Periodic Trends - NJCTL
Transcript of Periodic Trends - NJCTL
Slide 1 / 102
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AP Chemistry
Periodic Trends
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Recall that the periodic law states that the physical and chemical properties of the elements tend to recur in a systematic way when
arranged by increasing atomic number.
The Periodic Law
Slide 4 / 102
Over the course of this unit, we will use our knowledge of the atom to explain the periodic trends we see regarding the following
properties:
PROPERTY DEFINITION
Ionic Charge charge of common ion formed by that element
Atomic/Ionic Radii Distance from the nucleus to outermost electron
Density Ratio of Mass/Volume
Ionization Energy Energy required to remove valence electron
Metallic Character Disposition to have metallic characteristics - ie. conduct electricity
Electronegativity Measure of attraction for electrons when the atom is sharing electrons in a molecule.
The Periodic Law
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Recall that the periodic law states that the physical and chemical properties of the elements tend to recur in a systematic way when
arranged by increasing atomic number.
Let's look at the first eleven elements to illustrate this.
H He Li Be B C N O F Ne Na
Atomic Number
1 2 3 4 5 6 7 8 9 10 11
Ionic Charges
+1,-1 NA +1 +2 +3 +4 -3 -2 -1 NA +1
Notice that neither He or Ne form ions. Also, notice that in both cases the atom that precedes them can form a -1 ion
and the atom that succeeds them forms a +1 ion.
There is definitely a systemic pattern here!
The Periodic Law
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The pattern can be easily visualized on a graph, particularly as we move past the first 11 elements!
ion charge
+1
+2
+3
-1
-2
-3
atomic number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
+4
The Periodic Law
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The Periodic Law and the Quantum Model This trend in ionic charge can be easily explained if we apply the
quantum model of the atom.
Element
Principal Quantum
Number (N) of valence electrons
Electron Configuration
Lose/Gain
electrons
Ionic Charge
H 1 1s1 gain 1
lose 1
-1
+1He 1 1s2 NA NA Li 2 [He]2s1 lose 1 +1Be 2 [He]2s2 lose 2 +2
B 2 [He]2s22p1 lose 3 +3
C 2 [He]2s22p2 lose 4 +4
N 2 [He]2s22p3 gain 3 -3
O 2 [He]2s22p4 gain 2 -2 F 2 [He]2s22p5 gain 1 -1Ne 2 [He]2s22p6 NA NA
Na 3 [Ne]3s1 lose 1 +1
The pattern recurs with every increase in the
principal quantum number. This means
every time a new shell of electrons is filled, the
pattern repeats!
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Let's use to quantum model to answer some questions about these ionic charges.
Question 1: Why do both He and Ne not form ions?
Both have a full principal energy level
He = 1s2
Ne = [He]2s22p6
Question 2: Why do both Li and Na have the same charge?
Both require only a small amount of energy to lose 1 electron to become a noble gas with a full principal energy level.
The Periodic Law and the Quantum Model
move for answer
move for answer
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Question 3: Explain why P would be expected to have the same ionic charge as N?
Both have the same number of valence electrons (5) so both need to gain three electrons to fill their outer principal energy level.
N = [He]2s22p3 gain 3 e- --> Ne
P = [Ne]3s23p3 gain 3 e- --> Ar
Question 4: After sodium, which element would most likely form an ion with +1 charge and why?
Potassium (K), because it is beginning to fill the 4th principal energy level with 1 electron, just as sodium was beginning the
3rd with 1 electron.
The Periodic Law and the Quantum Model
move for answer
move for answer
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We have seen that the quantum model explains the periodic trend with regard to ionic charges for the main group elements in
the first three periods.
Quantum theory can also explain the periodic trends amongst the transition elements that are in the midst of filling their "d" orbitals.
d orbital
transition elements
+3+3
+4+4
+6+6
+7+7
+3+3
+2+2
+1+1
+2+2
+5+5
+3+3
The Periodic Law and the Quantum Model
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The charges increase from left to right as the atoms lose both their two valence "s" electrons and however many "d" electrons
they have also.
