prep 101 midterm #2 with sol.
©Prep101
Chem110MT2Solutions
Chapter 10 – Practice Problems Q 10.1 Solution:
-1
0 O S
-1 O
-1 O
O
+2 0 O
S O
+2
S O
-1
FC O in S = O
60
-1
O
0
1 u 2 1 2 1 64 u4 0 2
66
FC O in S - O
FC S
O
-1
+2
1 u 8 2 2
Q 10.2 Solution:
Correct Answer: B Two oxygen atoms will have a formal charge of –1 and there will also be 4 different resonant structures. O
O C O
C O
Bond order C-O = 1/2 (# bonding e- - # antibonding e-) = 3/2 Therefore statements 1 and 3 are true. Q 10.3 Solution:
Correct Answer: A There should be one pair of electrons on the central sulphur atom in A.
Solutionswillbepostedatwww.prep101.com/solutions 1
©Prep101
Chem110MT2Solutions
Q 10.4 Solution: Total # of valence electrons = 5 + 6 + 7 = 18 Formal charge on F = 7 – [6 + ½ (2)] = 0 Formal charge on N = 5 – [2 + ½ (6)] = 0 Formal charge on O = 6 – [4 + ½ (4)] = 0 Since the formal charge on N, O and F are 0, this is the most likely Lewis structure. Q 10.5 Solution:
X is Nitrogen (N).
Q 10.6 Solution:
There are 4 lone pairs of electrons O
C H2N
Q 10.7 Solution:
NH2
Correct answer: A A and C are the only neutral structures. A abides by octet rule.
Solutionswillbepostedatwww.prep101.com/solutions 2
©Prep101 Q 10.8 Solution:
Chem110MT2Solutions
Correct Answer: B Resonance structures occur when more than one Lewis structure, each with different electron distributions, can be drawn for a molecule. This usually results in molecules containing both double bonds and single bonds. (a) O3
O
O
O
O
O
O
(b) No resonance structure. CH4 cannot have resonance structures because no double bonds are present.
H
H
H
H (c) NO2-
O
-
N
O
O
N
O
(d) HCO3OH
O O
-1
OH
O
-1
O
Solutionswillbepostedatwww.prep101.com/solutions 3
©Prep101 Q 10.9 Solution:
(a) SCN-
Chem110MT2Solutions
Total # of valence electrons: 6 + 4 + 5 + 1 = 16 -1 (-)
(b) -1 (-)
Formal charge = (# of valence electron on a free atom) – (# of valence electrons assigned to the atom in the molecule) Valence electrons assigned = (# of lone pair electrons) + ½ (# of shared electrons) For S, formal charge = 6 – [4 + ½ (4)] = 0 For C, formal charge = 4 – [0 + ½(8)] = 0 For N, formal charge = 5 – [4 + ½(4)] = -1 For S, formal charge = 6 – [6 + ½ (2)] = -1 For C, formal charge = 4 – [0 + ½(8)] = 0 For N, formal charge = 5 – [2 + ½(6)] = 0
The first structure is preferred since N is more electronegative than S, and thus is more likely to have the negative charge. Q 10.10 Solution:
Correct Answer: C The formal charge of P in structures (i) and (iii) is = 5 – (5 + 0) = 0. Based on formal charge, structure (i) is the most important resonance structure. The formal charge of O in structure (ii) = 6 – (1 + 6) = -1. The most stable structure is that with the lowest formal charge.
Solutionswillbepostedatwww.prep101.com/solutions 4
©Prep101 Q 10.11 Solution:
Chem110MT2Solutions
Correct answer: C There are two lone pairs and four bonding pairs on the central S atom, giving AX4E2. The lone pairs will align to maximize the distance between them. Therefore, this atom would be square planar.
Q 10.12 Solution:
Correct answer: B CF4 is a tetrahedral molecule with all the same types of bonds at each point, so it will have a net dipole of zero.
