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Chemistry 1250 - Sp17
Solutions for Midterm 2
This material is copyrighted. Any use or reproduction is not allowed except with the expressed written
permission of Dr. Zellmer. If you are taking Chem 1250 you are allowed to print one copy for your own use
during the semester you are taking Chem 1250 with Dr. Zellmer. You are not allowed to disseminate this
material to anyone else during the semester or in the future.
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1)
a) T:
For an endothermic process ΔH is positive (Hf - Hi >0). This means the products are HIGHER
in energy than the reactants.
b) T: Enthalpy, H, is a state function. It is a property of a system which is determined by specifying
the system’s conditions or state. The value of a state function depends only on the present state
of the system and not how it got there. A change in a state function, ΔH, is path independent.
b) T: ΔE = q + w and work for chemical systems usually arises out of pressure-volume changes.
ΔE = q ! ΔPV (q = heat & w = work) Just another way to state the First Law of
Thermodynamics (Conservation of energy). At constant pressure ΔE = qP - PΔV, where qP is the
heat at constant pressure (ΔH), ΔE = ΔH - PΔV.
The heat at constant pressure is the enthalpy. (Heat at constant volume is ΔE).
c) T:
ΔH = qP
e) F:
q & w are not state functions. They depend on the path taken. It takes more work to push a cart
between two points if you go back and forth rather than in a straight. However, internal energy,
E, and enthalpy, H, are state functions (path independent).
E
2) heat stoichiometry problem
C
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3)
Standard Enthalpy (heat) of Formation: Change in enthalpy for a reaction which forms ONE mole of a
compound from its elements with all substances in their
standard states.
ΔHfE
Standard state: pure form at atmospheric pressure (and some temperature of interest). A specific temp.
is NOT part of the definition. Thus for H2O at 1 atm and 25 EC it’s standard state would
be a liquid. For H2O at 1 atm and 110 EC it’s standard state would be a gas.
The only reaction given which produces 1 mole of a substance in its standard state from it’s elements is
b) Na(s) + 1/2 Cl2(g) + 3/2 O2(g) 6
NaClO3(s)
Note: You may see fractions in these balanced eqns. for ΔHfE
B
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4)
For calorimetry problems you should know what heat capacity, C, is and how molar heat capacity, Cm,
and specific heat, Cs, are related to it and how they all related to heat transfer and temperature change.
Heat Capacity: The amount of heat required to raise the temperature of a substance by 1 EC. Often used
when you don’t have pure or uniform substances (such as a styrofoam cup in exp 6). This is an
extensive property (depends on how much is present, how big the cup is).
C = q/ΔT
Molar Heat Capacity: Heat Capacity of one mole of a substance (heat cap. per mol). The amount of heat
required to raise the temp. of 1 g of a substance by 1 EC. This is most often used for a pure substance
and is an intensive property (while the heat added is extensive).
Cm = C/mol
Specific Heat: Heat Capacity of one gram of a substance (heat cap. per gram). The amount of heat
required to raise the temp. of 1 g of a substance by 1 EC. This is most often used for a pure substance
and is an intensive property (while the heat added is extensive).
Cs = C/gram
This problem is very similar to what you did in experiment 6 in lab except the problem assumes no heat
is lost to the calorimeter (it has a heat capacity of zero). Heat was lost by the solution since it’s temp.
decreased (from 25.00 EC to 23.34 EC) and heat was gained by the dissolution process (NH4NO3
dissolving in H2O).
q = m × Cs × ΔT
Heat = (mass) @ (specific heat) @ (temperature change)
qgained by dissolution = ! qlost by H2O
Cs = 4.184 J/g@EC (the specific heat of solution to 4 s.f. - from the first sentence)
m = 75.0 g H2O + 1.60 g NH4NO3 = 76.6 g (need total mass of solution)
qH2O = m @ Cs @ ΔT = 76.6 g × (4.184 J/g@EC) × ( 23.34 EC ! 25.00 EC)
qH2O = ! 532.02 J
(3 s.f. since 23.34 ! 25.00 = ! 1.66, 3 s.f.)
qdissolution = !(qlost by H2O) = !(! 532.02 J) = 532.02 J
Δ Hdissolution
D
532.02 J x (1 kJ/103 J)
= ----------------------------------------------------- = 26.6 kJ/mol
(1.60 g NH4NO3 x 1 mol NH4NO3/80.05 g)
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5)
Similar to the practice exam problem. Use ΔHfE to determine the ΔHrxn, except in this case we are given
ΔHrxn and want to find the ΔHfE for one of the substances. Based on Hess’s Law.
