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Periodicity
Chapter 7
1
Atomic and
Ionic Radii
2
Atomic Radii
• Increase down the
chart
✓
because of increased
energy levels, each
one further from the
nucleus.
• Decrease across
(L-R)
✓
because of increased
effective nuclear
charge, Zeff, pulling on
electrons that are in
the same energy
level.
3
Based on atomic radii, which of the
following molecules would you expect
to have the shortest bonds?
1. F2
2. Cl2
3. Br2
4
Based on atomic radii, which of the
following molecules would you expect
to have the shortest bonds?
1. F2
•
Since fluorine is the smallest of
atoms (since it has the fewest
energy levels), the bonds of the
fluorine molecule would be the
smallest
2. Cl2
3. Br2
5
Which of the following is(are) expected
to be a correct order of atomic radius
from smallest to largest?
Select all that are correct
1.
2.
3.
4.
5.
6.
H<C<F
Mg < Ar < Al
S < Si < Na
Ne < B < H
K < Na < Li
P < As < Sb
6
Which of the following is(are) expected to
be a correct order of atomic radius from
smallest to largest? Select all that are correct.
1.
2.
3.
H<C<F
Mg < Ar < Al
S < Si < Na As you proceed from right to left,
• the effective nuclear charge is less pulling the electron cloud in
not so tightly.
4. Ne < B < H
5. K < Na < Li
6. P < As < Sb Down the table in a column,
• the number of energy levels increase making a longer distance
over which the attraction of the protons is acting on the
electrons
• and each full energy level provides more shielding thus
reducing the effective nuclear charge.
7
Which has larger radius:
1. Cl
−
2. Cl
8
Which has larger radius:
1. Cl
−
2. Cl
• All negative ions
are larger than
their parent
atom
• an e− has been
added and
would repel the
other e− in the
energy level
9
Which has larger radius:
1. Al
3+
2. Al
10
Which has larger radius:
1. Al
3+
2. Al
11
Which of the following gives the correct
ordering of their ionic size (from
smallest to largest)?
1.
2.
3.
4.
5.
6.
−
F
+
Na
< Ne <
−
+
F < Na < Ne
−
+
Ne < F < Na
+
−
Na < Ne < F
+
−
Na < F < Ne
They are all the same size
12
Which of the following gives the correct ordering
of their ionic size (from smallest to largest)?
F− < Ne < Na+
F− < Na+ < Ne
Ne < F− < Na+
Na+ < Ne < F−
•
These particles are all isoelectronic, but you must
consider the number of protons.
•
Na+ has greater effective nuclear charge pulling
in the 2s2 2p6 electrons.
5. Na+ < F− < Ne
6. They are all the same size
1.
2.
3.
4.
13
Order the following according to
increasing atomic/ionic radius.
3−
+
2−
N
Li
C O
1.
2.
3.
4.
5.
6.
7.
8.
+
Li
2−
O
3−
N
C<
<
<
3−
2−
+
C < N < O < Li
+
3−
2−
Li < N < O < C
+
2−
3−
Li < C < O < N
3−
2−
+
N < O < C < Li
3−
2−
+
N < O < Li < C
2−
3−
+
O < N < Li < C
2−
3−
+
O < N < C < Li
14
Order the following according to
increasing atomic/ionic radius.
3−
+
2−
N
Li
C O
1.
2.
3.
4.
•
•
•
5.
6.
7.
8.
C < Li+ < O2− < N3−
C < N3− < O2− < Li+
Li+ < N3− < O2− < C
Li+ < C < O2− < N3−
Li+ is smallest since it only has a 1st energy level of eC is smaller than the neg ions because when they add electrons, the
extra repulsion (p/e- ratio smaller) they become larger in size.
O2− is smaller than N3− because it has less protons holding the 10 eN3− < O2− < C < Li+
N3− < O2− < Li+ < C
O2− < N3− < Li+ < C
O2− < N3− < C < Li+
15
Size of Atoms and their Ions
• Cations
✓
✓
✓
are always smaller
than their parent atom
the atom lost electrons
the ion’s outermost e−
and the outer electrons
now one energy level
closer to nucleus
• Anions
✓
✓
are always larger than
their parent atom
the atom gains e−
increasing repulsion
16
Lattice
Energy
17
Lattice energy
• The enthalpy change (always positive) that
accompanies the separation of a solid
compound into separate gaseous ions.
• Coulomb’s Law shows that the force of
attraction between two oppositely charged
particles is directly proportional to the
magnitude of the charges and inversely
proportional to the distance between those
charges.
+
−
+
−
(Q )(Q )
(Q )(Q )
Fattraction α
F=k
2
2
d
d
18
Lattice Energy
+
−
(Q )(Q )
F=k
2
d
• Factors that affect lattice energy
✓ The magnitude of the ionic charge
✓ 3+ / 3− hold tighter than 1+/1−, thus 3+/3− has
higher lattice energy
✓ The ionic radius, although less important than the
charge factor since ionic radii do not vary greatly.
