© Pavel Sedach Chem201 Final Booklet Chemical Bonding ...learnfaster.ca/wp-content/uploads/chem201/final/Chem-201-Final-Booklet-Solutions...i) N(1) is sp3 hybridized; tetrahedral

© Pavel Sedach Chem201 Final Booklet

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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.



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© Prep101 Chem201 Final Booklet

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VSEPR Solutions


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.


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

©Prep101 Chem201 Final Booklet

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Practice Problems


Solution:

Non-zero zero zero with zero with non-zero

resonance resonance


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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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.


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)


(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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A) B) C)

H

O

O

O

NH2

O


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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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.


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


(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


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


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


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.


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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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.


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.


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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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


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


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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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


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


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

Documents

© Pavel Sedach Chem201 Final Booklet Chemical Bonding ...learnfaster.ca/wp-content/uploads/chem201/final/Chem-201-Final-Booklet-Solutions...i) N(1) is sp3 hybridized; tetrahedral