Q 10.13 Solution: Species
CS2
Lewis electron dot structure S
C
SO3
Linear
S
F
F
XeF4
Xe F
F S
O
O
Hybridizatio Polar (P) or n Nonpolar (N) on species central atom N sp
Square Planar
sp3d2
N
Trigonal planar
sp2
N
Trigonal Pyramidal
sp3
P
Octahedral
sp3d2
N
O
O
SO32
Shape (as given by nuclei)
O
S O
F
SF6
F F
F S
F
F
Solutionswillbepostedatwww.prep101.com/solutions 5
©Prep101
Chem110MT2Solutions
Q 10.14 Solution: H H
Solution:
sp2
Solution:
sp2
Solution:
Answer: 4
Solution:
Answer: 1
Solution:
Yes
Solution:
Yes
C
N
H
Solutionswillbepostedatwww.prep101.com/solutions 6
©Prep101
Chem110MT2Solutions
Q 10.15 Solution: Molecule
Lewis Structure O
H2CO
C H
Molecular Shape (VSEPR)
Hybridization of Central Atom
Polar (P) or Nonpolar (N)
Trigonal planar
sp2
Polar
H
sp3 OF2
Bent
O F
Polar
F
HCN H
C
N
linear
sp
Polar
tetrahedral
sp3
Non-polar
Square pyramidal
sp3d2
Polar
See-saw
sp3d
Polar
H
SiH4
Si H
H
H
IF5
F
F I
F
F F
SeF4
F F Se F F
Solutionswillbepostedatwww.prep101.com/solutions 7
©Prep101
Chem110MT2Solutions
Q 10.16 Solutions:
i)
l
N F
O
O
O
O
l
N F
O
N F
ii) Trigonal planar iii) 3 iv) sp2 v) X = C and Z = B
Solutionswillbepostedatwww.prep101.com/solutions 8
O
©Prep101
Chem110MT2Solutions
Q 10.17 Solution:
Species
Lewis Structure
Shape
Hybridization on Central atom
Polar/Nonpolar
PF5
AX5
Trigonalbipyramidal
sp3d
N
BF3
AX3
Trigonalplaner
sp2
N
PF3
AX3E
Tetrahedral
sp3
P
BrF3
AX3E2
T-shaped
sp3d
P
BrF2+
AX2E2
Angular
sp3
P
SeF4
AX4E
Seesaw
sp3d
P
BrF5
AX5E
Squarepyramidal
sp3d2
P
Solutionswillbepostedatwww.prep101.com/solutions 9
©Prep101 Q 10.18 Solution:
Solution:
Chem110MT2Solutions
C1 is sp3 with 109.5° (tetrahedral) C2 is sp2 with 120° (trigonal planar) OA is sp2 hybridized OB is sp3 hybridized Nitrogen is sp3 hybridized with a bond angle of 109.5°
Solution:
Q 10.19 Solution:
There are 9 ı bonds (single bonds) There is 1 ʌ bond (double bond)
Correct answer – C IF5 = 42 electrons. ĺ AB5E1 On the picture below, there is also a lone pair of electrons around the central iodine. This molecule is polar, because it has a net dipole towards the fluorine (in the plane). F
F I
F
F
F
Q 10.20 Solution:
Correct answer - E There are 14 sigma bonds (single bonds) and 3 pi bonds Recall that a double bond equals 1 sigma bond and 1 pi bond. Recall that a triple bond equals 1 sigma bond and 2 pi bonds.
Solutionswillbepostedatwww.prep101.com/solutions 10
©Prep101 Q 10.21 Solution:
Chem110MT2Solutions
Correct Answer: D D and E are the only answers that sum to give a -1 anion (eliminate A, B, C). For nitrogen: 5 – 4- 2 = -1, sulfur: 6- 4 -2 = 0, carbon: 4 – 4 =0
Q 10.22 Solution:
Answer: C Sodium iodide, NaI An ionic compound is generally the result of a compound composed of a non-metal and a metal.
Q 10.23 Solution:
Answer: D Iron is a metal, and has primarily metallic bonding – delocalization of electrons
Q 10.24 Solution:
Answer: B Fluorine is the most electronegative atom in the periodic table. Electronegative increases from bottom to top, and increases from left to right.
Q 10.25 Solution:
Correct answer – D There are 9 sigma bonds (single bonds) and 1 pi bond (double bond) Recall that a double bond equals 1 sigma bond and 1 pi bond.
Q 10.26 Solution:
Correct answer – C The oxygen is AB2E2 which is tetrahedral family and sp3
Solutionswillbepostedatwww.prep101.com/solutions 11
©Prep101 Q 10.27 Solution:
Chem110MT2Solutions
Correct answer – B This carbon behaves as trigonal planar, and therefore sp2 hybridization
Q 10.28 Solution:
Correct Answer – A Eliminate answers B,D (B has negative 2 charge, D has +2) C is incorrect because it puts negative formal charge on the less electronegative atom. In E, carbon octet is exceeded.
Q 10.29 Solution:
Correct answer – B (is false) NO3- has 24 electrons Lewis Structures (resonance):
O N O
O
O N O
O
HO N O
O
As a result, N-O bond length is identical in all N-O bonds (making B the false answer). Nitrogen atom has no lone pairs of electrons, and is an AB3 type molecule (trigonal planar), and hence sp2 hybridized.
Solutionswillbepostedatwww.prep101.com/solutions 12
©Prep101 Q*10.30 Solution:
Chem110MT2Solutions
Correct answer: C PCl2F3 can have several orientations; however one set cannot exist because we are told the molecule is polar! PCl2F3 = 5 + 14 + 21 = 40 electrons Cl
Cl F
P
F
Cl
P
F
F Cl
non-polar
F
F
polar
In the polar molecule, there is at least one F-P-F angle of 90°C Q 10.31 Solution:
Correct Answer: A Carbon = 4 – 4 – 0 = 0 Sulfur = 6 – 2- 4 = 0 Nitrogen = 5 – 4 = +1 Oxygen = 6 -1 – 6 = -1
Solutionswillbepostedatwww.prep101.com/solutions 13
©Prep101
Chem110MT2Solutions
Chapter 11 –Molecular Orbital Theory Chapter 11 – Practice Problems Q 11.1 Solution:
For O2 (total 12 electrons) Bond Order (O2) = (8 – 4)/2 = 2 Structure:
O=O
For F2 (total 14 electrons) Bond Order (F2) = (8 – 6)/2 = 1 Structure:
Q 11.2 Solution:
F – F
For N2 (total 10 electrons) Bond Order (N2) = (8 – 2)/2 = 3 Structure:
NŁN
For B2 (total 14 electrons) Bond Order (F2) = (4 – 2)/2 = 1 Structure:
B – B
Solutionswillbepostedatwww.prep101.com/solutions 14
©Prep101 Q 11.3 Solution:
Chem110MT2Solutions
H2 has a total of 2 electrons. In the MO diagram, both electrons reside in the bonding orbital giving H2 a bond order of 1. He2 has a total of 4 electrons. In the MO diagram, the first two electrons will occupy the bonding orbital however the next two will occupy antibonding orbitals. The bond order for He2 will be zero – and therefore non existent.