✓ Smaller ions get closer to each other and the +/− pull
felt is stronger thus higher lattice energy
19
Lattice Energy
• The energy required to separate a mole of solid
ionic compound into its gaseous ions.
ScF3
5096
20
Select the correct lattice energy values
comparisons. Choose all that apply.
1. LiF > LiI
2. RbF > LiF
3. MgO > NaF
4. CaS > BaS
5. K2S > K2O
6. MgF2 > AlF3
21
Select the correct lattice energy values
comparisons.
1. LiF > LiI
• F is a smaller size than I, greater force
2. RbF > LiF
3. MgO > NaF
• Mg is larger charge, greater force
4. CaS > BaS
• Ca is smaller size, greater force
5. K2S > K2O
6. BaF2 > AlF3
22
Each of the processes below are
measurable and represent some
process. Identify all of the ∆H’s below.
•
•
•
•
•
Na(s) + ½Cl2(g) → NaCl(s) ∆H_ = −411 kJ
+
−
Na(g) → Na (g) + 1e
∆H_ = 496 kJ
Cl2(g) → 2Cl(g)
∆H_ = 243 kJ
−
−
Cl(g) + 1e → Cl (g)
∆H_ = −349 kJ
Na(s) → Na(g)
∆H_ = 109 kJ
23
Each of the processes below are
measurable and represent some
process. Identify all of the ∆H’s below.
•
•
•
•
•
Na(s) + ½Cl2(g) → NaCl(s) ∆Hf = −411 kJ
+
−
Na(g) → Na (g) + 1e
∆HIE = 496 kJ
Cl2(g) → 2Cl(g)
∆HBDE = 243 kJ
−
−
Cl(g) + 1e → Cl (g)
∆HEA = −349 kJ
Na(s) → Na(g)
∆HSub = 109 kJ
24
Lattice energy can not be measured directly, it
must be calculated. Calculate the lattice
energy for NaCl(s) using Hess’ Law and the
following information.
NaCl(s) → Na+(g) + Cl−(g)
•
•
•
•
•
Na(s) + ½Cl2(g) → NaCl(s) ∆Hf = −411 kJ
+
−
Na(g) → Na (g) + 1e
∆HIE = 496 kJ
Cl2(g) → 2Cl(g)
∆HBDE = 243 kJ
−
−
Cl(g) + 1e → Cl (g)
∆HEA = −349 kJ
Na(s) → Na(g)
∆HSub = 109 kJ
25
Lattice energy can not be measured directly, it
must be calculated. Calculate the lattice
energy for NaCl(s) using Hess’ Law and the
following information.
NaCl(s) → Na+(g) + Cl−(g)
•
•
•
•
•
NaCl(s) → Na(s) + ½Cl2(g) ∆Hf = +411 kJ
+
−
Na(g) → Na (g) + 1e
∆HIE = 496 kJ
½Cl2(g) → Cl(g)
∆H½BDE = 121.5 kJ
−
−
Cl(g) + 1e → Cl (g)
∆HEA = −349 kJ
Na(s) → Na(g)
∆Hsub = 109 kJ
26
Lattice energy can not be measured directly, it
must be calculated. Calculate the lattice
energy for NaCl(s) using Hess’ Law and the
following information.
NaCl(s) → Na+(g) + Cl−(g)
•
•
•
•
•
•
NaCl(s) → Na(s) + ½Cl2(g) ∆Hf = +411 kJ
+
−
Na(g) → Na (g) + 1e
∆HIE = 496 kJ
½Cl2(g) → Cl(g)
∆H½BDE = 121.5 kJ
−
−
Cl(g) + 1e → Cl (g)
∆HEA = −349 kJ
Na(s) → Na(g)
∆Hsub = 109 kJ
+788 kJ
27
Magnetism
28
para- and dia- magnetism
• paramagnetism
✓
The tendency of a species to be attracted by a
magnetic field as a result of having unpaired
electrons.
• diamagnetism
✓
The tendency of a species not to be attracted (or even
slightly repelled) by a magnetic field as a result of all
of its electrons being paired.
• not to be confused with
✓
ferromagnetism
Materials that can maintain a magnetic field in the
absence of an external magnetic source (Fe,Co, Ni)
29
Evidence of electron configurations
• Paramagnetic studies are used to provide additional
evidence of proposed electron configurations
✓ Experimental evidence shows that both Ti and Ti2+ are
both paramagnetic, indicating which electrons the
atom has lost to beome an ion.
