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Chemical Bonding Solutions
Problem 1. Answer: B
Formal Charge = Valence electrons – lone pair electrons – bonds
FCF = 7 − 6 − 1 = 0
FCP = 5 − 2 − 3 = 0
FCS = 6 − 4 − 2 = 0
Problem 2. Answer: X is Carbon (C).
FCX = x − 0 − 4 = 0,
Therefore x = 4 and that is the number of valence electrons in the element. Giving us carbon.
Problem 3. Answer: C
There are 4 lone pairs of electrons.
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Problem 4. Answer: D
Resonance is defined as the movement of electrons through π (pi) bonds. π bonds are any bonds beyond a single bond. For instance a double bond has 1 π bond and a triple bond has 2 π bonds. If there are only single bonds, there are no π bonds, therefore there is no resonance.
a) 𝑁O2
b) 𝑆O3
2−
c) O3
d) CH4
Methane is the only compound with only single bonds. Moving electrons in this structure results in broken bonds – this is another useful way to think of structures that can and can’t have resonance.
e) O2
You may have thought oxygen only has one structure but, because it has a double bond, there is a minor structure that can occur.
Problem 5. Answer: D
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VSEPR Solutions
Problem 6. Answer: C
Solution: A triple bond corresponds to 2 bonds and 1 bond. The two single bonds consist of one
bond each. Therefore there is a total of three bonds and two bonds.
Problem 7. Answer:
a) N(1): sp3 N(2): sp2 O(1): sp2 O(2): sp3
b) i) 109 I ii) 120 iii) 6 lone pairs
Solution:
N(1) has 3 bonding pairs and 1 lone pairs of electrons sp3 hybridized
N(2) has 3 bonding pairs of electrons sp2 hybridized
O(1) has 1 bonding pair and 2 lone pairs of electrons sp2 hybridized
O(2) has 1 bonding pair and 3 lone pairs of electrons sp3 hybridized
i) N(1) is sp3 hybridized; tetrahedral structure; 109 bond angle
ii) N(2) is sp2 hybridized; trigonal planar; 120 bond angle
iii) By the Lewis Structure, there are 6 lone pairs of electrons.
Problem 8. Answer: D
Solution: From the Lewis diagram of the molecule we can see that the shape of the molecule is T-Shaped.
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Problem 9. Which of the following statements is/are correct for the formate ion HCO2 ?
Answer: C
Solution: the oxidation number of the C atom is +2, there are only 2 plausible contributing structures, and HCO2
= 18 e- = AX3 = trigonal planar
Problem 10. a) Answer:
b) Answer:
Sigma bonds = 4
Pi bonds = 1
c) Answer: A single bond contains one σ bond, whereas a double bond consists of one σ and one п bond. σ bonds are covalent bonds between electron pairs in the area between two atoms. п bonds are a covalent bond between parallel p orbitals and the electron pairs shared are below & above the line joining the 2 atoms.
C N
d) Answer: There are 3 electron pairs around the central C atom, and thus it is sp2 hybridized. Thus
the bond angle is 120.
п bond
σ bond
2p orbitals
sp2 orbitals
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Problem 11. Predict the geometric shape of ClO2- ion.
Answer: Bent
Solution:
Total valence electrons: 7 + 2(6) + 1 = 20 electrons
Center Cl has 2 bonding pairs and 2 lone pairs of electrons sp3 hybridized & tetrahedral formation. Since there are 2 lone pairs of electrons, the structure is bent-shaped.
Problem 12. Draw the Lewis structure for the peroxymonosulfate ion, H-O-O-SO3-, and estimate the H-O-O bond angle.
Answer: < 109
Solution:
The center oxygen in H-O-O has 4 pairs of electrons (2 lone pairs of electron and 2 bonding pairs of electrons); as such, this part of the molecule has a tetrahedral structure. The bond angle is less than
109.5 due to the 2 lone pairs of electrons that bend the structure more than bonding pairs of electrons. The tetrahedral formation means the electrons are distributed in sp3 orbitals.
