Download Chapter 2: Answers to Review Problems

Chapter 2: Answers to Review Problems Section 2.1 1) OH
O
C
N
H
O
H3C
NH2
O
O
2) Lewis Structure is: O
Cl
O
There are 4 non bonding electrons (2 pairs) on chlorine. 3) a) O (the most electronegative atom) has a 1 formal charge; N and C have no formal charge. b) N has a + 1 formal charge; C and O both have 1 formal charges. c) N has a 2 formal charge; O has a + 1 formal charge. d) N has a 1 formal charge; both C and O have no formal charge. e) C has a 1 formal charge; both N and O have no formal charge. The best Lewis structure has negative formal charges on the most electronegative atom; therefore a) is the best Lewis structure. 4) In the best Lewis structure of BrO3—, the Br has a formal charge of +2 and the single bonded O has a formal charge of 1. a) Incorrect; formal charges not as above. There are non bonding electron pairs missing on one O in this structure. b) Correct; the most electronegative atom bears the 1 formal charge, and all other formal charges have been minimized by the formation of double bonds. c) Not the best structure as the 1 formal charge has been placed on the Br, not on the O. d) Incorrect; the Br is missing a non bonding pair of electrons. e) Incorrect; the O which bears the 1 formal charge has been given 9 electrons, in violation of the octet rule. The other two O’s should bear 1 formal charges. The correct answer is b). Chapter 2 ‐ 1 5) a)
H
b)
O
C
H
H
H
N
H
H
S
H
d)
Cl
c)
As
Cl
Cl
B
Cl
Cl
Cl
e)
Cl
Te
Cl
f)
H
O
O
Cl
Cl
H
g)
H
h)
O
C
S
C
S
H
O
Chapter 2 ‐ 2 F
S
i)
F
O
j)
F
Br
F
F
F
F
k)
C
H
H
C
H
H
C
H
H
6) a) Correct; resonance structures differ only in the locations of the electrons. b) Incorrect; resonance structures do not differ in the locations of the atoms. c) Correct; delocalization of electrons tends to stabilize a molecule or an ion. 7) a) Incorrect; the formal charge on the O in the second structure is 0. b) Incorrect; the formal charge on the end N atom in the first structure is 0. c) Incorrect; the formal charge on the central N atom is +1 in all structures. d) Incorrect; the formal charge on the O atom in the third structure is +1. e) Correct; the end N atom in the second structure has a formal charge of 1. The correct answer is e). 8) a) Correct; there are a total of 4 equivalent resonance O
structures for this molecule. b) Incorrect; there are 11 non bonding pairs of electrons in this molecule. Sb
O
c) Correct; the central Sb atom bears no non bonding pairs of electrons. O
O
Both a) and c) are correct. Chapter 2 ‐ 3 9) d) The N has been given 9 electrons, in violation of the octet rule. 10) a)
O
Xe
O
b)
O
Se
O
O
O
2 more resonance structures;
average Se ‐ O bond order = 4/3
Only structure; average
Xe ‐ O bond order = 2
c)
Se
O
d)
O
O
Only structure; average
Se ‐ O bond order = 2
I
O
O
2 more resonance structures;
average I ‐ O bond order =5/3
e)
Cl
O
Cl
O
2 resonance structures; average Cl ‐ O bond order = 3/2
11) IF5 has one non bonding pair of electrons on the central atom, I. SF4 has one non bonding pair of electrons on the central atom, S. 12) O
O
3
Br
O
O
O
O
Br
O
O
3 more resonance structures;
average Br ‐ O bond order = 7/4
Chapter 2 ‐ 4 13) O
There are 3 other resonance
structures for this molecule;
4 in total.
O
As
O
O
Average As – O bond order in AsO43― is 1.25. 14) O
S
The average bond order is 2.
O
O
15)
O
N
O
N
N
O
O
O
O
a)
NO2+ has an average N ‐ O bond order of 2.