‣
Ti
‣
‣
‣
✓
[Ar] 4s23d2
⨂
Ti2+
⊘⊘〇〇〇
[Ar] 3d2
〇
⊘⊘〇〇〇
If it were the d orbital electrons that were removed,
from the Ti atom to produce the Ti ion, then the ion
would not be paramagnetic
30
Evidence of electron configurations
• Paramagnetic studies are used to provide additional
evidence of proposed electron configurations
✓ Iron atoms are paramagnetic
‣
Fe
‣
✓
⨂
⨂⊘⊘⊘⊘
Fe2+ ions do not demonstrate an increase or decrease in
paramagnetism
‣
Fe2+
[Ar]
3d6
〇 ⨂ ⊘⊘⊘⊘
‣
✓
[Ar] 4s23d6
Fe3+ ions demonstrate increased paramagnetism
‣
‣
Fe3+
[Ar]
〇
3d5
⊘⊘⊘⊘⊘
31
Effective
Nuclear Charge
32
The forces on electrons
•
•
•
Protons pull on electrons. Higher charges attract
more strongly than lower charges. Higher nuclear
charge lowers the energy of orbitals making
electrons in those orbital more stable.
In multi-electron atoms, a second force comes into
play, the repulsion from other electrons. This
repulsion counteracts the nuclear attractions
somewhat, making an electron easier to remove. We
call this repulsion shielding and it reduces the full
nuclear charge “felt” by an electron.
The greatest shielding effect is caused by inner core
full energy levels. Some small repulsion, shielding
does occur between electrons in the same orbital
sets and same energy levels
33
What is the approximate Zeff (effective
nuclear charge) “felt” by the 3p
electrons in the phosphorus atom?
• Input a value
34
What is the approximate Zeff (effective
nuclear charge) felt by the 3p electrons
in the phosphorus atom?
•
•
•
•
+5
The 10- core electrons blocks 10 of the
15+ from the protons resulting in +5
For atoms in the s and p valence
orbitals, we can consider the effective
nuclear charge to be equal to the
number of valence electrons.
sort of.....
35
Zeff
•
•
The effective nuclear charge experienced by the valence
electron in sodium depends mostly on the 11+ charge of the
nucleus and the 10− charge of the neon core. If the neon
core were totally effective in shielding the valence electron
from the nucleus, then the valence electron would
experience an effective nuclear charge of 1+.
But, the 3s1 electron has some probability of being inside
the Ne core. As a consequence of this "penetration," the
core is not totally effective in screening the 3s1 electron from
the nucleus. Thus the effective nuclear charge experienced
by the 3s1 electron is actually a teeny bit greater than 1+. 36
In a germanium atom, when comparing electrons
in the 3s, 3p and 3d orbitals, which electron(s) will
experience the greatest effective nuclear charge?
Electron(s) in the
1.
2.
3.
4.
5.
3s orbital
3p orbital
3d orbital
3s and 3p will feel the same
3s, 3p, and 3d will all feel the same
37
In a germanium atom, which electron(s) will
experience the greatest effective nuclear
charge? Electron(s) in the
1. 3s orbital
•
You might be inclined to think that 3s and 3p & 3d
would “feel” or “see” the same effective nuclear
charge, and it is nearly the same, however s orbital
electrons feel slightly more nuclear charge since the
s orbital penetrates deeper into the inner core of
electrons. (check the next slide)
2. 3p orbital
3. 3d orbital
4. 3s and 3p will feel the same
5. 3s, 3p, and 3d will all feel the same
38
Orbital Penetration & Zeff
•
•
•
•
The radial probability function for the
2s orbital of the hydrogen atom (red
curve) shows a "node" of probability
close to the nucleus, whereas the 2p
orbital (blue curve) does not.
As a result, an electron in the 2s
orbital for a many-electron atom
"sees" more of the nuclear charge
than does an electron in the 2p orbital.
In other words, the effective nuclear
charge experienced by the 2s electron
is greater than that for the 2p electron.
This difference leads to the
observation that in a many-electron
atom the orbitals for a given n value
increase in energy s < p < d < f, that
is, orbital ns is lower energy than np,
which is lower energy than nd.
39
Ionization
Energy
40
Ionization Energy
• The amount of energy required to forcibly
remove an electron from a gaseous atom.
✓
✓
Energy added as heat, light, or electricity
Equation: X + IE →
+
X
+
−
e
41
Which will have the highest first
ionization energy?
1. C
2. Si
42
Which will have the highest
first ionization energy?
1. C
2. Si
• C has fewer energy levels thus
the e− is closer and “feels”
greater nuclear attraction on the
electron, making it more difficult
to remove an electron
43
Is the first ionization energy of carbon
larger, smaller or the same as
nitrogen?
1. larger
2. smaller
3. the same as
44
Is the first ionization energy of carbon
larger, smaller or the same as
nitrogen?
1.
2.
•
3.
larger
smaller
extra proton provides
greater nuclear
attraction, making the
valence electron harder
to remove.
the same as
45
Is the first ionization energy of oxygen
larger, smaller or the same as
nitrogen?
1. larger
2. smaller
3. the same as
46
You would think that
the trend continues
for N and O, however
the ionization energy
of oxygen is lower
than for nitrogen.
Why?
47
You would think that the trend
continues for N and O, however
the ionization energy of oxygen
is lower than for nitrogen. Why?