-1
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Problem 13. a) Applying your knowledge of VSEPR, indicate the geometry of the three atoms indicated in acetic acid.
b) What is the O=C-O bond angle?
c) Indicate the orbital hybridization of the three atoms with the arrows.
Answer:
a) H3C- : 4 bonding pairs of electrons, thus it is tetrahedral
3 bonding pairs of electrons; thus it is trigonal planar
4 bonding pairs of electrons; thus it is tetrahedral.
b) The geometry around the central C carbon is trigonal planar structure. Thus, the bond angle is 120.
c)
H3C - C - OH Tetrahedral structure = sp3hybridization
H3C - C - OH Trigonal planar structure = sp2 planar
H3C - C - OH Tetrahedral structure = sp3 hybridization
CH3
OH
O
CH3
OH
O
CH3
OH
O
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Problem 14. a) Cl2CS (thiophosgene) (Carbon is central atom, Cl are equivalent).
Lewis:
*All formal charges are zero. Total valence electrons = 24
VSEPR
AX3
Electronic geometry: trigonal planar
Molecular geometry: trigonal planar
b) PF6
Lewis Structure
*Formal Charge on F = 0, Formal Charge on P = -1. Total valence electrons = 48
AX6
Electronic geometry: octahedral
Molecular geometry: octahedral
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Problem 15. a) HNO3
a) Total # of electrons = 1 + 5 + 3(6) = 24
Formal charge:
On O(1) 6 – [4 + ½ (4)] = 0
On O(2) 6 – [6 + ½ (2)] = -1
On O(3) 6 – [4 + ½ (4)] = 0
On N 5 – [0 + ½ (8)] = +1
On H 1 – [0 + ½ (2)] = 0
Molecular drawing:
b) O3
Total # of valence electrons: 3(6) = 18
Lewis structure:
Formal charge on:
O(1) 6 – [6 + ½ (2)] = -1
O(2) 6 – [2 + ½ (6)] = + 1
O(3) 6 – [4 + ½ (4)] = 0
Molecular structure around central O = bent
(1) (2) (3)
(2) (1)
(3)
Structure around central N: 3 electron pairs = trigonal planar
Structure around central O: 4 electron pairs = bent
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Practice Problems
Problem 1. Answer: B
Solution:
Non-zero zero zero with zero with non-zero
resonance resonance
Problem 2. Answer: B
Solution: The most ionic bond will have the most electronegative species and the least electronegative species. Electronegativity increases as you move across and up the periodic table. F is the most electronegative and Li, here, is the least E.N.
Problem 3. Answers:
NH3– Br: dipole – induced dipole, London dispersion
ClBr– ClBr: dipole – dipole, London dispersion
CO2– CO: induced dipole – dipole, London dispersion
b) All 3 molecules are non-polar, so only London dispersion forces
London dispersion forces increase in magnitude as MW increase
As IM forces increase, boiling point increases
Br2 has highest MW, so has strongest IM forces and highest boiling point
bp(N2) < bp(O2) < bp(Br2)
c) Compound Type of intermolecular forces
CH4 Non-polar; London dispersion forces
CH3OH Polar; London dispersion forces & H-bonding
CCl4 Non-polar; London dispersion forces that are greater
than those in CH4 or CH3OH
The greater the intermolecular forces present, the lower the vapor pressure.
(lowest vapor pressure) CCl4 < CH3OH < CH4 (highest vapor pressure)
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Problem 4. Answer: D
There is more interaction between the atoms of carbon tetrachloride due to stronger London dispersion forces, which increase the boiling point with respect to methane. Chlorine is more polarizable than hydrogen because it has MANY more electrons that are also further from the nucleus, resulting in stronger London forces.
Problem 5. Answer: B
Solution: HF is more likely to form hydrogen bonds with water because of the large dipole present in the HF molecule. This allows the hydrogen bonds to form more readily.
Problem 6. Solution: B
Methanol experiences hydrogen bonding due to the polar nature of the molecule.
Problem 7. Solution: D
The lowest boiling point will occur in non-polar compounds. The only non-polar compound above is N2.
