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(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about \(S i\) in terms of hydrid orbitals? Does it make a difference that Si lies in the row below \(\mathrm{C}\) in the periodic table? Explain.

Short Answer

Expert verified
In summary, the Lewis structures for ethane (C2H6), ethylene (C2H4), and acetylene (C2H2) have single, double, and triple bonds between their carbon atoms, respectively. The hybridization of the carbon atoms is sp3, sp2, and sp, respectively. Both ethylene (planar) and acetylene (linear) are planar molecules. Ethane has 7 σ bonds, ethylene has 5 σ bonds and 1 π bond, and acetylene has 3 σ bonds and 2 π bonds. Silicon analogs would have the same hybrid orbitals (sp3, sp2, and sp) as carbon analogs, but due to Si's larger atomic size and availability of 3d orbitals, it can have a higher coordination number and weaker π bonding than carbon-based molecules.

Step by step solution

01

Draw Lewis Structures

For each compound, start by counting the total number of valence electrons and then arrange the atoms for bonding. For ethane, ethylene, and acetylene, the Lewis structures are as follows: Ethane (C2H6): Each C atom contributes 4 valence electrons, and each H atom contributes 1 valence electron, for a total of 14 electrons. Ethane has a single bond between the two carbon atoms and single bonds connecting the other 6 hydrogen atoms. Ethylene (C2H4): Each C atom contributes 4 valence electrons, and each H atom contributes 1 valence electron, for a total of 12 electrons. Ethylene has a double bond between the two carbon atoms with the 4 hydrogen atoms single bonded to the carbon atoms. Acetylene (C2H2): Each C atom contributes 4 valence electrons, and each H atom contributes 1 valence electron, for a total of 10 electrons. Acetylene has a triple bond between the two carbon atoms with the 2 hydrogen atoms single bonded to the carbon atoms.
02

Determine hybridization of carbon atoms in each molecule

We determine the hybridization of carbon atoms by looking at the number of atomic orbitals participating in the formation of hybrid orbitals. Ethane (C2H6): Each carbon atom forms 1 bond with another carbon and 3 bonds with hydrogen, using a total of 4 orbitals. Therefore, the hybridization of the carbon atoms in ethane is sp3. Ethylene (C2H4): Each carbon atom forms 1 double bond with another carbon and 2 single bonds with hydrogen, using a total of 3 orbitals. Therefore, the hybridization of the carbon atoms in ethylene is sp2. Acetylene (C2H2): Each carbon atom forms 1 triple bond with another carbon and 1 single bond with hydrogen, using a total of 2 orbitals. Therefore, the hybridization of the carbon atoms in acetylene is sp.
03

Predict which molecules, if any, are planar

A molecule is planar if all its atoms lie in the same plane. Ethane (C2H6): The carbon atoms are surrounded by 4 electron groups with tetrahedral geometry, so ethane is not planar. Ethylene (C2H4): The carbon atoms are surrounded by 3 electron groups with trigonal planar geometry. Therefore, ethylene is planar. Acetylene (C2H2): The carbon atoms are surrounded by 2 electron groups with linear geometry. Therefore, acetylene is also planar.
04

Determine the number of σ and π bonds in each molecule

Sigma (σ) bonds involve head-on overlap of atomic orbitals, while pi (π) bonds involve side-by-side overlap of atomic orbitals. Ethane (C2H6): Ethane has one C-C σ bond and six C-H σ bonds, with no π bonds. A total of 7 σ bonds. Ethylene (C2H4): Ethylene has one C-C σ bond, one C-C π bond, and four C-H σ bonds. A total of 5 σ bonds and 1 π bond. Acetylene (C2H2): Acetylene has one C-C σ bond, two C-C π bonds, and two C-H σ bonds. A total of 3 σ bonds and 2 π bonds.
05

Describe the Si analogs in terms of hybrid orbitals and discuss the difference between Si and C

Supposing that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. For Si analogs, the bonding about Si in each structure would involve the same hybrid orbitals as in their carbon-based counterparts (sp3, sp2, and sp). However, there is a difference between Si and C, as Si lies in the row below C in the periodic table. Si has larger atomic size than C and has access to the 3d orbitals. These differences cause Si to be more comfortable with a higher coordination number (for example, silanes can have a higher number of Si-H bonds). Additionally, the π bonding in Si analogs of ethylene and acetylene would be weaker than in carbon-based molecules because the 3p orbitals' side-by-side overlap in Si is less effective than the 2p orbitals in C. Overall, these differences affect the stability and bonding strength of the silicon analogs compared to their carbon-based counterparts.

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Key Concepts

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

Chemical Bonding
At its core, chemical bonding is the fundamental force that holds atoms together within molecules. It arises from the mutual attraction between the positively charged nuclei of atoms and the negatively charged electrons they share. The most basic types of chemical bonds include ionic, covalent, and metallic bonds, each with distinct properties.

For instance, in the molecules ethane \( \mathrm{C}_{2}\mathrm{H}_{6} \), ethylene \( \mathrm{C}_{2}\mathrm{H}_{4} \), and acetylene \( \mathrm{C}_{2}\mathrm{H}_{2} \), covalent bonding is at play, where atoms share electrons to achieve stability. This sharing is seen in how carbon, with four valence electrons, forms single, double, or triple bonds with other carbon atoms and additional single bonds with hydrogen atoms to fulfill the octet rule, ensuring electrons are paired and stable.