Even though there is one added proton, the
added repulsion of 4th electron that is paired
in the p orbital is greater than the increase in
proton pull from the extra proton, this repulsion
makes the IE required to remove that p
electron lower in oxygen compared to nitrogen
N ⨂ ⊘⊘⊘ O ⨂ ⨂⊘⊘
In fluorine, the added proton overshadows the
increased repulsion. F ⨂ ⨂⨂⊘
48
Which will have the highest first
ionization energy?
1.
2.
3.
4.
5.
C
N
O
Al
Si
49
Which will have the highest first
ionization energy?
1.
2.
•
3.
4.
5.
C
N
While the O as it has greater nuclear
charge; the extra repulsion caused by
the paired p4 electron makes it a bit
easier to remove than one of the p3
electron in N
O
Al
Si
50
Ionization Energy
• The amount of energy required to forcibly remove an
electron from an atom.
✓
✓
Energy added as heat, light, or electricity
Equation: X + IE → X+1 + e-1
• Increases across the
chart (L-R)
✓
✓
Due to increased effective
nuclear charge
No increase in shielding
• Decreases down the
chart
✓
✓
due to outermost electrons
further from the nucleus
increased shielding
F ir st Ion iza t io n En er gie s ( kJ/mole )
H
1
1311
Li Be
2
521 899
Na Mg
3
496 737
K Ca
4
419 590
Rb Sr
5
403 550
Cs Ba
6
376 503
7 Fr Ra
He
1
2370
B
C
N
O
F
Ne
2
799 1087 1404 1314 1682 2080
Al Si
P
S
Cl
Ar
3
576 786 1052 1000 1245 1521
Ga Ge As
Se
Br
Kr
4
579 762 944 941 1140 1351
In Sn
Sb
Te
I
Xe
5
558 709 832 869 1009 1170
Tl Pb
Bi
Po
At
Rn
6
589 716 703 812
1037
51
7
Ionization Energy
F ir st Ion iza t io n En er gie s ( kJ/mole )
H
1
1311
Li Be
2
521 899
Na Mg
3
496 737
K Ca
4
419 590
Rb Sr
5
403 550
Cs Ba
6
376 503
7 Fr Ra
He
1
2370
B
C
N
O
F
Ne
2
799 1087 1404 1314 1682 2080
Al Si
P
S
Cl
Ar
3
576 786 1052 1000 1245 1521
Ga Ge As
Se
Br
Kr
4
579 762 944 941 1140 1351
In Sn
Sb
Te
I
Xe
5
558 709 832 869 1009 1170
Tl Pb
Bi
Po
At
Rn
6
589 716 703 812
1037
7
52
Zeff
•
•
The effective nuclear charge experienced by the valence
electron in sodium depends mostly on the 11+ charge of the
nucleus and the 10− charge of the neon core. If the neon
core were totally effective in shielding the valence electron
from the nucleus, then the valence electron would
experience an effective nuclear charge of 1+.
But, the 3s1 electron has some probability of being inside
the Ne core. As a consequence of this "penetration," the
core is not totally effective in screening the 3s1 electron from
the nucleus. Thus the effective nuclear charge experienced
by the 3s1 electron is actually a teeny bit greater than 1+. 53
Orbital Penetration & Zeff
•
•
•
•
The radial probability function for the
2s orbital of the hydrogen atom (red
curve) shows a "bump" of probability
close to the nucleus, whereas that for
the 2p orbital (blue curve) does not.
As a result, an electron in the 2s orbital
for a many-electron atom "sees" more
of the nuclear charge than does an
electron in the 2p orbital.
In other words, the effective nuclear
charge experienced by the 2s electron
is greater than that for the 2p electron.
This difference leads to the observation
that in a many-electron atom the
orbitals for a given n value increase in
energy with increasing l value, that is,
orbital ns is lower energy than np,
which is lower energy than nd.
54
Which ionization energy will be the
largest?
(Choose more than one if the same.)
1.
2.
3.
4.
5.
1st IE of Na
2nd IE of Na
1st IE of Mg
2nd IE of Mg
3rd IE of Mg
55
Which will be the largest?
(Choose more than one if the same.)
1.
2.
3.
4.
5.
•
1st IE of Na
2nd IE of Na
1st IE of Mg
2nd IE of Mg
3rd IE of Mg
6
The same 2p electron is being removed,
however the Mg has one extra proton
creating greater nuclear charge.
56
Successive Ionization Energy
• The amount of energy required to repeatedly remove
electrons.
✓
Energy could be added in the form of heat, light, or electricity.
X + IE → X+ + e−
Second: X+ + IE → X2+ + e−
Third: X2+ + IE → X3+ + e−
Etc, etc, etc.
There is a very large increased value when the electron
being removed is one more than the number of valence
electrons (in red on the next side)
✓ Compared to the previous, the removal of this electron is
from a full energy level that is one level closer to the
nucleus than the previous electron removed.
