Question 5.6: Predict the polarity of each of the following molecules. a. ......
Predict the polarity of each of the following molecules.
a. PCl_{3} (trigonal pyramidal) b. SCl_{2} (angular)
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\quad\quad\quad\quad\quad\quad \begin{matrix}:S:\\:\underset{.\ .}{Cl}:^{^{^{\diagup}}}\quad\quad^{^{^{\diagdown}}}:\underset{.\ .}{Cl}:\end{matrix}
c. SiBr_{4} (tetrahedral) d. C_{2}Cl_{2} (linear)
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\quad\quad\quad\quad\quad\quad\quad Cl-C\equiv C-Cl
Knowledge of a molecule’s geometry, which is given for each molecule in this example, is a prerequisite for predicting molecular polarity.
a. Noncancellation of the individual bond polarities in the trigonal pyramidal PF_{3} molecule results in its being a polar molecule.
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The bond polarity arrows all point toward fluorine atoms because fluorine is more electronegative than phosphorus.
b. For the bent SCl_{2} molecule, the shift in electron density in the polar sulfur–chlorine bonds will be toward the chlorine atoms because chlorine is more electronegative than sulfur. The SCl_{2} molecule as a whole is polar because of the noncancellation of the individual sulfur–chlorine bond polarities.
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c. The SiBr_{4} molecule is a tetrahedral molecule and all four atoms attached to the central atom (Si) are the same. This, and the highly symmetrical nature of a tetrahedral geometry (all bond angles are the same), means that the Si–Br bond polarities cancel each other and the molecule as a whole is nonpolar.
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d. The carbon–carbon bond is nonpolar. The two carbon–chlorine bonds are polar and are “equal but opposite” in terms of effect; that is, they cancel. The C_{2}Cl_{2} molecule as a whole is thus nonpolar.
\quad\quad\quad Cl-C\equiv C-Cl \quad \quad \overset{⇸}{Cl\ \ C}\ \overset{⇷}{C\ \ Cl} No net polarity