Unformatted text preview: ©Prep101 Chem110 Exam Solutions Chapter 8 – Electrons in Atoms
Q 1.1
Solution: E= λ= hc
λ → (6.626 x 10 λ= hc
E )( Js 2.998 x 10 8 ms −1
3.90 x 10 19 J
34 → λ = 5.09 x 10 5 cm ) → λ = 5.09 x 10 7 m λ = 509 nm Therefore, GREEN light will be emitted Q 1.2
Solution: We know the energy (499 kJ/mol). By using E = hν, we can solve for the
frequency (ν). The frequency is related to wavelength (λ) by the equation ν = λ/c ( ) E = 499 kJ mol = ( ) = 8.286 × 10 499 × 10 3 J mol
6.022 × 10 23 ( mol −1 ) −19 (J ) ()
(
)
( )
m
c 2.998 × 10 ( s)
c
m
= 2.397 × 10 ( )= 239.7 (nm)
ν= ⇒λ= =
ν 1.251 × 10 s
λ
8.203 × 10 J
E
= 1.251 × 10 15 s−1 or Hz
=
h 6.626 × 10−34 J ⋅ s
−19 E = hν ⇒ ν = 8 15 Q 1.3
Solution: E = hν → ( ) −7 −1 E = (6.626 x 1034 Js)(4.35 x 1014 s1) = 2.88 x 1019 J 1 ©Prep101 Q 1.4
Solution: Chem110 Exam Solutions First calculate the energy of a photon:
E = hν → E = 6.626 x 1034 Js (1.50 x 1014 s1) = 9.94 x 1020 J Now, let “a” be the number of photons (factor) to contribute 30.1 J since we know
that 1 photon emits 9.94 x 1020 J of energy
E(a) = 30.1 J → 9.94 x 1020 J (a) = 30.1 J a = 3.03 x 1020 photons Q 1.5
Solution: Divide by Avagadro’s number to give the energy per photon
2.58 J ⋅ mol1
= 4.28 4 x10 −24 J
−1
23
6.022 x 10 mol
Wavelength can now be determined by dividing the product of the speed of light
and Plank’s constant by the energy per photon.
E= hc
λ λ = 0.0464 m Q 1.6
Solution: → λ= hc
E → λ= (6.626x10 34 Js)(2.998 x10 8 ms 1 )
4.28x10 24 J or alternatively 4.64 x 102 m First convert the wavelength from μm → m
3.34 μm x (1 m/1E6 μm) = 3.34 x 106 m Now energy can be calculated by the standard equation: E = hc
λ E = (6.626x1034 Js)(2.998x108 ms1)/(3.34x106 m) = 5.95 x 1020 J 2 ©Prep101
Q 1.7
Solution: Chem110 Exam Solutions Amplitude is the maximum height of the wave above the centre line or the
maximum depth below.
Wavelength, λ – is the distance of one cycle of the wave. It is the distance
between the tops of two successive crests (or the bottoms of two troughs). http://www.users.mis.net/~pthrush/lighting/wave.jpg
Q 1.8
Solution: Correct answer: D
B has a higher frequency because the distance between the peaks of the waves are
smaller than A.
B also has a higher energy because Energy is inversely proportional to
wavelength. Since A has a longer wavelength, it will be of lower energy. Q 1.9
Solution: Correct answer: A
Speed of light = 3.0 x 108 m/sec.
Speed of electrons = (3.0 x 108 m/sec)0.005 = 1.5 x 106 m/sec
λ = h/mv = (6.626 x 1034Js)/[( 1.5 x 106 m/sec)(9.11 x 1031kg)
= 4.8 x 1010m = 0.48 nm 3 ©Prep101 Chem110 Exam Solutions Q 1.10
Solution: 1 1 ∆E = R H 2 − 2 n n f i RH = 2.179×10−18 J.
** your prof indicates that hcRH = 2.17 E18 J 1
1 =  4.58 × 1019 J
∆E = (2.179×10−18 J) − (2) 2 (5) 2 E= λ= hc λ , (EMISSION!) therefore hc (6.626 × 10 −34 Js)(3.0 × 108 m / s )
= 4.34 E7 m
=
E
4.58 × 10 −19 J 4.34 E7 m (1 E9 nm / 1 m) = 434 nm Q 1.11
Solution: 325 nm = 325 E9 m
Ephoton = hc/λ = (6.626 E34 Js)(2.998 E8 m)/(325 E9) = 6.11 E19 J/photon
6.11 E19 J/photon (6.022 E23 photon/mol) = 3.68 E5 J/mol = 368 kJ/mol Q 1.12
Solution: E = 400 E3 J/mol (1 mol / 6.022 E23 bonds) = 6.64 E19 J/bond
λ = hc/E → (6.626 E34 Js)(2.998 E8 m/s) / (6.63 E19 J/bond) → 3.00 E7 m 3.0 E7 m (109 nm / 1 m) = 300.0 nm
The visible region is considered in between 400800 nm. Therefore, 300 nm is
within the ultraviolet (UV) region.