• First:
•
•
•
•
57
Successive Ionization Energy
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
1st
1311
23 70
521
899
799
1087
1404
1314
1682
20 80
496
737
576
786
1052
1000
1245
1521
2nd
3rd
4th
5220
73 04 11752
1756 14 8 49 20899
2422 3657 25 0 19
2393 4622
6223
2856 4573
7468
3396 5297
7468
3367 6050
8423
3946 6165
9301
45 64 6918
9542
1447 77 38 10546
1814 2750 11 5 78
1582 3232
4361
1901 2914
4959
2258 3387
4544
2287 3850
5162
2653 3927
5886
5th
6th
7th
8th
32660
37 8 22
9446
10990
11028
12138
13373
13624
14820
16 0 07
6272
6947
6542
7526
46988
53 2 50
13325
15167
15148
16644
18033
18361
19693
21 5 16
8500
9359
8587
63970
71 3 12
17869
19972
20175
21767
23465
23658
25858
27 1 12
11028
11964
83652
91 9 50
22963
25501
25742
27575
29110
30489
31734
33 4 42
13778
58
Electron
Affinity
59
Electron Affinity
• The energy change accompanying addition
of electron to gaseous atom:
✓
✓
✓
✓
This is NOT the opposite of ionization energy.
Cl + e− → Cl−
It’s a measure of the attraction of an atom for an
electron.
The greater the attraction, the more negative its affinity
value will be.
‣
remember energy is released when bonds form...and an
incoming electron is like forming a bond between that
electron and the atom.
60
Electron Affinity
61
PES
Photon Electron
Spectroscopy
Evidence for the Validity
of the Shell Orbital
Model
62
PES − Measuring Binding Energy
• X-Rays of a particular energy (Ex-ray) are “shot” at atoms
• Electrons from each energy level of that atom leap away with
varying amounts of excess kinetic energy from the X-ray
• The charged hemisphere bends that spray of electrons
✓
faster (more KE) electrons bend less, slower (Less KE) electrons bend more
• Detector tells us how many of each electron of a particular
energy (KEdetected).
• Binding energy (BEelectron) can be calculated
• BEelectron = Ex-ray − KEdetected
Photo Electron Spectra
Relative Number of Electrons
for Scandium
1000
Scandiu
m
Electron Detector
100
10
Energy
1
0.1
Negatively
Charged
Hemisphere
(Constant V)
X-Rays
63
Which element is represented by the
PES black peaks shown below.
lithium
carbon
oxygen
neon
Photo Electron Spectra
Relative Number of Electrons
1.
2.
3.
4.
100
10
1
0.1
Energy
64
Which element is represented by the
PES black peaks shown below.
lithium
carbon
oxygen
neon
Photo Electron Spectra
2p4
Relative Number of Electrons
1.
2.
3.
4.
100
2s2
2
1s
10
1
0.1
Energy
•
Turn to your mate and decide what element is
represented by the red peaks.
65
Which element is represented by the
PES black peaks shown below.
lithium
carbon
oxygen
neon
Photo Electron Spectra
2p4
Relative Number of Electrons
1.
2.
3.
4.
100
2s2
2
1s
10
1
0.1
Energy
•
Explain to your mate why the first 2 peaks of
the red element are to the right of the first two
peaks of the black element.
66
Which elements are represented by the
PES black peaks shownbelow.
Photo Electron Spectra
Relative Number of Electrons
O
e
C
o
100
10
1
0.1
Energy
•
Explain to your mate why the first 2 peaks of
the red element are to the right of the first two
peaks of the black element.
67
Isotopes
Percent Abundance
68
Determine the average atomic mass for a
recently discovered element, Biggsium, Bg. Its
isotopes are found in nature according to the
chart below. Calculate the average atomic mass.
•
Report your answer to the
nearest 10ths place.
isotope
% abundance
300Bg
65%
30%
5%
303Bg
304Bg
69
Determine the average atomic mass for a
recently discovered element, Biggsium, Bg. Its
isotopes are found in nature according to the
chart below. Calculate the average atomic mass.
•
Report your answer to the nearest 10ths
place.
[(300 × 0.65) + (303 × 0.3) + (304 × 0.05)] = 301.1
isotope
% abundance
300Bg
65%
30%
5%
303Bg
304Bg
70
You would like to calculate the % abundance
found in nature of the newly discovered element
Maxogen, Mx. There are only two naturally
occurring isotopes: 324Mx and 327Mx. The
average atomic mass is 325.2
•
ths
10
Report the % to the nearest
place for the
isotope that occurs in highest quantity.
71
You would like to calculate the % abundance
found in nature of the newly discovered element
Maxogen, Mx. There are only two naturally
occurring isotopes: 324Mx and 327Mx. The
average atomic mass is 325.2
•
ths
10
Report the % to the nearest
place for the
isotope that occurs in highest quantity.
324x + 327(1 − x) = 325.2
324x + 327 − 327x = 325.2
−3x = −1.8
•
x = 0.6
60% is the 324Mx
72
Cesium is the most reactive of all the alkali
metals (Francium has not been studied
extensively due to its radioactivity.) What
atomic property is most responsible for its
high reactivity?
