helpsheet_16_103

CHEMISTRY 103 – Help Sheet #16 Chapter 7 (Part II): Sections 7.3, 7.4; Chapter 6 (Part II) (6.5a) Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob http://www.chem.wisc.edu/areas/clc (Resource page) Nuggets: Assign Hybrid Orbitals (HO) to Atoms; Assign HO that Make a Bond HYBRID ORBITALS/HYBRIDIZATION With molecules such as CH4, VSEPR predicts a tetrahedral geometry with bond angles of 109.5˚. This implies the orbitals used to create these bonds must lie at 109.5˚. In CH4, the central C atom has 2s and 2p atomic orbitals available to form bonds but the 2p orbitals lie at 90˚. Hence, the angles between the p-orbitals don’t align with the 109.5˚ molecular angles predicted. A new model is needed to explain the molecular bonding. Solution: Atomic orbitals are combined to form new hybrid orbitals (HO) that can be used to form molecular bonds at the proper angles. Tetrahedral Arrangement – sigma (σ ) bonds

z

z

+

sp3

sp3

sp3

sp3

s

+

sp 3

y

x

px

+

x

py

+

p

y

x

s

p

z

y

x

p

z

y

pz

+

+

sp 3

sp 3

sp 3

sp 3

Trigonal Planar Arrangement – sigma (σ ) bonds

z

z y

+

x

p

p sp2

+

x

px

p sp2

y

x

s

p

z y

+

pz

+

sp2

sp 2

s

sp 2

sp 2

sp 2

Linear Arrangement – sigma (σ ) bonds

z

z y

y

x

+

x

s

p

p

p

p sp

s

px

p +

sp

sp

sp

sp

Bonds are made from 2 electrons with opposite spins in two orbitals and those two orbitals overlap. Sigma (σ ) Bonds: The bonds/electrons lie between the nuclei of the two atoms on the internuclear axis. All bonds (single, double, or triple) contain one σ bond. Sigma (σ) bonds are commonly made from the overlap of two hybrid orbitals, HO–HO, (e.g., sp2–sp3) or 1s–hybrid orbital, 1s–HO (e.g., H–C, 1s–sp3). Sigma bonds can rotate, are symmetrical, and are stronger than pi (π) bonds. Note: Other bonds often involving halogen atoms have sigma (σ) bonds that are often described as being made from atomic orbitals rather than hybrid orbitals. For example, H–F can be described as atomic orbitals 1s–2pz overlapping to form a sigma (σ) bond. Likewise, F–F bonds are formed from 2pz–2pz atomic orbitals overlapping to form a sigma (σ) bond. H–H bonds are formed from the overlap of 1s–1s atomic orbitals to again form a sigma (σ) bond. (Note: Not all instructors present these non-HO sigma (σ) bonds in this fashion; some instructors will describe the H–F bond as 1s-sp3.)

Pi (π ) Bonds: Bonds/electrons do not lie on internuclear axis and usually occur with double or triple bonds. In the triangular planar (sp2) and linear (sp) arrangements, one p-orbital from each atom is used to make the pi (π) bond; the π bonds are made from the side-to-side overlap of two p-atomic orbitals, p–p, as shown below. 2 pz-orbitals before overlapping z

2 pz-orbitals overlapping to form π bond

z y

y

x

+

y y

x

x

pz

pz

π bond

How to make bonds: σ bonds: HO–HO, (e.g., sp2–sp3), 1s–HO (e.g., 1s–sp3), or 1s–1s (e.g., H–H) (sometimes bonds are described as being made as follows: H–F: 1s–2pz; F–F: 2pz–2pz)

π bonds: p–p (two p-atomic orbitals) Single bond = 1σ bond; Double bond = 1σ bond + 1π bond; Triple bond = 1σ bond + 2π bonds GEOMETRIC ISOMERS (Section 6.5a): cis versus trans configuration Does the molecule have geometric isomers? 1. Must contain a C=C (or another double bond) 2. On each C of the C=C, there must be 2 different groups Both criteria must be true for cis/trans isomers to exist. The molecule below does not have cis/trans isomers. The left C atom of the C=C bond does have 2 different groups (a –CH3 and a –CH2CH3 group) but the right C atom of the C=C has 2 H atoms which are the same (each C atom of the C=C needs two different groups attached). H

H3C C H3C

C

H2C

H

no cis or trans

If molecule has geometric isomers, is it cis or trans? Simple molecules have 2 groups (e.g., –CH3) that are the same attached to the 2 C’s of the C=C (these are the typical molecules seen in general chemistry) 1. Draw a line through C=C bond 2. Find the 2 groups that are the same on the 2 C of the C=C and circle them; if they’re on the same side of the line → cis isomer; different sides of the line → trans isomer H