H
o

v
products
2 HNO3(aq) + NO(g)
NO2(g)
NO(g)
H2O(R)
nH
0
f, products

3 NO2(g) + H2O(R)
mH
reactants
0
f, reactants
ΔHE = +138.5 kJ
ΔHfE = +33.84 kJ/mole
ΔHfE = +90.37 kJ/mole
ΔHfE = !285.85 kJ/mole
ΔHE = [(3 mol)(33.84 kJ/mol) + (1 mol)(!285.85 kJ/mol)] &
[(2 mol) ΔHfE(HNO3) + (1 mol)(90.37 kJ/mol)]
+138.5 kJ = [101.52 kJ) + (-285.85)] & [(2 mol) ΔHfE(HNO3) + 90.37 kJ]
+138.5 kJ = !274.70 kJ & [(2 mol) ΔHfE(HNO3)]
ΔHfE(HNO3) = ! 206.6 kJ
A
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6) Hess’s Law problem
Want ΔHrxn for a reaction given ΔHrxn for other reactions.
N2(g) + 4 H2O (R)
v
N2H4 (R) + 2 H2O2 (R)
ΔHrxn = ?
1a)
N2H4 (R) + O2(g) 6 N2(g) + 2 H2O (R)
ΔHE = -622.3 kJ
2a)
H2(g) + 1/2 O2(g) 6 H2O (R)
ΔHE = -285.8 kJ
3a)
H2(g) + O2(g) 6 H2O2 (R)
ΔHE = -187.8 kJ
Remember:
1) If you multiply an eqn. by some factor the ΔH gets multiplied by that factor.
2) If you reverse an eqn. it simply changes the sign of ΔH.
ΔHreverse = ! ΔHforward
To get the reverse of a reaction you multiply everything by !1, the coefficients and ΔH. Your
coefficients in the balanced eqn. would all be negative. To get rid of the negative signs you simple turn
the whole eqn. around (reverse it), just like in a math eqn. So in reality, this is #1 using a !1 as the
multiplying factor.
***** continued on next page *****
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(cont.)
1) Need 1 N2H4 molecule as a product. Take eqn 1a and reverse it (multiply by !1).
1b)
N2(g) + 2 H2O (R) 6 N2H4 (R) + O2(g)
ΔH1b = + 622.3 kJ
2) Need 2 H2O2 molecule as a product. Take eqn 3a multiply it by 2.
3b)
2 H2(g) + 2 O2(g) 6 2 H2O2 (R)
ΔH3b = (2) (-187.8 kJ) = ! 375.6 kJ
3) Need 2 more H2O as a reactant and 1 O2 on right (right now if you added 1b and 3b you would get
1 O2 on left. Can take eqn. 2a, multiply it by 2 and reverse it (multiply by !2).
2b)
2 H2O (R) 6 2 H2(g) + 1 O2(g)
ΔH2b = (-2) (-285.8 kJ) = + 571.6 kJ
Add the eqns. and delH values:
1b)
N2(g) + 2 H2O (R) 6 N2H4 (R) + O2(g)
ΔH1b = + 622.3 kJ
2b)
2 H2O (R) 6 2 H2(g) + 1 O2(g)
ΔH2b = + 571.6 kJ
3b)
2 H2(g) + 2 O2(g) 6 2 H2O2 (R)
ΔH3b = ! 375.6 kJ
N2(g) + 2 H2O (R) + 2 H2O (R) + 2 H2(g) + 2 O2(g) v
N2H4 (R) + 2 H2(g) + O2(g) + O2(g) + 2 H2O2 (R)
ΔHrxn = ΔH1b + ΔH2b + ΔH3b
Cancel things in eqn. which are the same on both sides, combine same things which appear on the same
side and add ΔH values:
N2(g) + 4 H2O (R)
v
N2H4 (R) + 2 H2O2 (R)
ΔHrxn = (+622.3 kJ) + (+571.6 kJ) + (!375.6 kJ)
= + 818.3 kJ
E
(desired eqn.)