1. Cs is the softest of the alkali metals.
2. Cs has the lowest ionization energy of the alkali
metals and the greatest metallic character.
3. Cs has the greatest density of the alkali metals.
4. Cs is the most conductive of the alkali metals.
5. Cs has the most electrons of the alkali metals.
73
Cesium is the most reactive of all the alkali metals
(Francium has not been studied extensively due to its
radioactivity.) What atomic property is most
responsible for its high reactivity?
1. Cs is the softest of the alkali metals.
2. Cs has the lowest ionization energy of the alkali
metals and the greatest metallic character.
• Remember that IE gives us a direct insight to the
ease with which electrons leave an atom, and the
easier an electron leaves, and the more metallic
character it has.
3. Cs has the greatest density of the alkali metals.
4. Cs is the most conductive of the alkali metals.
5. Cs has the most electrons of the alkali metals.
74
The electron configuration of germanium is
2
10
2
[Ar]4s 3d 4p This element is known to form
more than one oxide. Which pair of formulas
include the two most likely formulas?
1.
2.
3.
4.
5.
GeO and GeO2
GeO2 and Ge2O7
Ge2O3 and GeO2
GeO and Ge2O3
Ge2O3 and Ge2O7
75
The electron configuration of germanium is
2
10
2
[Ar]4s 3d 4p This element is known to form
more than one oxide. Which pair of formulas
include the two most likely formulas?
1. GeO and GeO2
• Ge like (Sn and Pb) forms oxidation
states +2 and +4
2. GeO2 and Ge2O7
3. Ge2O3 and GeO2
4. GeO and Ge2O3
5. Ge2O3 and Ge2O7
76
How does the electron structure of a
phosphorus atom differ from a
phosphorus ion?
1. The phosphide ion has more electrons.
2. The phosphorus atom has more unpaired
electrons.
3. The phosphide ion has more core electrons.
4. The phosphorus atom is smaller than the
phosphorus ion.
77
How does the electron structure of a
phosphorus atom differ from a
phosphorus ion?
•
1.
2.
3.
4.
Phosphorus forms a −3 oxidation state, gaining 3
electrons
The phosphide ion has more electrons.
The phosphorus atom has more unpaired
electrons.
The phosphide ion has more core electrons.
The phosphorus atom is smaller than the
phosphorus ion.
78
Which radiation below has the highest
energy?
1.
2.
3.
4.
5.
infrared
microwaves
ultraviolet
visible
x-rays
79
Which radiation below has the highest
energy?
1.
2.
3.
4.
5.
•
infrared
microwaves
ultraviolet
visible
x-rays
in order from lowest to highest it would
be: mw < IR < visible < UV < x-rays
80
Which radiation below can be used to
analyze the concentration of colored
solutions?
1.
2.
3.
4.
5.
infrared
microwaves
ultraviolet
visible
x-rays
81
Which radiation below can be used to
analyze the concentration of colored
solutions?
1.
2.
3.
4.
infrared
microwaves
ultraviolet
visible
• Since color is a manifestation of visible light.
We used the spectrophotometer to measure
+1
the intensity of color of a red FeSCN solution
to measure its concentration.
5. x-rays
82
Which radiation below is used in the
analysis of bond length and atomic
arrangement in crystal structures.
1.
2.
3.
4.
5.
•
infrared
microwaves
ultraviolet
visible
x-rays
x-ray crystallography is used to measure
bond lengths and radii of atoms since the
nucleus’s show up as more dense on the xray image
83
Which radiation below is used as a
detection beam across doorways and
windows?
1.
2.
3.
4.
5.
infrared
microwaves
ultraviolet
visible
x-rays
84
Which radiation below is used as a
detection beam across doorways and
windows
1. infrared
• IR is also used for motion detectors
2. microwaves
3. ultraviolet
4. visible
5. x-rays
85
The value of Plank’s constant is 6.63x10-34 J
8
-1
sec. The velocity of light is 3x10 m sec .
Which value in joules is closest to the energy of
15
-1
a photon with frequency of 8x10 sec
1.
2.
3.
4.
5.
-20
10
1x
-20
5 x 10
-19
5 x 10
-18
1 x 10
-18
5 x 10
86
The value of Plank’s constant is 6.63x10-34 J
8
-1
sec. The velocity of light is 3x10 m sec .
Which value in joules is closest to the energy of
15
-1
a photon with frequency of 8x10 sec
1.
2.
3.
4.
5.
•
-20
10
1x
-20
5 x 10
-19
5 x 10
-18
1 x 10
-18
5 x 10
E = hv, so the speed of light is a distractor
and not necessary to calculate the energy.
87
Calculate the wavelength of light that is
released when a hydrogen electron
transitions from n=4 to n=1
• E=
• wavelength =
88
Calculate the wavelength of light that is
released when a hydrogen electron
transitions from n=4 to n=1
• Pull equations off the blue sheet.
