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

Short Answer

Expert verified
(a) HCN: Trigonal planar electron-domain geometry and trigonal planar molecular geometry. (b) \(\mathrm{SO}_{3}^{2-}\): Tetrahedral electron-domain geometry and trigonal pyramidal molecular geometry. (c) \(\mathrm{SF}_{4}\): Trigonal bipyramidal electron-domain geometry and seesaw molecular geometry. (d) \(\mathrm{PF}_{6}^{-}\): Octahedral electron-domain geometry and octahedral molecular geometry. (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\): Tetrahedral electron-domain geometry and trigonal pyramidal molecular geometry. (f) \(\mathrm{N}_{3}^{-}\): Linear electron-domain geometry and linear molecular geometry.

Step by step solution

01

(a) HCN Electron-domain and molecular geometries

For HCN, we have carbon as the central atom. The Lewis structure is H – C ≡ N. Electron-domain geometry: There are three electron domains around the central atom (carbon): one single bond, one triple bond, and no lone pairs. Treat multiple bonds as a single electron domain; thus, we have three electron domains, which means the electron-domain geometry is trigonal planar. Molecular geometry: Since there are no lone pairs, the molecular geometry is the same as the electron-domain geometry, which is trigonal planar.
02

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

For \(\mathrm{SO}_{3}^{2-}\), we have sulfur as the central atom. The Lewis structure is: O=S=O | O²⁻ Electron-domain geometry: There are four electron domains around the central atom (sulfur): three double bonds and one lone pair. The electron-domain geometry is tetrahedral. Molecular geometry: Considering the three bonding domains and one lone pair, the molecular geometry is trigonal pyramidal.
03

(c) \(\mathrm{SF}_{4}\) Electron-domain and molecular geometries

For \(\mathrm{SF}_{4}\), we have sulfur as the central atom. The Lewis structure is: F | F - S - F | F Electron-domain geometry: There are five electron domains around the central atom (sulfur): four single bonds and one lone pair. The electron-domain geometry is trigonal bipyramidal. Molecular geometry: Considering the four bonding domains and one lone pair, the molecular geometry is 'seesaw'.
04

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

For \(\mathrm{PF}_{6}^{-}\), we have phosphorus as the central atom. The Lewis structure is octahedral with six single bonds to fluorine: F | F - P - F | F Electron-domain geometry: There are six electron domains around the central atom (phosphorus), all of which are single bonds, giving an electron-domain geometry of octahedral. Molecular geometry: Since there are no lone pairs, the molecular geometry is the same as the electron-domain geometry, which is octahedral.
05

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

For \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), we have nitrogen as the central atom. The Lewis structure is: H | H - N - H \ Cl⁺ Electron-domain geometry: There are four electron domains around the central atom (nitrogen): three single bonds and one lone pair. The electron-domain geometry is tetrahedral. Molecular geometry: Considering the three bonding domains and one lone pair, the molecular geometry is trigonal pyramidal.
06

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

For \(\mathrm{N}_{3}^{-}\), we have a linear arrangement of three nitrogen atoms. The Lewis structure is N ≡ N – N⁻ Electron-domain geometry: There are two electron domains around the central atom (middle nitrogen): one single bond and one triple bond. Treat multiple bonds as a single electron domain; thus, we have two electron domains, which means the electron-domain geometry is linear. Molecular geometry: Since there are no lone pairs, the molecular geometry is the same as the electron-domain geometry, which is 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 all about understanding how electrons are positioned around a central atom in a molecule. You can think of electron domains as regions where electrons are likely to be found. These regions can be bonds (single, double, or triple) or lone pairs. When determining electron-domain geometry:
  • Count each bond (whether single, double, or triple) as one domain.
  • Count each lone pair as one domain.
For example, in the case of \( ext{HCN}\), the central atom carbon has three electron domains (one single bond and one triple bond). This configuration gives it a trigonal planar geometry. Always remember, even if the number of bonds is higher (like a triple bond), they count as one domain.
Lewis structure
A Lewis structure is essentially a blueprint for the molecule, showing how atoms are arranged and where the valence electrons are. Drawing a correct Lewis structure can help in predicting molecular behavior and physical properties.
Steps for drawing a Lewis structure:
  • Determine the total number of valence electrons available from all atoms.
  • Draw a skeleton arrangement of the atoms, connecting them with single bonds.
  • Distribute the remaining electrons to achieve full outer shells, usually aiming for an octet (eight electrons) for main-group atoms.
  • Use double or triple bonds if necessary to satisfy the octet rule.
Consider \( ext{SO}_3^{2-}\); start by counting valence electrons: sulfur (6) + oxygen (6 per each × 3) + 2 (for the charge). Create bonds and distribute remaining electrons to complete octet structures for each atom. Understanding how to draw these structures helps to identify possible molecule shapes and stability.
VSEPR theory
Valence Shell Electron Pair Repulsion (VSEPR) theory explains the shapes of molecules based on electron repulsion. The theory uses the repulsion between electron pairs (bonding and lone pairs) to predict molecular spatial arrangement.Key principles:
  • Electrons around a central atom repel each other, and this repulsion drives them apart as much as possible.
  • Electron pairs arrange themselves to minimize repulsion, determining the molecular geometry.
  • VSEPR considers only the electron domains around the central atom.
For \( ext{SF}_4\), using VSEPR theory shows us that five domains (four bonds and one lone pair) adopt a trigonal bipyramidal electron-domain geometry. The lone pair creates variation, resulting in a 'seesaw' molecular shape. VSEPR is crucial in chemistry to understand the three-dimensional structure of molecules.

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Most popular questions from this chapter

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same? Explain.

(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 Si in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher- energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

Predict whether each of the following molecules is polar or nonpolar: (a) \(\mathrm{CCl}_{4},\) (b) \(\mathrm{NH}_{3}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{XeF}_{4}\), (e) \(\mathrm{CH}_{3} \mathrm{Br}\), (f) \(\mathrm{GaH}_{3}\)

(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}^{2-} ?\) For those ions that are paramagnetic, determine the number of unpaired electrons.

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