1. 1. Structure and Bonding Why this chapter?
• Review ideas from general chemistry: atoms, bonds, molecular geometry

2. What is Organic Chemistry?
Living things are made of organic chemicals
Proteins that make up hair
DNA, controls genetic make-up
Foods, medicines
Examine structures below

3. Origins of Organic Chemistry
Foundations of organic chemistry from mid-1700’s. Compounds obtained from plants, animals hard to isolate, and purify. Compounds also decomposed more easily. Torben Bergman (1770) first to make distinction between organic and inorganic chemistry. It was thought that organic compounds must contain some “vital force” because they were from living sources.

4. Because of “Vital force”, it was thought that organic compounds could not be synthesized in laboratory like inorganic compounds. 1816, Chervrut showed that not to be the case, he could prepare soap from animal fat and an alkali 1828, Wholer showed that it was possible to convert inorganic salt ammonium cyanate into organic substance “urea”

5. 90% of more than 30 million chemical compounds contain carbon.
• Organic chemistry is study of carbon compounds.
• Why is it so special?
90% of more than 30 million chemical compounds contain carbon.
Examination of carbon in periodic chart answers some of these questions.
Carbon is group 4A element, it can share 4 valence electrons and form 4 covalent bonds.

6. 1.1 Atomic Structure Structure of an atom
Positively charged nucleus (very dense, protons and neutrons) and small (10-15 m)
Negatively charged electrons are in a cloud (10-10 m) around nucleus
Diameter is about 2  m (200 picometers (pm)) [the unit angstrom (Å) is m = 100 pm]

7. X The Atomic Symbol A = Atomic mass Z = Atomic #
= # p + # n
C = Charge
= + or - values X A C Z #
Z = Atomic #
# p = # e
# = Number of atoms in a formula.

8. X - - - - - - The Atomic Symbol 6 A = Atomic mass 12 Z = Atomic number
= # protons + # neutrons 6 6 - - X 12 - - 6 + +
Z = Atomic number
= # protons = # electrons - + + + + -

9. C - - - - - - 6 The Atomic Symbol A = Atomic mass 12 Z = Atomic number
= # protons + # neutrons - - C 12 - - 6 + +
Z = Atomic number
= # protons = # electrons - + + + + -

10. Na A = Atomic mass Z = Atomic # = p = 11 C = Charge = +1
The Atomic Symbol
A = Atomic mass
= p + n = 23
C = Charge = +1 11 12 23 1+ Na 11
# = 1 atom in formula.
Z = Atomic # = p = 11
Sodium

11. Why is the atomic weight on the tables not a whole #?47
Atomic number
Name of the element
Elemental Symbol
Silver Ag
107.87
Atomic mass (weight)
Atomic weight = The average, relative mass of an atom in an element.

12. - - - Isotopes of Hydrogen
Isotopes = Atoms of the same element but having different masses. 1 2 1 3 1 H H H - - - + + +
Protium
99.99%
Tritium
Trace %
Deuterium
0.01%

13. Average Atomic weight of Hydrogen
Isotopes of Hydrogen
Isotopes = Atoms of the same element but having different masses. 1 2 1 3 1 H H H - - - + + +
Average Atomic weight of Hydrogen
= amu

14. Average Atomic weight of C= 12.011 amu
Isotopes of Carbon 12 13 14 C C C 6 6 6 - - - - - - - - + + + - + + + + + + + + + - - + + + + + + - - - - - - -
98.89%
1.11%
Trace %
Average Atomic weight of C= amu

15. C - - Radioactive Isotopes 6 14 H H-3 C-14 + + Nucleus is unstable
So falls apart (decays)
Giving radioactive particles

16. Electronic arrangement
A new layer is
added for each row or period in the table.

17. fill layers around nucleus
Electron arrangement 24 12 Mg
Electrons
fill layers around nucleus
Low  High 32 18 8 2
Shells = Energy levels

18. IA
IIA 1 H 4 2 He 7 3 Li 9 4 Be
2, 1
2, 2

19. IA
IIA
IIIA 1 H 11 5 B 7 3 Li 9 4 Be
2, 1
2, 2
2, 3

20. IIIA
IVA VA 11 5 B 12 6 C 13 7 N
2, 3
2, 4
2, 5

21. H He Be Li Ne Mg Ar Na 1 4 2 9 7 4 3 24 40 12 18 IA IIA VIIIA 20 10
2, 1
2, 2
2, 8 24 12 Mg 40 18 Ar 23 11 Na
2, 8, 8
2, 8, 1
2, 8, 2

