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Chapter 10: Problem 61

Sketch the bond moments and resultant dipole moments for these molecules: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{PCl}_{3}, \mathrm{XeF}_{4}, \mathrm{PCl}_{5}\) \(\mathrm{SF}_{6}\)

Short Answer

Expert verified
In H2O, the resultant dipole moment will be directed from the hydrogen side to the oxygen side due to its bent shape. For PCl3, XeF4, PCl5 and SF6, the resultant dipole moment will be zero because of the symmetric distribution of polar bonds which cancel each other out.

Step by step solution

01

Determine Electronegativity of Atoms

Find the electronegativity values for the atoms in each molecule. Electronegativity values can be found on a periodic table. Hydrogen (H) has an electronegativity of 2.1, Oxygen (O) has an electronegativity of 3.44, Phosphorus (P) has an electronegativity of 2.19, Chlorine (Cl) has an electronegativity of 3.16, Xenon (Xe) has an electronegativity of 2.6, Fluorine (F) has an electronegativity of 3.98 and Sulfur (S) has an electronegativity of 2.58.
02

Determine the Bond Moments

The bond moment is determined by the difference in electronegativity between the two atoms in a bond. For example, in H2O, oxygen is more electronegative than hydrogen, so the bond moments will be directed from the hydrogen atoms to the oxygen atom. Repeat this for each bond in the molecules PCl3, XeF4, PCl5 and SF6.
03

Determine the Resultant Dipole Moments

The resultant dipole moment is found by vector addition of the individual bond moments. If there is symmetry in the molecule, some of the bond moments may cancel out. For instance, the dipole moment in PCl3, XeF4, PCl5 and SF6 will be zero because of the symmetric distribution of polar bonds which cancel out each other. In H2O, the resultant dipole moment isn't zero due to its bent shape that makes bond moments not cancel out each other. So the direction of the resultant dipole moment will be from the hydrogen side to the oxygen side.

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Most popular questions from this chapter

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.

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