4 ©Prep101
Q 1.13
Solution: Chem110 Exam Solutions First calculate the energy ∆E from the transition from nf = 5 and ni = 2. 1
1 ∆E = 2.179 ×10 −18 J 2 − 2 → n i nf ∆E = 4.086 E19 J 1 1
∆E = 2.179 ×10 −18 J 2 − 2 4 2 (EMISSION) We can now calculate the frequency by the equation, E = hν
ν = E/h = 4.086 E19 J / (6.626 E34 Js) = 6.166 E14 s1
Now calculate the wavelength from the frequency, λ = c/ν
λ = 2.998 E8 ms1 / (6.166 E14 s1) = 4.86 E7 m
Q 1.14
Solution:
1 1
∆E = 2.179 ×10 −18 J 2 − 2 5 2 → ∆E = 4.576 E20 J We can now calculate the frequency by the equation, E = hν
ν = 4.576 E20 J / (6.626 E34 Js) = 6.906 E14 s1
Now calculate the wavelength from the frequency, λ = c/ν
λ = 2.998 E8 ms1 / (6.906 E14 s1) = 4.34 E7 m
4.23 E7 m → 434 nm 5 ©Prep101
Q 1.15
Solution: Chem110 Exam Solutions ni = 7, nf = 4 (Brackett series) 1
1 ∆E = 2.179 × 10 −18 J 2 − 2 7
4 → ∆E = 9.176 E20 J We can now calculate the frequency by the equation, E = hν
ν = 9.176 E20 J / (6.626 E34 Js) = 1.3848 E14 s1
Now calculate the wavelength from the frequency, λ = c/ν
λ = 2.998 E8 ms1 / (1.3848 E14 s1) = 2.16 E6 m
Radiation is emitted in the infrared region (because the frequency is 1014) Q 1.16
Answer Correct Answer: C
Paschen refers to nf = 3. We are also told the wavelength is from the emission of
an electron so ni > 3 (for this question). Eliminate answers A, B, E
1094.1 nm 1.094 E6 m
We can now calculate the frequency by the from the wavelength, λ = c/ν
ν = (2.998 E8 m·s1) / (1.094 E6 m) = 2.74 E14 s1
Energy can be calculated from the frequency, ∆E = hν,
∆E = 6.626E34 Js (2.74 E14 s1) = 1.8158 E19 J
** Energy is negative because energy was emitted 1 1
− 1.82 ×10 −19 J = 2.179 ×10 −18 J 2 − 2 x 3 → − 0.0833 + 1 1
=
9 x2 → → 0.02777 = 1 1
− 0.0833 = 2 − 2 x 3 1
→
x2 x2 = 36
x = ni = 6 6 ©Prep101
Q 1.17
Solution: Chem110 Exam Solutions D
Red has the longest wavelength. Wavelength is inversely proportional to
frequency. Therefore, red has a small frequency. Blue has short wavelength, so
frequency is larger than red. Q 1.18
Solution: B
A is false (see Bohr equation for orbit radius). C is false – emission occurs when
∆E is negative. Q 1.19
Solution:
Calculate the KEe from the excess light: (emitted – threshold)
νo = 5.57 E14 s1, ν = (3.0 E8 m·s1)/(5.1 E7 m) = 5.88 E14 s1
KEe= h(ννo) → KEe = 6.626 E34 Js ((5.88 E14 s1) – (5.57 E14 s1)) KEe = 6.626 E34 Js (3.12 E13 s1)
KEe = ½ mu2 → u= → 2 KEe
me KEe = 2.07 E20 J
→ u= 2 (2.07 E − 20 J )
(9.109 E − 31 kg ) u = 2.1 E5 m·s1 Q 1.20
Solution: Correct Answer  E
Calculate the KEe from the excess light (emitted – threshold)
KEe= h(ννo) → KEe = 6.626 E34 Js ((1.0 E15 s1) – (6.7 E14 s1)) KEe = 2.187 E19 J (or alternatively 2.187 E22 kJ) KEe = ½ mu2 → u= 2 KEe
me → u= 2 (2.2 E − 19 J )
(9.109 E − 31 kg ) u = 6.9 E5 m·s1 7 ©Prep101
Q 1.21
Solution: Chem110 Exam Solutions Correct answer C
KEe= h(ννo) where KEe = 1.54 E19 J
and
ν = c/λ = 3.0 E8 m·s1 / (4 E7 m) = 7.5 E14 s1 ( ) → → u= → E = 3.09 eV 1.54 ×10 −19 J = 6.626 ×10 −34 Js 7.5 ×1014 s −ν o → KEe = ½ mu2 u= 2 KEe
me νo = 5.18 E14 s1 2 (1.54 E − 19 J )
(9.109 E − 31 kg ) u = 5.81 E5 m·s1 Q* 1.22
Solution: B
Convert wavelength to energy (J → eV)
E = hc/λ → E = 4.97 E19 J KEe = h(EΦ) We know that electrons are emitted if the energy exceeds
the threshold frequency (or work function, 3.09 eV). So in
this case, electrons are emitted for Na and Cs and not Mg.
(Eliminate C, D). If the incident wavelength was increased,
the energy would be smaller and electrons would still not
be ejected. (Eliminate A) KEe = ½ mu2 KEe is proportional to velocity. KE for Cs > Ke for Na, so
velocity of Cs particle is greater. 8 ©Prep101
Q 1.23
Solution: Chem110 Exam Solutions A is correct
rn = n2ao where ao = 52.9 pm r5 = (5)2 (52.9 pm) = 1322.5 pm
This is the radius! The diameter is 2r. Therefore the diameter is 2645 pm
En = 2.179 E18 J / (52) = 8.716 E20 J Q 1.24
Solution: rn = (n)2(0.529 Å) → 4.00/0.529 = n2 → n = 2.75 Since n ≠ integer; then therefore such an orbit does not exist.