22. H Be Li B Mg Al Na 1 9 7 4 3 Valence electrons Where most chemical
Reactions occur. 1 H 2 3 9 4 Be 7 3 Li 11 5 B
2, 3
2, 1
2, 2 24 12 Mg 27 13 Al 23 11 Na
2, 8, 3
2, 8, 1
2, 8, 2

23. H He Be Li Ne Mg Ar Na Octet Rule 1 4 2 9 4 7 3 24 40 12 18 1 8 2 2010 Ne 7 3 Li
2, 1
2, 2
2, 8 24 12 Mg 40 18 Ar 23 11 Na
2, 8, 8
2, 8, 1
2, 8, 2

24. The octet rule
Atoms are most stable if they have a filled or empty outer layer of electrons.
Except for H and He, a filled layer contains 8 electrons - an octet.
Atoms gain, lose or share electrons to make a filled or empty outer layer.
Atoms gain, lose or share electrons based on what is easiest.

25. 1.2 Atomic Structure: Orbitals
Quantum mechanics: describes electron energies and locations by a wave equation
Wave function solution of wave equation
Each wave function is an orbital,

26. Orbitals A plot of y 2 describes where electron most likely to be
Electron cloud has no specific boundary so we show most probable area
p (3)
s (1)
d (5)
f (7)
An f orbital
4 different kinds of orbitals for e-s
s and p orbitals most important in organic & biochem

27. Orbitals and Shells
Orbitals are grouped in shells of increasing size and energy
Different shells contain different numbers and kinds of orbitals
Each orbital can be occupied by two electrons

28. Orbitals and Shells
1st shell contains one s orbital, denoted 1s, holds 2 e-s
2nd shell contains one s orbital (2s) and three p orbitals (2p), 8 electrons
3rd shell contains an s orbital (3s), three p orbitals (3p), and five d orbitals (3d), 18 electrons

29. p-Orbitals
In each shell there are three perpendicular p orbitals, px, py, and pz, of equal energy
Lobes of a p orbital are separated by region of zero electron density, a node

30. 1.3 Atomic Structure: Electron Configurations
Ground-state electron configuration (lowest energy arrangement) of an atom lists orbitals occupied by its electrons. Rules:
1. Lowest-energy orbitals fill first: 1s  2s  2p  3s  3p  4s  3d (Aufbau (“build-up”) principle)
2. Electrons act as if they were spinning around an axis. Electron spin can have only two orientations, up  and down . Only two electrons can occupy an orbital, and they must be of opposite spin (Pauli exclusion principle) to have unique wave equations
3. If two or more empty orbitals of equal energy are available, electrons occupy each with spins parallel until all orbitals have one electron (Hund's rule).

31. 1.3 Atomic Structure: Electron Configurations
The Aufbau (“build-up”) principle
Electrons fill from the low  high.
fill n = 1 before n = 2 ,
fill s before p ... s p d f
n = 4
Pauli Exclusion principle
n = 3
Each subshell contains orbitals which can hold a max of 2 e-s
e’s in same orbital must be of opposite spins: 1  & 1 .
n = 2
n = 1
Hund’s Rule
Electrons don’t share same orbital unless they need to.
(i.e. no pairing until each orbital of the set has an e-)

32. Major trends in electron filling5p 4d 4f 4s 4p 3d 3s 3p 2s 2p 1s
Major trends in electron filling
Exceptions:
Fill 4s before 3d
Fill 5s before 4 d
Fill 5p before 4f
This is why
transition metals
are assigned as
B group elements.

33. Electron Configuration
1s __
2s __
2p __ __ __
3s __
3p __ __ __
4s __
3d __ __ __ __ __
5s __
4p __ __ __
4d __ __ __ __ __
5p __ __ __
6s __
5d __ __ __ __ __
4f __ __ __ __ __ __ __ 7 3 Li
1s22s1 x y z x y z

34. Electron Configuration
1s __
2s __
2p __ __ __
3s __
3p __ __ __
4s __
3d __ __ __ __ __
5s __
4p __ __ __
4d __ __ __ __ __
5p __ __ __
6s __
5d __ __ __ __ __
4f __ __ __ __ __ __ __ 16 8 O
1s22s22p4 x y z