Q* 11.4 Solution:
A) CN- has 10 valence electrons Fill as appropriate: ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (2) < ʌ*2p < ı*2p CN+ = 8 electrons; BO = 2 CN = 9 electrons; BO = 2½ CN- = 10 electrons; BO = 3 Shortest bond order = longest bond
Solutionswillbepostedatwww.prep101.com/solutions 15
©Prep101
Chem110MT2Solutions
Q* 11.5 Solution:
BO = 8-8/2 = 0
Therefore, Ne2 does not exist
Solutionswillbepostedatwww.prep101.com/solutions 16
©Prep101
Chem110MT2Solutions
Q* 11.6 Solution:
Bond order for CO is larger than bond order for NO, therefore C-O bond is shorter.
Q* 11.7 Solution:
Q 11.8 Solution:
Correct answer: C (false) Orbitals are conserved: the number of MOs will always be the same as the number of AO used to construct them.
Solutionswillbepostedatwww.prep101.com/solutions 17
©Prep101 Q 11.9 Solution:
Chem110MT2Solutions
Correct answer – C Bond order = (8-2)/2 = 6/2 = 3 No unpaired electrons; therefore diamagnetic.
Q 11.10 Solution:
Correct answer – E Bond Order = (8-6)/2 = 2/2 = 1 No unpaired electrons, therefore diamagnetic
Solutionswillbepostedatwww.prep101.com/solutions 18
©Prep101 Q 11.11 Solution:
Chem110MT2Solutions
Correct answer – B O2+ has the same molecular orbital diagram as O2, except one electron is removed! The ʌ*2p therefore has 1 valence electron.
Solutionswillbepostedatwww.prep101.com/solutions 19
©Prep101 Q 11.12 Solution:
Chem110MT2Solutions
Correct answer – D B2 has 6 valence electrons, so B2- has 7 ı2s (2) < ı*2s (2) < ʌ2p (3)
Q 11.13 Solution:
Correct answer – D N2 has 10 valence electrons, so N2+ will have 9 valence electrons N2
ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (2)
BO = (8-2)/2 = 3
N2+
ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (1)
BO = (7-2)/2 = 2.5
As a result, bond order will decrease, and the unpaired electron makes it paramagnetic.
Solutionswillbepostedatwww.prep101.com/solutions 20
©Prep101 Q 11.14 Solution:
Q 11.15 Solution:
Chem110MT2Solutions
Correct answer – C B2
ı2s (2) < ı*2s (2) < ʌ2p (2)
BO = (4-2)/2 = 1
C2
ı2s (2) < ı*2s (2) < ʌ2p (4)
BO = (6-2)/2 = 2
N2
ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (2)
BO = (8-2)/2 = 3
O2
ı2s (2) < ı*2s (2) < ı2p (2)< ʌ2p (4) < ʌ*2p(2)
BO = (8-4)/2 = 2
F2
ı2s (2) < ı*2s (2) < ı2p (2)< ʌ2p (4) < ʌ*2p(4)
BO = (8-6)/2 = 1
Correct answer – B CN+ = 8 electrons; BO = 2 CN = 9 electrons; BO = 2½ CN- = 10 electrons; BO = 3 NO+ = 10 electrons BO = 3 Fill as appropriate: ı2s < ı*2s < ʌ2p < ı2p Shortest bond order = longest bond
Solutionswillbepostedatwww.prep101.com/solutions 21
©Prep101
Chem110MT2Solutions
Fenster – Extra Problems QI Solution:
Q II Solution: Q III Solution:
Q IV Solution:
Answer – D Corundum is a crystalline form of aluminum oxide (Al2O3) with traces amount of iron, titanium and chromium.
Answer – E
E There are over 500 chemicals found in natural apples
Answer –D N2O is dinitrogen oxide, and is referred to as laughing gas.
QV Solution:
Answer: A
Q VI Solution:
Answer – B
Solutionswillbepostedatwww.prep101.com/solutions 22
©Prep101
Chem110MT2Solutions
Chapter 24 – Coordination Chemistry Chapter 24 Problems Q 24.1 Solution:
Q 24.2 Solution:
Q 24.3 Solution:
Q 24.4 Solution:
Q 24.5 Solution:
Q 24.6 Solution:
Q 24.7 Solution:
Correct Answer: A H2O is neutral ligand, Cl is each -1, Cr = +3, so overall is -1
Correct Answer: D H2O is neutral ligand, Co = +2, so overall complex is +2.
