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Chapter 9: Problem 19

In which of these molecules or ions does the presence of nonbonding electron pairs produce an effect on molecular shape? (a) \(\mathrm{CO}_{2},(\mathbf{b}) \mathrm{CH}_{2} \mathrm{Br}_{2},(\mathbf{c}) \mathrm{OF}_{2},\) (d) \(\mathrm{BCl}_{3}\), (e) \(\mathrm{SF}_{6}\).

Short Answer

Expert verified
Nonbonding electron pairs affect the shape of OF2.

Step by step solution

01

Understanding Nonbonding Electron Pairs

Nonbonding electron pairs, also known as lone pairs, can influence the molecular geometry by occupying space and repelling bonded atoms. This repulsion often affects the overall shape of the molecule.
02

Analyzing Molecule CO2

The central atom in CO2 is carbon, which forms double bonds with two oxygen atoms. There are no lone pairs on the carbon atom, so the shape of CO2 remains linear without distortion due to nonbonding pairs.
03

Analyzing Molecule CH2Br2

In CH2Br2, the central atom is carbon, which forms single bonds with two hydrogen atoms and two bromine atoms. Carbon has no lone pairs, meaning the shape (tetrahedral) is not affected by nonbonding pairs at the central atom.
04

Analyzing Molecule OF2

Oxygen in OF2 acts as the central atom bonded to two fluorine atoms. Additionally, oxygen has two nonbonding electron pairs. These lone pairs cause a bent shape due to their repulsion, which distorts the molecular geometry from linear.
05

Analyzing Molecule BCl3

Boron in BCl3 is surrounded by three chlorine atoms with no lone pairs on boron. Thus, the molecular shape is a planar trigonal, and no effect is observed from nonbonding pairs as boron doesn't possess any.
06

Analyzing Molecule SF6

Sulfur is the central atom in SF6 and forms six equivalent bonds with fluorine atoms. Since no lone pairs are present on sulfur, the geometry is octahedral, and there is no distortion due to nonbonding electron pairs.

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Key Concepts

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

Nonbonding Electron Pairs
Nonbonding electron pairs, also known as lone pairs, are valence electrons that are not involved in bonding with other atoms. These electron pairs can strongly influence the shape of a molecule. Understanding the role of nonbonding electron pairs is crucial because they exert repulsive forces on the other bonded atoms in a molecule. This repulsion:
  • Changes the angles between the bonds.
  • Creates unique molecular shapes.
For example, in the molecule OF2, oxygen has two lone pairs that repel the bonded fluorine atoms, creating a bent shape. These lone pairs take up more space around the central atom compared to bonded pairs because they are only bounded by one nucleus instead of two, intensifying their repulsive effect.
Molecular Shape
The molecular shape or geometry of a molecule is defined by the positions of the nuclei of the atoms in space. The arrangement is determined by the number of bonds and lone pairs on the central atom. Key shapes to remember include:
  • Linear
  • Tetrahedral
  • Trigonal planar
  • Bent
  • Octahedral
Molecules such as CO2 and BCl3 have symmetrical shapes like linear and trigonal planar, respectively, because they lack lone pairs on the central atoms. However, in the presence of lone pairs, like in OF2, the shape changes to accommodate the added repulsion, resulting in a bent molecular shape. The shape adopted is the one that minimizes the repulsion between electron pairs.
Lone Pairs Effect
Lone pairs are unshared electrons that are crucial in defining the corners of a molecule's geometry, due to their ability to repel bonded electron pairs. This effect changes the perceived shape and angular displacement of bonds. Lone pairs possess greater repulsive power in comparison to bonding electron pairs. This increase in repulsion
  • reduces the bond angle between bonded pairs,
  • pushes bonded atoms closer together,
modifying the expected geometric shapes. As a result, molecules like OF2, diverge from expected linear shapes to a bent shape because the lone pairs on the oxygen push the fluorine atoms closer together by occupying more space around the oxygen atom. This effect is absent in molecules such as CO2 and SF6 where nonbonding electron pairs are not present on the central atom, thus preserving the predicted geometries.

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Most popular questions from this chapter

How would we describe a substance that contains only paired electrons and is weakly repelled by a magnetic field? Which of the following ions would you expect to possess similar characteristics: \(\mathrm{H}_{2}^{-}, \mathrm{Ne}_{2}^{+}, \mathrm{F}_{2}, \mathrm{O}_{2}^{2+}\) ?

Which of the following statements about hybrid orbitals is or are true? (i) After an atom undergoes sp hybridization, there is one unhybridized \(p\) orbital on the atom, (ii) Under \(s p^{2}\) hybridization, the large lobes point to the vertices of an equilateral triangle, and (iii) The angle between the large lobes of \(s p^{3}\) hybrids is \(109.5^{\circ}\).

The oxygen atoms in \(\mathrm{O}_{2}\) participate in multiple bonding, whereas those in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the oxygen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{O}-\mathrm{O}\) bond?

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure \(9.46,\) how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on F interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) \((\mathbf{e})\) Look at the Lewis structure for HF. Where are the nonbonding electrons?

Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.
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