35. Electron Configuration
1s __
2s __
2p __ __ __
3s __
3p __ __ __
4s __
3d __ __ __ __ __
5s __
4p __ __ __
4d __ __ __ __ __
5p __ __ __
6s __
5d __ __ __ __ __
4f __ __ __ __ __ __ __ 30 Zn x y z
1s22s22p63s23p64s23d10 x y z
[Ar] 4s23d10
[Ar] 3d104s2

36. Classification by sublevelsp H He d Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Hf Zr Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba Ls Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu f Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
3 - 25

37. Using the periodic table to find sublevels1 H Li Na Cs Rb K Ba Fr He Be Mg Sr Ca Ra p 2 2 Tl Rn At Po Bi Pb In Xe I Te Sb Sn Ga Kr Br Se As Ge Al Ar Cl S P Si B Ne F O N C 3 d 3 4 3 Sc Ti V Cr Mn Fe Co Ni Cu Zn 4 5 4 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd 5 5 Ls Hf Ta W Re Os Ir Pt Au Hg 6 6 6 Ac 7 4 Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Lu Lr Er Fm Tm Md Dy Cf Ho Es f 5

38. Learning Check: 1s22s22px22py02pz0 1s22s22px12py12pz0
Which of the following represents the ground-state electronic configuration for carbon (atomic number 6)?
1s22s22px22py02pz0
1s22s22px12py12pz0
1s22s12px12py1 2pz1
1s12s22px12py1 2pz1
1s22s22px02py02pz2

39. Solution: 1s22s22px22py02pz0 1s22s22px12py12pz0 1s22s12px12py1 2pz1
Which of the following represents the ground-state electronic configuration for carbon (atomic number 6)?
1s22s22px22py02pz0
1s22s22px12py12pz0
1s22s12px12py1 2pz1
1s12s22px12py1 2pz1
1s22s22px02py02pz2

40. H He Be Li Ne Ar Mg Na Octet Rule 1 4 2 9 7 4 3 1s2 1s1 20 10 1s2, 2s2
1s2, 2s2 2p6
1s2, 2s1 40 18 Ar 24 12 Mg 23 11 Na
1s2, 2s2 2p6, 3s2 3p6
1s2, 2s2 2p6, 3s1
1s2, 2s2 2p6, 3s2
[Ne] 3s1

41. H Li Na K H Li Lewis Structures Show only Valence Electrons Na 1 7 323 11 Na K

42. He O F N Si S P Ca H Li C Na K Se B Ne Be Al Cl Ar Mg Kr Ga As Br Ge 1
Nonmetals Share e-s
with other nonmetals 1 8 H He 2 3 4 5 6 7 C O Li B N F Ne Be Si Al S Cl Ar Mg P Na
Metals give e-s to nonmetals Kr Ca Ga As Se Br K Ge

43. Common ions Representative Elements Transition Elements 1+ 4+ 4- H He2+ 3+ 3- 2- 1-
Transition Elements Li Be B C N O F Ne
Variable Na Mg Al Si P S Cl Ar K Ca Hg Au Hf Ls Pt Ir Os Re W Ta Cd Ag Zr Y Pd Rh Ru Tc Mo Nb Ac Zn Cu Ti Sc Ni Co Fe Mn Cr V Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Lu Lr Er Fm Tm Md Dy Cf Ho Es Ga Ge As Se Br Kr Rb Sr In Sn Sb Te I Xe Cs Ba Tl Pb Bi Po At Rn Fr Ra 12 12

44. Na + Cl Na+ + Cl _ Formation of NaCl e- moves from Metal  Nonmetal
Stable octets _
Na + Cl Na+ + Cl
Metal
Cation
Nonmetal
Anion
+ and - ions attract to form an ionic bond. 15 15

45. Ionic compounds NaCl sodium chloride Not individual molecules
Form crystal arrays
Ions touch many others
Formula represents the average ion ratio
NaCl
sodium chloride Na Cl Cl Na Cl Na 18 18

46. He O F N Si S P Ca H Li C Na K Se B Ne Be Al Cl Ar Mg Kr Ga As Br Ge 1
Nonmetals Share e-s
with other nonmetals 1 8 H He 2 3 4 5 6 7 C O Li B N F Ne Be Si Al S Cl Ar Mg P Na
Metals give e-s to nonmetals Kr Ca Ga As Se Br K Ge

47. Covalent Bonds H H H + Cl Cl + Cl O O O + N N N +

48. Covalent Bonds H
H-H H2 Cl
Cl-Cl Cl2 O
O=O O2 N2 N
N N

49. May modify rules to improve sound. ie - monoxide not monooxide.
Covalent Bonds CO
Carbon monoxide C O
C O
CO2
Carbon dioxide O C O
O=C=O
May modify rules to improve sound.
ie - monoxide not monooxide.