Correct answer: A NH3is neutral, Br = -1, Pt = +2, therefore overall = zero
Correct answer: A Coordination number = # ligands = 4 SO4 = -2, ammonia = 0, therefore Cu = +2
Correct Answer: E Coordination number = # ligands = 2 CN = -1, overall complex = -1, therefore, Ag = +1
Correct Answer: C Coordination number = # ligands = 4+2 = 6 H2O = neutral, Cl = -1, counterion = -1, therefore Cr = +3
Correct Answer: B Coordination number = # ligands = 4 CN = -1(4) = -4, counterion = +1(4) = +4, therefore Ni = 0
Solutionswillbepostedatwww.prep101.com/solutions 23
©Prep101 Q 24.8 Solution:
Q 24.9 Solution:
Q 24.10 Solution:
Q 24.11 Solution:
Q 24.12 Solution:
Q 24.13 Solution:
Q 24.14 Solution:
Q 24.15 Solution:
Chem110MT2Solutions
Correct Answer: E Correct name is tetrachlorocobaltate(II) ion, (underline for emphasis only)
Correct Answer: A Chloride because counterion, 4 = tetra, di = 2
Correct Answer: E Nickelate because anionic, oxidation number = 0, tetracyano
Correct answer: A B is not by alphabetical order, oxidation number is incorrect in C and D
Correct answer: D Bromide = counterion, (en) = neutral, so cobalt = +3. must be balanced by 3 bromide ions
Correct Answer: A CN = -1, Ni = 0, therefore 4 K are required. Ligands on inside, counterion (K) on outside K (counterion) is positive (cation), so goes on left hand side.
Correct Answer: A Coordination number = # Ligands = 2+2 = 4 Cl = -1, CN = -1, Therefore Cd = +4, and counterion is -1(2).
[FeCl6] must equal -3. Balance charge 3(+2) = 6; 2(x) = -6 to give neutral charge. x = -3.
Solutionswillbepostedatwww.prep101.com/solutions 24
©Prep101 Q 24.16 Solution:
Q 24.17 Solution:
Q 24.18 Solution:
Q 24.19 Solution:
Q 24.20 Solution:
Q 24.21 Solution:
Chem110MT2Solutions
Correct Answer: -1 NH4 is +1 and counterion to give neutral charge, therefore complex must equal -1.
Correct answer: B V = +4, O = 2-, CN = -1, therefore overall charge must be equal to -2
Correct answer: A There are two cyanide ions complexed with copper(I). This gives the complex a charge of 1-, which means there must be one potassium ion.
Correct Answer: E Rh = +1, CO = 0, Br = -1(2); therefore overall ion complex = -1, so 1 potassium ions are required to make it neutral.
Correct answer: B A is incorrect (Coordination number is 3 of Ni(en)3) en is bidentate (making C incorrect) hexaamminenickel(3) ion is incorrect, it needs roman numerals 3 sulphate groups are required; cross charges, to give [Ni(en)3]2(SO4)3
Correct Answer: C Ionization isomer is one where the counter-ion is switched. So switch the bromo for the chloro in (c).
Solutionswillbepostedatwww.prep101.com/solutions 25
©Prep101 Q 24.22 Solution:
Chem110MT2Solutions
Correct answer: D Cu2+ is the oxidation state. Cu2+ is d9. Irrevalent whether it is high spin or low spin system. 1 unpaired electron
Q 24.23 Solution:
Correct answer: B Name tells you oxidation state of Mn(II) = 2 Shape = octahedral Mn2+ = d5 CN = strong field; t2g5eg0 Æ 1 unpaired electron
Solutionswillbepostedatwww.prep101.com/solutions 26
©Prep101 Q 24.24 Solution:
Chem110MT2Solutions
Correct answer: E [CoCl(NH3)5] Cl2 Æ pentaamminechlorocobalt(III) chloride Co3+ = d6 Coordination number = 6 = octahedral t2g6eg0 Æ 0 unpaired electrons (strong field)
Q 24.25 Solution:
Correct answer: D Cs[FeCl4]
ĺ
[FeCl4] = -1
ĺ
Fe = +3
Fe3+ ĺ d5 Cl is a weak-field ligand; so 5 unpaired electrons
Solutionswillbepostedatwww.prep101.com/solutions 27
©Prep101 Q 24.26 Solution:
Chem110MT2Solutions
Correct answer: A Cl = -1, en = neutral, NH3= neutral; therefore Ni = +2 Coordination number = 6 (en is bidentate); Ni2+ĺ d8 It does not matter whether it is high field or low field, d8 gives 2 unpaired electrons in either situation.