Problem 8. Solution: C
Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure.
Problem 9. Solution: A) H2O
The boiling point depends on the type of intermolecular forces present.
H-bonding > dipole-dipole > London dispersion forces.
CH4 has the lowest boiling point as there are only London dispersion forces present.
OHH H
SH F
H NHH
HH H
H
H
H-bonding H-bonding H-bonding H-bonding London forces
2 strong dipoles 2 weak dipoles 1 very strong 3 weak dipoles
dipole
The order of intermolecular bonding strength & hence boiling temp is:
(highest BP & bond strength) H2O > HF > H2S > NH3 > CH4 (lowest BP & bond strength)
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Problem 10. Solution: A
The higher the intermolecular forces, the higher the boiling point of the solution will be.
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Practice Problems
Problem 11. Answers:
O
CH3
CH3
H3C
CO
O
C CH3CH2C Cl
H3C C CH
CH3H2C C
HC
CH3
CH3
(i)
(ii)
(iii)K
(v)
(iv)
Problem 12. Answers:
(i) butanal
(ii) Cyclopentane
(iii) 3-chlorobutanoic acid
(iv) butylamine
(v) 2-ethyl-1-pentene
(vi) (cis)-2-hexene
(vii) methoxyethane
(viii) benzoic acid
(ix) 2-pentyne
(x) 1,3-dichorobenzene
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Problem 13. Answers:
A) B) C)
H
O
O
O
NH2
O
Problem 14. Answer: D
D has the formula C7H14. All the above structures have 7 carbons; therefore, you can just count the H atoms.
Problem 15. Answer: 9
Solution:
heptane 2-methylhexane 3-methylhexane 2,4-dimethylpentane
2,3-dimethylpentane 3-ethylpentane 3,3-dimethylpentane 2,2,3-trimethylbutane
2,2-dimethylpentane
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Problem 16. Answer: C
Solution:
A chiral carbon is one that has 4 different groups attached to it via σ bonds. A carbon containing double bonds does not count as asymmetric carbons since there are only 3 σ bonds.
A) None B) None C) D) None
Br NH2
OH
O
NH2
O
OH
NH2
COOH
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Problem 17. Answer: Alkene, alkyne, ketone, alcohol
Notes:
Alkene (double bond), alkyne (triple bond),
ketone (contains a carbonyl), alcohol (contains a hydroxyl)
NOTE: Hydroxyl (-OH) and carbonyl (C=O) ARE NOT functional groups. They are pieces of functional groups. For instance, an alcohol (R-OH) contains a hydroxyl. A ketone (R2C = O) contains a carbonyl. A carboxylic acid (R-COOH) contains BOTH a hydroxyl and carbonyl.
Problem 18. Answer: B
An ester group is one that contains RCOOR ( ) which is not present
Problem 19. Answer:
CH3CH2CHCH2CH2CH3
OH
H3C Cl
C
CC
C
CC
Cl
OH
H
H
H
H
C C OH3C
H
NH3
O
**
*The middle two structures are achiral
Problem 20. Consider the following molecules A, B, C, D and E.
Answers:
i. A and D or B and D
ii. A and B
iii. NONE
Problem 21. Answers:
(i) A and F
(ii) C and E
(iii) B, C, D and E are not optically active, thus, no rotation of plane polarized light. A and F are optically active but form racemic mixture. Thus, no rotation of plane polarized light.
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Problem 22. Answer:
OCH3
CH3
HO
Problem 23. Answer:
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Problem 24. Answer:
On reason nylon is so strong is that it is a high density polymer meaning it has long non-branching chains of high molecular weight. In the absence of the special amid functional group these chains could pack together tightly in a uniform structure and be fairly strong. However, nylon’s amide functional groups enable it to form hydrogen bonds between the nitrogen hydrogen of one strand and the carbonyl oxygen of an adjacent strand. These hydrogen bonds allow for tight packing and greatly increase its strength:
N
N
O
O
H
H
N
N
O
O
H
H
N
N
O
O
H
H
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Periodic Properties Practice Problems
Problem 16. Answer: B
Solution: H will have the highest 1s orbital energy because it has the least positive charge on the nucleus.