50. Naming covalent compounds
Review:
Naming covalent compounds CO
CO2
N2O5
SiO2
ICl3
P2O5
CCl4
carbon monoxide
carbon dioxide
dinitrogen pentoxide
silicon dioxide
iodine trichloride
diphophorous pentoxide
carbon tetrachloride
May modify rules to improve the sound.
Example - use monoxide not monooxide. 34 34

51. Shape
CO2 O C O
O=C=O
Linear Shape (180o)

52. Shape
BF3 F B F F F B
(120o)
Trigonal Planar

53. 1.4 Development of Chemical Bonding Theory
Kekulé and Couper independently observed that carbon always has four bonds
van't Hoff and Le Bel proposed that the four bonds of carbon have specific spatial directions
Atoms surround carbon as corners of a tetrahedron

54. O=C=O Shape Linear Trigonal planar e-’s in 2 directions = 180o

55. Shape N H N H Pyramidal Trigonal Pyramid Configuration of Atoms
e-’s in 4 directions = 109.5o N H N H
Pyramidal
(109.5o)
Tetrahedral Configuration of Electrons
Trigonal Pyramid Configuration of Atoms

56. Tetrahedral electron-pair Geometries
Pyramidal
Bent

58. Non-bonding electrons
Valence electrons not used in bonding are called nonbonding electrons, or lone-pair electrons
Nitrogen atom in ammonia (NH3)
Shares six valence electrons in three covalent bonds and remaining two valence electrons are nonbonding lone pair

59. Learning Check
How many hydrogen atoms does phosphorus bond to in forming phosphine, PH?
Hint: Draw the electron dot structure and then use Hydrogen to provided the needed bonded electrons

60. Solution
How many hydrogen atoms does phosphorus bond to in forming phosphine, PH? H P H 3 P H H
Hint: Draw the electron dot structure and then use Hydrogen to provided the needed bonded electrons

61. Learning Check
How many hydrogen atoms does Carbon need in forming chloroform, CH?Cl C Cl H

62. Solution 3 C Cl H H C Cl H Cl C H H
How many hydrogen atoms does Carbon need in forming chloroform, CH?Cl 3 C Cl H H C Cl H Cl C H H

63. 1.5 The Nature of Chemical Bonds: Valence Bond Theory
Covalent bond forms when two atoms approach each other closely so that a singly occupied orbital on one atom overlaps a singly occupied orbital on the other atom
Two models to describe covalent bonding.
Valence bond theory, Molecular orbital theory
Valence Bond Theory:
Electrons are paired in the overlapping orbitals and are attracted to nuclei of both atoms
H–H bond results from the overlap of two singly occupied hydrogen 1s orbitals
H-H bond is cylindrically symmetrical, sigma (s) bond

64. Bond Energy Reaction 2 H·  H2 releases 436 kJ/mol
Product has 436 kJ/mol less energy than two atoms:
H–H has bond strength of 436 kJ/mol.
(1 kJ = kcal; 1 kcal = kJ)

65. Bond Length Distance between nuclei that leads to maximum stability
If too close, they repel because both are positively charged
If too far apart, bonding is weak

66. 1.6 sp3 Orbitals and the Structure of Methane
Carbon has 4 valence electrons (2s2 2p2)
In CH4, all C–H bonds are identical (tetrahedral)
sp3 hybrid orbitals: s orbital and three p orbitals combine to form four equivalent, unsymmetrical, tetrahedral orbitals (sppp = sp3), Pauling (1931)

67. The Structure of Methane
sp3 orbitals on C overlap with 1s orbitals on 4 H atoms to form four identical C-H bonds
Each C–H bond has a strength of 436 (438) kJ/mol and length of 109 pm
Bond angle: each H–C–H is 109.5°, the tetrahedral angle.

