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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.

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

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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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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