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Chapter 9: Problem 20

In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. (a) \(\mathrm{H}_{2} \mathrm{~S}\), (b) \(\mathrm{BCl}_{3}3,\) (c) \(\mathrm{CH}_{3} \mathrm{I},(\mathrm{d}) \mathrm{CBr}_{4}\), (e) \(\mathrm{TeBr}_{4}\)

Short Answer

Expert verified
For the given molecules: (a) \(H_2S\) has a bent molecular geometry with bond angles close to 104.5°; we can confidently predict this angle. (b) \(BCl_3\) has a trigonal planar molecular geometry with bond angles of 120°, which can be confidently predicted. (c) \(CH_3I\) and (d) \(CBr_4\) both have tetrahedral molecular geometries with bond angles around 109.5°, which can be confidently predicted. (e) \(TeBr_4\) has a see-saw molecular geometry, and we can be uncertain about the bond angles due to the effect of the lone pair on the central Tellurium atom, causing deviations from ideal bond angles.

Step by step solution

01

(a) H2S

For \(\mathrm{H}_{2} \mathrm{S}\), the central atom is Sulfur (S) with 6 valence electrons. There are two bonded atoms (Hydrogen), and two lone pairs of electrons. The total number of electron domains is 4, which gives the electron domain geometry of tetrahedral. Since there are two bonded atoms and two lone pairs, the molecular geometry is bent. In the case of H2S, we can confidently predict the bond angles as being close to 104.5°, although the presence of lone pairs may somewhat decrease the angle.
02

(b) BCl3

In \(\mathrm{BCl}_{3}\), the central atom is Boron (B) with 3 valence electrons. There are three bonded atoms (Chlorine) and no lone pairs. This gives a total of 3 electron domains, which leads to a trigonal planar electron domain geometry. The molecular geometry is also trigonal planar since there are no lone pairs. In this case, we can confidently predict the bond angles to be 120°.
03

(c) CH3I

In \(\mathrm{CH}_{3} \mathrm{I}\), the central atom is Carbon (C) with 4 valence electrons. There are three bonded Hydrogen atoms and one bonded Iodine atom, with no lone pairs on the central Carbon atom. The total number of electron domains is 4, which provides a tetrahedral electron domain geometry. The molecular geometry is also tetrahedral, as there are no lone pairs. Here, we can confidently predict the bond angles to be around 109.5°.
04

(d) CBr4

For \(\mathrm{CBr}_{4}\), the central atom is Carbon (C) with 4 valence electrons. There are four bonded Bromine atoms and no lone pairs. The total number of electron domains is 4, giving a tetrahedral electron domain geometry. The molecular geometry is also tetrahedral due to the absence of lone pairs. In this case, we can confidently predict the bond angles to be around 109.5°.
05

(e) TeBr4

In \(\mathrm{TeBr}_{4}\), the central atom is Tellurium (Te) with 6 valence electrons. There are four bonded atoms (Bromine) and one lone pair on the central Tellurium atom. The total number of electron domains is 5, which gives the electron domain geometry of trigonal bipyramidal. The molecular geometry is see-saw due to the presence of one lone pair. In this case, we can be uncertain about the bond angles since they may deviate from the ideal bond angles expected for a trigonal bipyramidal geometry. The axial bond angle can be close to 180° and equatorial bond angles can be close to 120°, but the exact bond angles may be distorted due to the lone pair affecting the bond angles between the Bromine atoms.

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Most popular questions from this chapter

A compound composed of \(2.1 \% \mathrm{H}, 29.8 \% \mathrm{~N},\) and \(68.1 \% \mathrm{O}\) has a molar mass of approximately \(50 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if \(\mathrm{H}\) is bonded to \(\mathrm{O} ?\) (c) What is the geometry of the molecule? (d) What is the hybridization of the orbitals around the \(\mathrm{N}\) atom? (e) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Are any valence atomic orbitals of \(\mathrm{B}\) left unhybridized? If so, how are they oriented relative to the hybrid orbitals?

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a double or triple bond as a single electron domain. Why is this justified?

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH},\) gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

(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.)
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