NO2‐ has an average N ‐ O bond order of 1.5
b)
Between the same atoms, the higher the bond order, the shorter the bond
length; 110 pm in NO2+ and 124 pm in NO2‐.
Molecular formula of ATP is C10H12N5 O13P34–. 16) Chapter 2 ‐ 5 Section 2.2 1)
a)
b)
Cl
F
S
F
Al
Cl
Cl
F
120o
AX3, Bond Angles are
Shape is triangular or trigonal planar
AX5; Shape is trigonal bipyramidal
Cl
d)
c)
O
H
H
P
H
AX3E, Bond angles <
Shape is triangular pyramidal
Cl
AX4, Bond angles are 109.5o
Shape is tetrahedral
Cl
e)
f)
Cl
Cl
Si
Cl
Cl
Cl
Cl
AX4, Bond angles are 109.5o
Shape is tetrahedral
AX6, Bond angles are 90o
Shape is octahedral
g)
h)
S
O
F
Br
F
O
O
F
Cl
Cl
Cl
Sb
Cl
Cl
109.5o
F
120o
F
90o
AX3E, Bond angles < 109.5o
Shape is triangular pyramidal
AX3E2, Bond angles are 90o
Shape is T-shaped
Chapter 2 ‐ 6 i)
F
B
Cl
F
F
k)
l)
~180o
Se
F
F
~120o
Cl
AX3, Bond angles are 120o
Shape is triangular or trigonal planar
AX5E, Bond angles are 90o
Shape is square pyramid
Cl
F
S
j)
F
F
F
~90o
B
F
F
F
F
AX4E, Shape is see-saw
AX4, Bond angles are 109.5o
Shape is tetrahedral
n)
m)
O
Se
O
O
O
N
O
O
AX3, Bond angles are ~ 120o
Shape is triangular or trigonal planar
AX3E, Bond angles < 109.5o
Shape is triangular pyramidal
p)
o)
O
Br
O
Br
O
O
O
AX4, Bond angles are 109.5o
Shape is tetrahedral
O
AX2E2, Bond angle < 109.5o
Shape is bent
Chapter 2 ‐ 7 q)
r)
P
C
F
H
F
H
F
H
AX3E, Bond angles < 109.5o
Shape is triangular pyramidal
s)
AX3E, Bond angles < 109.5o
Shape is triangular pyramidal
H
t)
H
H
C
C
H
O
H
H
H
o
AX3, Bond Angles are 120
Shape is triangular or trigonal planar
u)
N
Carbon is AX4; tetrahedral, 109.5o
Oxygen is AX2E2; bent, <109.5o
F
v)
N
AXE, Linear
90o
F
F
I
F
H
w)
F
AX5E; Shape is square planar
C
O
C
H
H
H
H
x)
O
109.5o
Carbon 1 is AX4; tetrahedral,
Carbon 2 is AX3; triangular planar, 120o
Oxygen is AX2E2; bent, <109.5o
y)
Cl
O
O
C
C
H
H
N
Carbon 1 is AX4; tetrahedral, 109.5o
Carbon 2 is AX2; linear, 180o
Nitrogen is AXE; linear
O
AX3E, Bond angle < 109.5o
Shape is triangular pyramidal
Chapter 2 ‐ 8 2)
AX2E geometry
Molecule not symmetrical
Theref ore molecule is polar
S
O
O
AX 4 geometry
Molecule is symmetrical
Theref ore molecule is not polar
H
N
H
H
H
AX4 geometry
Molecule is symmetrical due to resonance
Theref ore molecule is not polar
O
P
O
O
O
3) i) Non polar molecules containing only non polar bonds: Cl2 ii) Non polar molecules containing polar bonds: iii) Polar molecules: SF4, OF2, BrF5, IF, PF3, HCF3 CF4, BeF2, XeCl4, TeCl6, AlF3, SO42―, NO3― 4) AsF5 is AX5; trigonal bipyramidal, SiF5― is AX5; trigonal bipyramidal The correct answer is a). Chapter 2 ‐ 9 5) a) CH2F2 is tetrahedral, but not symmetrical, so polar; PCl5 is trigonal bipyramidal, non polar b) BeF2 is linear, non polar; ClF3 is T‐shaped, polar c) SF6 is octahedral, non polar; ClF3 is T‐shaped, polar d) CH2F2 is tetrahedral, but not symmetrical, so polar; ClF3 is T‐shaped, polar e) PCl5 is trigonal bipyramidal, non polar ; SF6 is octahedral, non polar The correct answer is e). 6) “Main Group” means Groups 13 to 18. If the atom “A” has 3 unpaired electrons, it must have the configuration s2 p3, which is Group 15. “A” contributes 5 electrons to the total count. If there is one non bonding pair of electrons on “A” in AFn, there are 3 electrons remaining to form bonds. The formula of AFn is AF3. This gives an AX3E geometry, which is triangular pyramidal. The correct answer is e). 