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7)
The frequency of radiation is related to wavelength,
c
ν = -----λ
Within this distance of 1.6 x 10!7 m:
wave A has 4.5 wavelengths and a single wavelength is 3.6 x 10!8 m and the frequency is 8.4 x 1015 Hz
wave B has two wavelengths and a single wavelength is 8.0 x 10!8 m and the frequency is 3.8 x 1015 Hz
c
3.00 x 108 m/s
ν = ------ = -------------------- = 3.75 x 1015 s-1 = 3.8 x 1015 Hz
λ
8.0 x 10!8 m
Can see that wave A has a shorter wavelength than wave B (there’s more waves in same distance) and
since frequency is inversely proportional to the wavelength it means wave A has a higher frequency.
The energy of a wave is directly proportional to the frequency of the wave.
E=hν
A
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8)
The rules for the values for the quantum numbers is given in section 6.5 (and 6.7 concerning the spin q.n.).
n = 1, 2, 3, ..., 4
(integer values starting at 1) principal quantum number (shell number)
R = 0, 1, 2, ... , (n-1) angular momentum (azimuthal) q.n. (subshell q.n. and defines shape of orbitals
w/in a subshell)
mR = -R, ..., 0, ..., +R
(values increase by 1)
ms = + ½ and - ½
(only one with non-integer values)
magnetic q.n. (describes orientation of orbital in
space - orbitals are degenerate unless
in an applied magnetic field)
spin magnetic q.n. (electrons “spin” around
an axis, only has two directions of spin, “up”
or “down”)
Correct q.n.:
For (2) when n = 3 and R = 2 the possible values of mR are mR = -2, -1, 0, 1, 2 and ms = - ½ is fine
For (4) when n = 4, can have R = 0, 1, 2, 3 (can’t have R = 4 when n = 4)
Incorrect q.n.:
For (1) when n = 3 and R = 0 is okay and the possible values of mR is mR = 0 and ms = - ½ is fine
For (3) when n = 4 and R = 3 the possible values of mR are mR = -3, -2, -1, 0, 1, 2, 3 and ms = + ½ is fine
For (5) when n = 5 and R = 1 the possible values of mR are mR = -1, 0, 1 and ms = + ½ is fine
E
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9)
For energy changes in a hydrogen atom we have the following eqn.,
Δ EHydrogen
1
1
= ! (hcRH) ( ----- ! ----- )
nf2
n i2
Δ EHydrogen = ! (2.18 x 10
!18
1
1
J) ( ----- ! ----- )
nf2
n i2
Δ EHydrogen = ! (2.18 x 10!18 J) (1/nf2 - 1/ni2)
This eqn. gives the correct sign for ΔE for both emission (nf < ni, ΔE < 0, energy released) and for excitation
(nf > ni, ΔE > 0 , energy required.
Then you could use the following eqn to calc. energy associated with this and the frequency or wavelength:
|ΔE| = hν = hc/λ
= (2.18 x 10!18 J) (1/2f2 - 1/6i2) = 4.8444 x 10!19 J/electron
? J/mol = 4.8444 x 10!19 J/electron x (6.02 x 1023 electrons/mol)
= 2.9163 x 105 J/mol = 292 kJ/mol
Instead you could just use the Rydberg eqn. to first calculate the wavelength and then the energy
1/λ = RH (1/nf2 - 1/ni2)
=
1.097 x 107 m!1 (1/nf2 - 1/ni2)
Technically one uses the absolute value of the (1/nf2 - 1/ni2) since λ has to be positive. It’s understood, when
nf > ni it corresponds to an absorption of a photon (excitation) and when
nf < ni it corresponds to an emission of a photon.