After the Mn group, the charges decrease, one of the reasons being that the stability of the "d" orbital increases as
it becomes full.
+3
d orbital
transition elements
+3 +4+4
+6+6
+7+7
+3+3
+2+2
+1+1
+2+2
+5+5
+3+3
The Periodic Law and the Quantum Model
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Let's use quantum theory to explain the trends we see amongst the charges of the transition elements.
Question 1: Elements within the Fe group can form ions of both +2 and +3 charges. Explain why the +3 charge is more common:
Fe = [Ar]4s23d6
The 4s electrons are readily lost yielding the +2 ion.
A half-full "d" orbital is quite stable so Fe will lose 1 d orbital electron as well to yield the +3 ion.
The Periodic Law and the Quantum Model
move for answer
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Let's use quantum theory to explain the trends we see among the charges of the transition elements.
Question 2: Why do the elements in the zinc group tend to only form ions with a +2 charge?
Zn = [Ar]4s23d10
The "d" orbital is full so only the outer "s" electrons are lost.
The Periodic Law and the Quantum Model
move for answer
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1 The trends in chemical and physical properties tend to recur as atoms…
A Fill a new principal energy level
B Gain more neutrons
C Decrease in mass
D Increase in atomic number
E Both A and D
Ans
wer
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2 An atom with a +2 charge must be in the same group as barium.
True
False
Ans
wer
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3 Which of the following BEST explains why O and S both form ions with a -2 charge?
A They both have the same atomic number
B They are both in the same period
C They both have the same electron configuration
D They both have the same number of valence electrons
E They both have the same mass
Ans
wer
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4 An atom with the electron configuration of [Kr]5s24d2 would be in the same group as _____ and have a likely charge of ____.
A Sc, +1
B Hf, +4
C Ti, +3
D Zn, +2
E Y, +1
Ans
wer
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5 Atoms on the right side of the chart tend to form negative ions because...
A Their principal energy level is almost empty
B Their principal energy level is almost full
C Their atomic number is less than other elements in that period
D Both B and C
E A, B, and C
Ans
wer
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The Periodic Law and Atomic/Ionic RadiiThe atomic/ionic radii of an atom can be measured and or
calculated a number of different ways. We will be using values calculated via the Clementi method (E. Clementi,
D.L.Raimondi, and W.P. Reinhardt, J. Chem. Phys. 1963, 38, 2686.)
The atomic radius of an atom or ion can be thought of as the distance
between the nucleus and the region of space where the outermost valence electrons would be most likely found.
radius
**Note: Remember an electron is not in orbit round the nucleus like a planet. The radius therefore is determined out to the point
where the electron charge density starts to diminish
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Let's examine the trend in atomic radii for the first 18 elements.
atomic number
radius (pm)
0
200
100
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
H
LiNa
HeNe
Ar
We clearly see two trends!
1. As atomic number increases down a group, the radii increase.
H < Li < Na
2. As atomic number increases across a period, the radii decrease.
Li > Be > B > C > N > O > F > Ne
The Periodic Law and Atomic Radii
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The Periodic Law and Atomic Radii
The distance the electrons can be from the nucleus is governed by Coulomb's law of attraction. The greater the charge, the greater the
attraction between the charges, and the shorter the distance.
As atomic number increases across a period, so does the nuclear charge (Z) resulting in a greater attraction and a smaller
distance between the nucleus and the outermost electrons.
Lithium (Z=3)
radii = 167 pm
Carbon (Z=6)
radii = 67 pm
Neon (Z=10)
radii = 38 pm
**Note: The size of an atom is NOT determined by the size of the nucleus. It is the electron cloud that contains most of the volume
of an atom and therefore determines the radii.
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The Periodic Law and Atomic Radii
Lithium (Z=3)
1s22s1
radii = 167 pm
Only a certain number of electrons are permitted within
a given energy level, so additional ones must be added to higher energy levels farther from the
nucleus.
Hydrogen (Z=1)
1s1
radii = 53 pm
Sodium (Z=11)
1s22s22p63s1
radii = 190 pm
Why don't the radii continue to get smaller as the atomic number and nuclear charge increase. The quantum model explains why.