H3C C

H3C

C

H

C CH3

trans

CH3 C

H

H

cis

Example: a. Draw the Lewis dot structure for CF2=CH2. b. Identify the hybridization for each atom. c. Identify what orbitals make up each bond. 4 regions = sp3

3 regions = sp2

Single bond = σ bond σ bond = HO–HO = sp2(C)–sp 3(F)

3 regions = sp2 1s

F

H C

C

F

H 1s

4 regions = sp3

Answer:

Single bond = σ bond σ bond = HO–1s = sp2(C)–1s(H)

Double bond = σ + π bonds σ = HO–HO = sp2(C)–sp 2(C) π = p–p = p(C)–p(C)

1. For each molecule below, determined the electron region geometry and hybridization around the central atom. a. CO3-2 b. SO2 c. SO3-2 d. BCl3 2. I. a. For the molecule, C2H4, what hybrid orbitals are used to make the C=C bonds? (Hint: Draw the Lewis dot structure first.) b. What orbitals are used to make the C–H bonds? II. a. For the molecule, Cl2CO, what hybrid orbitals are used to make the C=O bonds? (Hint: Draw the Lewis dot structure first.) b. What orbitals are used to make the C–Cl bonds? a

H H H 3. a. Draw in the lone pairs in the structure that are missing. b. What is the bond order for the carbon-carbon bond labeled “a”? H C C C C C H C C c. What is the hybridization on the N labeled “b”? d d. What is the angle for the H–C–C labeled “c”? C C c e. What is the electron region geometry for the O atom labeled “f”? H C O H f. What orbitals overlap to form the bond labeled “e”? f N b g. How many σ and π bonds are in the molecule below? H H h. What is the molecular geometry around the N labeled “b”? i. What orbitals overlap to form the σ bond labeled “d”? What orbitals are used to form the π bond labeled “d”?

4. Use the structure to answer the questions below about the structure and bonding in this molecule. a. What is the angle between the H-C-H labeled “c”? H d b. What is the electron region geometry around the carbon atom labeled “b”? H C c. What is the hybridization of the carbon atom labeled “a”? H C C C d. What bonds make up the triple bond between carbon atoms labeled “d”? C e. What orbitals overlap to make the bonds between the C atoms labeled “d”? H b f. How many σ and π bonds are there in the molecule? H

e

a

H

H C

C

H C

C C

C H

c

H

5. How many π and σ bonds does the molecule below have? HO

C

C

H

H

H

H

O

H

C

C

C

C

C

C

NH2

H

a. 16σ, 5π

b. 16σ, 3π

c. 16σ, 4π

d. 19σ, 5π

e. 19σ, 4π

6. (Do if cis/trans isomers are covered) Which of these molecules can have geometric isomers? Not a multiple choice question; there may be more than one molecule. H

CH2 C

a.

CH3

H

CH2

CH3 CH2

C H3C

H2C

CH2 C

H

d.

H

C

b. CH3

CH3 CH2 C

e.

c.

H2C

C H

H3C

H3C

CH3 C

H3C H3C

CH3

C H3C

C CH3

CH3

C CH3

ANSWERS 1. a. triangular planar, sp2 b. triangular planar, sp2 c. tetrahedral, sp3 d. triangular planar, sp2 2. I. a. σ bond = sp2(C)–sp2(C); π bond = p(C)–p(C) b. σ bond: sp2(C)–1s(H) II. a. σ bond = sp2(C)–sp2(O); π bond = p(C)–p(O) b. σ bond = sp2(C)–sp3(Cl) 3. a. See figure right. a H H H b. 1.5 H C C C C C H c. sp3 C C e d d. 120˚ C C c e. tetrahedral H C O H f. 1s(H)-sp(C) f N b g. 21σ bonds; 6π bonds H H h. triangular pyramid i. σ: sp2(C)–sp2(C); π: p(C)–p(C) 4. a. 109.5˚ b. linear c. sp2 d. 1σ + 2π e. σ bond: sp(C)–sp(C); π bond: px(C)–px(C); π bond: py(C)–py(C) assumes z-axis is internuclear axis f. 22σ, 5π 5. d {π bonds: each double bond = 1π bond; each triple bond = 2π bonds; π bonds = 1 + 1 + 1 + 2 = 5π bonds; each bond (single, double, or triple) has 1σ bond; be careful not to miss the O–H single bond (1σ bond) and the 2 N–H single bonds (2σ bonds)}

6. b, d, e