Q* 1.25
Solution: A
rn (n)2 (5.29 E11 m) / (Z) where Z = 2 protons for He+ rn = (2)2(5.29 E11 m) / (2) → Use de Broglie equation λ = h/mu rn = 1.058 E10 m = λ u = (6.626 E34 J·s) / ( 9.109 E31 kg)(1.058 E10 m) = 6.88 E6 m/s Q 1.26
Solution: B Q 1.27
Solution: E
The Bohr model applies to all hydrogenlike atoms (1 electron systems). 9 ©Prep101
Q 1.28
Solution: Chem110 Exam Solutions Answer E
Recall that rn = n2(ao)/Z where Z = # protons So as ↑Z, then radius decreases; eliminate D
Examine E → For n = 3, Z = 1, n2/Z = 9/1 = 9(ao)
Examine A → For n = 4, Z = 2, n2/Z = 16/2 = 8(ao)
E>A Q* 1.29
Solution: A
Orbitals of same ‘n’ value can overlap in space (See Fig 834 in text). The Ψ2 is
the finding the electron in space. Q* 1.30
Solution: B
The frequency is different for a hydrogen atom than a hydrogenlike atom, by the
equation ∆E = Z2hcRH((1/n12)(1/nf2)) and Z = # protons. The photon involved in
the H transition lies in the visible spectrum (recall the emission lines). For Be3+ (Z
= 4), so the photon emitted by hydrogen is 16 times the wavelength absorbed by
Be3+ because Z2 = 16 and energy is inversely proportional to wavelength. When nf
= 2, this is a Balmer emission series. Q 1.31
Solution: Each electron in an atom has a unique set of quantum numbers. When one
electron in an atom is described by a particular set of quantum numbers, no other
electron in the atom is described by the same set.
• No two electrons can have the same set of four quantum numbers Q 1.32
Solution:
n
2
2
2
2
2
2 ℓ
1
1
1
1
1
1 mℓ
1
1
0
0
1
1 ms
+½
½
+½
½
+½
½
10 ©Prep101 Chem110 Exam Solutions Q 1.33
Solution:
Set of quantum #’s
(n, l, ml, ms)
1, 0, 0, ½
3, 1, 2, 0
3, 2, 1, ½
4, 4, 3, ½
5, 0, 0, ½ Q 1.34
Solution: Q 1.35
Solution: Allowed?
(yes or no)
Y
N
N
N
Y Designation
1s
5s An electron can either be spin up or spin down, designated by either ms = + ½ or –
½ respectively Electron #1 n = 3, ℓ = 0, mℓ = 0, ms = +½ Electron #1 n = 3, ℓ = 0, mℓ = 0, ms =  ½ Q 1.36
Solution: Q 1.37
Solution: Those not possible are: (5, 1, 2, ½) & (4, 4, 0, ½)
The remaining three, of increasing order, are: (1, 0, 0, ½), (4, 3, 0, ½),
(5, 1, 0, ½)
11 ©Prep101
Q 1.38
Solution: Chem110 Exam Solutions Degenerate refers to energy equivalent atomic orbitals (equal in energy).
The 2px, 2py, 2pz are the 3 atomic orbitals of the 2p level. These are all equivalent
in energy and thus termed degenerate. Q 1.39
Solution: A) [He]2s22p5
B) [Ar]4s1
C) [Xe]6s24f6
D) [Rn]7s26d1
E) [Xe]6s24f145d6
F) [Kr]5s14d8 Important exception G) [Ar]4s13d10 Important exception H) [Xe]6s24f145d1
I) [Ar]4s13d5
J) [Ar]4s23d5
K) [Kr]5s24d105p2
L) [Rn]7s25f146d1 Q 1.40
Solution: A) Indium
B) Argon
C) Tellurium (Te) 12 ©Prep101
Q 1.41
Solution: Chem110 Exam Solutions 2 unpaired electrons F = [He]2s22p5 1 unpaired electron Mo = [Kr]5s14d5 6 unpaired electrons (1s+5d) Co = [Ar]4s23d7 Q 1.42
Solution: Si = [Ne]3s23p2 3 unpaired electrons E
ℓ cannot be negative. Series AD are all acceptable quantum numbers Q* 1.43
Solution: D
Mn → 4s2d5 (5 unpaired electrons)
P → 3s23p3 (3 unpaired electrons)
Rb → 5s1 (1 unpaired electron)
Mo → 5s14d5 (6 unpaired electrons)
Cu → 4s13d10 (1 unpaired electron) Q* 1.44
Solution: E
Look up Os (Z = 76) → d6 → ℓ must equal 2; eliminate A, B, C
We have 6 electrons that have to be situated in the dorbital. The first five are
placed singly in each of the 5 degenerate dorbitals (by Hund’s Rule). The sixth
(and last electron) is of opposite spin and starts at the ℓ orbital. Q 1.45
Solution: Count the number of electrons
[Xe]f145d106s1 = 54 + 14 + 10 + 1 = 79
Au has Z = 79 (neutral state) 13 ©Prep101
Q 1.46
Solution: Chem110 Exam Solutions E
Te → [Kr]5s24d105p4 → 2 + 10 + 4 = 16 Q 1.47
Solution: D
Principle quantum number = n
If mℓ = 7, then ℓ ≥ 7; therefore minimum ‘n’ must be 8. 14 ©Prep101 Chem110 Exam Solutions Chapter 9 – The Periodic Table and Some Atomic Properties
Q 2.1
Solution: V5+ < Ti4+ < Sr2+ < BrThe electron configurations are...
# of protons
Sr+2: [Kr]
38
Br : [Kr]
35
V+5: [Ar]
23
Ti+4: [Ar]
22
Sr+2 and Br_ have the same number of electrons; however, Zeff is greater for
Sr+2 due to a greater # of protons, resulting in a greater + charge that pulls the
electrons closer to the nucleus. Thus, Sr+2 is smaller than Br.