The core electrons shield the valence electrons from the nucleus thus diminishing the coulombic attraction and
increasing the atomic radii.
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The Periodic Law and Ionic RadiiWhen electrons are gained or lost, the effect on the radii can be
dramatic or slight but there are some certainties.
If an atom loses electrons, the radii will
decrease.
Ca --> Ca2+ + 2e-
194 pm 99 pm
When electrons are lost, the remaining electrons feel a
stronger coulombic attraction from the nucleus.
If an atom gains electrons, the radii will
increase.
F + e- --> F-
42 pm 136 pm
When electrons are gained, the nuclear charge is spread
over a larger number of electrons, resulting in a
weaker coulombic attraction.
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The Periodic Law and Ionic RadiiLet's rank a series of atoms and ions in order of increasing radii.
Al3+ Al Mg Mg2+
Whenever comparing radii, use the following procedure:
1. Determine the energy level of the atom/ion.
2. For atoms in the same energy level, use the nuclear charge (Z) to determine the radii.
Al3+ Al Mg Mg2+
Energy Level 2 3 3 2
"Z" 13 13 12 12
Al3+ < Mg2+ < Al < Mg
radius (pm) 50 < 65 < 118 < 145
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In this case, Na+, Mg2+, Al3+, O2-, and F- are all isoelectronic with Ne. As a result, they all experience the same core shielding.
The ionic radii then decreases with an increasing nuclear charge.
Al3+ < Mg2+ < Na+ < F- < O2-
Z = 13 12 11 9 8
The Periodic Law and Ionic Radii
Recall that in an isoelectronic series, the atoms/ions have the same number of electrons.
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The Periodic Law and Ionic RadiiLet's try a few more together.
1. Explain why Si has an atomic radii of 111 pm while C has an atomic radii of 67 pm despite Si having a higher nuclear charge (Z)?
Si has an additional energy level, so the valence electrons are farther away and more shielded than those
of C resulting in a smaller coulombic attraction. move for answer
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The Periodic Law and Ionic RadiiLet's try a few more together.
2. Explain why iron (Fe) has a smaller atomic radii - 154 pm - than does scandium (Sc) - 184 pm.
Although both have the same amount of shielding, Fe has a larger Z creating a stronger coulombic attraction and a
smaller radii. move for answer
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6 Which of the following influences the atomic/ionic radii?
A the number of neutrons
B the amount of core electrons between the nucleus and the valence electrons
C the number of protons
D A and B
E B and C
Ans
wer
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7 The atomic radius of main-group elements generally increases down a group because __________.
A effective nuclear charge increases down a group
B effective nuclear charge decreases down a group
C effective nuclear charge zigzags down a group
D the principal quantum number of the valence orbitals increases
Eboth effective nuclear charge increases down a group and the principal quantum number of the valence orbitals increases
Ans
wer
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8 Of the following, which gives the correct order for atomic radius for Be, Li, N, C and Ne?
A Be > Li > N > C > Ne
B Ne > C > N > Li > Be
C C > N > Ne > Li > Be
D Li > Be > C > N > Ne
E Ne > N > C > Be > Li
Ans
wer
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9 Which of the following atoms would have a smaller atomic radii than Ar and why?
A Fe - It has more core electrons
B Si - It has fewer core electrons
C O - It has fewer core electrons
D Ne - it has a higher nuclear charge (Z)
E Ca - it has a higher nuclear charge (Z)
Ans
wer
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10 Which ion below has the largest radius?
A O2-
B Li+
C I-
D N3-
E K+
Ans
wer
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11 Which of the following pairs correctly shows the proper relationship between the two atoms/ions in terms of atomic/ionic radii?
A Ca < Ca2+
B F < F-
C V < Mn
D Ca < Be
E He > Li
Ans
wer
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12 Which of the following correctly states why the atomic radii do not consistently decrease as the atomic number rises throughout the periodic table?
