SeS}_{3}, \operatorname{SeS}_{2}, \operatorname{PCl}_{5}, \operatorname{TeCl}_{4},\) ICl \(_{3}\), and \(\mathrm{XeCl}_{2}\) all exhibit at least one bond angle which is approximately \(120^{\circ} .\) b. The bond angle in \(\mathrm{SO}_{2}\) should be similar to the bond angle in \(\mathrm{CS}_{2}\) or \(\mathrm{SCl}_{2}\) c. Of the compounds \(\mathrm{CF}_{4}, \mathrm{KrF}_{4},\) and \(\mathrm{SeF}_{4},\) only \(\mathrm{SeF}_{4}\) exhibits an overall dipole moment (is polar). d. Central atoms in a molecule adopt a geometry of the bonded atoms and lone pairs about the central atom in order to maximize electron repulsions.

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions: $$ \begin{aligned} & \mathrm{NCl}_{3}(g) \longrightarrow \mathrm{NCl}_{2}(g)+\mathrm{Cl}(g) & \Delta E &=375 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{ONCl}(g) & \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}(g) & & \Delta E=158 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Rationalize the difference in the values of \(\Delta E\) for these reactions, even though each reaction appears to involve only the breaking of one \(\mathrm{N}-\mathrm{Cl}\) bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with 25 -nm light. a. Using the following information, indicate what species can form in the flask during irradiation. $$\begin{array}{ll} \mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g) & \Delta E=941 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-} & \Delta E=1501 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-} & \Delta E=1402 \mathrm{kJ} / \mathrm{mol} \end{array}$$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of \(\mathrm{N}_{2}(1501 \mathrm{kJ} /\) mol) is greater than the first ionization energy of atomic nitrogen (1402 kJ/mol).
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