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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

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

(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.

Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: (a) 3, (b) 4 , (c) 5, (d) 6

(a) Explain why the following ions have different bond angles: \(\mathrm{C} 1 \mathrm{O}_{2}^{-}\) and \(\mathrm{NO}_{2}^{-}\). Predict the bond angle in each case. (b) Explain why the \(\mathrm{XeF}_{2}\) molecule is linear and not bent.

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a sigma bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals make a \(\pi\) bond? Explain.
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