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Chapter 10: Problem 11

Predict the geometry of the following molecules using the VSEPR method: (a) \(\mathrm{HgBr}_{2}\), (b) \(\mathrm{N}_{2} \mathrm{O}(\mathrm{ar}-\) rangement of atoms is \(\mathrm{NNO}\) ), (c) \(\mathrm{SCN}^{-}\) (arrangement of atoms is \(\mathrm{SCN}\) ).

Short Answer

Expert verified
The molecular geometry of \(\mathrm{HgBr}_{2}\) and \(\mathrm{SCN}^{-}\) are both linear while the geometry of \(\mathrm{N}_{2}\mathrm{O}\) is bent or V-shaped.

Step by step solution

01

Predict the geometry of \(\mathrm{HgBr}_{2}\)

1. Total valence electrons: \(\mathrm{Hg}\) (2e-) + 2 \(\mathrm{Br}\) (14e-) = 16e- \n 2. Lone pairs: \(\mathrm{Hg}\) (central atom) doesn't have any lone pair.\n 3. Electron pair geometry is linear since there are two bonding atoms and no lone pairs.\n Therefore, the molecular geometry of \(\mathrm{HgBr}_{2}\) is linear.
02

Predict the geometry of \(\mathrm{N}_{2}\mathrm{O}\)

1. Total valence electrons: 2\(\mathrm{N}\) (10e-) + \(\mathrm{O}\) (6e-) = 16e-\n 2. Lone pairs: Central atom \(\mathrm{N}\) has one lone pair.\n 3. The electron pair geometry is trigonal planar as there are three bonding atoms and one lone pair.\n Therefore, the molecular geometry of \(\mathrm{N}_{2}\mathrm{O}\) is bent or V-shaped (due to the existence of a lone pair).
03

Predict the geometry of \(\mathrm{SCN}^{-}\)

1. Total valence electrons: \(\mathrm{S}\) (6e-) + \(\mathrm{C}\) (4e-) + \(\mathrm{N}\) (5e-) + 1e- (for 1- charge) = 16e-\n 2. Lone pairs: Central atom \(\mathrm{C}\) has no lone pair.\n 3. The electron pair geometry is linear as there are two bonding atoms and no lone pairs.\n Therefore, the molecular geometry of \(\mathrm{SCN}^{-}\) is linear.

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Most popular questions from this chapter

Specify which hybrid orbitals are used by carbon atoms in the following species: (a) \(\mathrm{CO}\), (b) \(\mathrm{CO}_{2}\), (c) \(\mathrm{CN}^{-}\).

The molecule benzyne \(\left(\mathrm{C}_{6} \mathrm{H}_{4}\right)\) is a very reactive species. It resembles benzene in that it has a sixmembered ring of carbon atoms. Draw a Lewis structure of the molecule and account for the molecule's high reactivity.

The geometries discussed in this chapter all lend themselves to fairly straightforward elucidation of bond angles. The exception is the tetrahedron, because its bond angles are hard to visualize. Consider the \(\mathrm{CCl}_{4}\) molecule, which has a tetrahedral geometry and is nonpolar. By equating the bond moment of a particular \(\mathrm{C}-\mathrm{Cl}\) bond to the resultant bond \(\mathrm{mo}-\) ments of the other three \(\mathrm{C}-\mathrm{Cl}\) bonds in opposite directions, show that the bond angles are all equal to \(109.5^{\circ}\)

Draw the Lewis structure for the \(\mathrm{BeCl}_{4}^{2-}\) ion. Predict its geometry and describe the hybridization state of the Be atom.

What is valence bond theory? How does it differ from the Lewis concept of chemical bonding?
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