89
Which is a list of elements in order of
increasing first ionization energy?
Select all that apply.
1. Si < P < Cl
2. N < P < As
3. Sr < Ca < Mg
4. Cl < Br < I
5. F < Ne < Na
90
Which correctly list elements in order of
increasing first ionization energy?
1. Si < P < Cl due to greater effective
nuclear charge as you proceed L>R
across the chart
2. N < P < As
3. Sr < Ca < Mg due to less energy levels
as you proceed up the chart.
4. Cl < Br < I
5. F < Ne < Na
91
The overall electron configuration of the sulfide
ion is most similar to (isoelectronic with) the
electron configuration of the
1.
2.
3.
4.
5.
oxide ion
chlorine atom
oxygen atom
sodium ion
potassium ion
92
The overall electron configuration of the sulfide
ion is most similar to (isoelectronic with) the
electron configuration of the
1.
2.
3.
4.
5.
•
oxide ion
chlorine atom
oxygen atom
sodium ion
potassium ion
-1
Since potassium ion has 18e (since it
-1
lost 1e ) as does the sulfide ion (since it
-1
gained two e )
93
Which describes the behavior of
potassium metal during a chemical
reaction? Select all that apply.
1. K atoms become ions with a 1+ charge.
2. K atoms take on protons.
3. K atoms become ions with a
corresponding decrease in radius.
4. When K atoms bond with sulfur, the
bond is stronger than the bond that
forms when K atoms bond with
selenium.
94
Which describes the behavior of potassium metal
during a chemical reaction?
1.
K atoms become ions with a 1+ charge.
• due to the loss of 1e-1
2. K atoms take on protons.
3. K atoms become ions with a corresponding decrease in
radius.
• cations lose e-1 and the p/e ratio increases, and the
valence e-1 are one energy level closer.
4. When K atoms bond with sulfur, a stronger ionic bond
results than when K atoms bond with selenium.
• Coulombs law says that when distance is shorter, the
force is greater (and the bond is stronger). S is smaller
than Se.
95
-34
10
The value of Planck’s constant is 6.63 x
17
-1
Jsec. The velocity of light is 3.0 x 10 nm sec .
Which value is closest to the wave length in
nanometers of a quantum of light with frequency
15
-1
of 6 x 10 sec ?
1.
2.
3.
4.
5.
10
25
50
75
100
96
-34
10
The value of Planck’s constant is 6.63 x
17
-1
Jsec. The velocity of light is 3.0 x 10 nm sec .
Which value is closest to the wave length in
nanometers of a quantum of light with frequency
15
-1
of 6 x 10 sec ?
1. 10
2. 25
3. 50
•
c = wavelength*frequency
•
Planck’s constant is a distractor
4. 75
5. 100
97
What is the number of half-filled
orbitals in an atom of phosphorus?
1.
2.
3.
4.
5.
none
one
three
five
seven
98
What is the number of half-filled
orbitals in an atom of phosphorus?
1.
2.
3.
4.
5.
none
one
three
five
seven
99
Which description of electron configuration
applies to atoms of potassium, chromium,
and copper?
1.
2.
3.
4.
5.
only one half-filled orbital
only one filled sublevel
only three half-filled orbitals
only nine filled orbitals
only five filled sublevels
100
Which description of electron configuration
applies to atoms of potassium, chromium,
and copper?
1. only one half-filled orbital
• they all have a half filled s1 orbital
(sublevel)
2. only one filled sublevel
3. only three half-filled orbitals
4. only nine filled orbitals
5. only five filled sublevels
101
The most abundant isotopes of hydrogen and
oxygen are 11H, 21H, 168O, and 178O, respectively.
Using these isotopes only, what is the number of
different possible values for the molar mass of
water in grams?
•
input your numeric answer
102
The most abundant isotopes of hydrogen and
oxygen are 11H, 21H, 168O, and 178O, respectively.
Using these isotopes only, what is the number of
different possible values for the molar mass of
water in grams?
•
4
•
•
•
there are 6 combinations, but 2 of
them repeat
18(16+1+1), 19(16+1+2), 20(16+2+2)
19(17+1+1), 20(17+1+2), 21(17+2+2)
103
Which set of formulas or symbols best
illustrates the Law of Multiple Proportions?
12
14
1. 6C and 6C
14
14
2. 6C and 7N
3. CO and CO2
4. CH4 and CCl4
5. C2H5OH and CH3OCH3
104
Which set of formulas or symbols best
illustrates the Law of Multiple Proportions?
12
14
1. 6C and 6C
14
14
2. 6C and 7N
3. CO and CO2
4. CH4 and CCl4
5. C2H5OH and CH3OCH3
105
All of the following can be inferred from the
Lewis dot diagram of a neutral atom below
EXCEPT
X
1.
2.
3.
4.