V+5 and Ti+4 have the same number of electrons; however, Zeff is greater for V+5
due to a greater # of protons. Thus V+5 is smaller than Ti+4.
Sr+2 & Br_ are larger than V+5 & Ti+4 because there are more electrons held in a
larger subshell. Q 2.2
Solution: Cl > Br > I
Electron affinity decreases from top to bottom.
EA (Cl) = 349 kJ/mol
EA (Br) = 324 kJ/mol
EA (I) = 295 kJ/mol
Chlorine is most likely to gain an electron. Q 2.3
Solution: Electron affinity becomes more negative from left to right because Cl has higher
Zeff than S.
EA(Cl) = 349 kJ/mol, EA (S) = 200.4 kJ/mol 15 ©Prep101
Q 2.4
Solution: Q 2.5
Solution: Q 2.6
Solutions: Chem110 Exam Solutions For K, valence electron is in 4s orbital while valence electron in Li is in 2s orbital.
4s orbital is larger than 2s. Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron.
1
Ionization energy ∝ 2
n A) Ca2+ < K+ < Cl− < S2−
B) B < Be < O < N Q 2.7
Solution: Correct Answer: Mg
Ionization energy increase as you move up a group and it also increases as you
move right across a period. Q 2.8
Solution: Correct answer: Mg
Radius increase as you move left along a period and as you move down a group. 16 ©Prep101
Q 2.9
Solutions: Chem110 Exam Solutions A) [Ar] or [Ne]3s23p6
B) S2
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).
C) Ca+2
Ca+2 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.
D) S2 < Ar < Ca+2
Ionization energy is the energy required to remove an electron. It is most difficult
to remove an electron from Ca+2, as this will involve the removal of a core
electron instead of a valence electron, and Ca+2 has the greatest Zeff. It is easier to
remove an electron from S2 versus Ar because S2 has a smaller Zeff. Q 2.10
Solutions: A) IE1(K) < IE1 (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 (K) > IE2
(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) → K2+ (g) + e−
Ca+(g) → Ca2+(g) + 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. 17 ©Prep101 Chem110 Exam Solutions Q 2.11
1s22s22p6 = Ne
1s22s22p63s1 = Na
1s22s22p63s2 = Mg Solution: Ne has the largest first ionization energy as an electron is removed from the n = 2
shell, compared to n = 3 shell for Na and Mg. An electron from Na or Mg requires
less energy to remove because the n =3 shell is further from the nucleus, and is
better shielded by core electrons in the n = 2 shell.
Mg has a smaller second ionization energy because the second electron to be
removed is a valence electron (n = 3), whereas the second electron removed from
Na is taken from a core electron (n = 2). Q 2.12
Solution: A
All isoelectronic, so therefore greater number of electrons gives more pull
towards nucleus (smaller radius). Anion > neutral > cation Q 2.13
Solution:
Se < As < Ge < Sn < Pb
Left to right:
Top to bottom: Q 2.14
Solution: Ge > As > Se
Pb > Sn > Ge Correction Answer:
We know that IE is the ability (energy required) to remove an electron. Therefore,
IE will increase from L → R (as the atom becomes more electronegative). It will
be easier to remove an electron from a bigger atom than a smaller atom, so IE will
decrease from top to bottom.
L→R
T→B Al < Si < S
Al > Ga > In Therefore (smallest) In < Ga < Al < Si < S (largest)
18 ©Prep101
Q 2.15
Solution: Chem110 Exam Solutions Answer – C
Paramagnetic – has unpaired electrons
Ne is noble gas – Eliminate E
F → 1s22s22p6
Ca2+ and S2 → [Ne]3s23p6
Fe2+ → [Ar]3d6 (all paired)
(all paired)
(contains unpaired electrons) 19 ©Prep101 Chem110 Exam Solutions Chapter 10 – Chemical Bonding I (Basic Concepts)
Q 10.1
Solution:
SO3 = 6 + 18 = 24 electrons S S S
O O O O O O O O O 1
× 2 = −1
2
1
FC O in S = O = 6 − 4 − × 4 = 0
2
1
FC S = 6 − 0 − × 8 = +2
2 ( ) ( ) FC O in S  O = 6 − 6 − () 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 CO = 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.
20 ©Prep101 Chem110 Exam Solutions 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. 21 ©Prep101
Q 10.8
Solution: Chem110 Exam Solutions 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 1
OH O
O 22 ©Prep101 Chem110 Exam Solutions Q 10.9
Solution:
(a) SCN 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. 23 ©Prep101
Q 10.11
Solution: Chem110 Exam Solutions 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 Shape
(as given by
nuclei) Hybridizatio
n
on
central atom Linear sp Square
Planar sp3d2 N Trigonal
planar sp2 N Trigonal
Pyramidal sp3 P Octahedral Lewis
electron dot
structure sp3d2 N Polar (P) or
Nonpolar (N)
species
N CS2 S S C F F XeF4 Xe
F F O SO3 S O O SO32− O O S
O F SF6 F F
S
F F
F 24 ©Prep101 Chem110 Exam Solutions Q 10.14
Draw its Lewis Dot structure.
Solution:
H
C N H H What is the hybridization at the carbon atom?
Solution: sp2 What is the hybridization at the nitrogen atom?
Solution: sp2 How many σ bonds does H2CNH have?
Solution: Answer: 4
How many π bonds does H2CNH have? Solution: Answer: 1 Is H2CNH planar?
Solution: Yes Is H2CNH polar?
