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Chapter 9: Problem 71

Complete the Lewis structures of the following molecules. Predict the molecular structure, polarity, bond angles, and hybrid orbitals used by the atoms marked by asterisks for each molecule.

Short Answer

Expert verified
To analyze the given molecule, follow these steps: 1. Draw the Lewis structure: Count the valence electrons, connect the central atom with single bonds to surrounding atoms, distribute the remaining valence electrons as lone pairs, and form multiple bonds if required. 2. Determine the electron domain geometry: Identify electron domains (bonds and lone pairs) using VSEPR theory. 3. Determine the molecular geometry: Use VSEPR theory to predict the molecular geometry considering only the position of the atoms. 4. Predict the polarity: Compare electronegativity values of bonded atoms to determine if there are polar bonds. Determine if there's a net dipole moment in the molecule. 5. Determine bond angles: Identify ideal bond angles based on the electron domain geometry and account for any deviations due to lone pairs or multiple bonds. 6. Identify hybrid orbitals: Determine the type of hybrid orbitals used by the marked atoms based on the electron domain geometry (e.g., sp for two electron domains, sp2 for three, and sp3 for four). Using these steps, you can complete the Lewis structures, molecular structure, polarity, bond angles, and hybrid orbitals used by the marked atoms for the given molecules.

Step by step solution

01

Draw the Lewis structure

To draw the Lewis structure of the molecule M, we need to follow these steps: 1. Count the total number of valence electrons in the molecule. 2. Write the symbol of the central atom (usually the least electronegative) and connect it with single bonds to surrounding atoms. 3. Distribute the remaining valence electrons as lone pairs to satisfy the octet rule for each atom (except hydrogen, which should have 2 electrons). 4. Check if the central atom has an octet. If not, form multiple bonds (double or triple) to complete the octet.
02

Determine the electron domain geometry

Identify the number of electron domains (regions of high electron density) around the central atom. This includes bonds (single, double, or triple) and lone pairs. Use the VSEPR (Valence Shell Electron Pair Repulsion) theory to determine the electron domain geometry.
03

Determine the molecular geometry

Based on the electron domain geometry, determine the molecular geometry by considering only the position of the atoms (ignoring lone pairs). Use the VSEPR theory to predict this geometry.
04

Predict the polarity of the molecule

Check if the molecule has polar bonds by comparing electronegativity values of bonded atoms. If there are polar bonds, determine if they are symmetrically arranged, canceling each other's effects. If there's a net dipole moment in the molecule, it's polar; otherwise, it's nonpolar.
05

Determine bond angles

Bond angles are determined based on the electron domain geometry, which was found in step 2. Check the ideal bond angles for the specific electron domain geometry and see if there are any deviations due to the presence of lone pairs or multiple bonds.
06

Identify hybrid orbitals used by the marked atoms

Based on the electron domain geometry (step 2), determine the type of hybrid orbitals used by the marked atoms. Look for the number of electron domains and find the corresponding hybridization (e.g., two electron domains correspond to sp hybridization, three correspond to sp2, and four correspond to sp3). Using these steps, complete the Lewis structures of the given molecules, predict their molecular structure, polarity, bond angles, and hybrid orbitals used by the marked atoms.

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Most popular questions from this chapter

An unusual category of acids known as superacids, which are defined as any acid stronger than \(100 \%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with \(\mathrm{SbF}_{5}\) produces the superacid \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) : $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of \(\left[\mathrm{H}_{2} \mathrm{~F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) can be prepared when \(2.93 \mathrm{~mL}\) anhydrous HF (density \(=0.975 \mathrm{~g} / \mathrm{mL}\) ) and \(10.0 \mathrm{~mL} \mathrm{SbF}_{5}\) (density \(=3.10 \mathrm{~g} / \mathrm{mL}\) ) are allowed to react?

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C} \equiv \mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43 .\) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and NO than the hybrid orbital model?

For each of the following molecules or ions that contain sulfur, write the Lewis structure(s), predict the molecular structure (including bond angles), and give the expected hybrid orbitals for sulfur. a. \(\mathrm{SO}_{2}\) b. \(\mathrm{SO}_{3}\) e. \(\mathrm{SO}_{3}^{2-}\) f. \(\mathrm{SO}_{4}^{2-}\) g. \(\mathrm{SF}_{2}\) h. \(\mathrm{SF}_{4}\) i. \(\mathrm{SF}_{6}\) j. \(\mathrm{F}_{3} \mathrm{~S}-\mathrm{SF}\) k. \(\mathrm{SF}_{5}{ }^{+}\)

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lower-energy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of \(\mathrm{OH}\). Explain.
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