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. \((\mathbf{a}) \mathrm{H}_{2} \mathrm{S},(\mathbf{b}) \mathrm{BCl}_{3},(\mathbf{c}) \mathrm{CH}_{3} \mathrm{I},(\mathbf{d}) \mathrm{CBr}_{4,}(\mathbf{e}) \mathrm{TeBr}_{4}\)

Short Answer

Expert verified
We can confidently predict the bond angles for BCl3 (trigonal planar), CBr4 (tetrahedral), and TeBr4 (trigonal bipyramidal) using the VSEPR theory. However, we cannot confidently predict the bond angles for H2S (bent) and CH3I (tetrahedral) due to the presence of lone pairs and differing electronegativities of surrounding atoms, respectively.

Step by step solution

01

Identify central atom and electron pairs

First, we need to identify the central atom in each molecule and determine the number of bonding and non-bonding electron pairs around it: (a) H2S - Central atom: S; 2 bonding pairs (H-S); 2 lone pairs (b) BCl3 - Central atom: B; 3 bonding pairs (B-Cl); 0 lone pairs (c) CH3I - Central atom: C; 3 bonding pairs (C-H); 0 lone pairs; 1 bonding pair(C-I) (d) CBr4 - Central atom: C; 4 bonding pairs (C-Br); 0 lone pairs (e) TeBr4 - Central atom: Te; 4 bonding pairs (Te-Br); 1 lone pair
02

Predict molecular geometry using VSEPR

Now, we'll determine the molecular geometry based on the VSEPR theory: (a) H2S - 2 bonding pairs and 2 lone pairs - Bent molecular geometry (b) BCl3 - 3 bonding pairs and 0 lone pairs - Trigonal planar molecular geometry (c) CH3I - 3 C-H bonding pairs and 1 C-I bonding pair - Tetrahedral molecular geometry (d) CBr4 - 4 bonding pairs and 0 lone pairs - Tetrahedral molecular geometry (e) TeBr4 - 4 bonding pairs and 1 lone pair - Trigonal bipyramidal molecular geometry
03

Identify molecules with uncertain bond angles

We can now identify the molecules with uncertain bond angles: (a) H2S - As there are two bonding pairs and two lone pairs, the electron pair repulsion is unequal, and the bond angle might be slightly less than the ideal angle for a bent geometry, which is \(104.5^{\circ}\). (c) CH3I - In the case of CH3I, there are unequal electron pair repulsion forces because of the difference in electronegativity of H and I. This makes the bond angle prediction uncertain.
04

Conclusion

Based on the VSEPR theory and molecular geometry, we can confidently predict the bond angles for BCl3 (trigonal planar), CBr4 (tetrahedral), and TeBr4 (trigonal bipyramidal). However, we cannot confidently predict the bond angles for H2S (bent) and CH3I (tetrahedral) due to the presence of lone pairs and differing electronegativities of surrounding atoms, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Understanding molecular geometry is crucial for predicting properties like polarity, reactivity, and color of different compounds. Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The Valence Shell Electron Pair Repulsion (VSEPR) theory guides us in determining these shapes by assuming that electron pairs around a central atom will arrange themselves to minimize repulsion, giving us a predictable molecular structure.

For example, a molecule like BCl3, which has three bonding pairs and no lone pairs, adopts a trigonal planar geometry to minimize repulsion. On the other hand, H2S with two bonding pairs and two lone pairs forms a bent shape due to the additional space lone pairs require compared to bonding pairs. Predicting the geometry not only helps in visualizing the molecule but is also foundational in understanding its interaction with other molecules and its overall chemistry.
Bond Angles Prediction
Predicting bond angles is a bit like solving a spatial puzzle. Using the VSEPR theory, we can estimate the angles between bonded electron pairs. In a molecule with a tetrahedral layout such as CBr4, with equal repulsion among the four bonding electron pairs, the bond angles are roughly 109.5 degrees each. However, the presence of lone pairs or differences in the size and electronegativity of surrounding atoms can alter these angles.

Lone pair electrons occupy more space and can push bonding pairs closer together, resulting in smaller bond angles than in an ideal tetrahedral. This can be seen in a molecule like TeBr4, where the lone pair forces the Br-Te-Br bond angles in the equatorial plane to be less than the ideal 120 degrees. Predictions can become uncertain when these and other factors, like hybridization or the presence of multiple bonds, come into play.
Electron Pair Repulsion
Electron pair repulsion is the fundamental idea behind the VSEPR theory. It's the proverbial 'social distancing' at the atomic level. Electrons, being negatively charged, repel each other. This repulsion between valence electron pairs—whether bonding or non-bonding (lone pairs)—affects molecular geometry.

Bonding pairs, consisting of two electrons each, are shared between two atoms and are generally closer together than lone pairs, which are not shared and are only influenced by the nucleus of one atom. In our case of H2S, the lone pairs are a bit like extra guests at a dinner table—they need more room, leading to a bending of the structure away from the ideal linear geometry. The theory helps explain not only the shapes of molecules but also the angles between bonds, providing a framework for predicting many molecular properties and behaviors.

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Most popular questions from this chapter

There are two compounds of the formula Pt \(\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} :\) The compound on the right is called cisplatin, and the compound on the left is called transplatin. (a) Which compound has a nonzero dipole moment? (b) One of these compounds is an anticancer drug, and one is inactive. The anticancer drug works by its chloride ions undergoing a substitution reaction with nitrogen atoms in DNA that are close together, forming a \(\mathrm{N}-\mathrm{Pt}-\mathrm{N}\) angle of about \(90^{\circ} .\) Which compound would you predict to be the anticancer drug?

Draw a picture that shows all three 2\(p\) orbitals on one atom and all three 2\(p\) orbitals on another atom. (a) Imagine the atoms coming close together to bond. How many \(\sigma\) bonds can the two sets of 2\(p\) orbitals make with each other? (b) How many \(\pi\) bonds can the two sets of 2\(p\) orbitals make with each other? (c) How many antibonding orbitals, and of what type, can be made from the two sets of 2\(p\) orbitals?

Name the proper three-dimensional molecular shapes for each of the following molecules or ions, showing lone pairs as needed: \((\mathbf{a}) \mathrm{ClO}_{2}^{-}(\mathbf{b}) \mathrm{SO}_{4}^{2-}(\mathbf{c}) \mathrm{NF}_{3}(\mathbf{d}) \mathrm{CCl}_{2} \mathrm{Br}_{2}(\mathbf{e}) \mathrm{SF}_{4}^{2+}\)

In ozone, \(\mathrm{O}_{3}\) , the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plust (rotate) around a single \(\sigma\) bond same?

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