Solutionswillbepostedatwww.prep101.com/solutions 28
©Prep101 Q 24.27 Solution:
Chem110MT2Solutions
Correct Answer: C Each are Ni2+ ĺ d8 Cl is a weak-field ligand, while CN is a strong field ligand
Q 24.28 Solution:
Correct answer: E Revisit the spectrochemical series. AĺD are all strong field. Br is weak field, and would result in a high spin system with unpaired electrons (paramagnetic). Co3+ = d6
Solutionswillbepostedatwww.prep101.com/solutions 29
Chem110MT2Solutions
Chapter 10 – Practice Problems Q 10.1 Solution:
-1
0 O S
-1 O
-1 O
O
+2 0 O
S O
+2
S O
-1
FC O in S = O
60
-1
O
0
1 u 2 1 2 1 64 u4 0 2
66
FC O in S - O
FC S
O
-1
+2
1 u 8 2 2
Q 10.2 Solution:
Correct Answer: B Two oxygen atoms will have a formal charge of –1 and there will also be 4 different resonant structures. O
O C O
C O
Bond order C-O = 1/2 (# bonding e- - # antibonding e-) = 3/2 Therefore statements 1 and 3 are true. Q 10.3 Solution:
Correct Answer: A There should be one pair of electrons on the central sulphur atom in A.
Solutionswillbepostedatwww.prep101.com/solutions 1
©Prep101
Chem110MT2Solutions
Q 10.4 Solution: Total # of valence electrons = 5 + 6 + 7 = 18 Formal charge on F = 7 – [6 + ½ (2)] = 0 Formal charge on N = 5 – [2 + ½ (6)] = 0 Formal charge on O = 6 – [4 + ½ (4)] = 0 Since the formal charge on N, O and F are 0, this is the most likely Lewis structure. Q 10.5 Solution:
X is Nitrogen (N).
Q 10.6 Solution:
There are 4 lone pairs of electrons O
C H2N
Q 10.7 Solution:
NH2
Correct answer: A A and C are the only neutral structures. A abides by octet rule.
Solutionswillbepostedatwww.prep101.com/solutions 2
©Prep101 Q 10.8 Solution:
Chem110MT2Solutions
Correct Answer: B Resonance structures occur when more than one Lewis structure, each with different electron distributions, can be drawn for a molecule. This usually results in molecules containing both double bonds and single bonds. (a) O3
O
O
O
O
O
O
(b) No resonance structure. CH4 cannot have resonance structures because no double bonds are present.
H
H
H
H (c) NO2-
O
-
N
O
O
N
O
(d) HCO3OH
O O
-1
OH
O
-1
O
Solutionswillbepostedatwww.prep101.com/solutions 3
©Prep101 Q 10.9 Solution:
(a) SCN-
Chem110MT2Solutions
Total # of valence electrons: 6 + 4 + 5 + 1 = 16 -1 (-)
(b) -1 (-)
Formal charge = (# of valence electron on a free atom) – (# of valence electrons assigned to the atom in the molecule) Valence electrons assigned = (# of lone pair electrons) + ½ (# of shared electrons) For S, formal charge = 6 – [4 + ½ (4)] = 0 For C, formal charge = 4 – [0 + ½(8)] = 0 For N, formal charge = 5 – [4 + ½(4)] = -1 For S, formal charge = 6 – [6 + ½ (2)] = -1 For C, formal charge = 4 – [0 + ½(8)] = 0 For N, formal charge = 5 – [2 + ½(6)] = 0
The first structure is preferred since N is more electronegative than S, and thus is more likely to have the negative charge. Q 10.10 Solution:
Correct Answer: C The formal charge of P in structures (i) and (iii) is = 5 – (5 + 0) = 0. Based on formal charge, structure (i) is the most important resonance structure. The formal charge of O in structure (ii) = 6 – (1 + 6) = -1. The most stable structure is that with the lowest formal charge.
Solutionswillbepostedatwww.prep101.com/solutions 4
©Prep101 Q 10.11 Solution:
Chem110MT2Solutions
Correct answer: C There are two lone pairs and four bonding pairs on the central S atom, giving AX4E2. The lone pairs will align to maximize the distance between them. Therefore, this atom would be square planar.
Q 10.12 Solution:
Correct answer: B CF4 is a tetrahedral molecule with all the same types of bonds at each point, so it will have a net dipole of zero.
