Edward J. Neth; Paul Flowers; William R. Robinson, PhD; Klaus Theopold; Richard Langley

1.

Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).

3.

Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side-by-side to form the π bond.

This figure shows the orbitals of two atoms being added together as they form bonds. The two atoms are shown separately on the right, each having two peanut-shaped orbitals lying perpendicularly to one another. A right-facing arrow shows that the two have moved closer together and now the upper and lower portions of the vertical peanut-shaped orbitals are shown as merging together above and below the plane of the molecule while the horizontal peanut-shaped orbitals are overlapping between the two nuclei.

5.

No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.

Two nitrogen atoms are shown both in a Lewis structure and as a diagram. The Lewis structure depicts the two nitrogen atoms bonded by a triple bond. The diagram shows two nitrogen atoms and their three peanut-shaped p-orbitals. One of the orbitals lies horizontally and overlaps between the two nuclei. It is labeled, “sigma bond.” The other two lie vertically and in the z-plane of the page. They overlap above and below and into and out of the page in relation to the nuclei. They are each labeled, “pi bond.”

7.

Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.

9.

There are no d orbitals in the valence shell of carbon.

11.

trigonal planar, sp²; trigonal pyramidal (one lone pair on A) sp³; T-shaped (two lone pairs on A sp³d, or (three lone pairs on A) sp³d²

13.

(a) Each S has a bent (109°) geometry, sp³

A Lewis structure is shown in which eight sulfur atoms, each with two lone pairs of eletrons, are single bonded together into an eight-sided ring.

(b) Bent (120°), sp²

Two Lewis structure are shown, connected by a double-ended arrow. The left structure shows a sulfur atom with one lone pair of electrons double bonded to an oxygen atom with two lone pairs of electrons on the left and single bonded to an oxygen atom with three lone pairs of electrons on the right. The right structure shows the same molecule, except that the double bonded oxygen is on the right side of the sulfur and the single bonded oxygen is to the left of the sulfur.

(c) Trigonal planar, sp²

A Lewis structure of a sulfur atom singly bonded to two oxygen atoms, each with three lone pairs of electrons, and double bonded to a third oxygen atom with two lone pairs of electrons is shown.

(d) Tetrahedral, sp³

A Lewis structure is shown in which a sulfur atom is single bonded to four oxygen atoms. Two of the oxygen atoms have three lone pairs of electrons while the other two each have two lone pairs of electrons and are each singly bonded to a hydrogen atom.

15.

(a) XeF₂
(b)

(c) linear (d) sp³d

17.

(a)

Two Lewis structure are shown, the left of which depicts three phosphorus atoms single bonded together to form a triangle. Each phosphorus is bonded to a sulfur atom by a vertical single bond and each of those sulfur atoms is then bonded to a single phosphorus atom so that a six-sided ring is created with a sulfur in the middle. Each sulfur atom in this structure has two lone pairs of electrons while each phosphorus has one lone pair. The second Lewis structure shows a chlorine atom with one lone pair of electrons single bonded to three oxygen atoms, each of which has three lone pairs of electrons.

(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp³; (d) Oxidation states P +1, S $- 1 \frac{1}{3},$ Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1

19.

Two Lewis structures are shown. The left structure shows a nitrogen atom with one lone pair of electrons single bonded to three fluorine atoms, each of which has three lone pairs of electrons. The right structure shows a phosphorus atoms single bonded to five fluorine atoms, each of which has three lone pairs of electrons.

Phosphorus and nitrogen can form sp³ hybrids to form three bonds and hold one lone pair in PF₃ and NF₃, respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp³d hybrid orbitals to bind five fluorine atoms in NF₅. Phosphorus has d orbitals and can bind five fluorine atoms with sp³d hybrid orbitals in PF₅.

21.

A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.

23.

(a)

A Lewis structure is shown in which a carbon atom is attached by single bonds to three hydrogen atoms. It is also attached by a single bond to a carbon atom that is triple bonded to a nitrogen atom with one lone electron pair. Below the structure is a right facing arrow with its head near the nitrogen and its tail, which looks like a plus sign, located near the carbon atoms. The arrow is labeled, “dipole moment.”

(b) The terminal carbon atom uses sp³ hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2p orbital on carbon and a nitrogen 2p orbital.

25.

(a) sp²; (b) sp; (c) sp²; (d) sp³; (e) sp³; (f) sp³d; (g) sp³

27.

(a) sp², delocalized; (b) sp, localized; (c) sp², delocalized; (d) sp³, delocalized

29.

A diagram is shown in two parts, connected by a right facing arrow labeled, “Hybridization.” The left diagram shows an up-facing arrow labeled, “E.” To the lower right of the arrow is a short, horizontal line labeled, “2 s,” that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, “2 p.” Above both sets of lines is the phrase, “Orbitals in an isolated C atom.” There are two upward facing arrows on two of these lines. The right side of the diagram shows two short, horizontal lines placed halfway up the space and each labeled, “s p.” An upward-facing half arrow is drawn vertically on each line. Above these lines are two other short, horizontal lines, each labeled, “2 p,” and which have two upward facing arrows on them. Above both sets of lines is the phrase, “Orbitals in the s p hybridized C in C O subscript 2.”

Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.

31.

(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.

33.

An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.

35.

Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.

37.

The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.

39.

(a) H₂ bond order = 1, $H_{2}^{+}$ bond order = 0.5, $H_{2}^{−}$ bond order = 0.5, strongest bond is H₂; (b) O₂ bond order = 2, $O_{2}^{2+}$ bond order = 3; $O_{2}^{2−}$ bond order = 1, strongest bond is $O_{2}^{2+};$ (c) Li₂ bond order = 1, ${Be}_{2}^{+}$ bond order = 0.5, Be₂ bond order = 0, strongest bond is ${Li}_{2}$ ;(d) F₂ bond order = 1, $F_{2}^{+}$ bond order = 1.5, $F_{2}^{−}$ bond order = 0.5, strongest bond is $F_{2}^{+};$ (e) N₂ bond order = 3, $N_{2}^{+}$ bond order = 2.5, $N_{2}^{−}$ bond order = 2.5, strongest bond is N₂

41.

(a) H₂; (b) N₂; (c) O; (d) C₂; (e) B₂

43.

Yes, fluorine is a smaller atom than Li, so atoms in the 2s orbital are closer to the nucleus and more stable.

45.

2+

47.

N₂ has s-p mixing, so the π orbitals are the last filled in $N_{2}^{2+} .$ O₂ does not have s-p mixing, so the σ_p orbital fills before the π orbitals.

Chapter 5