For a transition from ni = 6 to nf = 2 (corresponding to an emission):
1/λ = 1.097 x 107 m!1 |(1/22 ! 1/62)| = 2.437 x 106 m!1
λ = 4.1020 x 10!7 m = 410.2 nm (visible region)
c
3.00 x 108 m/s
ν = ------ = ---------------------- = 7.31 x 1014 s-1 = 7.31 x 1014 Hz
λ
4.1020 x 10!7 m
B
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10)
a) T:
According the Heisenberg uncertainty principle there is a fundamental limitation on how
precisely one can simultaneously measure both the location (x) and momentum (mv) of an
object. According to this principle the uncertainty in the position times the uncertainty in the
momentum is given by the following formula:
Δx C Δp $ h/4π
x = position
p = momentum (mCv, m = mass, v = velocity)
h = Planck’s constant
C
b) T:
Lower energy orbitals are always filled with electrons first to give the ground state (lowest
energy state).
c) F:
When filling degenerate orbitals (such as the three p orbitals in a p-subshell) each orbital is singly
occupied, with the same spin for each electron, before electron pairing occurs. This is Hund’s
rule: “for degenerate orbitals, the lowest energy is attained when the number of electrons having
the same spin is maximized”. This can only occur when the electrons are added one a time to
each orbital until all the orbitals in the subshell have one electron (all with the same spin) before
the electrons are paired up (two in an orbital).
d) T:
The energy levels for the subshells in a shell are different (not degenerate) in multi-electron
atoms (e.g. the 2s is lower in energy than the 2p). This is due to the fact the multiple electrons
interact with each other and the nucleus. This causes the energies of the subshells to be different.
The orbitals within a subshell are still degenerate (have the same energy, e.g. the three 2p orbitals
have the same energy). The orbitals will split (have different energies) in the presence of an
external magnetic field).
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The correct statements are 2, 3 & 4:
2) A d subshell can have a maximum of 10 electrons. (See below.)
3) Each p orbital within a subshell consists of two lobes. (figure eight, 4; dumbbell)
4) Only 2 electrons can occupy an orbital and they must have opposite spins.
The corrected answers for 1 & 5 would be:
1) There are nine orbitals in an g subshell. (See below.)
5) Not every subshell w/in a shell is filled prior to putting e& in the next shell (4s filled before the 3d).
l=
0
1
2
3
4
5. . .
subshell letter
s
p
d
f
g
h. . .
# orbitals in subshell
1
3
5
7
9
11. . .
max. # e- in subshell
2
6
10
14
18
22. . .
E
(2, 3 & 4 are correct)
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[Rn] 7s 2 5f 7 6d 1
13
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[Xe] 5f 7 6d 1 7s 2
If the # e& is ˜ # e& in a half-filled subshell, then;
the number of unpaired e& = the number of e& in the subshell.
If the # e& is > half filled (or filled), then;
# of unpaired e& = (the maximum # of e& in the subshell) - (the # of e& present in the subshell).
There are 7 unpaired e& in the 5f, 1 unpaired e& in the 6d & 0 unpaired e& in the 7s. This gives a total of 8
unpaired e&.
¼ ¼ ¼ ¼ ¼ ¼ ¼
¼
5f
C
¼¿
6d
7s
(8 unpaired e&-)
13)
Use the periodic table to get the electron configurations. Remember, for the shorthand notation you use
the noble gas from the previous period in square brackets to signify the core electrons and just show the
subshells and electrons outside the core.
Correct ones:
1) Sc: [Ar]4s23d1
2) P: [Ne]3s23p3
5) Ag: [Kr]5s14d10
Incorrect ones:
3) Cr: [Ar]4s13d5
==>
Cr2+ : [Ar]3d4
Cr is an exception to how the orbitals are filled (would expect [Ar]4s23d4 ). When forming cations the
electrons are removed first from the valence shell. In this case, that would be the 4s. Once those are lost
then electrons from the (n-1)d orbitals are removed. Thus, Cr2+ would have the configuration given
above.