7) a) CH2F2 is tetrahedral, but is not symmetrical. This molecule is polar. b) SO42― is tetrahedral and has 4 equivalent resonance structures. This molecule is non polar. c) BF3 is triangular planar. This molecule is non polar. d) CH3OH is tetrahedral, but is not symmetrical. This molecule is polar. e) SF4 is a see‐saw. The shape is irregular so this molecule is polar. The molecules in answers b) and c) are non polar. 8) a) ClF3 is T‐shaped, polar; ClF5 is a square pyramid, polar. b) ClF2― is linear, non polar; ClF4― is square planar, non polar. c) ClF2― is linear, non polar; ClF3 is T‐shaped, polar. d) ClF3 is T‐shaped, polar; ClF4― is square planar, non polar. e) ClF2― is linear, non polar; ClF5 is a square pyramid, polar. The correct answer is a); both species are polar. 9) a) H2O is bent; polar b) F2O is bent; polar c) HOF is bent; polar d) FBeCl is linear, but not symmetrical; polar e) HCl is linear, but the one bond is polar, therefore the molecule is polar. f) PCl6― is octahedral; non polar g) BF3 is triangular planar; non polar h) CH4 is tetrahedral; non polar i) PCl4+ is tetrahedral; non polar Chapter 2 ‐ 10 10) XeF3― has 12 electrons or 6 pairs of electrons. The geometry of this molecule is AX3E3. The arrangement of the electron pairs is octahedral, but there are 3 non bonding pairs. The correct answer is e). 11) SF4 is a see‐saw, SeF5+ is a trigonal bipyramid, AsO43― is tetrahedral, PH3 is trigonal pyramidal The correct answer is b). 12) O
O
H
H
C
C
O
O
a) Correct; there are two equivalent resonance structures as shown above. b) Correct; the molecule is polar. c) Correct. All statements are correct. 13) SF42– is square planar in shape; AX4E2 14) BrF3 is T‐shaped; BeCl2 is linear; BF4― is tetrahedral The correct answer is a). 15) BF3 is triangular planar; non polar SnCl2 is bent; polar OCl2 is bent; polar 16) a) CO2 is linear b) HCN is linear c) H2S is bent d) NF3 is trigonal pyramidal e) H2O is bent The molecules in a) and b) are linear. 17) NO3― is trigonal planar in shape with an average bond order of 1.33 (See 1m) Answer d) is correct. Chapter 2 ‐ 11 Section 2.3 1) c) is impossible; the square pyramidal geometry requires sp3d2 hybridization. 2) The geometry of the ICl5 molecule is square pyramidal. (AX5E) There are 6 electron pairs, therefore the hybridization is sp3d2. This molecule is polar. Answer a) is incorrect. 3) a) The N in NH4+ is sp3 hybridized. b) The B in BF4 is sp3 hybridized. c) The Xe in XeF4 is sp3d2 hybridized. d) The O in H2O is sp3 hybridized. e) The C in CH4 is sp3 hybridized. The hybridization is different in molecule c). 4) a) Triangular pyramid is AX3E; 4 electron pairs (sp3 hybrid) but only 3 bonding pairs. Therefore, 5 electrons must come from the central atom, which must be in Group 15. Examples include NH3, PF3, AsF3, SbH3, BiCl3. b) Octahedral shape is AX6; 6 electron pairs (sp3d2 hybrid). All 6 electron pairs are bonding; one electron comes from each terminal atom. The remaining 6 electrons come from the 1 charge and the valence electrons of the central atom, of which there are 5. The central atom must be in Group 15 . Examples include PCl6 , AsF6 , SbH6 , BiF6 . c) Tetrahedral shape is AX4; 4 electron pairs (sp3 hybrid). All 4 electron pairs are bonding; one electron comes from each terminal atom. The remaining 4 electrons are the sum of the +1 charge and the valence electrons on the central atom. There must be 5 valence electrons on the central atom, which is in Group 15. Examples include NH4+, PF4+, AsCl4+, SbH4+, BiCl4+. d) T‐shaped molecule is AX3E2; 5 electron pairs (this must be an sp3d hybrid). Only 3 electron pairs are bonding; one electron comes from each of the terminal atoms. This leaves 7 electrons around the central atom, so the central atom must be in Group 17. Examples include ClF3, BrCl3, IF3 Chapter 2 ‐ 12 H
