/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 (a) Does \(\mathrm{SCl}_{2}\) ha... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 37

(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?

Short Answer

Expert verified
(a) Yes, SCl2 has a dipole moment that points from the S atom to the midpoint between the two Cl atoms due to its bent molecular geometry and the significant difference in electronegativity between S and Cl atoms. (b) No, BeCl2 does not have a dipole moment because the linear molecular geometry causes the polarities of its individual bonds to cancel each other out, despite the electronegativity differences between Be and Cl atoms.

Step by step solution

01

Determine the molecular geometry of SCl2

First, let's determine the molecular geometry of SCl2. Sulfur has 6 valence electrons, and each chlorine atom contributes 7 valence electrons. Thus, the total number of valence electrons is 6 + (2 脳 7) = 20. These 20 electrons form 2 single bonds between sulfur and the two chlorine atoms, and a pair of lone pair electrons on the sulfur. In this arrangement, we have a bent molecular geometry for SCl2.
02

Determine electronegativity differences for SCl2

Now, let's check the electronegativity difference between S and Cl. The electronegativity of S is 2.58, and that of Cl is 3.16. Thus, the difference in electronegativity between sulfur and chlorine is 3.16 - 2.58 = 0.58. This difference is significant and implies that SCl2 has a polar bond.
03

Determine the dipole moment direction of SCl2

Since we have a bent molecular geometry, the asymmetrical distribution of electron density leads to a net dipole moment. The net dipole points towards the more electronegative atom, which is chlorine. So, the direction of the net dipole in SCl2 is from S to the midpoint between the two Cl atoms. Now let's analyze BeCl2:
04

Determine the molecular geometry of BeCl2

First, let's determine the molecular geometry of BeCl2. Beryllium has 2 valence electrons, and each chlorine contributes 7 valence electrons. This makes a total of 2 + (2 脳 7) = 16 valence electrons. Beryllium uses its 2 valence electrons to form single bonds with both chloride atoms. The molecular geometry for BeCl2 is linear.
05

Determine electronegativity differences for BeCl2

Now, let's check the electronegativity difference between Be and Cl. The electronegativity of Be is 1.57, and the value for Cl stays at 3.16. The difference in electronegativity between beryllium and chlorine is 3.16 - 1.57 = 1.59. This indicates that the Be-Cl bond is polar.
06

Determine the dipole moment direction of BeCl2

The molecular geometry of BeCl2 is linear, causing the individual bond dipoles to cancel each other out, resulting in no net dipole moment for BeCl2. In summary: (a) SCl2 has a dipole moment that points from the S atom to the midpoint between the two Cl atoms. (b) BeCl2 does not have a dipole moment, as the polarity of its individual bonds cancel each other out due to the linear molecular geometry.

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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) Explain why \(\mathrm{BrF}_{4}{ }^{-}\)is square planar, whereas \(\mathrm{BF}_{4}{ }^{-}\)is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Suppose that silicon could form molecules that are precisely the analogs of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). How would you describe the bonding about \(\mathrm{Si}\) in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

Give the electron-domain and molecular geometries for the following molecules and ions: (a) \(\mathrm{HCN}\), (b) \(\mathrm{SO}_{3}^{2-}\), (c) \(\mathrm{SF}_{4}\), (d) \(\mathrm{PF}_{6}^{-}\), (e) \(\mathrm{NH}_{3} \mathrm{Cl}^{+}\), (f) \(\mathrm{N}_{3}^{-}\).

Consider the following \(\mathrm{XF}_{4}\) ions: \(\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+}\), and \(\mathrm{AlF}_{4}^{-}\). (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electrondomain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry? (d) Which of the ions will exhibit a see-saw molecular geometry?
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