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

Short Answer

Expert verified
In molecules (b) PF3 and (e) SO2, the presence of nonbonding electron pairs affects the molecular shape due to repulsion between the lone pairs and bonding pairs of electrons. In the other molecules, there are no nonbonding electron pairs on the central atom, so they do not have an effect on the molecular shape.

Step by step solution

01

(a) SiH4

: Silicon is the central atom with 4 hydrogen atoms bonded to it, creating a tetrahedral arrangement. The silicon atom has a total of 4 valence electrons which are entirely involved in bonding with the hydrogens. There are no lone pairs on silicon, so the molecular geometry is also a tetrahedron. The effect of nonbonding electron pairs is not seen.
02

(b) PF3

: In PF3, phosphorus is the central atom with 3 fluorine atoms bonded to it. Phosphorus has 5 valence electrons, 3 of which form bonds with the fluorine atoms, leaving behind 1 lone pair (2 nonbonding valence electrons). These lone pairs can distort the molecular shape due to their repulsion with the bonding pairs. The presence of nonbonding electron pairs has an effect on the molecular shape.
03

(c) HBr

: HBr is a linear diatomic molecule with no central atom. Hence nonbonding electron pairs do not have an effect on molecular shape.
04

(d) HCN

: In this linear molecule, carbon is the central atom with 2 bonding electron pairs: one with hydrogen and one with nitrogen. Carbon has 4 valence electrons, all of which are involved in bonding, leaving no nonbonding electron pairs. Hence, there is no effect due to nonbonding electron pairs on the molecular shape.
05

(e) SO2

: In SO2, sulfur is the central atom, and it is bonded to 2 oxygen atoms. Sulfur has 6 valence electrons, 4 of which are involved in the bonding with 2 oxygen atoms, leaving behind 1 lone pair (2 nonbonding valence electrons). These lone pairs lead to repulsion with the bonding pairs, distorting the molecular shape. The presence of nonbonding electron pairs has an effect on the molecular shape. In conclusion, the presence of nonbonding electron pairs has an effect on the molecular shape of (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
Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used in chemistry to predict the shape of individual molecules based on the number of electrons surrounding their central atoms. It operates under the principle that electron pairs around a central atom will organize themselves to minimize repulsion between each other. This repulsive force mainly arises because electron pairs are negatively charged and like charges repel.

The arrangement of electron pairs is determined first by totaling the number of bonding (shared with other atoms) and nonbonding (lone pairs) electron pairs. Then, VSEPR theory predicts the geometric distribution of these pairs to minimize repulsion and determine the molecular shape. The shapes can range from linear, trigonal planar, tetrahedral, to more complex geometries such as trigonal bipyramidal and octahedral, depending on the number of electron pairs.
Lone Pair Repulsion
Lone pair repulsion is a concept within the VSEPR theory that explains how nonbonding electron pairs can influence the shape of a molecule. When electrons are in lone pairs, they occupy more space around the central atom than bonding pairs because they are solely under the influence of one nucleus, as opposed to bonding pairs which are 'shared' between two nuclei.

As a result, lone pairs exert greater repulsive forces on each other and on bonding pairs of electrons. This increased repulsion can lead to distortions in the idealized molecular shape. For instance, in a molecule with one lone pair, you might expect the bonds to deviate from their positions when compared to a similar molecule with all bonding pairs. Understanding how the presence and number of lone pairs affect molecular structure is crucial in predicting how molecules will look and behave.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The geometry of a molecule significantly influences its physical and chemical properties, including reactivity, polarity, and interaction with other molecules.

The VSEPR theory helps in understanding these geometries by considering the number of bonding and nonbonding electron pairs. Some common geometries are linear, bent, trigonal planar, pyramidal, and tetrahedral. Lone pair repulsions can cause deviations from these basic geometries, leading to 'bent' or 'distorted' shapes like those seen in PF3 and SO2 molecules from the original exercise. It's essential to distinguish between the electron-pair geometry, which describes the arrangement of all electron pairs (bonding and nonbonding), and the molecular shape, which describes the spatial arrangement of just the atoms.
Valence Electrons
Valence electrons are the outermost electrons of an atom and are significant in determining how the atom can bond with others. These electrons have the highest energy level and are usually the only electrons involved in chemical bonds. The number of valence electrons an atom has influences its chemical properties and the type of bonds it can form.

Atoms tend to follow an 'octet rule,' where they are most stable with eight valence electrons. In bonding, atoms will share, lose, or gain electrons to achieve a full valence electron shell. During this process, lone pairs can be formed when atoms have extra valence electrons that aren't shared in bonds. By analyzing these electrons through the VSEPR theory, we can predict the molecular shape and understand the structural reasons behind the molecule's behavior and reactivity.

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Most popular questions from this chapter

Consider the Lewis structure for glycine, the simplest amino acid: (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygens and the nitrogen atom, and what are the approximate bond angles at the nitrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}{ }^{-}\)ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\)ion to be stable? (e) Which of the following statements about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{4}{ }^{-}\right)\)are both described as tetrahedral. What does this indicate about their bond angles? (b) The \(\mathrm{NH}_{3}\) molecule is trigonal pyramidal, while \(\mathrm{BF}_{3}\) is trigonal planar. Which of these molecules is flat?

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?

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\)-axis is vertical in the plane of the paper and the \(x\)-axis horizontal. Write " \(\mathrm{M}^{"}\) at the origin to denote a metal atom. (b) Now, on the \(x\) axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(M\). The CO bond axis should be on the \(x\)-axis. (c) Draw the \(\mathrm{CO} \pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2}^{*}\) orbital of \(\mathrm{CO}\) ? (e) What kind of bond is being made with the orbitals between \(M\) and \(\mathrm{C}_{,} \sigma\) or \(\pi\) ? (f) Predict what will happen to the strength of the CO bond in a metal\(\mathrm{CO}\) complex compared to \(\mathrm{CO}\) alone.

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