Problem 17. Answer: a) V5+ < Ti4+ < Sr2+ < Br-
The electron configurations are...
# of protons
Sr2+: [Kr] 38
Br-: [Kr] 35
V5+: [Ar] 23
Ti4+: [Ar] 22
Sr2+ and Br- has the same number of electrons; however, Zeff is greater for Sr2+ due to a greater # of protons, resulting in a greater + charge that pulls the electrons closer to the nucleus. Thus, Sr2+ is smaller than Br-.
V5+ and Ti4+ has the same number of electrons; however, Zeff is greater for V5+ due to a greater # of protons. Thus V5+ is smaller than Ti4+.
Sr2+ and Br- are larger than V5+ and Ti4+ because there are more electrons held in a larger subshell.
(smallest) V5+ < Ti4+ < Sr2+ < Br- (largest)
b) Cl > Br > I
Problem 18. Solutions
a) Electron affinity becomes more negative from left to right because Cl has higher Zeff than S.
b) For k, valence electron is in 4s orbital while valence electron in Li is in 2s orbital. 4s orbital is larger than 2s.
c) Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. Ionization energy
2
1
n
d) Zeff is larger in O than in B.
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Problem 19. a) Mg, Ionization energy increase as you move up a group and it also increases as you move right across a period.
b) Mg, Radius increase as you move left along a period and as you move down a group.
c) Ca<Be<P<Cl<O
Electronegativity increases as you up a group and as you move right across a period. F is the most electronegative element and Cs is the least electronegative element.
Problem 20. Answer:
a) [Ar] or [Ne]3s23p6 b) S2- c) S2- d) Ca2+ e) S2- < Ar < Ca2+
Solution:
a) [Ar] or [Ne]3s23p6 b) Ar, Ca2+ and S2- all have the same number of electrons; however, Ar has 18 protons, Ca2+ has
20, and S2- has 16. Thus S2- has the least Zeff since it has the smallest charge pulling on the electrons.
c) The species with the least favorable electron affinity is S2- because it has the smallest Zeff (a smaller positive charge to pull electrons towards the nucleus).
d) Ca2+ has the greatest Zeff because it has the greatest number of protons pulling on the 18 electrons, resulting in the electrons being closer to the nucleus.
Ionization energy is the energy required to remove an electron. It is most difficult to remove an electron from Ca2+, as this will involve the removal of a core electron instead of a valence electron, and Ca2+ has the greatest Zeff. It is easier to remove an electron from S2- versus Ar because S2- has a smaller Zeff. S2- < Ar < Ca2+
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Problem 21.
has a noble gas electron configuration? Kr
has the smallest ionization energy? Cs
has an atomic number Z = 13? Al
has a half-filled sub-shell with l = 2? Cr
is a hydrogen-like species? He+
has only one 4s electron? Cr
have two unpaired electrons? O, C, V3+
has only two d electrons with n = 3? V3+
is diamagnetic? Kr
has the largest radius? Cs
has only one electron with l = 1? Al
has the largest number of unpaired electrons?
Cr
are transition metal species? V3+, Cr
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Problem 22. Answer:
a) IE1 for K > IE1 for Ca because Zeff increases from left to right across a period, so Ca has a higher Zeff and it is therefore harder to remove an electron. IE2 for K > IE2 for Ca because the ionization process for Ca leads to the formation of the stable noble gas configuration (Ca2+) while the ionization reaction for K requires the destruction of a stable noble gas configuration. The former will always be lower in energy.
b)
K(g)+ → K(g)
2+ + e
Ca(g)+ → Ca(g)
2+ + e
c) The larger the negative charge, the greater the repulsion between electrons, the larger the radius. These are all atoms that have the configuration of Kr, so only charge influences radius. Therefore, Rb+
< Br < Se2
The ionization energy decreases as you go down Group 1 because as n increases it is easier to remove the outer electron. Radius increases as you go down Group A because as n increases, the radius increases.