68. 1.7 sp3 Orbitals and the Structure of Ethane
Two C’s bond to each other by s overlap of an sp3 orbital from each
Three sp3 orbitals on each C overlap with H 1s orbitals to form six C–H bonds
C–H bond strength in ethane 423 kJ/mol
C–C bond is 154 pm long and strength is 376 kJ/mol
All bond angles of ethane are tetrahedral

69. 1.8 sp2 Orbitals and the Structure of Ethylene
sp2 hybrid orbitals: 2s orbital combines with two 2p orbitals, giving 3 orbitals (spp = sp2). This results in a double bond.
sp2 orbitals are in a plane with120° angles
Remaining p orbital is perpendicular to the plane

70. Bonds From sp2 Hybrid Orbitals
Two sp2-hybridized orbitals overlap to form a s bond
p orbitals overlap side-to-side to formation a pi () bond
sp2–sp2 s bond and 2p–2p  bond result in sharing four electrons and formation of C-C double bond
Electrons in the s bond are centered between nuclei
Electrons in the  bond occupy regions are on either side of a line between nuclei

71. Structure of Ethylene H atoms form s bonds with four sp2 orbitals
H–C–H and H–C–C bond angles of about 120°
C–C double bond in ethylene shorter and stronger than single bond in ethane
Ethylene C=C bond length 134 pm (C–C 154 pm)

72. 1.9 sp Orbitals and the Structure of Acetylene
C-C a triple bond sharing six electrons
Carbon 2s orbital hybridizes with a single p orbital giving two sp hybrids
two p orbitals remain unchanged
sp orbitals are linear, 180° apart on x-axis
Two p orbitals are perpendicular on the y-axis and the z- axis

73. Orbitals of Acetylene
Two sp hybrid orbitals from each C form sp–sp s bond
pz orbitals from each C form a pz–pz  bond by sideways overlap and py orbitals overlap similarly

74. Bonding in Acetylene Sharing of six electrons forms C ºC
Two sp orbitals form s bonds with hydrogens

76. 1.10 Hybridization of Nitrogen and Oxygen
Elements other than C can have hybridized orbitals
H–N–H bond angle in ammonia (NH3) 107.3°
C-N-H bond angle is °
N’s orbitals (sppp) hybridize to form four sp3 orbitals
One sp3 orbital is occupied by two nonbonding electrons, and three sp3 orbitals have one electron each, forming bonds to H and CH3.

77. 1.11 Molecular Orbital Theory
A molecular orbital (MO): where electrons are most likely to be found (specific energy and general shape) in a molecule
Additive combination (bonding) MO is lower in energy
Subtractive combination (antibonding) MO is higher energy

78. Molecular Orbitals in Ethylene
The  bonding MO is from combining p orbital lobes with the same algebraic sign
The  antibonding MO is from combining lobes with opposite signs
Only bonding MO is occupied

79. 1.12 Drawing Structures
Drawing every bond in organic molecule can become tedious.
Several shorthand methods have been developed to write structures.
Condensed structures don’t have C-H or C-C single bonds shown. They are understood.
e.g.

80. 3 General Rules:
1) Carbon atoms aren’t usually shown. Instead a carbon atom is assumed to be at each intersection of two lines (bonds) and at the end of each line.
2) Hydrogen atoms bonded to carbon aren’t shown.
3) Atoms other than carbon and hydrogen are shown (See table 1.3).

81. Summary Organic chemistry – chemistry of carbon compounds
Atom: positively charged nucleus surrounded by negatively charged electrons
Electronic structure of an atom described by wave equation
Electrons occupy orbitals around the nucleus.
Different orbitals have different energy levels and different shapes
s orbitals are spherical, p orbitals are dumbbell-shaped
Covalent bonds - electron pair is shared between atoms
Valence bond theory - electron sharing occurs by overlap of two atomic orbitals
Molecular orbital (MO) theory, - bonds result from combination of atomic orbitals to give molecular orbitals, which belong to the entire molecule

82. Summary (cont’d)
Sigma (s) bonds - Circular cross-section and are formed by head-on interaction
Pi () bonds – “dumbbell” shape from sideways interaction of p orbitals
Carbon uses hybrid orbitals to form bonds in organic molecules.
In single bonds with tetrahedral geometry, carbon has four sp3 hybrid orbitals
In double bonds with planar geometry, carbon uses three equivalent sp2 hybrid orbitals and one unhybridized p orbital
Carbon uses two equivalent sp hybrid orbitals to form a triple bond with linear geometry, with two unhybridized p orbitals
Atoms such as nitrogen and oxygen hybridize to form strong, oriented bonds
The nitrogen atom in ammonia and the oxygen atom in water are sp3-hybridized

