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Chapter 9: Problem 55

(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.

Short Answer

Expert verified
(a) A \(\sigma\) bond formed by two \(p\) orbitals is created through the head-on overlap of their dumbbell-shaped lobes. (b) A \(\pi\) bond formed by two \(p\) orbitals occurs when the orbitals overlap side-by-side, with electron density spread out above and below the plane of the atoms. (c) A \(\sigma\) bond is generally stronger than a \(\pi\) bond due to the greater extent of overlap and concentrated electron density. (d) Two \(s\) orbitals cannot form a \(\pi\) bond as they have a spherical shape, lacking distinct lobes for side-by-side overlap; they can only form \(\sigma\) bonds.

Step by step solution

01

(a) Drawing a \(\sigma\) bond formed by \(p\) orbitals

To draw a \(\sigma\) bond formed by two \(p\) orbitals, we must first understand that \(p\) orbitals have a dumbbell shape with two lobes. These orbitals can overlap end-to-end, with their lobes pointing towards each other. This is an example of a head-on overlap, which forms a \(\sigma\) bond. Step 1: Draw two atoms with their \(p\) orbitals, both having dumbbell shapes and pointing towards each other. Step 2: Show the overlapping region of two \(p\) orbitals in a different shade or color, if possible. Step 3: Indicate that this overlap is forming a \(\sigma\) bond in the molecule.
02

(b) Sketching a \(\pi\) bond formed by \(p\) orbitals

To sketch a \(\pi\) bond formed by two \(p\) orbitals, keep in mind that \(p\) orbitals have a dumbbell shape with two lobes. These orbitals can overlap side-by-side, with their lobes parallel to each other. This is a parallel overlap, which forms a \(\pi\) bond. Step 1: Draw two atoms with their \(p\) orbitals, both having dumbbell shapes and oriented parallel to each other. Step 2: Show the overlapping region of two \(p\) orbitals on the top and bottom, in a different shade or color, if possible. Step 3: Indicate that this side-by-side overlap is forming a \(\pi\) bond in the molecule.
03

(c) Comparing \(\sigma\) and \(\pi\) bond strengths

Generally, a \(\sigma\) bond is stronger than a \(\pi\) bond. The reason for this difference in strength can be explained by the extent of the overlap between orbitals. In a \(\sigma\) bond, the orbitals overlap end-to-end, which means that the electron density is concentrated between the two atoms. This leads to a greater extent of overlap and stronger bond formation. On the other hand, in a \(\pi\) bond, the orbitals overlap side-by-side, and the electron density is spread out above and below the plane of the atoms, resulting in a lesser extent of overlap and a weaker bond.
04

(d) Forming a \(\pi\) bond with two \(s\) orbitals

No, two \(s\) orbitals cannot combine to form a \(\pi\) bond. The reason for this is that \(s\) orbitals have a spherical shape, while \(\pi\) bonds are formed by the side-by-side overlap of orbitals with distinct lobes, such as \(p\) orbitals. Since \(s\) orbitals do not have distinct lobes and their electron density is spread uniformly around the nucleus, they cannot participate in the formation of a \(\pi\) bond. Instead, when two \(s\) orbitals overlap, they form a \(\sigma\) bond.

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Most popular questions from this chapter

(a) An \(\mathrm{AB}_{6}\) molecule has no lone pairs of electrons on the \(\mathrm{A}\) atom. What is its molecular geometry? (b) An \(\mathrm{AB}_{4}\) molecule has two lone pairs of electrons on the A atom (in addition to the four \(\mathrm{B}\) atoms). What is the electron-domain geometry around the A atom? (c) For the \(\mathrm{AB}_{4}\) molecule in part (b), predict the molecular geometry.

In which of the following AF \(_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: $\mathrm{PF}_{6}^{-}, \mathrm{SbF}_{\mathrm{s}}, \mathrm{SF}_{4} ?$

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated \(\mathrm{B}\) atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B-F bonds in \(\mathrm{BF}_{3}\) ? (d) What valence orbitals, if any, remain unhybridized on the \(\mathrm{B}\) atom in \(\mathrm{BF}_{3} ?\)

(a) Which geometry and central atom hybridization would you expect in the series $\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}{ }^{+} ?(\mathbf{b})$ What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?
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