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Chapter 6: Problem 16

Consider the methane molecule. (a) Draw the methane molecule showing its tetrahedral shape, using lines, solid wedges, and dashed wedges to show three-dimensionality. (b) Draw the methane molecule inscribed inside ? tetrahedron such that the \(\mathrm{H}\) atoms touch the vertices of the tetrahedron. (c) Why is it better for methane to have \(109.5^{\circ}\) bond angles rather than \(90^{\circ}\) bond angles?

Short Answer

Expert verified
To represent the methane molecule (\(CH_4\)), depict the carbon atom at the center of a tetrahedron and the hydrogen atoms at the vertices, with lines and wedges indicating three-dimensional bonds. The methane molecule is inscribed inside a tetrahedron, with vertices shared between the molecule and the polyhedron. The \(109.5^{\circ}\) bond angles are energetically more favourable than \(90^{\circ}\) angles, due to the valence shell electron pair repulsion (VSEPR) theory, which minimizes repulsion and maintains stability in the molecule.

Step by step solution

01

(Drawing the methane molecule)

To draw the methane molecule (\(CH_4\)), first, place the carbon atom (\(C\)) at the center and the four hydrogen atoms (\(H\)) at the vertices of a tetrahedron. Represent the bonds between the atoms using lines, solid wedges, and dashed wedges to show three-dimensionality. Solid wedges indicate the bonds coming out of the plane towards you, while dashed wedges indicate the bonds going behind the plane.
02

(Inscribing methane inside a tetrahedron)

Draw a tetrahedron, placing its vertices at the four hydrogen atoms of the methane molecule. The edges of this tetrahedron will connect the hydrogen atoms. This way, you'll have both, the methane molecule and the tetrahedron, sharing the same vertices. Consequently, the methane molecule can be regarded as inscribed inside the tetrahedron.
03

(Explaining the bond angles)

The bond angles in methane are approximately \(109.5^{\circ}\), and not \(90^{\circ}\), because of the tetrahedral shape. This shape allows for the maximum possible separation between the hydrogen atoms, which is essential for minimizing electron-electron repulsion, as per the valence shell electron pair repulsion (VSEPR) theory. If the bond angles were equal to \(90^{\circ}\), that would mean the hydrogen atoms would be closer together, leading to increased repulsion and instability of the molecule. Therefore, having a bond angle of \(109.5^{\circ}\) is energetically more favourable for methane.

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Most popular questions from this chapter

The dot diagram \(\ddot{\mathrm{Q}}=\mathrm{S}=\ddot{\mathrm{O}}\) for \(\mathrm{SO}_{2}\) is incorrect. (a) Draw the correct dot diagram. (b) How do the correct and incorrect dot diagrams differ in their prediction of molecular shape and polarity?

Consider the molecule \(\mathrm{SiCl}_{4}\). (a) Draw the dot diagram. (b) Draw the molecule's three-dimensional shape, and label the numeric value of all bond angles. (c) What is the shape of this molecule? (d) Draw in the individual bond dipole moments. (e) Is the molecule polar? If yes, draw the molecular dipole moment vector.

Sulfur dioxide, \(\mathrm{SO}_{2}\), a pollutant that comes from burning coal contaminated with sulfur.

Draw a combined Lewis dot, molecular-shape diagram for each of the following species. Name each shape, and indicate whether the molecule or ion has an overall dipole moment. If so, draw the dipole moment vector. (a) \(\mathrm{Cl}_{2} \mathrm{O}\) (b) NOF (c) \(\mathrm{PF}_{3}\) (d) \(\mathrm{ICl}_{2}^{+}\) (Hint: See problem statement and hint for Problem 6.69. Hint for \((b): \mathrm{N}=\mathrm{O}\) bonds are common.)

Consider \(\mathrm{SO}_{2}\) and \(\mathrm{CO}_{2}\). Both have polar covalent bonds. One of these molecules is polar and the other is nonpolar. Which is which and why?
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