4) As: [Ar]4s23d104p5
E
==>
[Ar]4s23d104p3
(Third element in p-block, 3 electrons.)
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14)
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Again, look at the periodic table (opposite of metallic character and size):
Inc. Ionization Energy
Bi
Te
Se
Cl
O
&
Electronegativity
E
15)
Need to write out the e- configuration for each atom and then look at which electrons will come
out when it forms an ion. Remember, the valence e! come out first when forming ions.
Hg
[Xe] 6s2 4f14 5d10
Hg2+
[Xe] 4f14 5d10
Pb
[Xe] 6s2 4f14 5d10 6p2
Pb2+
[Xe] 6s2 4f14 5d10
6s is the valence shell (removed first)
6p is the valence shell (removed first)
A
16)
The radius of 31Ga is greater than 32Ge. Size increases from right to left across a period.
For b) The ionic radius of S2! is greater than Ca2+.
S2! & Ca2+ are isoelectronic (they have the same number of e& and look like Ar). In an isoelectronic series,
anions are bigger than cations. The cations have more protons and the same # e& as the anions so “extra”
protons pull the e& in closer.
C
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17)
The following is the proportionality eqn. for LE and it’s dependence on charge and distance:
Q1 Q2
LE % ---------d
Q = charges on ions
d = distance between cation and anion
(sum of ionic radii)
This eqn. shows the LE is proportional to the charges on the ions. The bigger the charges the greater
the LE. (Use the magnitude, absolute value, of the charges).
The eqn also shows the LE is inversely proportional to the distance between the charges. The
smaller the distance (the smaller the ions) the greater the LE.
The numerator is more important then the denominator. (i.e. has a bigger effect).
NaCl: +1 on Na+ and !1 on Cl!
CaS +2 on Ca2+ and !2 on S2!
Al2O3 +3 on Al3+ and !2 on O2!
The Al2O3 has +3 and -2 charges so it’s numerator is larger than CaS with +2/-2 charges which as a
much larger numerator than NaCl. Based on this one would expect the LE for Al2O3 would be the
greatest and that for the CaS to be next and that for NaCl to be the smallest since the charges are only +1
and -1.
How about the sizes of the ions? One would expect the sizes of the cations to increase in the order
Al3+ < Na+ < Ca2+. The Cl! and S2! are isoelectronic so they inc. in size Cl! < S2! . The O2! is smaller
than these so one expects the sizes of the anions to increase in the order, O2! < Cl! < S2!. The LE inc. as
the ions get smaller. However, the sizes of ions aren’t that different and the numerator in the above eqn
is the more important factor.
We would expect the Al2O3 to have the largest LE since it has the largest charges and smallest ions.
We would expect the CaS to have the next largest LE even though the ions are a little larger than those
in NaCl the numerator for CaS is four times larger.
The expected lattice energies increase in the order,
NaCl < CaS < Al2O3
D
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18)
Simply look at the electronegativities of the atoms. Remember, EN increases bottom to top and left to right in
the periodic table.
Inc. ionization energy
&
electronegativity
Also remember the ordering of a number of the most used atoms and the fact there is a big drop in EN going
from the 2nd row to the 3rd row and then from the 3rd row down the EN decreases but not by much.
The EN of some of the most common elements inc. in the following order:
B, As < H, P < C,S,I < Br < N, Cl < O < F
Based on this the H-O bond would be the most polar (largest dipole moment) of those listed.
B
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19) CO32&
C (4A) O (6A) !2 chg
1) A = 1(4e!) + 3(6e!) + (2e!) = 24 val e!
2) Draw skeleton structure. The more
EN O atoms attached to C. This
accounts for 6 e-.
3) Put 6 e- on each O to fulfill octet.
This uses 18 e-.
