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Chapter 9: Problem 43

Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) \(\mathrm{CS}_{2}\) (c) \(\mathrm{SO}_{3}\) (d) \(\mathrm{PCl}_{3}\) (e) \(\mathrm{SF}_{6}\) (f) \(\mathrm{IF}_{5}\)

Short Answer

Expert verified
(a) IF is polar, (b) \(\mathrm{CS}_{2}\) is nonpolar, (c) \(\mathrm{SO}_{3}\) is nonpolar, (d) \(\mathrm{PCl}_{3}\) is polar, (e) \(\mathrm{SF}_{6}\) is nonpolar, and (f) \(\mathrm{IF}_{5}\) is polar.

Step by step solution

01

a) IF molecule

1. Difference in electronegativity (EN) between I and F: Since Iodine (I) has an EN of 2.66 and Fluorine (F) has an EN of 3.98, there is a significant difference in electronegativity, making the IF bond polar. 2. Shape of the molecule: IF is a diatomic molecule, consisting of just one bond between Iodine and Fluorine. There are no other dipoles to cancel out or interact with. Since the bond is polar and there are no other dipole interactions to consider, the IF molecule is polar.
02

b) \(\mathrm{CS}_{2}\) molecule

1. Difference in EN between C and S: Carbon (C) has an EN of 2.55 and Sulfur (S) has an EN of 2.58. The small difference in electronegativity indicates that the CS bonds are polar. 2. Shape of the molecule: \(\mathrm{CS}_{2}\) has a linear molecule shape, meaning that the dipoles of the two bonds cancel each other out. Although the bonds are polar, the linear shape of the molecule results in a nonpolar overall structure. Therefore, \(\mathrm{CS}_{2}\) is nonpolar.
03

c) \(\mathrm{SO}_{3}\) molecule

1. Difference in EN between S and O: Sulfur (S) has an EN of 2.58 and Oxygen (O) has an EN of 3.44. The significant difference in electronegativity makes the SO bonds polar. 2. Shape of the molecule: \(\mathrm{SO}_{3}\) has a trigonal planar shape. The arrangement of the SO bonds in this shape results in the dipoles canceling each other out. Even though the SO bonds are polar, the overall structure of the molecule is nonpolar due to the symmetric arrangement of the dipoles. Hence, \(\mathrm{SO}_{3}\) is nonpolar.
04

d) \(\mathrm{PCl}_{3}\) molecule

1. Difference in EN between P and Cl: Phosphorus (P) has an EN of 2.19 and Chlorine (Cl) has an EN of 3.16. The difference in electronegativity makes the PCl bonds polar. 2. Shape of the molecule: \(\mathrm{PCl}_{3}\) has a trigonal pyramidal shape, which causes the dipoles of the P-Cl bonds not to cancel each other out. Since the bonds are polar and the dipoles do not cancel each other out due to the shape of the molecule, \(\mathrm{PCl}_{3}\) is a polar molecule.
05

e) \(\mathrm{SF}_{6}\) molecule

1. Difference in EN between S and F: Sulfur (S) has an EN of 2.58 and Fluorine (F) has an EN of 3.98. The significant difference in electronegativity makes the SF bonds polar. 2. Shape of the molecule: \(\mathrm{SF}_{6}\) has an octahedral shape. This arrangement causes the dipoles of the SF bonds to cancel each other out, making the overall structure nonpolar. Although the SF bonds are polar, the symmetric arrangement of the dipoles in the octahedral shape results in a nonpolar overall structure. Thus, \(\mathrm{SF}_{6}\) is nonpolar.
06

f) \(\mathrm{IF}_{5}\) molecule

1. Difference in EN between I and F: Iodine (I) has an EN of 2.66 and Fluorine (F) has an EN of 3.98. The considerable difference in electronegativity makes the IF bonds polar. 2. Shape of the molecule: \(\mathrm{IF}_{5}\) has a square pyramidal shape, which does not allow the dipoles of the IF bonds to cancel each other out. Since the bonds are polar and the dipoles do not cancel each other out due to the shape of the molecule, \(\mathrm{IF}_{5}\) is a polar molecule.

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Most popular questions from this chapter

The three species \(\mathrm{NH}_{2}^{-}, \mathrm{NH}_{3}\), and \(\mathrm{NH}_{4}^{+}\) have \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angles of \(105^{\circ}, 107^{\circ}\), and \(109^{\circ}\), respectively. Explain this variation in bond angles.

(a) Does \(\mathrm{SCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{BeCl}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

What is the distinction between a bond dipole and a molecular dipole moment?

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 \(\mathrm{N}-\mathrm{N}\) bond?

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of \(\mathrm{NO} ?\) (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+},\) and \(\mathrm{NO}^{-},\) and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?
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