Q 10.13 Solution: Species
CS2
Lewis electron dot structure S
C
SO3
Linear
S
F
F
XeF4
Xe F
F S
O
O
Hybridizatio Polar (P) or n Nonpolar (N) on species central atom N sp
Square Planar
sp3d2
N
Trigonal planar
sp2
N
Trigonal Pyramidal
sp3
P
Octahedral
sp3d2
N
O
O
SO32
Shape (as given by nuclei)
O
S O
F
SF6
F F
F S
F
F
Solutionswillbepostedatwww.prep101.com/solutions 5
©Prep101
Chem110MT2Solutions
Q 10.14 Solution: H H
Solution:
sp2
Solution:
sp2
Solution:
Answer: 4
Solution:
Answer: 1
Solution:
Yes
Solution:
Yes
C
N
H
Solutionswillbepostedatwww.prep101.com/solutions 6
©Prep101
Chem110MT2Solutions
Q 10.15 Solution: Molecule
Lewis Structure O
H2CO
C H
Molecular Shape (VSEPR)
Hybridization of Central Atom
Polar (P) or Nonpolar (N)
Trigonal planar
sp2
Polar
H
sp3 OF2
Bent
O F
Polar
F
HCN H
C
N
linear
sp
Polar
tetrahedral
sp3
Non-polar
Square pyramidal
sp3d2
Polar
See-saw
sp3d
Polar
H
SiH4
Si H
H
H
IF5
F
F I
F
F F
SeF4
F F Se F F
Solutionswillbepostedatwww.prep101.com/solutions 7
©Prep101
Chem110MT2Solutions
Q 10.16 Solutions:
i)
l
N F
O
O
O
O
l
N F
O
N F
ii) Trigonal planar iii) 3 iv) sp2 v) X = C and Z = B
Solutionswillbepostedatwww.prep101.com/solutions 8
O
©Prep101
Chem110MT2Solutions
Q 10.17 Solution:
Species
Lewis Structure
Shape
Hybridization on Central atom
Polar/Nonpolar
PF5
AX5
Trigonalbipyramidal
sp3d
N
BF3
AX3
Trigonalplaner
sp2
N
PF3
AX3E
Tetrahedral
sp3
P
BrF3
AX3E2
T-shaped
sp3d
P
BrF2+
AX2E2
Angular
sp3
P
SeF4
AX4E
Seesaw
sp3d
P
BrF5
AX5E
Squarepyramidal
sp3d2
P
Solutionswillbepostedatwww.prep101.com/solutions 9
©Prep101 Q 10.18 Solution:
Solution:
Chem110MT2Solutions
C1 is sp3 with 109.5° (tetrahedral) C2 is sp2 with 120° (trigonal planar) OA is sp2 hybridized OB is sp3 hybridized Nitrogen is sp3 hybridized with a bond angle of 109.5°
Solution:
Q 10.19 Solution:
There are 9 ı bonds (single bonds) There is 1 ʌ bond (double bond)
Correct answer – C IF5 = 42 electrons. ĺ AB5E1 On the picture below, there is also a lone pair of electrons around the central iodine. This molecule is polar, because it has a net dipole towards the fluorine (in the plane). F
F I
F
F
F
Q 10.20 Solution:
Correct answer - E There are 14 sigma bonds (single bonds) and 3 pi bonds Recall that a double bond equals 1 sigma bond and 1 pi bond. Recall that a triple bond equals 1 sigma bond and 2 pi bonds.
Solutionswillbepostedatwww.prep101.com/solutions 10
©Prep101 Q 10.21 Solution:
Chem110MT2Solutions
Correct Answer: D D and E are the only answers that sum to give a -1 anion (eliminate A, B, C). For nitrogen: 5 – 4- 2 = -1, sulfur: 6- 4 -2 = 0, carbon: 4 – 4 =0
Q 10.22 Solution:
Answer: C Sodium iodide, NaI An ionic compound is generally the result of a compound composed of a non-metal and a metal.
Q 10.23 Solution:
Answer: D Iron is a metal, and has primarily metallic bonding – delocalization of electrons
Q 10.24 Solution:
Answer: B Fluorine is the most electronegative atom in the periodic table. Electronegative increases from bottom to top, and increases from left to right.
Q 10.25 Solution:
Correct answer – D There are 9 sigma bonds (single bonds) and 1 pi bond (double bond) Recall that a double bond equals 1 sigma bond and 1 pi bond.
Q 10.26 Solution:
Correct answer – C The oxygen is AB2E2 which is tetrahedral family and sp3
Solutionswillbepostedatwww.prep101.com/solutions 11
©Prep101 Q 10.27 Solution:
Chem110MT2Solutions
Correct answer – B This carbon behaves as trigonal planar, and therefore sp2 hybridization
Q 10.28 Solution:
Correct Answer – A Eliminate answers B,D (B has negative 2 charge, D has +2) C is incorrect because it puts negative formal charge on the less electronegative atom. In E, carbon octet is exceeded.
Q 10.29 Solution:
Correct answer – B (is false) NO3- has 24 electrons Lewis Structures (resonance):
O N O
O
O N O
O
HO N O
O
As a result, N-O bond length is identical in all N-O bonds (making B the false answer). Nitrogen atom has no lone pairs of electrons, and is an AB3 type molecule (trigonal planar), and hence sp2 hybridized.
Solutionswillbepostedatwww.prep101.com/solutions 12
©Prep101 Q*10.30 Solution:
Chem110MT2Solutions
Correct answer: C PCl2F3 can have several orientations; however one set cannot exist because we are told the molecule is polar! PCl2F3 = 5 + 14 + 21 = 40 electrons Cl
Cl F
P
F
Cl
P
F
F Cl
non-polar
F
F
polar
In the polar molecule, there is at least one F-P-F angle of 90°C Q 10.31 Solution:
Correct Answer: A Carbon = 4 – 4 – 0 = 0 Sulfur = 6 – 2- 4 = 0 Nitrogen = 5 – 4 = +1 Oxygen = 6 -1 – 6 = -1
Solutionswillbepostedatwww.prep101.com/solutions 13
©Prep101
Chem110MT2Solutions
Chapter 11 –Molecular Orbital Theory Chapter 11 – Practice Problems Q 11.1 Solution:
For O2 (total 12 electrons) Bond Order (O2) = (8 – 4)/2 = 2 Structure:
O=O
For F2 (total 14 electrons) Bond Order (F2) = (8 – 6)/2 = 1 Structure:
Q 11.2 Solution:
F – F
For N2 (total 10 electrons) Bond Order (N2) = (8 – 2)/2 = 3 Structure:
NŁN
For B2 (total 14 electrons) Bond Order (F2) = (4 – 2)/2 = 1 Structure:
B – B
Solutionswillbepostedatwww.prep101.com/solutions 14
©Prep101 Q 11.3 Solution:
Chem110MT2Solutions
H2 has a total of 2 electrons. In the MO diagram, both electrons reside in the bonding orbital giving H2 a bond order of 1. He2 has a total of 4 electrons. In the MO diagram, the first two electrons will occupy the bonding orbital however the next two will occupy antibonding orbitals. The bond order for He2 will be zero – and therefore non existent.
