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Chapter 9: Problem 36

Consider a molecule with formula \(\mathrm{AX}_{3}\) . Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{3}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle increases from \(100^{\circ}\) to \(120^{\circ}\)

Short Answer

Expert verified
As the X-A-X bond angle in an AX3 molecule with trigonal planar geometry increases from 100° to 120°, the net dipole moment's magnitude changes. At the ideal 120° bond angle, the net dipole moment becomes zero, as the individual A-X bond dipole moments are evenly distributed and cancel each other out in the plane of the molecule.

Step by step solution

01

Consider the molecular geometry of AX3

In general, the molecular geometry of an AX3 molecule can be trigonal planar, without a lone pair on the central atom A, or trigonal pyramidal, with a lone pair on the central atom A. For this exercise, we'll consider the trigonal planar geometry.
02

Calculate the dipole moment vector components

Since the A-X bond is polar, there will be a net dipole moment in each A-X bond. We can consider each bond's dipole moment as a vector. The net dipole moment of the AX3 molecule will be the vector sum of these three individual bond dipole moment vectors. To find the net dipole moment, let's split each dipole moment vector's components into their x and y-components, considering the molecule in a two-dimensional plane. If the bond angle X-A-X increases, the molecule remains in the plane.
03

Determine how the net dipole moment changes as the bond angle increases

Now, let's investigate the net dipole moment as the bond angle X-A-X increases from 100° to 120°. In the XY plane, each A-X bond's dipole moment contributes to both the x- and y-components. When the bond angle is 100°, the molecule is less symmetrical, leading to a non-zero net dipole moment. As the bond angle increases to 120°, the AX3 molecule becomes more symmetrical (the ideal bond angle in a trigonal planar molecule is 120°). In this case, the dipole moment vectors become evenly distributed and cancel each other out, resulting in a net dipole moment of zero.
04

Conclusion

As the X-A-X bond angle increases from 100° to 120° in an AX3 molecule with a trigonal planar geometry, the net dipole moment's magnitude changes. At the ideal 120° bond angle, the net dipole moment becomes zero, as the individual A-X bond dipole moments are evenly distributed and cancel each other out in the plane of the molecule.

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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 \((\sigma\) plust (rotate) around a single \(\sigma\) bond same?

Consider the molecule \(\mathrm{PF}_{4} \mathrm{Cl}\) (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl} .\) How did your answer for part (b) influence your answer here in part \((\mathrm{c}) ?(\mathbf{d})\) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

The nitrogen atoms in \(\mathrm{N}_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has the stronger \(N-N\) bond?

(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 \(\mathrm{H}_{2}^{-} .(\mathbf{d})\) 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? (e) Which of the following statements about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{\star}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{4} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of M, draw the Lewis structure of a CO molecule, with the carbon nearest the M. The CO bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M},\) with phases. Can you see how they will overlap with the \(\pi_{2 p}^{\star}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?(\mathrm{f})\) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.
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