A The nuclear charge (Z) does not always increase with atomic number
B The number of neutrons start to influence the atomic radii
C Filled energy levels shield the nucleus and diminish coulombic forces
D Electrons become less negative the more there are
E A higher atomic number increases the size of the radii, not decreases it.
Ans
wer
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13 Which of the following would correctly rank the following in order of decreasing atomic/ionic radii?
A V4+ > V5+ > F > F-
B V4+ > V5+ > F- > F
C V5+ > V4+ > F- > F
D V5+ > V4+ > F > F-
E F > F- > V4+ > V5+
Ans
wer
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14 Isotopes of an element, like C-12 and C-13, are likely to have different atomic radii?
Yes
NoA
nsw
er
Slide 37 / 102
The Periodic Law and Ionization Energy
Ionization energy is the amount of energy required to remove an electron from an atom. This creates an ion, hence the name!
The stronger the Coulombic attraction between the valence electron and the nucleus, the greater the energy required
to remove an electron.
Element Ionization Energy
Li + IE --> Li+ + e- 520 kJ/mol
Na + IE --> Na+ + e- 496 kJ/mol
Less energy is required to remove sodium's electron than lithium's because sodium has a full energy level more of core
electrons shielding the nuclear charge.
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The Periodic Law and Ionization Energy
Unless you're hydrogen, you've got multiple electrons that can be lost. As a result we have to distinguish between 1st, 2nd, 3rd, etc.
ionization energies.
Ionization Ionization Energy
1st: Na + IE --> Na+ + e- 496 kJ/mol
2nd: Na+ + IE --> Na2+ + e- 4560 kJ/mol3rd: Na2+ + IE --> Na3+ + e- 6,900 kJ/mol
4th: Na3+ + IE --> Na4+ + e- 9540 kJ/mol
Note the huge jump in ionization energy from the 1st to the 2nd. After sodium loses it's first electron, it is isoelectronic with [Ne], with
an extremely stable full s and p orbital and minimal shielding.
Each successive ionization energy is always higher than the previous. This is due to the higher nuclear charge felt by the
remaining electrons.
Slide 39 / 102
The Periodic Law and Ionization EnergyThe chart below clearly shows the impact of being isoelectronic
with a noble gas on the ionization energy.
Ionization Energy (kJ/mol)
Na+
Mg2+
Al3+
Si4+
P5+
S6+
Slide 40 / 102
The Periodic Law and Ionization Energy
The trend in first ionization energies mostly matches what we would expect.
The ionization energy increases across a period with increasing atomic number. ( Li < Ne)
The ionization energy decreases down a group with increasing atomic number due to additional core electrons from each filled energy level
shielding the nucleus. ( He > Ne)
Slide 41 / 102
The Periodic Law and Ionization Energy
There are however a few hiccups that need to be explained. Let's look carefully at the ionization energies of Be and B as well as N
and O indicated in the circles.
Shouldn't the ionization energy increase with increasing atomic number across a period? Quantum theory will explain.
Slide 42 / 102
The Periodic Law and Ionization Energy
Be: [He]2s2 N: [He]2s22p3
B: [He]2s22p1 O: [He]2s22p4
More energy is required to remove an electron from Be's full "s" orbital
More energy is required to remove an electron from
N's 1/2 full "p" orbital
Slide 43 / 102
The Periodic Law and Ionization EnergyLet's look at another hiccup in the trend.
Notice that a lot less energy is required to remove an electron from Ga (Z=31) than from Zn (Z=30). How can this be?
Zinc has a full "s" and "d" orbital conferring extra stability while in gallium, the electron is being taken from a "p" orbital which is
heavily shielded from the nucleus by the "d" orbital itself.
Slide 44 / 102
The Periodic Law and Ionization EnergyLet's practice ranking atoms/ions in terms of ionization energy:
1. Rank the following in terms of increasing ionization energy:
C Al Na+ Ne Na
As with atomic radii, determine their outermost principal energy level and nuclear charge.
