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Chapter 9: Problem 21

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN}\), (b) \(\mathrm{SO}_{3}{ }^{2-}\), (c) \(S F_{4}\) (d) \(\mathrm{PF}_{6}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).

Short Answer

Expert verified
(a) HCN: Electron-domain geometry is linear, molecular geometry is linear. (b) \(\mathrm{SO}_{3}{ }^{2-}\): Electron-domain geometry is tetrahedral, molecular geometry is trigonal planar. (c) \(SF_{4}\): Electron-domain geometry is trigonal bipyramidal, molecular geometry is seesaw. (d) \(\mathrm{PF}_{6}\): Electron-domain geometry is octahedral, molecular geometry is octahedral. (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\): Electron-domain geometry is trigonal planar, molecular geometry is trigonal planar. (f) \(\mathrm{N}_{3}^{-}\): Electron-domain geometry is linear, molecular geometry is linear.

Step by step solution

01

(a) Electron-domain and molecular geometries for HCN

1. Identify the central atom: In HCN, the central atom is Carbon (C). 2. Determine the number of electron domains: Carbon has 4 valence electrons, 1 forms a single bond with Hydrogen (H), and 3 form a triple bond with Nitrogen (N). Thus, there are 2 electron domains around carbon. 3. Determine the electron-domain geometry: With 2 electron domains, the electron-domain geometry is linear. 4. Determine the molecular geometry: Since there are no lone pairs on the central atom, the molecular geometry is also linear.
02

(b) Electron-domain and molecular geometries for \(\mathrm{SO}_{3}{ }^{2-}\)

1. Identify the central atom: In \(\mathrm{SO}_{3}{ }^{2-}\), the central atom is Sulfur (S). 2. Determine the number of electron domains: Sulfur has 6 valence electrons, 3 form single bonds with Oxygen (O) atoms, and there are 2 additional electrons due to the 2- charge. Thus, there are 4 electron domains around sulfur. 3. Determine the electron-domain geometry: With 4 electron domains, the electron-domain geometry is tetrahedral. 4. Determine the molecular geometry: Since there are no lone pairs on the central atom, the molecular geometry is trigonal planar.
03

(c) Electron-domain and molecular geometries for \(SF_{4}\)

1. Identify the central atom: In \(SF_{4}\), the central atom is Sulfur (S). 2. Determine the number of electron domains: Sulfur has 6 valence electrons, and 4 form single bonds with Fluorine (F) atoms. Thus, there are 5 electron domains around sulfur. 3. Determine the electron-domain geometry: With 5 electron domains, the electron-domain geometry is trigonal bipyramidal. 4. Determine the molecular geometry: Since there is only 1 lone pair on the central atom, the molecular geometry is seesaw.
04

(d) Electron-domain and molecular geometries for \(\mathrm{PF}_{6}\)

1. Identify the central atom: In \(\mathrm{PF}_{6}\), the central atom is Phosphorus (P). 2. Determine the number of electron domains: Phosphorus has 5 valence electrons, and 6 form single bonds with Fluorine (F) atoms, which means an expanded octet. Thus, there are 6 electron domains around phosphorus. 3. Determine the electron-domain geometry: With 6 electron domains, the electron-domain geometry is octahedral. 4. Determine the molecular geometry: Since there are no lone pairs on the central atom, the molecular geometry is also octahedral.
05

(e) Electron-domain and molecular geometries for \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\)

1. Identify the central atom: In \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), the central atom is Nitrogen (N). 2. Determine the number of electron domains: Nitrogen has 5 valence electrons, 3 form single bonds with Hydrogen (H) atoms, and there are 1 fewer electrons due to the 1+ charge. Thus, there are 3 electron domains around nitrogen. 3. Determine the electron-domain geometry: With 3 electron domains, the electron-domain geometry is trigonal planar. 4. Determine the molecular geometry: Since there are no lone pairs on the central atom, the molecular geometry is trigonal planar.
06

(f) Electron-domain and molecular geometries for \(\mathrm{N}_{3}^{-}\)

1. Identify the central atom: In \(\mathrm{N}_{3}^{-}\), the central atom is the middle Nitrogen (N) atom. 2. Determine the number of electron domains: The central Nitrogen atom has 5 valence electrons, 2 form a single bond with the first Nitrogen (N) atom, and the remaining 3 form a double bond with the last Nitrogen (N) atom. Thus, there are 2 electron domains around the central nitrogen. 3. Determine the electron-domain geometry: With 2 electron domains, the electron-domain geometry is linear. 4. Determine the molecular geometry: Since there are no lone pairs on the central atom, the molecular geometry is also linear.