X belongs to the same family as sulfur
X has two half filled p-orbitals
X has at least ten kernel (core) electrons
X can accept two electrons to become an ion with
a charge of 2-
5. X has four electrons with l quantum number = 1
106
All of the following can be inferred from the
Lewis dot diagram of a neutral atom below
EXCEPT
X
1.
2.
3.
4.
X belongs to the same family as sulfur
X has two half filled p-orbitals
X has at least ten kernel (core) electrons
X can accept two electrons to become an ion with
a charge of 25. X has at least four electrons from l quantum
number of 1
107
Which gives a list of species with the same
electron configuration; that is, species that
are isoelectronic?
1. Mg, Ca, Sr
2+
2+
2+
2. Mg , Ca , Sr
2−
3−
3. F , S , As
4. Fe, Co, Ni
2−
+
5. S , Cl , K
108
Which gives a list of species with the same
electron configuration; that is, species that
are isoelectronic?
1. Mg, Ca, Sr
2+
2+
2+
2. Mg , Ca , Sr
2−
3−
3. F , S , As
4. Fe, Co, Ni
2+
5. S , Cl , K
109
Which color in the visible spectrum is
associated with the lowest frequency?
1.
2.
3.
4.
5.
blue
green
violet
yellow
red
110
Which color in the visible spectrum is
associated with the lowest frequency?
1.
2.
3.
4.
5.
blue
green
violet
yellow
red
111
According to quantum mechanics, what is the
maximum number of electrons that can occupy
the third energy level in a ground state atom?
•
input your numeric answer
112
According to quantum mechanics, what is the
maximum number of electrons that can occupy
the third energy level in a ground state atom?
•
•
18
s, d, p electrons only (no f’s etc)
113
Which pair of atoms represents nuclei
that have the same number of
neutrons?
1.
2.
3.
4.
5.
55Co
and 58Co
57Mn and 57Fe
58Ni and 57Fe
58Ni and 57Co
59Ni and 56Fe
114
Which pair of atoms represents nuclei
that have the same number of
neutrons?
55Co
1.
2.
3.
4.
•
5.
and 58Co
57Mn and 57Fe
58Ni and 57Fe
58Ni and 57Co
both have 30 neutrons
59Ni and 56Fe
115
Quantum
Numbers
Quantum Numbers are no longer
part of the AP curriculum.
We will review them before the
SAT subject test in June.
116
Describe a possible set of quantum
numbers for the electron removed for
either the 2nd IE of Na or the 3rd IE of
Mg
117
Describe a possible set of quantum
numbers for the electron removed for
either the 2nd IE of Na or the 3rd IE of
Mg
•
•
2, 1, 1, +½
OR 2, 1, (or -1, or 0), +½(or -½)
118
What is the number of electrons in 23V
that have an L quantum number of 2?
•
Input your answer
119
What is the number of electrons in 23V
that have an L quantum number of 2?
•
Quantum number L = 2 represents d electrons,
and Vanadium has d3 or 3 electrons in the d
orbitals
120
What is the number of electrons in 16S
that have an L quantum number of 1?
•
Input your answer
121
What is the number of electrons in 16S
that have an L quantum number of 1?
•
Quantum number L = 1 represents p electrons,
and sulfur has p4 or 4 electrons in the 3p orbitals
AND p6 or 6 electrons in the 2p orbitals
122
Consider the set of quantum number numbers:
3, 2, -1, -1/2
Which set of quantum numbers represents an
electron with the same energy (degenerate
orbitals) but different angular orientation in space
as the electron represented above?
1.
2.
3.
4.
5.
3, 2, -1, +½
3, 1, -1, -½
3, 2, 0, +½
4, 1, 0, +½
2, 2, -1, -½
123
Consider the set of quantum number numbers:
3, 2, -1, -1/2
Which set of quantum numbers represents an
electron with the same energy but different
orientation in space as the electron represented
above?
1. 3, 2, -1, +1/2
2. 3, 1, -1, -1/2
3. 3, 2, 0, +1/2
• both d electrons, but different orientation.
4. 2, 1, 0, +1/2
5. 2, 2, -1, -1/2
124
Consider the set of quantum number numbers:
3, 2, -1, -1/2
Which set of quantum numbers represents a
different electron in the same (exact) orbital.
1.
2.
3.
4.
5.
3, 2, -1, +1/2
3, 1, -1, -1/2
3, 2, 0, +1/2
2, 1, 0, +1/2
2, 2, -1, -1/2
125
Consider the set of quantum number numbers:
3, 2, -1, -1/2
Which set of quantum numbers represents a
different electron in the same orbital.
1.
2.
3.
4.
5.
3, 2, -1, +1/2
3, 1, -1, -1/2
3, 2, 0, +1/2
2, 1, 0, +1/2
2, 2, -1, -1/2
126
Write three possible groups of quantum numbers
4
that could describe any one of the 4f electrons
in the Neodymium atom.
127
Write a set of three possible sets of quantum
numbers to describe the ground state all three of
the 4p3 electrons in the Arsenic atom.
128