a)
5)
N: Geometry of electron pairs is triangular
planar. Hybridization is sp2.
O: Geometry of electron pairs is tetrahedral.
Hybridization is sp3.
N
O
O
H
b)
C
C: Geometry of electron pairs is linear.
Hybridization is sp.
N
c)
S: Geometry of electron pairs is tetrahedral.
Hybridization is sp3.
C: Geometry of electron pairs is tetrahedral.
Hybridization is sp3.
S
H
H
C
C
H
H
H
H
F
F
d)
C: Geometry of electron pairs is tetrahedral.
Hybridization is sp3.
In this structure, each F atom has 3 non
bonding pairs of electrons.
F
C
F
C
F
C
F
F
F
6) sp2 hybridization results in a triangular planar arrangement of electron pairs. (AX3) a) CO2 is linear; sp hybridized. b) OCS is linear; sp hybridized. c) K2CO is the CO32 ion with K atoms bonded to the negatively charged O atoms. d) The H2C=O molecule is triangular planar; sp2 hybridized. e) In C2H4, there is a double bond between the two C atoms. The arrangement of electron pairs The CO32 molecule is triangular planar; sp2 hybridized. around each C is triangular planar; sp2 hybridized. f) The C2H2 molecule has a triple bond between the two C atoms. The arrangement of electron pairs around each C is linear; sp hybridized. g) The HCCl3 molecule is tetrahedral; sp3 hybridized. The molecules in c), d) and e) contain a C atom which is sp2 hybridized. Chapter 2 ‐ 13 7) a) Incorrect; a double bond consists of a δ bond and a π bond. b) Incorrect; a triple bond consists of a δ bond and two π bonds. c) Correct; rotation about the δ bond axis cannot occur when a π bond is present. d) Incorrect; a π bond consists of only one pair of electrons. e) Incorrect; a triple bond consists of a δ bond and two π bonds. Only answer c) is correct. 8) H – C N: is the structure of HCN. The hybridization of the C atom is sp, the CN bond is a triple bond and the molecule is linear. The correct answer is c). 9) a) CO2 is a linear molecule and therefore planar. b) CH3NH2 is tetrahedral about the C atom and triangular planar about the N atom. This molecule is not planar. c) CH3+ is triangular planar. d) HCOO― is triangular planar about the C atom. Molecules a), c) and d) have all their atoms in the same plane, or are planar molecules. 10) A σ bond is a single bond, formed by the overlap of two atomic or hybrid orbitals between the atoms. A π bond is formed by sideways overlap of pure p orbitals, both above and below the plane of the σ bond. Due to the sideways overlap, the atoms involved cannot rotate. 11) A double bond consists of one σ bond and one π bond. As there are 4 double bonds in this molecule, there must be 4 π bonds. Chapter 2 ‐ 14 Section 2.4 1) The following chart lists the limitations of VB Theory and how they are addressed by MO Theory. Valence Bond Theory (Atomic Orbitals) Molecular Orbital Theory (Atomic Orbitals) Bonding electrons are confined to the Bonding electrons are delocalized and are area between the atoms; does not associated with the entire molecule. explain resonance structures. No prediction of paramagnetism. Accounts for unpaired electrons and properties resulting from these unpaired electrons. Does not describe the energies of the There are two types of MOs and the lowest electrons. energy ones are always filled first. 