4) # e- left = 24 e- - (6 e- + 18 e-) = 0 eNo more e- to put on C to fulfill octet. Must move a pair from an O atom to form double bond
between C and O. There are 3 oxygen atoms for which you could do this. Thus the double bond
could be drawn in 3 different positions leading to 3 resonance structures (see problem 19).
3 resonance structures ( double bond can be in 1 of the 3 different positions)
trigonal planar, angles of exactly 120E
B
20)
To fulfill the octet rule on sulfur there has to be a double bond between the S and an O atom. The double bond
could be drawn in 2 different positions leading to 2 resonance structures. The double bond isn’t moving or
flipping back and forth between the two regions. There aren’t two separate molecules. The two bonds are
identical (have the same bond length and electron distribution). There’s simply no easy way to draw one single
Lewis Structure for this molecule. What it represents is that the “extra” pair of electrons represented in the
double bond is actually being shared equally in both regions (actually shared by all three atoms in a molecular
orbital which spans all three atoms). It’s like the “extra” bond is shared equally to give a bond order of 1.5.
2 resonance structures ( double bond can be in 1 of the 2 different positions)
Bent, angles of approx. 120E
B
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21)
18
Formal Charge on the atoms in SCN!
Divide e- in bonds equally between atoms
(each atom gets ½ the e- involved in the bond).
lpe- assigned to the atom they’re on.
Subtract this total from the valence e- on the atom.
Ex:
For S: 6 - (2 + 3) = + 1
For C: 4 - (4) = 0
For N: 5 - (6 + 1) = !2
FC should add up to give overall chg., in this case !1 (as they do)
D
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22) ICl3
I: 7A Cl: 7A
I
Cl
1) A = 1(7e-) + 3(7e-) = 28 val e!
2) Draw skeleton structure. The more
EN Cl atoms attached to I This
accounts for 6 e-.
Iodine is in the middle
3) Put 6 e- on each Cl to fulfill octet.
This uses 18 e-.
4) # e- left = 28 e- - (6 e- + 18 e-) = 4 eThese go on the I atom. This gives 10 eon the I atom, more than an octet.
Iodine has 3 bonds & 2 lpe- in this case since it is in the middle
(when halogens are on the outside they have only 1 bond)
The ED geometry around the I is trigonal bipyramidal and the molecular geometry is T-shaped with
bond angles of - 90E
B
(2 lone pairs of electrons on the Iodine)
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23)
BrF3 (see ICl3 above) and AlCl3
There are several instances when the octet rule is not obeyed.
There are cases when there is less than an octet and there’s an even number of electrons.
This occurs for Be (group 2A) and B and Al (group 3A).
Be gets only 4 e! around it and forms 2 bonds (group 2A)
B and Al get only 6 e! around them and form 3 bonds (group 3A)
C Be C
C
C B C (Al like B since they are in the same group, 3A)
2 bonds
4 e!
3 bonds
6 e!
There’s also the cases when an atom can get more than an octet of electrons. This occurs
for atoms from the 3rd row down. It does NOT occur for atoms in row 2. Examples are
PCl5, SF6, XeF2 (3 lone pairs and 2 atoms), XeF4 (2 lone pairs and 4 atoms) and others.
There are some systems with an odd # of electrons. These will not get an octet on one of
the atoms and there will be an unpaired electron. This occurs most often with Nitrogen
containing compounds (group 5A) and halogens (other than F and when they’re in the
middle surrounded by other atoms, particularly O atoms).
NO2 has 17 valence electrons and it winds up with an unpaired e! in the molecule
and it’s usually shown on the N atom.
While H has less than an octet it conforms to the “octet” rule, what I also called the “noble-gas” rule. It
winds up with 2 e! around it to look like He, a noble gas. Some books will state H conforms to the octet
rule even though it doesn’t get eight electrons around it and some state it doesn’t.
D
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24)
Use bond enthalpies to determine the ΔHrxn. You will see bond enthalpies as BE(bond) or D(bond).