Such electron sharing and bond formation must be represented accurately in Lewis structures, which depict the valence electron distribution in molecules, and it serves as a visual representation of how atoms bond chemically.
Molecular Hybridization
Molecular hybridization refers to the concept in which atomic orbitals within an atom mix to form new, identical hybrid orbitals. This theory provides insight into the molecular shape and bond properties that can't be explained by simply observing the electron configuration of atoms.

Examining the hybridization in our example molecules, ethane's carbon atoms undergo sp3 hybridization to accommodate four sigma bonds, while ethylene and acetylene's carbon atoms exhibit sp2 and sp hybridizations, respectively, to account for their unique bonding situations. sp3 gives a tetrahedral geometry, perfect for four equivalent bonds in ethane, while sp2 and sp result in planar and linear shapes, matching the double and triple bonding scenarios seen in ethylene and acetylene.
Molecular Geometry
The shape or molecular geometry of a molecule significantly influences its physical and chemical properties. The geometry is determined by the spatial arrangement of the electron pairs around the central atom. The Valence Shell Electron Pair Repulsion (VSEPR) theory postulates that electron pairs repel each other and, so, adopt an orientation that minimizes repulsion, leading to the molecular geometry.

For example, according to VSEPR theory, ethane has a tetrahedral geometry due to four regions of electron density (sp3 hybridization) around each carbon atom, whereas ethylene and acetylene are planar and linear due to sp2 and sp hybridizations, which correspond to three and two regions of electron density, respectively. These geometries affect many properties, such as boiling point, reactivity, and the possible formation of isomers.
Sigma and Pi Bonds
Sigma (\(\sigma\)) and pi (\(\pi\)) bonds are types of covalent bonds that form due to the overlap of atomic orbitals. A sigma bond is the strongest type of covalent bond and is formed by the head-on overlap of orbitals. It allows for free rotation of the bonded atoms, given its cylindrical symmetry.

In contrast, pi bonds result from the lateral or side-by-side overlap of orbitals, and they usually accompany sigma bonds in double and triple bonds, as seen in ethylene and acetylene. Pi bonds restrict the rotation due to their electron cloud's shape, contributing to the rigidity seen in molecules with multiple bonds. Ethane, with only single bonds, has sigma bonds only, while ethylene and acetylene feature both \(\sigma\) and \(\pi\) bonds, reflecting their different bonding complexities.
Periodic Table Differences
The periodic table is organized by atomic number and electron configuration, which systematically affect an element's chemical properties. One key aspect is that elements in the same group or column generally exhibit similar valence electron configurations, leading to analogous chemical behaviors.

However, moving down a group as from carbon (C) to silicon (Si), noticeable differences arise. Silicon, for instance, has a larger atomic size and can utilize its 3d orbitals. These differences cause Si-based molecules to behave differently than their C-based analogs. While Si can form similar looking compounds, such as silane analogs to ethane, the bond strength, angle, and reactivity can be markedly different due to less effective p-orbital overlap for pi bonding, and a preference for different coordination numbers, as Si is less electronegative and more shielded by its inner electron shells.

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Most popular questions from this chapter

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\mathrm{a}} ; \mathrm{PCl}_{3}\) \(100.3^{\circ} ; \mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{Pl}_{3}, 102.0^{\circ} .\) The trend is generally at- tributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(X-P-X\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCI}_{4}\)

(a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular orbital theory, would either \(\mathrm{Be}_{2}\) or \(\mathrm{Be}_{2}{ }^{7}\) be expected to exist? Explain.

For both atoms and molecules, ionization energies (Section 7.4) are related to the energies of orbitals: The lower the energy of the orbital, the greater the ionization energy. The first ionization energy of a molecule is therefore a measure of the energy of the highest occupied molecular orbital (HOMO). See the "Chemistry Put to Work" box on Orbitals and Energy. The first ionization energies of several diatomic molecules are given in electron-volts in the following table: \begin{tabular}{ll} \hline Molecule & \(I_{1}(\mathrm{eV})\) \\ \hline \(\mathrm{H}_{2}\) & \(15.4\) \\ \(\mathrm{~N}_{2}\) & \(15.6\) \\ \(\mathrm{O}_{2}\) & \(12.1\) \\ \(\mathrm{~F}_{2}\) & \(15.7\) \\ \hline \end{tabular} (a) Convert these ionization energies to \(\mathrm{kJ} / \mathrm{mol} .(\mathrm{b})\) On the same plot, graph \(I_{1}\) for the \(\mathrm{H}, \mathrm{N}, \mathrm{O}\), and \(\mathrm{F}\) atoms (Figure 7.11) and \(I_{1}\) for the molecules listed. (c) Do the ionization energies of the molecules follow the same periodic trends as the ionization energies of the atoms? (d) Use molecular orbital energy-level diagrams to explain the trends in the ionization energies of the molecules.

A compound composed of \(2.1 \% \mathrm{H}, 29.8 \% \mathrm{~N}\), and \(68.1 \% \mathrm{O}\) has a molar mass of approximately \(50 \mathrm{~g} / \mathrm{mol} .\) (a) What is the molecular formula of the compound? (b) What is its Lewis structure if \(\mathrm{H}\) is bonded to \(\mathrm{O}\) ? (c) What is the geometry of the molecule? (d) What is the hybridization of the orbitals around the \(\mathrm{N}\) atom? (e) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \mathrm{~A}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ}\). The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

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