Solution: Yes 25 ©Prep101 Chem110 Exam Solutions Q 10.15 Solution:
Molecule Lewis
Structure Molecular
Shape
(VSEPR) Polar (P) or
Nonpolar
(N) Trigonal
planar O H2CO Hybridization
of Central
Atom
sp2 Polar C
H H sp3
OF2 Bent O
F Polar F HCN
N sp Polar sp3 Nonpolar Square pyramidal C linear tetrahedral H sp3d2 Polar Seesaw sp3d Polar H SiH4 Si
H H
H IF5 F F I F F
F SeF4 F
F
Se
F
F 26 ©Prep101 Chem110 Exam Solutions Q 10.16
Solutions: i) ↔ N
F O O O O ↔ N
F O N
O F ii) Trigonal planar
iii) 3
iv) sp2
v) X = C and Z = B 27 ©Prep101 Chem110 Exam Solutions 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 Tshaped sp3d P BrF2+ AX2E2 Angular sp3 P SeF4 AX4E Seesaw sp3d P BrF5 AX5E Squarepyramidal sp3d2 P 28 ©Prep101 Chem110 Exam Solutions Q 10.18
What are the approximate bond angles about each of the two carbon
atoms, and what are the hybridizations of the orbitals on each of
them?
Solution: C1 is sp3 with 109.5° (tetrahedral)
C2 is sp2 with 120° (trigonal planar)
What are the hybridizations of the orbitals on the two oxygen
molecules and the nitrogen atom? What is the bond angle of
nitrogen? Solution: OA is sp2 hybridized
OB is sp3 hybridized
Nitrogen is sp3 hybridized with a bond angle of 109.5°
What is the total number of σ bonds and π bonds in the molecule? Solution: Q 10.19
Solution: There are 9 σ bonds (single bonds)
There is 1 π bond (double bond)
Correct answer – C
IF5 = 42 electrons. → 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 29 ©Prep101
Q 10.20
Solution: Chem110 Exam Solutions 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. Q 10.21
Solution: 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
nonmetal 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. 30 ©Prep101
Q 10.26
Solution: Chem110 Exam Solutions Correct answer – C
The oxygen is AB2E2 which is tetrahedral family and sp3 Q 10.27
Solution: 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
O
O N O As a result, NO bond length is identical in all NO 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. 31 ©Prep101
Q*10.30
Solution: Chem110 Exam Solutions 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 nonpolar F F polar In the polar molecule, there is at least one FPF 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 32 ©Prep101 Chem110 Exam Solutions Chapter 11 –Molecular Orbital Theory
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: F–F 33 ©Prep101 Chem110 Exam Solutions Q 11.2 Solution: For N2 (total 10 electrons)
Bond Order (N2) = (8 – 2)/2 = 3
Structure: N≡N For B2 (total 14 electrons)
Bond Order (F2) = (4 – 2)/2 = 1
Structure: Q 11.3
Solution: B–B 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: σ2s (2) < σ*2s (2) < π2p (4) < σ2p (2) < π*2p < σ*2p
CN+ = 8 electrons; BO = 2
CN = 9 electrons; BO = 2½
CN = 10 electrons; BO = 3
Shortest bond order = longest bond 34 ©Prep101 Chem110 Exam Solutions Q 11.5
Solution: BO = 88/2 = 0 Therefore, Ne2 does not exist 35 ©Prep101 Chem110 Exam Solutions Q 11.6
Solution: Bond order for CO is larger than bond order for NO, therefore CO bond is
shorter. Q 11.7
Solution: 36 ©Prep101 Chem110 Exam Solutions 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. Q 11.9
Solution: Correct answer – C
Bond order = (82)/2 = 6/2 = 3
No unpaired electrons; therefore diamagnetic. 37 ©Prep101
Q 11.10
Solution: Chem110 Exam Solutions Correct answer – E
Bond Order = (86)/2 = 2/2 = 1
No unpaired electrons, therefore diamagnetic 38 ©Prep101
Q 11.11
Solution: Chem110 Exam Solutions Correct answer – B
O2+ has the same molecular orbital diagram as O2, except one electron is
removed! The π*2p therefore has 1 valence electron. 39 ©Prep101
Q 11.12
Solution: Chem110 Exam Solutions Correct answer – D
B2 has 6 valence electrons, so B2 has 7
σ2s (2) < σ*2s (2) < π2p (3) Q 11.13
Solution: Correct answer – D
N2 has 10 valence electrons, so N2+ will have 9 valence electrons
N2 σ2s (2) < σ*2s (2) < π2p (4) < σ2p (2) BO = (82)/2 = 3 N2+ σ2s (2) < σ*2s (2) < π2p (4) < σ2p (1) BO = (72)/2 = 2.5 As a result, bond order will decrease, and the unpaired electron makes it
paramagnetic. 40 ©Prep101 Chem110 Exam Solutions Q 11.14
Solution: Correct answer – C
B2 BO = (42)/2 = 1 C2 σ2s (2) < σ*2s (2) < π2p (4) BO = (62)/2 = 2 N2 σ2s (2) < σ*2s (2) < π2p (4) < σ2p (2) BO = (82)/2 = 3 O2 σ2s (2) < σ*2s (2) < σ2p (2)< π2p (4) < π*2p(2) BO = (84)/2 = 2 F2 Q 11.15