C Al Na+ Ne NaValence "N" 2 3 2 2 3
"Z" 6 13 11 10 11
Na < Al < C < Ne < Na+
IE(kJ/mol) 496 578 1086 2081 4560
move for answer
Slide 45 / 102
15 What is the ionization energy?
A Energy change associated with the gain of an electron
B Amount of energy that is required to move an electron from an s to a p orbital
C Measure of the attraction of an atom for electrons when in a compound
D Pull of the neutrons on the electrons
E Amount of energy required to remove an electron from an atom or ion
Ans
wer
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16 Which of the following would NOT influence the ionization energy?
A The shielding from core electrons
B The extent to which an orbital is full
C The nuclear charge
D The number of principal energy levels between the valence electrons and the nucleus
E All of these influence the ionization energy
Ans
wer
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17 Which of the following elements would be expected to have a higher ionization energy than magnesium (Mg)?
A Al
B Ca
C Na
D K
E B
Ans
wer
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18 Which of the following correctly ranks the elements below in order of decreasing ionization energy?
A Ne > O > N
B Ne > N > O
C H > He > Ne
D Li > Mg > Ga
E Zn > Ga > Br
Ans
wer
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19 Which of the following elements best fits the data provided below?
A Li
B C
C Be
D Ne
E O
Ionization Ionization Energy
1st: X + IE --> X+ + e- 900 kJ/mol
2nd: X+ + IE --> X2+ + e- 1757 kJ/mol3rd: X2+ + IE --> X3+ + e- 14,850 kJ/mol
Ans
wer
Slide 50 / 102
20 Which of the following pairs are correct in terms of relative first ionization energy and why?
A O2- < Ne , due to smaller nuclear charge on oxide ion
B Li > Na , due to increased shielding in the Na atom
C Zn > Cu , due to a higher nuclear charge in zinc
D Cl > S , due to the smaller nuclear charge in sulfur
E All of these
Ans
wer
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21 The second ionization energy will always be higher than the first.
True
False
Ans
wer
Slide 52 / 102
22 _________ have the lowest first ionization energies of the groups listed.
A Alkali metals
B Transition elements
C Halogens
D Alkaline eath metals
E Noble gases
Ans
wer
Slide 53 / 102
23 Of the choices below, which gives the order for decreasing first ionization energies?
A Cl > S > Al > Ar > Si
B Ar > Cl > S > Si > Al
C Al > Si > S > Cl > Ar
D Cl > S > Al > Si > Ar
E S > Si > Cl > Al > Ar
Ans
wer
Slide 54 / 102
Ionization Energy and PESIonization energy data can be determined from PES (photoelectron spectroscopy). Recall that PES looks at the energy of light required
to remove electrons from an atom. Each orbital appears as a peak on the spectrum.
The PES spectrum clearly shows that the core electrons require the most energy to remove. It also shows that Be has a higher 1st IE for the removal of the valence electrons than does Li. This is expected
as Be has a higher "Z".
Li (1s)
Be (1s)Be (2s)
Li (2s)
Intensity
binding energy
Slide 55 / 102
Ionization Energy and PES
Let's interpret another PES spectra, this one of nitrogen and oxygen.
Intensity
binding energy
N (2s) N (1s)N (2p)
O (2p)O (2s) O (1s)
Why is the N (2p) peak greater than the O (2p) peak?
N has a half-full "p" orbital increasing
the ionization energy
Why is the N(2s) peak less than the O (2s) peak?
O has the higher nuclear charge
move for answer
move for answer
Slide 56 / 102
Ionization Energy and PES
Click to go to an interactive PES spectra database
and answer the questions.
Why is the binding energy of the electrons greater in He than H?
Which peak in the Li spectra represents the valence electrons?
Why is the valence peak binding energy less in Li than in H?
Why is the core peak (1s) binding energy greater in Li than in H?
Similar shielding but greater "Z"
Peak with lower binding energy
Increased shielding due to core 1s electrons, lessens coulombic force
Lithium has a higher nuclear charge "Z" so higher coulombic attractionsmove for answer
move for answer
move for answer
move for answer
Slide 57 / 102
24 The following PES spectrum shows the valence "p" orbital peaks for Si and for C. Which of the following would be TRUE?