H
4H-N-H
|
H
+
N - H = 391 kJ
O - H = 463 kJ
 BE

broken
3 O2 ÷ 6 H - O - H

+
 BE
formed
2N/N
O = O = 495 kJ
N / N = 941 kJ
ΔHE = [(12) D(N - H) + (3) D(O = O)] & [(12) D(O - H) + (2) D(N / N)]
ΔHE = [(12) (391 kJ) + (3) (495 kJ)] & [(12) (463 kJ) + (2) (941 kJ)]
= [6177 kJ] & [7438 kJ]
= ! 1261 kJ
A
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BrO2&
Br: 7A
O: 6A &1chg
Br
O
1e&
1) A = 1(7e-) + 2(6e-) + (1e-) = 20 val e!
2) Draw skeleton structure. The more
EN O atoms attached to Br This
accounts for 4 e-.
Bromine is in the middle
3) Put 6 e- on each O to fulfill octet.
This uses 12 e-.
4) # e- left = 20 e- - (4 e- + 12 e-) = 4 eThese go on the Br atom to fulfill octet.
-
Br has 2 bonds & 2 lpe in this case since it is in the middle (when halogens are on the outside
they have only 1 bond, F is always on the outside, never in the middle)
-
O has 1 bond & 3 lpe
(Oxygen and other group 6A elements usually have 2 bonds and 2 lpe- or 1
bond 3 lpe-, particularly when surrounding another atom, like Cl, Br, I, N
and S)
Shape and bond angle
2 atoms (bonds) & 2 lpe& on Bromine
bent, angle of . 109.5E
There are 4 “things” (electron domains) around Br (2 O’s and 2 lpe&). These 4 “things” arrange themselves in a
tetrahedral structure around Br and bond angles start out at 109.5E. However, since all 4 “things” around Br are
not identical the O & Br & O angle is < 109.5E or . 109.5E but not exactly 109.5E. Two atoms and 2 lpe& on Br
gives a bent shape around the Br with a bond angle of . 109.5E. Ignore the lpeG when giving the molecular
shape but they are important since they help determine the bond angle.
B
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26) SO42! and SiCl4
In order to be tetrahedral there must be 4 atoms surrounding the central atom. (NO lpe&)
Tetrahedral: 4 atoms surrounding A and no lpe&.
angle = 109.5E (if all surrounding atoms identical)
1) SF4
4 atoms and 1 lpe& on S (4 F and 1 lpe!)
Seesaw
angle . 90E & . 120E
Polar
4) AlCl3
3 atoms and no lpe& (3 things total)
trigonal planar
angle = 120E (exactly)
Nonpolar
*3) SO42!
4 atoms and no lpe& on S (4 O and 0 lpe&)
tetrahedral
angle = 109.5E (exactly)
(neither polar nor nonpolar since it’s an ion)
4) NF3
3 atoms and 1 lpe& on N (3 F and 1 lpe&)
trigonal pyramidal
angle . 109.5E
Polar
*5) SiCl4
4 atoms and no lpe& on Si (4 Cl and 0 lpe&)
tetrahedral
angle = 109.5E (exactly)
Nonpolar
B
(3 & 5, SO42! and SiCl4, are tetrahedral)
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27)
1) AsBr3
trigonal pyramidal: 4 things
around As and they are not
all identical (3 Br and 1 lpe!)
lpe& on As (. 109.5E angles)
Polar (like :NH3, :NF3, :PH3, :PF3, etc.)
*2) BCl 3
trigonal planar (120E angles, exactly)
3 things around B and all identical
(and no lpe&)
Nonpolar (like BH3, BF3, AlCl3 etc.)
3) SCl2
bent: 4 things around S and not
all identical (2 H and 2 lpe!)
lpe& on O (. 109.5E angles)
..
..
Polar (like H2O:, H2S:, etc.)
4) CH2F2
tetrahedral (.109.5E angles)
all 4 atoms on C are not identical
Polar (like CHCl3, CH3Cl, CH2Cl2, ect.)
