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In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape, assuming they are all in the gaseous state? (a) \(\mathrm{SiH}_{4}\) (b) \(\mathrm{PF}_{3},\) (c) \(\mathrm{HBr}\), (d) \(\mathrm{HCN},\) (e) \(\mathrm{SO}_{2}\)

Short Answer

Expert verified
In the gaseous state, the presence of nonbonding electron pairs produce an effect on molecular shape for (b) PF3 and (e) SO2. PF3 has a trigonal pyramidal shape and SO2 has a bent (V-shaped) shape, both affected by their nonbonding electron pairs.

Step by step solution

01

Identify the central atom of each molecule/ion

For each molecule/ion, we will identify the central atom which is usually the least electronegative atom (except for hydrogen). (a) SiH4: The central atom is Si. (b) PF3: The central atom is P. (c) HBr: The central atom is Br. (d) HCN: The central atom is C. (e) SO2: The central atom is S.
02

Determine the number of nonbonding electron pairs on each central atom

We will now find the number of nonbonding electron pairs on each central atom: (a) SiH4: Si has 4 valence electrons, which are all bonded to four H atoms; thus, there are 0 nonbonding electron pairs. (b) PF3: P has 5 valence electrons, three of which are bonded to three F atoms. Thus, there remains 1 lone pair (nonbonding electron pair). (c) HBr: Br has 7 valence electrons, one of which is bonded to the H atom. Thus, there are 3 nonbonding electron pairs. (d) HCN: C has 4 valence electrons, two of which are bonded to the H atom and two others forming a triple bond with the N atom. Thus, there are 0 nonbonding electron pairs. (e) SO2: S has 6 valence electrons, four of which are bonded to two O atoms (with double bonds). Thus, there remains 1 nonbonding electron pair.
03

Use VSEPR theory to predict the molecular shape and the effect of nonbonding electron pairs

We will now use VSEPR theory to predict the molecular shape of each species and determine if the nonbonding electron pairs affect the shape: (a) SiH4: 0 nonbonding electron pairs; the shape is tetrahedral and unaffected by nonbonding electron pairs. (b) PF3: 1 nonbonding electron pair; the shape is trigonal pyramidal and affected by the nonbonding electron pair. (c) HBr: 3 nonbonding electron pairs; the shape is linear and unaffected by nonbonding electron pairs due to Br and H forming single bond with no other atoms present. (d) HCN: 0 nonbonding electron pairs; the shape is linear and unaffected by nonbonding electron pairs. (e) SO2: 1 nonbonding electron pair; the shape is bent (V-shaped) and affected by the nonbonding electron pair.
04

Conclusion

Among the given molecules/ions, the presence of nonbonding electron pairs produce an effect on the molecular shape for (b) PF3 and (e) SO2.

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Key Concepts

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

VSEPR Theory
VSEPR (Valence Shell Electron Pair Repulsion) theory is a fundamental concept used to predict the molecular geometry of molecules based on the repulsion between electron pairs around a central atom. It operates under the principle that electron pairs, both bonding and nonbonding, exert repulsive forces on each other. This is due to the negative charges of the electrons. These repulsions cause the electron pairs to arrange themselves as far apart as possible around the central atom.

By estimating the angles and distances between electron pairs, chemists can predict the shape of the molecule. For instance, when a molecule like \( ext{SiH}_4\) forms, the electron pairs distribute themselves in a way that minimizes repulsion, resulting in a tetrahedral shape. This theory is crucial as it aids in understanding the 3D arrangement of atoms in a molecule, which in turn affects molecular properties such as polarity and reactivity.
Nonbonding Electron Pairs
Nonbonding electron pairs, also known as lone pairs, are pairs of valence electrons that are not shared with another atom or involved in bonding. They occupy space around the central atom and can significantly impact the molecular geometry. Unlike bonding pairs, which are involved in a covalent bond, nonbonding pairs are fully localized on an atom and exert a greater repulsive force than bonding pairs.

This increased repulsion affects the overall shape of the molecule. The presence of lone pairs can lead to smaller bond angles. For example, in \( ext{PF}_3\), the phosphorus atom has one lone pair. This lone pair pushes the bonding pairs of electrons closer together, altering the shape from what might have been tetrahedral to a trigonal pyramidal shape. Similarly, in \( ext{SO}_2\), the lone pair on sulfur results in a bent shape, which differentiates it from a linear structure. Understanding nonbonding pairs is essential as they contribute significantly to the prediction of molecular shapes.
Molecular Shape Prediction
Predicting the shape of a molecule involves understanding both the number of bonding pairs and the number of nonbonding pairs (lone pairs) using VSEPR theory. The molecular shape is significantly dictated by the repulsions between electron pairs, which aim to be as far apart as possible. The process usually begins by examining the central atom of the molecule.

Consider \( ext{PF}_3\) and \( ext{SO}_2\) as examples. In \( ext{PF}_3\), we identified a trigonal pyramidal shape due to the lone pair on the phosphorus atom repelling the bonding pairs. This adjustment is necessary to minimize electron pair repulsion and leads us away from a typical trigonal planar shape. For \( ext{SO}_2\), a lone pair on the sulfur atom results in a bent structure, which deviates from linear geometry primarily due to the lone pair repulsion.

By understanding how nonbonding electron pairs influence bond angles and molecular shapes, students can predict the 3D arrangements of molecules with greater accuracy, an essential skill in both theoretical and practical chemistry applications.

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Most popular questions from this chapter

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a sigma bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two \(s\) orbitals combine to form a \(\pi\) bond? Explain.

The \(\mathrm{PF}_{3}\) molecule has a dipole moment of \(1.03 \mathrm{D}\), but \(\mathrm{BF}_{3}\) has a dipole moment of zero. How can you explain the difference?

What property of the electron causes electron domains to have an effect on molecular shapes?

Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: \((a) 3,(b) 4,(c) 5,(d) 6\)

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}:\) The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is ineffective for cancer therapy. Both compounds have a square- planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently, cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the \(\mathrm{Cl}^{-}\) ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

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