Solution: σ2s (2) < σ*2s (2) < π2p (2) σ2s (2) < σ*2s (2) < σ2p (2)< π2p (4) < π*2p(4) BO = (86)/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: σ2s < σ*2s < π2p < σ2p
Shortest bond order = longest bond 41 ©Prep101 Chem110 Exam Solutions Fenster – Extra Problems
QI
Solution: Answer – D
Corundum is a crystalline form of aluminum oxide (Al2O3) with traces
amount of iron, titanium and chromium. Q II
Solution: Answer – E Q III
Solution: E
There are over 500 chemicals found in natural apples Q IV
Solution: Answer –D
N2O is dinitrogen oxide, and is referred to as laughing gas. QV
Solution: Answer: A Q VI
Solution: Answer – B 42 ©Prep101 Chem110 Exam Solutions Chapter 24 – Coordination Chemistry
Q 24.1
Solution: Correct Answer: A
H2O is neutral ligand, Cl is each 1, Cr = +3, so overall is 1 Q 24.2
Solution: Correct Answer: D
H2O is neutral ligand, Co = +2, so overall complex is +2. Q 24.3
Solution: Correct answer: A
NH3is neutral, Br = 1, Pt = +2, therefore overall = zero Q 24.4
Solution: Correct answer: A
Coordination number = # ligands = 4
SO4 = 2, ammonia = 0, therefore Cu = +2 Q 24.5
Solution: Correct Answer: E
Coordination number = # ligands = 2
CN = 1, overall complex = 1, therefore, Ag = +1 43 ©Prep101
Q 24.6
Solution: Chem110 Exam Solutions Correct Answer: C
Coordination number = # ligands = 4+2 = 6
H2O = neutral, Cl = 1, counterion = 1, therefore Cr = +3 Q 24.7
Solution: Correct Answer: B
Coordination number = # ligands = 4
CN = 1(4) = 4, counterion = +1(4) = +4, therefore Ni = 0 Q 24.8
Solution: Correct Answer: E
Correct name is tetrachlorocobaltate(II) ion, (underline for emphasis only) 44 ©Prep101
Q 24.9
Solution: Chem110 Exam Solutions Correct Answer: A
Chloride because counterion, 4 = tetra, di = 2 Q 24.10
Solution: Correct Answer: E
Nickelate because anionic, oxidation number = 0, tetracyano Q 24.11
Solution: Correct answer: A
B is not by alphabetical order, oxidation number is incorrect in C and D Q 24.12
Solution: Correct answer: D
Bromide = counterion,
(en) = neutral, so cobalt = +3. must be balanced by 3 bromide ions Q 24.13
Solution: 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. Q 24.14
Solution: Correct Answer: A
Coordination number = # Ligands = 2+2 = 4
Cl = 1, CN = 1, Therefore Cd = +4, and counterion is 1(2).
45 ©Prep101
Q 24.15
Solution: Chem110 Exam Solutions [FeCl6] must equal 3. Balance charge
3(+2) = 6; 2(x) = 6 to give neutral charge. x = 3. Q 24.16
Solution: Correct Answer: 1
NH4 is +1 and counterion to give neutral charge, therefore complex must equal 1. Q 24.17
Solution: Correct answer: B
V = +4, O = 2, CN = 1, therefore overall charge must be equal to 2 Q 24.18
Solution: 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. Q 24.19
Solution: 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. Q 24.20
Solution: Correct answer: B
A is incorrect (Coordination number is 6 of Ni(en)3 for both complexes)
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 Q 24.21
Solution: Correct Answer: C
Ionization isomer is one where the counterion is switched. So switch the bromo
for the chloro in (c). 46 ©Prep101
Q 24.22
Solution: Chem110 Exam Solutions 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 1 unpaired electron 47 ©Prep101
Q 24.24
Solution: Chem110 Exam Solutions Correct answer: E
[CoCl(NH3)5] Cl2 pentaamminechlorocobalt(III) chloride
Co3+ = d6
Coordination number = 6 = octahedral
t2g6eg0 0 unpaired electrons (strong field) Q 24.25
Solution: Correct answer: D
Cs[FeCl4] → [FeCl4] = 1 → Fe = +3 Fe3+ → d5
Cl is a weakfield ligand; so 5 unpaired electrons 48 ©Prep101
Q 24.26
Solution: Chem110 Exam Solutions Correct answer: A
Cl = 1, en = neutral, NH3= neutral; therefore Ni = +2
Coordination number = 6 (en is bidentate); Ni2+→ d8
It does not matter whether it is high field or low field, d8 gives 2 unpaired
electrons in either situation. 49 ©Prep101
Q 24.27
Solution: Chem110 Exam Solutions Correct Answer: C
Each are Ni2+ → d8
Cl is a weakfield ligand, while CN is a strong field ligand Q 24.28
Solution: Correct answer: E
Revisit the spectrochemical series. A→D are all strong field. Br is weak field, and
would result in a high spin system with unpaired electrons (paramagnetic).