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Problem 23. a)Answer:
Ca2+ < K+ < Cl < S2
b) Answer:
B < Be < O < N
c) Answer:
Sr < Mg < S < F
Problem 24. Answer:
a) When AB is put into aqueous solution, it will dissociate into A- and B+ ions rather than A+ and B-. A has a greater ionization energy, and thus is less able to form A+ than B forming B+. Further, electron affinity for A is larger, and thus the reaction A A- is more favorable than B B-.
b) A has a greater electronegativity; A has a larger electron affinity value and thus releases more energy when it gains an electron, making the reaction more favorable than B gaining an electron.
c) A will be more to the right on the periodic table than B, since it has a higher ionization energy and larger electron affinity. Because A is further to the right, it has a greater Zeff and thus will be smaller than B.
d) Since nonmetals are on the right hand side of the periodic table, and thus have more favorable electron affinity due to a greater Zeff, element A is the nonmetal.
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Quantum Numbers Practice Problems
Problem 25. Answer: C
Solution:
This gives the probability of finding an electron in a region of space.
Problem 26. Answer: O2- has 10 electrons therefore Ne is isoelectronic with it.
The electron configuration is 1s22s22p6
Problem 27. Answer: D
Solution: The electron configuration for N is 1s22s22p3. The 1s and 2s levels will have be full, each consisting of 2 electrons with opposite spins (which rules out diagram A). The 2p level will have 3 electrons, one in each orbital (which rules out diagram E) and C). The best diagram will show all three electrons in the 2p orbitals with the same spin (i.e. either all facing up or down). Therefore, diagram D) is the best representation of an N atom in its ground state.
Problem 28. Answer: C
Solution: Sodium, Na, should be 1s2 2s2 2p6 3s1
Problem 29. Answer:
a) [Ar]3d1. The electronic configuration of Ti is [Ar]4s23d2. Electrons are removed from the s subshell first when transition metals, such as Ti, changes to an ionic state. Thus, the electron configuration of Ti3+ is [Ar]3d1.
b) n = 3, l = 2, ml = -2,-1,0,1,2, ms = + ½, ½
c)
Radial probability
r
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Problem 30. Answer: D
The electron configuration for As is 1s22s22p63s23p64s23d104p3
The valence electrons for the element As all reside in the 4th level so n = 4. The valence electrons only exist in s or p sublevels so l = 0 or 1. In the s sublevel there is only one orbital while in the p sublevel, there are 3 orbitals, therefore the value of ml = 0 for the s sublevel and ml = -1, 0, +1 for the p sublevel. ms = +1/2 or –1/2 since the spin of the electron in any orbital can only be up or down. The only answer that falls in range of all the possible choices is D.
Problem 31. Answer: C
Solution: 𝑙 = 0, 1, … (𝑛 − 1), therefore if n = 3, then 𝑙 cannot equal 3.
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Problem 32. Answer: E
a) n is the principal quantum number and can be any whole integer number. l is the secondary quantum number = 0 to n-1. ml is the third quantum number = -l to +l. Set II) is invalid because ml cannot equal -2 if l is 1. Set III) is invalid because ml cannot equal 3 if l is 2. Set IV) is invalid because l cannot equal 1 if n is 1.
Problem 33. Answer: D
Solution: These quantum numbers correspond to 4p
The atom could be Br.
Problem 34.
K [Ar]4s1
V3+ [Ar]3d2
Mo [Kr]5s14d5
Ru [Kr]5s24d6
Y3+ [Kr]
b) Sn = Tin
Problem 35.
[Ne]3s2 Magnesium (Mg)
[Ne]3s23p1 Aluminum (Al)
[Ar]4s13d5 Chromium (Cr)
[Kr]5s24d105p4 Tellurium (Te)
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Problem 36. Answers
a) 7, 𝑚𝑙 = −𝑙 … 0 … + 𝑙 where 𝑙 = 3 as we are dealing with the f shell and 𝑚𝑙 =−3, −2, −1,0,1,2,3
b) 6, [Ar] 4s 3d
c) 50, 𝑙 = 0,1,2, to a maximum of 𝑛 − 1 i.e. 4. Therefore
5𝑠 = 2𝑒− (𝑙 = 0), 5𝑝 = 6𝑒− (𝑙 = 1), 5𝑑 = 10𝑒− (𝑙 = 2), 5𝑓 = 14𝑒− (𝑙 = 3),
5𝑔 = 18𝑒− (𝑙 = 4): 𝑡𝑜𝑡𝑎𝑙 = 50.