Q* 11.4 Solution:
A) CN- has 10 valence electrons Fill as appropriate: ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (2) < ʌ*2p < ı*2p CN+ = 8 electrons; BO = 2 CN = 9 electrons; BO = 2½ CN- = 10 electrons; BO = 3 Shortest bond order = longest bond
Solutionswillbepostedatwww.prep101.com/solutions 15
©Prep101
Chem110MT2Solutions
Q* 11.5 Solution:
BO = 8-8/2 = 0
Therefore, Ne2 does not exist
Solutionswillbepostedatwww.prep101.com/solutions 16
©Prep101
Chem110MT2Solutions
Q* 11.6 Solution:
Bond order for CO is larger than bond order for NO, therefore C-O bond is shorter.
Q* 11.7 Solution:
Q 11.8 Solution:
Correct answer: C (false) Orbitals are conserved: the number of MOs will always be the same as the number of AO used to construct them.
Solutionswillbepostedatwww.prep101.com/solutions 17
©Prep101 Q 11.9 Solution:
Chem110MT2Solutions
Correct answer – C Bond order = (8-2)/2 = 6/2 = 3 No unpaired electrons; therefore diamagnetic.
Q 11.10 Solution:
Correct answer – E Bond Order = (8-6)/2 = 2/2 = 1 No unpaired electrons, therefore diamagnetic
Solutionswillbepostedatwww.prep101.com/solutions 18
©Prep101 Q 11.11 Solution:
Chem110MT2Solutions
Correct answer – B O2+ has the same molecular orbital diagram as O2, except one electron is removed! The ʌ*2p therefore has 1 valence electron.
Solutionswillbepostedatwww.prep101.com/solutions 19
©Prep101 Q 11.12 Solution:
Chem110MT2Solutions
Correct answer – D B2 has 6 valence electrons, so B2- has 7 ı2s (2) < ı*2s (2) < ʌ2p (3)
Q 11.13 Solution:
Correct answer – D N2 has 10 valence electrons, so N2+ will have 9 valence electrons N2
ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (2)
BO = (8-2)/2 = 3
N2+
ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (1)
BO = (7-2)/2 = 2.5
As a result, bond order will decrease, and the unpaired electron makes it paramagnetic.
Solutionswillbepostedatwww.prep101.com/solutions 20
©Prep101 Q 11.14 Solution:
Q 11.15 Solution:
Chem110MT2Solutions
Correct answer – C B2
ı2s (2) < ı*2s (2) < ʌ2p (2)
BO = (4-2)/2 = 1
C2
ı2s (2) < ı*2s (2) < ʌ2p (4)
BO = (6-2)/2 = 2
N2
ı2s (2) < ı*2s (2) < ʌ2p (4) < ı2p (2)
BO = (8-2)/2 = 3
O2
ı2s (2) < ı*2s (2) < ı2p (2)< ʌ2p (4) < ʌ*2p(2)
BO = (8-4)/2 = 2
F2
ı2s (2) < ı*2s (2) < ı2p (2)< ʌ2p (4) < ʌ*2p(4)
BO = (8-6)/2 = 1
Correct answer – B CN+ = 8 electrons; BO = 2 CN = 9 electrons; BO = 2½ CN- = 10 electrons; BO = 3 NO+ = 10 electrons BO = 3 Fill as appropriate: ı2s < ı*2s < ʌ2p < ı2p Shortest bond order = longest bond
Solutionswillbepostedatwww.prep101.com/solutions 21
©Prep101
Chem110MT2Solutions
Fenster – Extra Problems QI Solution:
Q II Solution: Q III Solution:
Q IV Solution:
Answer – D Corundum is a crystalline form of aluminum oxide (Al2O3) with traces amount of iron, titanium and chromium.
Answer – E
E There are over 500 chemicals found in natural apples
Answer –D N2O is dinitrogen oxide, and is referred to as laughing gas.
QV Solution:
Answer: A
Q VI Solution:
Answer – B
Solutionswillbepostedatwww.prep101.com/solutions 22
©Prep101
Chem110MT2Solutions
Chapter 24 – Coordination Chemistry Chapter 24 Problems Q 24.1 Solution:
Q 24.2 Solution:
Q 24.3 Solution:
Q 24.4 Solution:
Q 24.5 Solution:
Q 24.6 Solution:
Q 24.7 Solution:
Correct Answer: A H2O is neutral ligand, Cl is each -1, Cr = +3, so overall is -1
Correct Answer: D H2O is neutral ligand, Co = +2, so overall complex is +2.