Co3+ = d6 50 ©Prep101 Chem110 Exam Solutions Chapter 23 – The Transition Metals
Q 23.1
Solution: D (is false)
Froth flotation is used to extract light ores from heavy impurities. Q 23.2
Solution: Correct answer: C
A → incorrect, roasting requires air
C → incorrect, calcination is heating in the absence of air
D → electrolytic refining Q 23.3
Solution: Correct answer: D
Ex: HgS + O2 (reducing agent) → Hg(l) + SO2
ZnO + C (reducing agent) → Zn(l) + CO Q 23.4
Solution: Correct answer: A
B → incorrect (used to make steel from iron) Q 23.5
Solution: Correct answer: C Q 23.6
Solution: Correct answer: C
Steel is an alloy of iron and carbon 51 ©Prep101
Q 23.7
Solution: Chem110 Exam Solutions Correct answer: A
Iron coordinates to the heme protein in the body 52 ©Prep101 Chem110 Exam Solutions Chapter 12  Intermolecular Forces
Q 12.1
Solution: Correct answer: C
NO has the highest boiling point since it is polar, even though it has a
slightly lower molar mass than O2. O2 > N2 by LDF (increased MW) Q 12.2
Solution: Correct answer: D
The CO molecule is polar; thus, it has dipoledipole interactions, giving it a
higher boiling point. Q 12.3
Solution: Correct Answer: E
The butane molecule is more spread out so that it interacts better with the
other butane molecules. Q 12.4
Solution: Correct answer: E
The C3H8 is nonpolar, while the CH3OCH3 is polar and has the higher
boiling point. Q 12.5
Solution: Correct answer: B
CO2 is a linear molecule (no net dipole). Only LDF exist 53 ©Prep101
Q 12.6
Solution: Chem110 Exam Solutions Correct answer: E
Both covalent and hydrogen bonding have directional bonding (have
constraints as to how the molecules are oriented). Q 12.7
Solution: Correct answer: C
HF has LDF, Dipoledipole and Hydrogen bonding!
Then: HI > HBr > HCl due to increased LDF (LDF dominate over dipoledipole for this case because big difference in molecular weight) Q 12.8
Solution: Yes
Consider HI > HBr > HCl
All have dipoledipole, one would expect dipole for HCl to be greatest
(because Cl is most electronegative). Q 12.9
Solution: Correct answer: D Q 12.10
Solution: Correct answer: B
An allptrope (such as carbon in graphite or diamond), where the element
can have more than structural forms. 54 ©Prep101 Chem110 Exam Solutions Chapter 21 – Main Group Elements
Q 21.1
Solution: E
Potassium is group 1 → Alkali Q 21.2
Solution: B
LiX and MgX2 have similar chemical and physical properties (“diagonal
relationship)
Alkali metals (Grp 1) have one valence electron, not alkaline earth
Reactivity increases from T → B for both Grp 1 and Grp 2
Alkali metals have low densities Q 21.3
Solution: C
Alkali metals are stored in oil because they do not react with oil and do
react with water and oxygen. They also react with halogens. Q 21.4
Solution: Correct answer: A
2Li+ + O2 → 2LiO Q 21.5
Solution: (normal oxide), where M = Li for this case Correct answer: E
K+ + O2 → KO2 (superoxide); both answers B and D are correct 55 ©Prep101
Q 21.6
Solution: Chem110 Exam Solutions Correct answer: E
Group 1 metals and some group 2 metals will react with heat (flame) to
give a coloured flame that is indicative of that element.
Group 1 metals react violently with water. Group 1 metals react differently
with oxygen. Calcium (not magnesium) is found in lime and used for
cement. Q 21.7
Solution: Correct answer: A
Temporary water hardness can be removed by boiling or addition of lime
(answer A). Permanent water hardness has calcium and magnesium
sulfates (SO42) and/or chlorides, cannot be removed by boiling, and is
removed by a water softener. 56 ©Prep101 Chem110 Exam Solutions Chapter 25  Nuclear Chemistry
Q 25.1
Solutions: These questions are straightforward. Just make sure the numbers of
protons and the molecular weights are same for both the reactants and
the products. The chemical symbol of an element is dictated by the
number of protons.
A) 4
2 B) 214
82 C) 40
18 Q 25.2
Solution: 212
84 Q 25.3
Solution: D) 32
15 Pb E) 99
43 Ar F) 3 He P
Tc E
238
92 Q 25.4
Solution: Po → 208 Pb + 4 He
82
2 U → 4 He +
2 234
90 Th Correct Answer: E
High atomic number (Z>83); likely alpha emitter 57 ©Prep101
Q 25.5
Solution: Chem110 Exam Solutions Correct answer: B
A beta particle has a penetrating power greater than an alpha particle. Q 25.6
Solution: 14
6 Q 25.7
Solution: Correct Answer: D C → 14 N +
7 131
53 Q 25.8
Solution: β 0
−1 I → 131 Xe +
54 β 0
−1 Correct answer: B
A beta particle is an electron, Q 25.9
Solution: e (no mass, and 1 charge) Correct answer: A
40
19 Q 25.10
Solution: 0
−1 40
0
K → 18 Ar + 1 e Correct answer: B
A – incorrect – it is often referred to as a “positive electron”
C – incorrect  no effect on mass
D – incorrect  Likely for nuclei that have low neutron to proton ratio Q 25.11
Solution: 11
6 0