A The Si peak is of lower energy due to it's higher nuclear charge
B The Si peak is of higher energy due to the increased shielding from core electrons
C The Si peak is of lower energy due to the increased shielding from core electrons
D The Si peak is of higher energy due to its higher nuclear charge
Intensity
binding energy
Ans
wer
Slide 58 / 102
25 The 3s peak for magnesium should have a higher binding energy than that of the 4s peak in calcium due to calcium's higher amount of shielding by core electrons?
True
False
Ans
wer
Slide 59 / 102
26 Below is an actual PES spectrum of palladium (Pd). Which of the following would be TRUE? (Note: the outer 5s and 4d peaks are not shown)
A Compared to Pd, the 3d peak in Cd would be found to the left of the 3d Pd peak
B Compared to Pd, the 3d peak in Rb would be of a higher binding energy due to lower nuclear charge
C Compared to Pd, the 3p peak in Kr should be found to the left of the 3p peak in Pd
3s 3p
3d
4p 4s
Ans
wer
Slide 60 / 102
27 Based on the PES data below, what would be TRUE regarding atoms 1 and 2?
A I only B II and III only C 1 and III only
D II and IV only E I, II, III, and IV
Binding Energy
Inte
nsity
0 10
10 100
28.6
1.091.72
Binding Energy
Inte
nsity
0 10
10 100
39.6
1.402.45
1 2
I. Atom 1 has a smaller atomic radii II. Atom 2 has a larger first ionization energy
III. Both atoms are in the same period
IV. Both atoms are in the same group
Ans
wer
Slide 61 / 102
Ionization Energy and Metallic CharacterMetals are generally described as being able to lose
electrons readily which promotes conductivity.
Since metals lose electrons easily, they must have low ionization energies compared to non-metals.
Element Metal or Non-metal1st Ionization
Energy (kJ/mol)Na metal 496
O non-metal 1314
Slide 62 / 102
Ionization Energy and Metallic CharacterWe can predict, based on ionization energies, where the
metals and non-metals are on the periodic table.
semi-metals or metalloids
Notice that an element becomes more metallic as the shielding increases and as the nuclear charge - for a given
level of shielding - decreases.
Slide 63 / 102
Ionization Energy and Metallic CharacterLet's answer a few questions regarding metallic character.
1. Why is lead considered a metal and carbon a non-metal
despite being in the same group?
Pb has much more shielding due to more levels of core electrons
thereby causing it's electrons to be lost far more easily than that of C.
CSi
GeSnPbmove for answer
Slide 64 / 102
Let's answer a few questions regarding metallic character.
Cu
Ag
2. Which metal would we expect to be a better conductor of electricity? Ag or Cu
Ag due to the higher amount of shielding, causing it to ionize more easily, thereby
creating mobile electrons. move for answer
Ionization Energy and Metallic Character
Slide 65 / 102
Application: Elements of Life
The most common elements in living things are C,H,N,O,P, and S. Interestingly, these are all non-metals.
Interestingly, all metal atoms found in living
things are in their ionic form (Mg2+, Ca2+,
Zn2+, etc.)
In order to form large stable, yet complex,
molecules, the elements must not be able to lose electrons
easily.
Serotonin - brain hormone
2+ 2+
Ionization Energy and Metallic Character
Slide 66 / 102
28 Which of the following is the LEAST metallic of those below?
A F
B At
C Ne
D Xe
E Ba
Ans
wer
Slide 67 / 102
29 Which of the following would be TRUE?
A The higher the ionization energy, the less metallic an element will be
B The lower the ionization energy, the less metallic an element will be
C For a given amount of core electron shielding, the higher the nuclear charge, the more metallic an element will be
D Both A and C
E Both B and C
Ans
wer
Slide 68 / 102
30 Which of the following has the elements correctly ordered by increasing metallic character?
A Li < Be < B
B Ca < K < Ga
C Ga < Ca < K
D Rb < Cs < As
E Ga < As < Ba
Ans
wer
Slide 69 / 102
Ionization Energy and LightAs we have seen, EM radiation can provide the necessary energy to
ionize an electron from an atom.