Problem 37. Answer
a) Many solutions. Some examples are: anion = S2- and cation K+
K+ is [Ar] and S2- is [Ar]
b) Fe=26, for Fe: [Ar] 4s23d6, for Fe2+: [Ar] 3d6
c) It is the ground state for copper.
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Problem 39. Answer: B
Solution:
C has 2 unpaired electrons
N has 3 unpaired electrons
O has 2 unpaired electrons
F has 1 unpaired electrons
Ne has 0 unpaired electrons
Therefore, N will more strongly pulled into an in-homogeneous magnetic field.
Problem 40.
Atomic Number Configuration Species State Paramagnetic Or Diamagnetic
1 1s2 H Ground Diamagnetic
7 1s22s22p3 N Ground Paramagnetic
17 1s22s22p63s23p6 Cl Ground Diamagnetic
11 1s22s22p63p1 Na Excited Paramagnetic
22 1s22s22p63s23p64s2 Ti2+ Excited Diamagnetic
13 1s22s22p6 Al3+ Ground Diamagnetic
Note: Ti2+ is in an excited state because it would normally lose its 4s electrons first before its 3d electrons. The given configuration is correct for the neutral atom but NOT the cation.
Problem 41.
Species Si Cr V3- P Zn2+
Number of unpaired electrons 2 4 4 3 2
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+ +
+
+
Problem 42.
2 2i) 3p ii) 3d iii) 3s iv) 3dy xyx y
Answer:
i) ii)
iii) iv)
Problem 43.
Orbital Number of nodal planes
Number of additional
degenerate orbitals
2 23dx y
2 4
2py 1 2
y
x x
x
y y
x
y
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Z
X
Problem 44.
Cross – Section
Name of Orbital
3dz2 3dxz 2pz orbital
Value of n 3 3 Likely 2 (nodes will show
up if 3 or higher)
Value of l 2 2 1
Total Number of Nodal Planes
& Surfaces 2 nodal surfaces 2 nodal surfaces
2 nodal surfaces (one plane, one sphere)
Problem 45.
Orbital l All values of ml Atom
1s 0 0 H
2p 1 -1, 0, 1 B or F
3d 2 -2, -1, 0, 1, 2 Sc
3f X X X
4p 1 -1, 0, 1 Ga or Br
6s 0 0 Cs
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Electromagnetic Radiation Practice Problems
Problem 46. Answer: D
Solution: Statements (a), (b), and (c) are all true.
Problem 47. What is the energy of one mole of photons with a wavelength of 285 nm?
Answer: A
Solution:
34 8
9
19(6.626 10 )(3.0 10 / )
(285 10 ) 6.98 10
hc Js m
mJ
sE
This is the energy per photon; we want the molar energy
6.98×10−19𝐽
𝑝ℎ𝑜𝑡𝑜𝑛×
6.022×1023𝑝ℎ𝑜𝑡𝑜𝑛𝑠
𝑚𝑜𝑙= 4.20×105 J
Problem 48. Answer: D
Solution: Energy per photon = h = hc/
-34 -1 8 -1-19
-9
(6.626 10 J s )(2.998 10 m s )7.73 10 J
257 10 m
Problem 49. a) Answer: (iii)
Solution: There are six possible transitions-
b) Answer: (iii)
Solution: Going from n=4 to n=1 will have the largest E.