Correct answer: A NH3is neutral, Br = -1, Pt = +2, therefore overall = zero
Correct answer: A Coordination number = # ligands = 4 SO4 = -2, ammonia = 0, therefore Cu = +2
Correct Answer: E Coordination number = # ligands = 2 CN = -1, overall complex = -1, therefore, Ag = +1
Correct Answer: C Coordination number = # ligands = 4+2 = 6 H2O = neutral, Cl = -1, counterion = -1, therefore Cr = +3
Correct Answer: B Coordination number = # ligands = 4 CN = -1(4) = -4, counterion = +1(4) = +4, therefore Ni = 0
Solutionswillbepostedatwww.prep101.com/solutions 23
©Prep101 Q 24.8 Solution:
Q 24.9 Solution:
Q 24.10 Solution:
Q 24.11 Solution:
Q 24.12 Solution:
Q 24.13 Solution:
Q 24.14 Solution:
Q 24.15 Solution:
Chem110MT2Solutions
Correct Answer: E Correct name is tetrachlorocobaltate(II) ion, (underline for emphasis only)
Correct Answer: A Chloride because counterion, 4 = tetra, di = 2
Correct Answer: E Nickelate because anionic, oxidation number = 0, tetracyano
Correct answer: A B is not by alphabetical order, oxidation number is incorrect in C and D
Correct answer: D Bromide = counterion, (en) = neutral, so cobalt = +3. must be balanced by 3 bromide ions
Correct Answer: A CN = -1, Ni = 0, therefore 4 K are required. Ligands on inside, counterion (K) on outside K (counterion) is positive (cation), so goes on left hand side.
Correct Answer: A Coordination number = # Ligands = 2+2 = 4 Cl = -1, CN = -1, Therefore Cd = +4, and counterion is -1(2).
[FeCl6] must equal -3. Balance charge 3(+2) = 6; 2(x) = -6 to give neutral charge. x = -3.
Solutionswillbepostedatwww.prep101.com/solutions 24
©Prep101 Q 24.16 Solution:
Q 24.17 Solution:
Q 24.18 Solution:
Q 24.19 Solution:
Q 24.20 Solution:
Q 24.21 Solution:
Chem110MT2Solutions
Correct Answer: -1 NH4 is +1 and counterion to give neutral charge, therefore complex must equal -1.
Correct answer: B V = +4, O = 2-, CN = -1, therefore overall charge must be equal to -2
Correct answer: A There are two cyanide ions complexed with copper(I). This gives the complex a charge of 1-, which means there must be one potassium ion.
Correct Answer: E Rh = +1, CO = 0, Br = -1(2); therefore overall ion complex = -1, so 1 potassium ions are required to make it neutral.
Correct answer: B A is incorrect (Coordination number is 3 of Ni(en)3) en is bidentate (making C incorrect) hexaamminenickel(3) ion is incorrect, it needs roman numerals 3 sulphate groups are required; cross charges, to give [Ni(en)3]2(SO4)3
Correct Answer: C Ionization isomer is one where the counter-ion is switched. So switch the bromo for the chloro in (c).
Solutionswillbepostedatwww.prep101.com/solutions 25
©Prep101 Q 24.22 Solution:
Chem110MT2Solutions
Correct answer: D Cu2+ is the oxidation state. Cu2+ is d9. Irrevalent whether it is high spin or low spin system. 1 unpaired electron
Q 24.23 Solution:
Correct answer: B Name tells you oxidation state of Mn(II) = 2 Shape = octahedral Mn2+ = d5 CN = strong field; t2g5eg0 Æ 1 unpaired electron
Solutionswillbepostedatwww.prep101.com/solutions 26
©Prep101 Q 24.24 Solution:
Chem110MT2Solutions
Correct answer: E [CoCl(NH3)5] Cl2 Æ pentaamminechlorocobalt(III) chloride Co3+ = d6 Coordination number = 6 = octahedral t2g6eg0 Æ 0 unpaired electrons (strong field)
Q 24.25 Solution:
Correct answer: D Cs[FeCl4]
ĺ
[FeCl4] = -1
ĺ
Fe = +3
Fe3+ ĺ d5 Cl is a weak-field ligand; so 5 unpaired electrons
Solutionswillbepostedatwww.prep101.com/solutions 27
©Prep101 Q 24.26 Solution:
Chem110MT2Solutions
Correct answer: A Cl = -1, en = neutral, NH3= neutral; therefore Ni = +2 Coordination number = 6 (en is bidentate); Ni2+ĺ d8 It does not matter whether it is high field or low field, d8 gives 2 unpaired electrons in either situation.
Solutionswillbepostedatwww.prep101.com/solutions 28
©Prep101 Q 24.27 Solution:
Chem110MT2Solutions
Correct Answer: C Each are Ni2+ ĺ d8 Cl is a weak-field ligand, while CN is a strong field ligand
Q 24.28 Solution:
Correct answer: E Revisit the spectrochemical series. AĺD are all strong field. Br is weak field, and would result in a high spin system with unpaired electrons (paramagnetic). Co3+ = d6
Solutionswillbepostedatwww.prep101.com/solutions 29
