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Chapter 9: Problem 17

What is the difference between the electron-domain geometry and the molecular geometry of a molecule? Use the water molecule as an example in your discussion.

Short Answer

Expert verified
The difference between electron-domain geometry and molecular geometry lies in the consideration of electron pairs in determining the geometries. Electron-domain geometry considers both bonding and nonbonding electron pairs surrounding the central atom, while molecular geometry only takes the positions of bonded atoms into account. Using the water molecule (H2O) as an example, its electron-domain geometry is tetrahedral, as there are 4 electron domains (2 bonding pairs and 2 lone pairs) around the central oxygen atom. However, its molecular geometry is bent or V-shaped, considering only the positions of the bonded hydrogen atoms with an O-H bond angle of approximately 104.5掳.

Step by step solution

01

Define Electron-Domain Geometry

Electron-domain geometry refers to the arrangement of electron pairs around the central atom in a molecule, including both bonding electron pairs and any nonbonding electron pairs (lone pairs). In determining the electron-domain geometry, we don't distinguish between bonding and nonbonding electron pairs.
02

Define Molecular Geometry

Molecular geometry refers to the arrangement of atoms in a molecule based on the positions of the bonding electron pairs and does not take into account the nonbonding electron pairs. Only the positions of the bonded atoms matter in determining the molecular geometry.
03

Determine the Electron-Domain Geometry of Water

Water molecule has the chemical formula H2O. The central atom, oxygen, has two bonding electron pairs (bonded with two hydrogen atoms) and two lone pairs of electrons. Thus, there are a total of 4 electron domains around the central oxygen atom. Since these electron domains are arranged in a tetrahedral manner to minimize electron repulsion, the electron-domain geometry of the water molecule is tetrahedral.
04

Determine the Molecular Geometry of Water

To determine the molecular geometry of the water molecule, we only consider the positions of the bonded hydrogen atoms and ignore the lone electron pairs on the oxygen atom. In water, the two O-H bonds (hydrogen atoms) are bent with an angle of approximately 104.5掳. Due to this bent shape of the two bonding electron pairs, the molecular geometry of the water molecule is described as bent or V-shaped.
05

State the Difference between Electron-Domain Geometry and Molecular Geometry

The electron-domain geometry of a molecule takes into account both bonding and nonbonding electron pairs, while the molecular geometry only considers the positions of atoms bonded to the central atom. In the case of the water molecule, its electron-domain geometry is tetrahedral, while its molecular geometry is bent or V-shaped.

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Most popular questions from this chapter

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?

(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{6}\right)\) are both described as tetrahedral. What does this indicate about their bond angles? (b) The \(\mathrm{NH}_{3}\) molecule is trigonal pyramidal, while \(\mathrm{BF}_{3}\) is trigonal planar. Which of these molecules is flat?

Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN}\), and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of NO? (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+}\), and \(\mathrm{NO}^{-}\), and describe the magnetic properties of each (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6 r}\) you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) Suggest a hybridization scheme for the Xe atom in \(\mathrm{XeF}_{6}\). (e) The molecule \(\mathrm{IF}_{7}\) has a pentagonal- bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\)
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