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Chapter 6: Problem 41

Consider the following molecules. For those that are polar, draw the molecular dipole moment. (a) \(\mathrm{CHBr}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{Br}\) (c) \(\mathrm{H}_{2} \mathrm{~S}\) (d) \(\mathrm{NOC}\) (e) \(\mathrm{C}_{2} \mathrm{Cl}_{2}\) (connected \(\left.\mathrm{Cl}-\mathrm{C}-\mathrm{C}-\mathrm{Cl}\right)\)

Short Answer

Expert verified
(a) CHBr3: Polar, net dipole moment points towards the three bromine atoms. (b) CH3Br: Polar, net dipole moment points towards the bromine atom. (c) H2S: Polar, net dipole moment points towards the sulfur atom. (d) NOC: Polar, net dipole moment points towards the oxygen atom. (e) C2Cl2: Non-polar, no net dipole moment due to the cancellation of dipole moments.

Step by step solution

01

(a) Analyzing and Drawing CHBr3

In CHBr3, carbon is bonded with three bromine atoms and one hydrogen atom. The electronegativity difference between C-Br bond will make this bond polar. The molecular geometry of CHBr3 is tetrahedral. Thus, there will be a net dipole moment pointing towards the three bromine atoms.
02

(b) Analyzing and Drawing CH3Br

In CH3Br, carbon is bonded with three hydrogen atoms and one bromine atom. The C-Br bond is polar due to the electronegativity difference between C and Br. The molecular geometry of CH3Br is also tetrahedral. The net dipole moment will point towards the bromine atom.
03

(c) Analyzing and Drawing H2S

In H2S, sulfur is bonded with two hydrogen atoms. The electronegativity difference between S and H atom will make S-H bonds polar. The molecular geometry of H2S is bent. These two polar bonds create a net dipole moment pointing towards the sulfur atom.
04

(d) Analyzing and Drawing NOC

In NOC, nitrogen is bonded with oxygen and carbon. The N-O bond is polar due to the electronegativity difference between N and O, while the N-C bond is also polar because of the electronegativity difference between N and C. The molecular geometry of NOC is linear, with the least electronegative atom, nitrogen, in the center. The net dipole moment will point towards the oxygen atom, since it is the most electronegative.
05

(e) Analyzing C2Cl2

In C2Cl2, two carbon atoms are bonded with two chlorine atoms (one at each end). The electronegativity difference between C-Cl bond will make this bond polar. The molecular geometry of C2Cl2 is linear. As both chlorine atoms are at equal distance and the shape is linear, the two polar bonds' dipole moments will cancel each other out, resulting in no net dipole moment. Hence, C2Cl2 is not a polar molecule.

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Most popular questions from this chapter

Why is it reasonable to treat a multiple bond as a single electron group in VSEPR shape determination?

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.)

Are dipole-dipole forces between molecules as strong as the forces between oppositely charged ions? Explain.

There are exceptions to the predictions of VSEPR. Consider \(\mathrm{CH}_{3}\), known as a methyl radical. (a) Create a dot diagram for the methyl radical. How is it fundamentally different from other dot diagrams you have done? (b) Use VSEPR to predict the shape of the methyl radical and draw it with that shape (treat the odd electron as a single electron group). (c) The methyl radical is known to be planar with \(120^{\circ} \mathrm{H}-\mathrm{C}-\mathrm{H}\) angles. What steric number is being employed here, and what is the carbon atom doing with respect to the odd electron in determining molecular shape? (d) The \(\mathrm{CF}_{3}\) radical does obey VSEPR. Draw it according to its VSEPR-predicted shape. What steric number is being employed here? (e) The \(\mathrm{C}-\mathrm{H}\) bond is shorter than the \(\mathrm{C}-\mathrm{F}\) bond. When bonds are short, the atoms at the ends of the bonds can bang into each other (this is called steric congestion) unless a geometry is adopted to get around this. Use this knowledge to explain why \(\mathrm{CH}_{3}\) violates VSEPR, but \(C F_{3}\) does not.

What do we mean by intermolecular forces? What evidence is there that they exist?
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