C → 11 B + 1 e
5 58 ©Prep101
Q 25.12
Solution: Chem110 Exam Solutions Correct Answer: B
C14 has 6 protons, 8 neutrons. High N:P ratio → favour beta decay
(converts neutron to proton) Q 25.13
Solution: Correct Answer: B
A → incorrect → It has (somewhat) high N:P ratio, but lies below the belt
of stability
B → incorrect → both electron capture and position emission decay to
give Xe118 to I118
C → incorrect → mass number remains unchanged
D → correct → both EC and position form a neutron
E → incorrect (see C) Q 25.14
Solution: Q 25.15
Solution: 59 ©Prep101
Q 25.16
Solution: Chem110 Exam Solutions Correct Answer: C
It takes 82 minutes to go from 1.0 g → 0.125 g, so half life must be less
(eliminate B)
Find decay rate constant, k N
ln t
N 0 = −kt → 1 Nt =k − ln t N0 → 1 0.125 g − 82 min ln 1.000 g = k k = 0.025 min1 Find half life, t1/2
k= 0.693
t1/ 2 → 0.02536 year −1 = 0.693
t1/ 2 → t1/2 = 27 min 27 min / 60 min(hour)1 = 0.45 hour
Q 25.17
Solution: Correct Answer: A
We know that it will be less than 5.26 years
k= 0.693
t1/ 2 N
ln t
N 0 → = −kt k= 0.693
5.26 year → k = 0.132 year1 where Nt/No = 75% = 0.75 t = 2.178 = 2.2 year
2.2 years (12 months/year) = 26.4 months → 2 years and about 2 months 60 ©Prep101
Q*25.18
Solution: Chem110 Exam Solutions Correct answer: D
k= 0.693
t1/ 2 N
ln t
N 0 → = −kt k= → k = 0.132 year1 N → ln t = −(0.132 yr −1 )(26.5 yr )
N 0 N t −3.498 = 0.03026
N =e 0 Q 25.19
Solution: 0.693
5.26 year → closest to answer D Correct Answer: A
Find decay rate constant, k N
ln t
N 0 = −kt → 1 Nt =k − ln t N0 → 1 0.953 g − 2.00 y ln 1.000 g = k k = 0.0241 year1
Now, use same equation, (with same rate constant, but change time) N
ln t
N 0 = −kt → N ln t = −0.120
N 0 N ln t = −(0.0241 year −1 )(5.0 year )
N 0 → N t −0.120 = 0.887
N =e 0 N0 = 1.000 g, find for Nt
Nt = N0(0.887) = 1.000g (0.887) = 0.887 g 61 ©Prep101
Q 25.20
Solution: Chem110 Exam Solutions Correct Answer: A
Find decay rate constant, k N
ln t
N 0 = −kt → 1 Nt =k − ln t N0 → 1 0.953 g − 2.00 y ln 1.000 g = k k = 0.0241 year1 Find half life, t1/2
k= Q 25.21
Solution: 0.693
t1/ 2 → Mass of products 0.0241 year −1 = 0.693
t1/ 2 → t1/2 = 28.8 yr = (2 x 1.00728 amu) + (2 x 1.00866 amu)
= 4.03188 amu Mass of reactants = 4.00150 amu (Given)
∆m = (products) – (reactants) = 4.03188 – 4.00150 = 0.03038 amu
∆E = c2 (∆m) ( ∆E = 2.998 ×108 ms −1 c2 = (2.998E8 ms1)2 = 8.988E16 m2s2
∆m = 0.03038 amu 1g
1 kg ) (0.03038 amu ) 6.022E23 amu 1000 g =4.533 x 1012 J 2 To obtain the energy change per mol, multiply by Avagadro’s number
∆E = 4.533 x 1012 J (6.022 x 1023 mol1) = 2.730 x 1012 Jmol1 62 ©Prep101
Q 25.22
Solution: Chem110 Exam Solutions ∆m = ∆E/c2 = (393500 kgm2s1)/(2.998 x 108 ms1)2 = 4.378 x 1012 kg
∆m = 4.378 x 1012 kg (1000 g/kg) = 4.378 x 109 g Q 25.23
Solution: 56
26 Fe → 261 p + 301 n
1
0 Mass of products = (26)(1.00728 amu) + (30)(1.00866 amu)
Mass of products = 56.44914 amu
Mass of reactants = 55.92068 amu (Given)
∆m = (products) – (reactants) = 56.44914 – 55.92068 = 0.52846 amu ( ∆E = 2.998 × 10 8 ms −1
∆E = 7.87405x10 11 1g
1 kg ) (0.52846 amu) 6.022E23 amu 1000 g 2 J (also use 0.52866 amu x 1.49 E10 J/amu = 7.87 x1011 J)
To obtain binding energy per nucleon, divide by mass of nucleus
∆E = 7.87x1011 J / (55.92068) = 1.41 x 1012 J/nucleon Q 25.24
Solution: 238
92 1
U → 921 p + 1460 n
1 Mass of products = (92)(1.00728 amu) + (146)(1.00866 amu)
Mass of products = 239.93412 amu
Mass of reactants = 238.00031 amu (Given)
∆m = (products) – (reactants) = 239.93412 – 238.00031 = 1.93381 amu ( ∆E = 2.998 × 10 8 ms −1 1g
1 kg ) (1.93381 amu) 6.022E23 amu 1000 g 2 ∆E = 2.886310 J
To obtain binding energy per nucleon, divide by mass of nucleus
∆E = 2.886310 J / (238.00031) = 1.2127 x 1012 J/nucleon 63 ©Prep101 Chem110 Exam Solutions Q 25.25
Solution:
Mass defect = products  reactants
Mass defect = (13.00335 amu + 1.00783 amu) – (10.01294 amu +
4.00260 amu)
Mass defect = 0.00436 amu
Energy = 0.00436 amu (931.5 MeV/amu) = 4.06 MeV
Given the question asks “how much energy is released”, one can assume
exothermic and report 4.06 MeV (omit negiatve) * Negative energy implies energy is released
* Petrucci textbook may use ∆m = reactants  products
Q 25.26
Solution:
Mass defect = products – reactants
Mass defect = (231.0363 amu + 4.0026 amu) – (235.0429 amu)
Mass defect = 0.0050 amu
Energy = 0.0050 amu (931.5 MeV/amu) = 4.7 MeV Given the question asks “how much energy is released”, one can assume
exothermic and report 4.7 MeV (omit negiatve)
* Negative energy implies energy is released
* Petrucci textbook may use ∆m = reactants  products 64 ...
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 Fenster
 Chemistry, Atom, Electron, Correct Answer, Chemical bond

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