The higher the ionization energy, the higher the frequency of light needed to ionize the electron.
e-photon
Slide 70 / 102
Ionization Energy and LightWhich of the following elements would require the shortest
wavelength to lose an electron?
Si C N
Short wavelength means high energy so this would be the element with the largest ionization energy.
Si C N "N" 3 2 2
"Z" 14 6 8
N has similar shielding as carbon but a higher nuclear charge so it would require the shortest wavelength to ionize an
electron.
move for answer
Slide 71 / 102
Ionization Energy and LightWhat would be the necessary wavelength required to remove
one of Neon's outermost p electrons?
1. Look up 1st IE of Neon = 2081 kJ/mol
2. Convert to kJ/atom = 2081 kJ x 1 mol = 3.46 x 10-21 kJ
mol 6.022 x 1023 atoms
3. Convert to J = 3.46 x 10-18 J
4. Convert to v via E=hv --> v = E/h = 3.46 x 10-18 J = 5.2 x 1015 1/s
6.3 x 10-34 J*s
5. Convert to wavelength via v = c --> = c/v
3 x 108 m*s = 5.77 x 10-8 m = 57.7 nm
5.2 x 1015 s
move for answer
Slide 72 / 102
31 Which of the following orbitals of calcium would require the highest frequency of light to ionize?
A 2s
B 2p
C 3s
D 3p
E 4s
Ans
wer
Slide 73 / 102
32 Based on the table of 1st ionization energies below, which element is likely to ionized by light with wavelength of 214 nm?
A I
B Ga
C In
D He
E Rb
Element Ionization Energy (kJ/mol)
I 1009
Ga 579
In 558
He 2372
Rb 403
Ans
wer
Slide 74 / 102
33 What frequency of light would be required to ionize the first electron of cesium (1st IE = 376 kJ/mol)?
Ans
wer
Slide 75 / 102
Periodic Law and ElectronegativityAs we know, atoms do not often exist in isolation. They form bonds with other atoms to make molecules and compounds.
Recall that electronegativity is defined as a measure of an atom's attraction for electrons in a bond.
The greater the nuclear charge and the smaller the shielding, the greater the electronegativity.
waterOH H
Slide 76 / 102
Periodic Law and ElectronegativityLet's compare the electronegativities of H and O within the
water molecule.
OH H
O has more shielding but a much higher nuclear charge so it will have the higher electronegativity. Therefore the electrons
get pulled unevenly toward the oxygen atom.
OH H
Slide 77 / 102
Periodic Law and ElectronegativityTrends in electronegativity for periods 2-4.
SO
Se
Li Na K
What is the trend in electronegativity down a group?
What is the trend in electronegativity across a period from left to right?
increases, due to increasing nuclear charge with steady amount of shielding
decreases, due to additional shielding from each new energy level
move for answer
move for answer
Slide 78 / 102
Periodic Law and ElectronegativityTrends in electronegativity for periods 2-4.
Why do the noble gases not have published electronegativity values?
They have a full outer "s" and "p" system and do not form compounds.
move for answer
Slide 79 / 102
Periodic Law and ElectronegativityThe following electronegativity values will need to be
memorized as this will aid in understanding bonding later on.
H
2.2
C
2.5
N
3.0
O
3.5
F
4.0S
2.6
Cl
3.2Br
3.0
Slide 80 / 102
34 Of the atoms below, __________ is the most electronegative.
A SiB ClC Rb
D CaE S
Ans
wer
Slide 81 / 102
35 Which of the following BEST explains why fluorine has a higher electronegativity than oxygen?
A F has a higher nuclear charge and less shielding than O
B F has a higher nuclear charge and similar shielding of O
C F has the equivalent nuclear charge and less shielding than O
D F has the equivalent nuclear charge and more shielding than O
E None of these
Ans
wer
Slide 82 / 102
36 Which of the following groups of elements are ranked properly from lowest to highest electronegativity?
A H < Li < Na
B H < C < Li
C C < Si < Ge
D I < Br < Cl
E F < S < As
Ans
wer
Slide 83 / 102
37 An element with a small electronegativity value is likely to have...
A Valence shell PES peaks with high binding energies
B A high nuclear charge and a low amount of shielding
C A low nuclear charge and a high amount of shielding
D Both A and B
E Both A and C
Ans
wer
Slide 84 / 102
Specific Groups of Periodic Table
Group 1: Alkali Metals
Group 2: Alkaline Earth Metals
Group 3-12: Transition Metals
Group 13/14/15: Metalloids
Group 17: Halogens
Group 18: Noble Gases
We will now examine six groups of the periodic table in more detail.