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Key Concepts

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

Electron-Domain Geometry
Electron-domain geometry is a way to visualize how different pairs of electrons are arranged around a central atom in a molecule. The number of electron domains determines the geometry. These domains include both bonding pairs (shared between two atoms) and lone pairs (not shared).
  • If there are 2 electron domains, the geometry is linear.
  • With 3 domains, it becomes trigonal planar.
  • 4 domains lead to a tetrahedral shape.
  • 5 domains create a trigonal bipyramidal structure.
  • 6 domains result in an octahedral geometry.
Understanding electron-domain geometry helps predict molecular geometry, which describes the actual shape of the molecule based on the location of the atoms.
Valence Electrons
Valence electrons are the outermost electrons of an atom. They play a crucial role in forming chemical bonds. Knowing how many valence electrons an atom has helps in predicting how it will bond with other atoms.
For example:
  • Carbon has 4 valence electrons, allowing it to form 4 bonds.
  • Oxygen has 6 valence electrons, typically forming 2 bonds.
  • Phosphorus has 5 valence electrons, forming multiple bonds depending on the molecule.
Valence electrons are used to determine the number of electron domains around a central atom, which, in turn, influences the molecule's shape.
Central Atom
In a molecule, the central atom is usually the one with the highest bonding capacity or the least electronegative element that can form multiple bonds. It's the atom around which other atoms are arranged.
Here are some examples:
  • In \(\text{HCN}\), Carbon is the central atom because it can form the most bonds.
  • In \(\text{SO}_3^{2-}\), Sulfur is central, having multiple bonding options.
  • In complex ions like \(\text{PF}_6^-\), Phosphorus acts as the central atom due to its ability to expand its valence shell.
Identifying the central atom is a key step in determining the overall structure and geometry of the molecule.
Lone Pairs
Lone pairs are non-bonding pairs of electrons that belong to a single atom. They can influence the shape and geometry of a molecule significantly.
  • Lone pairs occupy space and can repel bonding pairs, affecting angles between atoms.
  • In \(\text{SF}_4\), the lone pair causes a seesaw shape instead of a trigonal bipyramidal geometry.
  • In molecules without lone pairs like \(\text{PF}_6^-\), the electron-domain and molecular geometries are identical.
Understanding lone pairs is vital in predicting not just the shape but also the reactivity and properties of the molecule.

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Most popular questions from this chapter

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{\prime} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The \(\mathrm{CO}\) bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the Closer Look box on phases) in the plane of the paper. Two lobes should be pointing toward \(\mathrm{M}\). (d) Now draw the \(\mathrm{d}_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

Consider the bonding in an \(\mathrm{MgH}_{2}\) molecule. (a) Draw a Lewis structure for the molecule, and predict its molecular geometry. (b) What hybridization scheme is used in \(\mathrm{MgH}_{2} ?\) (c) Sketch one of the two-electron bonds between an \(\mathrm{Mg}\) hybrid orbital and an \(\mathrm{H} 1 \mathrm{~s}\) atomic orbital.

(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.

Consider the \(\mathrm{H}_{2}{ }^{+}\) ion. (a) Sketch the molecular orbitals of the ion, and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}{ }^{+}\) ion? (c) Draw the electron configuration of the ion in terms of its MOs (d) What is the bond order in \(\mathrm{H}_{2}{ }^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higherenergy MO. Would you expect the excitedstate \(\mathrm{H}_{2}{ }^{+}\) ion to be stable or to fall apart? Explain.

Predict the trend in the \(\mathrm{F}(\) axial \()-\mathrm{A}-\mathrm{F}\) (equatorial) bond angle in the following \(\mathrm{AF}_{n}\) molecules: \(\mathrm{PF}_{5}, \mathrm{SF}_{4}\) and \(\mathrm{ClF}_{3}\)
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