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Problem 50. Answer: B
Solution: 1Watt = 1J/sec,
therefore, Energy for all photons = 72 / sec 5sec 360J J
Energy for one photon = 19
20
3603.63 10
9.91 10
JJ
photons
hcE
, therefore
34 8
19
(6.626 10 )(3 10 )547
3.63 10
hcnm
E
Problem 51. Answer: B
Solution: E = h
For one atom of helium:
1st Ionization energy = 23
2370 1
6.022 10
kJ mol
mol atoms
= 3.94 10-21kJ
3
1518
4
3.94 105.95
6.6210
6 10
E J
h Js
Problem 52.
a) Answer:
𝐸 =ℎ𝑐
𝜆= ℎ𝜈,
𝑐
𝜆= 𝜈
𝑐
𝜈= 𝜆 =
3.00108 ms1
6.911014 s1= 4.342×107 m = 434.2 nm
b) Answer:
E = hν = 6.262×1034 Js×6.91×1014 s1 =4.579×1019 J
photon
4.579×1019 J
photon×
6.022×1023𝑝ℎ𝑜𝑡𝑜𝑛𝑠
𝑚𝑜𝑙=
275.6 kJ
mol
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Molecular Orbital Theory Solutions
Problem 53.
(a) Answer: (iii)
Solution: Bond Order =Bonding electrons – Antibonding electrons
2
O2− has 13 valence electrons 2s
2 s∗22p
2 2py2 2pz
2 2py∗2 2pz
∗1
Bond order =(8 − 5)
2= 3×½ = 1.5
(b) Answer: O2− is paramagnetic (meaning there are unpaired electrons); thus it is attracted towards a
magnetic field.
Problem 54. Answer: A
Solution: The larger the bond order, the stronger the bond.
B. O. for Li2 =4 − 2
2 = 1
B. O. for Be2 =6 − 2
2= 0
B. O. for H2− =
2 − 1
2= 0.5
B. O. for He2+ =
2 − 1
2= 0.5
B. O. for He2 =2 − 2
2= 0
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Problem 55. For the homo-diatomic B2 provide the following:
a) A full and labelled molecular orbital diagram including bond order
b) A representation of the highest energy orbital that is occupied by at least one electron
c) Determine whether B2− would have a longer or shorter bond than B2. Explain your answer.
Answers:
a) Bond order =2−0
2= 1
b) It is an p2 orbital
c) B
2 has one more bonding electron, so the bond order of B
2 would be greater than for B2.
Therefore, B
2 will have a shorter bond than B2.
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Problem 56. Complete the following table:
Ion Paramagnetic
or Diamagnetic
Bond
Order
Bond Length:
Longer or
Shorter than
that of O2
A neutral diatomic molecule with the same MO electron configuration using any combination of the atoms C, N, O, F
O2+ Para 2½ S NO or CF
O22 + Dia 3 S N2, CO
O2 Para 1½ L OF
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Problem 57.
a) Answer: 2
2C
MO configuration: 1s2 1s
∗22s2 2s
∗22pz2 2py
2 2pz2
B. O. = 3
diamagnetic
b) Answer: 2
2O
MO configuration: 1s2*1s22s2*2s22pz22px22py2*2px2*2py2
B. O. = 1
diamagnetic
Problem 58. Benzene (C6H6)
Answer: it is in resonance
Problem 59. Answer: D
Solution: Levels “b” and “a” are anti-bonding orbitals and levels “d” and “c” are bonding orbitals.
Bond order = 2
MOs gantibondinin e of no. MOs bondingin e of no. , therefore adding an electron to
level “b”, an anti-bonding orbital, decreases the bond order.
Level “c” is a 2p orbital, “d” are orbitals therefore D is true.
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Problem 60. The
Lewis Structure
VSEPR Valence Bond Theory
Molecular Orbital Theory
Bohr Theory
Which model can account for the bond order of ½ for H2
+?
X
Which model can predict the detailed shapes of molecules using simple electrostatic arguments?
X
Which model can rationalize the fact that liquid oxygen “sticks” to the poles of a strong magnet?
X
Which model invokes the use of hybrid orbitals?
X
Which model can most easily calculate the ionization energy of U91
+ ?
X
Which model involves the concept of anti-bonding orbitals?
X
Which model uses “resonance” to describe bonds that are not localized between two atoms?
X X