Slide 85 / 102
Alkali Metals
They are highly reactive due to their extremely low ionization energies. As a result, they are found only in compounds in nature,
not in their elemental forms.
They have low densities and melting points. In fact Li, Na, and K have densities so low, they'll float on water!
Slide 86 / 102
Alkaline earth metals have higher densities and melting points than alkali metals.
Their ionization energies are low, but not as low as those of alkali metals so they are slightly less reactive.
Alkaline Earth Metals
Slide 87 / 102
Alkaline Earth Metals
Beryllium does not react with water and magnesium reacts only with steam, but the others react readily with water.
Reactivity tends to increase as you go down the group.
Can you explain why that would be?
Slide 88 / 102
Transition MetalsThe transition metals vary somewhat in properties but we can simplify to say that they are less reactive than either of the first two groups. In fact, the least reactive metals (Au, Pt, Ag) are in
this group.
Transition metals tend also to have higher densities and melting points than the first two groups.
Due to their "d" orbitals, they can form ions with much higher charges than the first two groups which will allow them to form
colored complex ions with water and other species.
Slide 89 / 102
Transition MetalsSome complex ions formed from transition metals and their colors.
Slide 90 / 102
Metalloids
These have some characteristics of metals and some of nonmetals.
For instance, silicon looks shiny like a metal, but is brittle and a fairly poor conductor.
There are six elements that are classified as metalloids:Boron (B)Silicon (Si)Arsenic (As)Tellurium (Te)Germanium (Ge)Antimony (Sb)
Slide 91 / 102
Metalloids
Metalloids like Si, although they are not particularly conductive due to higher ionization energies than metals can be made to be by
"doping" them with certain elements to increase their conductivity.
Circuits that form the basis for modern electronics are composed of doped metalloids like Si and Ge.
Slide 92 / 102
Halogens
The halogens are prototypical nonmetals. They only require one more electron to have a full "s" and "p" and are therefore highly
reactive.
The name comes from the Greek words halos and gennao: “salt formers”.
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Halogens (at standard temp and pressure)
Flourine is a colorless gas
Chlorine is a greenish gas
Bromine is a brownish liquid
Iodine is a purplish solid
Slide 94 / 102
Noble Gases
The noble gases have very high ionization energies.
Therefore, they are relatively unreactive.
As a result, unlike the diatomic halogens, they are found as monatomic gases
Slide 95 / 102
38 An atom with a very high ionization energy and is a liquid at room temperature is most likely a:
A Alkali metal
B Alkaline earth metal
C transition metal
D Halogen
E Noble gas
Ans
wer
Slide 96 / 102
39 Which of the following ranks the metals in order of increasing reactivity?
A Li < Na < Mg < K
B Mg < Li < Na < K
C K < Li < Na < K
D Li < Fe < Zn < Au
E None of these
Ans
wer
Slide 97 / 102
40 Which of the following elements would form colored complex ions?
A F
B Co
C Ca
D Al
E Na
Ans
wer
Slide 98 / 102
41 Which of the following elements would serve as a semiconductor?
A Ge
B C
C F
D Pb
E Y
Ans
wer
Slide 99 / 102
42 What would be the alkaline earth metal with the highest ionization energy?
A Li
B Al
C Be
D B
E Ra
Ans
wer
Slide 100 / 102
43 Which would be the halogen with the smallest atomic radii?
A Ne
B F
C At
D Pb
E Fr
Ans
wer
Slide 101 / 102
Now that we have a good understanding of some of the properties of various elements, we will now examine how
they react and what they produce when they do in the next chapter.
Slide 102 / 102