83. Learning Check:
Based on the octet rule, which line-bond structures is/are correct?
A and B
A and C
B and D
Only B
Only C

84. Solution:
Based on the octet rule, which line-bond structures is/are correct?
A and B
A and C
B and D
Only B
Only C

85. Which of these statements concerning p-orbitals is false?
Learning Check:
Which of these statements concerning p-orbitals is false?
They consist of two equivalent lobes.
They are absent from the first shell of atomic orbitals.
They can form π bonds.
They only participate in bonding on carbon atoms.
They can hold a maximum of two electrons.

86. Which of these statements concerning p-orbitals is false?
Solution:
Which of these statements concerning p-orbitals is false?
They consist of two equivalent lobes.
They are absent from the first shell of atomic orbitals.
They can form π bonds.
They only participate in bonding on carbon atoms.
They can hold a maximum of two electrons.

87. Learning Check: s p sp sp2 sp3
What is the hybridization of the oxygen-bonded carbon atom in the following molecule? s p sp
sp2
sp3

88. Solution:
What is the hybridization of the oxygen-bonded carbon atom in the following molecule? s p sp
sp2
sp3

89. What is the hybridization of nitrogen atom in trimethylamine :N(CH3)3?
Learning Check:
What is the hybridization of nitrogen atom in trimethylamine :N(CH3)3? sp
sp2
sp3
dsp2
unhybridized

90. What is the hybridization of nitrogen atom in trimethylamine :N(CH3)3?
Solution:
What is the hybridization of nitrogen atom in trimethylamine :N(CH3)3? sp
sp2
sp3
dsp2
unhybridized

91. Learning Check: s p sp sp2 sp3
What type of orbital is not used in constructing the molecule shown below? s p sp
sp2
sp3

92. Solution:
What type of orbital is not used in constructing the molecule shown below? s p sp
sp2
sp3

93. What is the molecular formula of the molecule shown below?
Learning Check:
What is the molecular formula of the molecule shown below?
C8H12O
C8H11O
C8H7O
C7H7O
C7H10O

94. What is the molecular formula of the molecule shown below?
Solution:
What is the molecular formula of the molecule shown below?
C8H12O
C8H11O
C8H7O
C7H7O
C7H10O

95. Electrons cannot occupy an antibonding molecular orbital.
Learning Check:
Electrons cannot occupy an antibonding molecular orbital.
True
False

96. Electrons cannot occupy an antibonding molecular orbital.
Solution:
Electrons cannot occupy an antibonding molecular orbital.
True
False

97. Learning Check:
How many hydrogen atoms are present in the naturally-occurring terpene α-terpinene shown below? 14 15 16 17 18

98. Solution:
How many hydrogen atoms are present in the naturally-occurring terpene α-terpinene shown below? 14 15 16 17 18

99. Learning Check:
Select the best condensed structural formula for the following molecule:
(CH3)2CHCH2COHOHCOH
CH3CH3CHCH2C(OH)2CHO
(CH3)2CHCH2C(OH)2CHO
(CH3)2CHCH2C(OH)2COH
CH3CHCH3CH2C(OH)2CHO

100. Solution:
Select the best condensed structural formula for the following molecule:
(CH3)2CHCH2COHOHCOH
CH3CH3CHCH2C(OH)2CHO
(CH3)2CHCH2C(OH)2CHO
(CH3)2CHCH2C(OH)2COH
CH3CHCH3CH2C(OH)2CHO

101. Learning Check:
What is the correct order of carbon-carbon bond lengths in ethane, ethylene and acetylene?
ethane < ethylene < acetylene
ethane < acetylene < ethylene
ethylene < ethane < acetylene
ethylene < acetylene < ethane
acetylene < ethane < ethylene
acetylene < ethylene < ethane

102. Solution:
What is the correct order of carbon-carbon bond lengths in ethane, ethylene and acetylene?
ethane < ethylene < acetylene
ethane < acetylene < ethylene
ethylene < ethane < acetylene
ethylene < acetylene < ethane
acetylene < ethane < ethylene
acetylene < ethylene < ethane

103. Learning Check:
What kind of orbital and with how many nodal planes is shown in the picture below?
σ antibonding, two nodal planes
σ antibonding, one nodal plane
π bonding, one nodal plane
π antibonding, two nodal planes
π antibonding, one nodal plane

