The molecule shown here is difluoromethane (CH_2F2), which is used as a refrigerant called R-32. (a) Based on the structure, how many electron domains surround the \(\mathrm{C}\) atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, which of the following describes the direction of the overall dipole moment vector in the molecule: (i) from the carbon atom toward a fluorine atom, (ii) from the carbon atom to a point midway between the fluorine atoms, (iii) from the carbon atom to a point midway between the hydrogen atoms, or (iv) from the earbon atom toward a hydrogen atom? [Sections 9.2 and 9.3\(]\)

Short Answer

Expert verified
The difluoromethane (CH2F2) molecule has 4 electron domains surrounding its carbon atom. It also has a nonzero dipole moment, making it a polar molecule. The overall dipole moment vector points from the carbon atom to a point midway between the fluorine atoms (option ii).

Step by step solution

01

Determine the number of electron domains surrounding the carbon (C) atom

The number of electron domains surrounding an atom is the total number of lone pairs and bonded atoms. In the difluoromethane (CH2F2) molecule, the carbon atom is bonded to two hydrogen atoms and two fluorine atoms. Since carbon has no lone pairs, the electron domains surrounding the carbon atom are those formed by the two hydrogen and two fluorine atoms. Number of electron domains = Number of bonded atoms = 2 hydrogen atoms + 2 fluorine atoms = 4 electron domains
02

Determine if the molecule has a nonzero dipole moment

To determine if the molecule has a nonzero dipole moment, we need to consider the individual bond dipoles and their orientation in the molecular structure. The bond between carbon and fluorine (C-F) is polar, as fluorine is highly electronegative. Hence, there will be a bond dipole moment vector pointing from the carbon atom to each fluorine atom. On the other hand, the bond between carbon and hydrogen (C-H) is less polar, but still has a small bond dipole moment vector pointing from the carbon atom to each hydrogen atom. Due to CH2F2's tetrahedral shape, the molecule is asymmetric, so the dipoles don't cancel each other out, resulting in a nonzero net dipole moment. Hence, the molecule has a nonzero dipole moment and is polar.
03

Determine the overall dipole moment vector direction

Since CH2F2 is polar, we now need to describe the overall dipole moment vector direction. Given the options in the exercise: (i) from the carbon atom toward a fluorine atom (ii) from the carbon atom to a point midway between the fluorine atoms (iii) from the carbon atom to a point midway between the hydrogen atoms (iv) from the carbon atom toward a hydrogen atom The bond dipoles of the C-F bonds are larger than the bond dipoles of the C-H bonds. Therefore, the overall dipole moment vector is influenced more by the C-F bond dipoles and points in the direction closest to the fluorine atoms. Since there are two fluorine atoms, the overall dipole moment vector will point from the carbon atom to a point midway between the fluorine atoms. Thus, the correct option is (ii) from the carbon atom to a point midway between the fluorine atoms.

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Most popular questions from this chapter

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as \(\pi_{2 \mathrm{p}}-\pi_{2 \mathrm{p}}{ }^{*}\), $\sigma_{2 s}-\sigma_{2 s}{ }^{*},$ and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (a) Write out the molecular orbital valence electron configurations for the ground state and first excited state for \(\mathrm{N}_{2}\). (b) Is \(\mathrm{N}_{2}\) paramagnetic or diamagnetic in its first excited state? (c) The electronic absorption spectrum of the \(\mathrm{N}_{2}\) molecule has the lowest energy peak at \(170 \mathrm{nm}\). To what orbital transition does this correspond? (d) Calculate the energy of the HOMO-LUMO transition in part (a) in terms of \(\mathrm{kJ} / \mathrm{mol}\). (e) Is the \(\mathrm{N}-\mathrm{N}\) bond in the first excited state stronger or weaker compared to that in the ground state?

Sodium azide is a shock-sensitive compound that releases \(\mathrm{N}_{2}\) upon physical impact. The compound is used in automobile airbags. The azide ion is \(\mathrm{N}_{3}\). (a) Draw the Lewis structure of the azide ion that minimizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central \(\mathrm{N}\) atom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?

(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 \(\mathrm{H}_{2}^{-}\). (d) Suppose that theion is excited by light, sothat 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?

(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 \((\sigma\) plus \(\pi)\) bond, or would they be the same?

Predict whether each of the following molecules is polar or nonpolar: (a) \(\mathrm{CCl}_{4},(\mathbf{b}) \mathrm{NH}_{3},(\mathbf{c}) \mathrm{SF}_{4}\) (d) \(\mathrm{XeF}_{4}\), (e) \(\mathrm{CH}_{3} \mathrm{Br},\) (f) \(\mathrm{GaH}_{3}\).

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