2) A node is an area where there is zero probability of finding an electron. If there are no electrons shielding the nuclei from each other, repulsion between the nuclei occurs. This does not lead to bond formation, so reduces the stability of the molecule. 3) A paramagnetic molecule is attracted to a magnetic field due to its unpaired electrons. A diamagnetic molecule with paired electrons is not attracted to a magnetic field. 4) There is no Figure 2.4.10; use Figure 2.4.8. a) O22― : Bond order = 1, diamagnetic b) O2― : Bond order = 1.5, paramagnetic c) O2 : Bond order = 2, paramagnetic d) O2+ : Bond order = 2.5, paramagnetic e) O22+ : Bond order = 3, diamagnetic f) Ne2 : Bond order = 0, diamagnetic g) C2 : Bond order = 2, paramagnetic The most stable molecules have the highest bond order. From most stable to least stable: e), d), c) = g), b), a), f) 5) The π2py and the π2pz are degenerate because they are both formed by sideways overlap of py or pz atomic orbitals. They are higher in energy than the π2px which is formed by head on overlap of px atomic orbitals. Chapter 2 ‐ 15 Section 2.5 1)
a) Correct; the greater the surface area, the greater the van der Waals forces. b) Incorrect; the strength of van der Waals forces increases as surface area increases. c) Incorrect; dipole‐dipole interactions are attractive forces between permanent dipoles of polar molecules. d) Correct; although this attraction can also occur within a molecule. 2) These three molecules, although they have the same molecular formula, have different shapes due to the different bonding sequences of their atoms. Due to the different bonding sequences, different shapes result and therefore, different surface areas of the three molecules. Boiling point depends on the strength of van der Waals interactions, and the smaller the surface area, the lower will be the boiling point. Melting point is more affected by the symmetry of the molecule, as symmetrical molecules can pack together tightly, resulting in a much higher melting point than molecules which do not pack together tightly, due to lack of symmetry. 3) The second molecule will have a higher boiling point as it has greater surface area. The first molecule will have a higher boiling point as it has both dipole‐dipole interactions and hydrogen bonding interactions compared to the second molecule which has hydrogen bonding only. Chapter 2 ‐ 16 The first molecule will have the higher boiling point due to hydrogen bonding. The first molecule will have the higher boiling point due to its larger surface area. The second molecule will have the higher boiling point due to hydrogen bonding, which is a stronger interaction than the dipole‐dipole interaction present in the first molecule. The second molecule exhibits a dipole‐dipole interaction which is stronger than the van der Waals forces in the first molecule. The second molecule has the higher boiling point. The second molecule will have hydrogen bonding interactions, resulting in a higher boiling point. The first molecule has ion‐ion interactions, which will result in a higher boiling point than the hydrogen bonding interactions in the second molecule. Chapter 2 ‐ 17 The second molecule will exhibit dipole‐dipole interactions resulting in a higher boiling point than the first molecule. The first molecule shows both hydrogen bonding and dipole‐dipole interactions, resulting in a higher boiling point than the second molecule which has only dipole‐dipole interactions. Chapter 2 ‐ 18