*5) CS2
linear (180E angles)
the 2 atoms on C are identical
Nonpolar (as is CO2)
Note: The symmetric shapes (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral and
square planar) are the only ones that can give nonpolar. For linear, trigonal planar and tetrahedral
all surrounding atoms have to be identical to each other for the molecule to be nonpolar.
For CH2F2 (#4) all surrounding atoms are NOT identical so the molecule is polar.
Also, the molecules (1 & 3) with lpe! are polar. Generally, if there are 1 or more lpe- on the central
atom, the molecule will be polar (at least for molecular shapes coming from ED geometries involving
2, 3 or 4 ED). The exceptions to this are for the linear molecular geometry arising from the trigonal
bipyramidal ED geometry and the square planar molecular geometry arising from the octahedral ED
geometry. These can be nonpolar if all the atoms surrounding the central atom are identical.
A
(2 & 5 are NONpolar)
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28)
Shape around:
C
C atoms:
tetrahedral - 4 atoms attached to the C atoms and no lone pairs, bond angles around C
are -109.5E and it is polar around the C atoms.
O atom:
bent - 2 atoms and 2 lone pairs of electrons on the O atom. Since the electron-pair
arrangement around the O atom is tetrahedral but all 4 things on the O atom are not
identical the molecule will have a bond angle (C-O-H) of -109.5E and it will be polar
around the O atom. Around the O atom it looks like H2O but with one of the H atoms
replaced with the CH3CH2- group.
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sp3
sp3 sp sp sp3 sp3 sp2 sp sp2
CH2 - CH - C / C - CH2 - CH2 - CH = C = CH2
|
|
CH = CH
sp2 sp2
For carbon atoms with one double bond the carbon forms 3 sp2 hybrid orbitals each with a p orbital
remaining. The sp2 orbitals form a σ bond with hydrogen and/or with carbon atoms (total of 3 σ
bonds). The fourth bond between the carbon atoms results from a side-by-side overlap of the p orbitals
of each carbon atom to form one pi (π) bond. The molecule above has 4 sp3, 4 sp2 and 3 sp hybridized
C atoms.
The molecule above has 3 double bonds. The double-bonded C atoms in the ring each have only 1
double bond to them so there are 2 sp2 hybridized carbon atoms (Remember, each “normal” double
bond has 2 sp2 hybridized carbon atoms; see exception below). There are 2 more double bonds. The C
atom on the right end has only 1 double bond to it so it is sp2 hybridized. The 3rd carbon from the right
has only 1 double bond to it so it is sp2 hybridized. That’s a total of 4 sp2 hybridized carbon atoms.
Normally carbon atoms with double bonds are sp2 hybridized. However, there is an exception when a
carbon atom has 2 double bonds to it (the 2nd carbon from the right). In this case the carbon atom is sp
hybridized (see below).
!C/C!
sp hybrid carbon atoms
\
*
/
C=C=C
/
\
*
/
sp hybrid carbon atom (= C = in middle, = C is sp2)
\
\
/
C=C
/
\
sp2 hybridized carbon (2 single bonds and double bond to carbon)
*
!C !
D
sp3 (4 single bonds)
(4 sp2 hybridized C atoms)
2 things attached to
a central atom
sp hybridization
linear, 180E
3 things attached to
a central atom
sp2 hybridization
trigonal planar, 120E
4 things attached to
a central atom
sp3 hybridization
tetrahedral, 109.5E
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C
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sp3 hybrid orbitals give an overall tetrahedral geometry with angles of 109.5E. When an atom
has 4 things around it the atom generally is using sp3 hybrid orbitals for bonding. The trigonal
pyramidal shape arises when an atom is bonded to 3 other atoms and has 1 lpe& on it (4 things
total). An example is :NH3. The N atom is sp3 hybridized and uses three of the hybrid orbitals
to form the bonds to the H atoms and 1 sp3 hybrid orbital is used for the lpe& on the N atom.
sp
linear, 180E
sp2
trigonal planar, 120E (molecular shapes: trigonal planar, bent)
sp3
tetrahedral, 109.5E (molecular shapes: tetrahedral, trigonal pyramidal, bent)