104. Solution:
What kind of orbital and with how many nodal planes is shown in the picture below?
σ antibonding, two nodal planes
σ antibonding, one nodal plane
π bonding, one nodal plane
π antibonding, two nodal planes
π antibonding, one nodal plane

105. What is the best description of bond length?
Learning Check:
What is the best description of bond length?
a distance between nuclei that yields the best orbital overlap
a distance between nuclei that yields the smallest nuclear-nuclear repulsion
a distance between nuclei that yields the smallest electron-electron repulsion
a distance between nuclei that yields the largest electron-nuclei attraction
a distance between nuclei that is a compromise of all electrostatic interactions

106. What is the best description of bond length?
Solution:
What is the best description of bond length?
a distance between nuclei that yields the best orbital overlap
a distance between nuclei that yields the smallest nuclear-nuclear repulsion
a distance between nuclei that yields the smallest electron-electron repulsion
a distance between nuclei that yields the largest electron-nuclei attraction
a distance between nuclei that is a compromise of all electrostatic interactions

107. What is the O-C-O bond angle in potassium carbonate, K2CO3?
Learning Check:
What is the O-C-O bond angle in potassium carbonate, K2CO3?
60°
90°
109.5°
120°
180°

108. What is the O-C-O bond angle in potassium carbonate, K2CO3?
Solution:
What is the O-C-O bond angle in potassium carbonate, K2CO3?
60°
90°
109.5°
120°
180°

109. What is the relative orientation of two π orbitals in acetylene?
Learning Check:
What is the relative orientation of two π orbitals in acetylene?
60°
90°
109.5°
120°
180°

110. What is the relative orientation of two π orbitals in acetylene?
Solution:
What is the relative orientation of two π orbitals in acetylene?
60°
90°
109.5°
120°
180°

111. In relation to σ bonds, which statement about π bonds is correct?
Learning Check:
In relation to σ bonds, which statement about π bonds is correct?
π bonds are weaker and π electrons have higher energy
π bonds are stronger and π electrons have higher energy
π bonds are weaker and π electrons have lower energy
π bonds are stronger and π electrons have lower energy
π bonds and σ bonds have equal strengths, and their electrons have the same energies

112. In relation to σ bonds, which statement about π bonds is correct?
Solution:
In relation to σ bonds, which statement about π bonds is correct?
π bonds are weaker and π electrons have higher energy
π bonds are stronger and π electrons have higher energy
π bonds are weaker and π electrons have lower energy
π bonds are stronger and π electrons have lower energy
π bonds and σ bonds have equal strengths, and their electrons have the same energies

113. Which of the orbitals shown below can be occupied in a hydrogen atom?
Learning Check:
Which of the orbitals shown below can be occupied in a hydrogen atom? A B C D
Any one of these

114. Which of the orbitals shown below can be occupied in a hydrogen atom?
Solution:
Which of the orbitals shown below can be occupied in a hydrogen atom? A B C D
Any one of these

115. Which statement about the antibonding orbital in H2 is false?
Learning Check:
Which statement about the antibonding orbital in H2 is false?
It is higher in energy than the corresponding bonding orbital
It is a molecular orbital
It has a nodal plane (a node for short)
It can hold up to two electrons
The two lobes have different charges

116. Which statement about the antibonding orbital in H2 is false?
Solution:
Which statement about the antibonding orbital in H2 is false?
It is higher in energy than the corresponding bonding orbital
It is a molecular orbital
It has a nodal plane (a node for short)
It can hold up to two electrons
The two lobes have different charges

117. What is the hybridization of nitrogen in :NCCH3?
Learning Check:
What is the hybridization of nitrogen in :NCCH3? sp
sp2
sp3
sp3d
nitrogen is not hybridized

118. What is the hybridization of nitrogen in :NCCH3?
Solution:
What is the hybridization of nitrogen in :NCCH3? sp
sp2
sp3
sp3d
nitrogen is not hybridized

119. Each shell (floor of the Hotel)
Hotel Model
Each shell (floor of the Hotel)
Has subshells (s,p,d,f) f
n = 4 (4th floor) d p
32 e- s
n = 3 (3rd floor) d
18 e- p s
n = 2 (2nd floor) p
8 e- s
n = 1 (1st floor) s
2 e-