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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)

• f valence

electrons Electron Configuration Lose/ Gain electrons Ionic Charge H 1 1s1 gain 1 lose 1

• 1

+1 He 1 1s2 NA NA Li 2 [He]2s1 lose 1 +1 Be 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

• 1

Ne 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

• f 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 Slide 12 / 102

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

Answer

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2 An atom with a +2 charge must be in the same group as barium. True False

Answer

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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

Answer

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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

Answer

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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

Answer

Slide 19 / 102 The Periodic Law and Atomic/Ionic Radii

The 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) 200 100 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 H Li Na He Ne 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

Slide 21 / 102 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
• f an atom and therefore determines the radii.

Slide 22 / 102 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.

Slide 23 / 102 The Periodic Law and Ionic Radii

When 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

• ver a larger number of

electrons, resulting in a weaker coulombic attraction.

Slide 24 / 102 The Periodic Law and Ionic Radii

Let'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.

Slide 26 / 102 The Periodic Law and Ionic Radii

Let'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

• f C resulting in a smaller coulombic attraction.

move for answer

Slide 27 / 102 The Periodic Law and Ionic Radii

Let'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

Answer

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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 E both effective nuclear charge increases down a group and the principal quantum number of the valence

• rbitals increases

Answer

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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

Answer

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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)

Answer

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10 Which ion below has the largest radius? A O2- B Li+ C I- D N3- E K+

Answer

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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

Answer

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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.

Answer

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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+

Answer

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14 Isotopes of an element, like C-12 and C-13, are likely to have different atomic radii? Yes No

Answer

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/mol 3rd: 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 Energy

The 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+

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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)

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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.

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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 Energy

Let'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 Energy

Let'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 Na Valence "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

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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

Answer

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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

Answer

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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

Answer

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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

Answer

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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/mol 3rd: X2+ + IE --> X3+ + e- 14,850 kJ/mol

Answer

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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

Answer

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21 The second ionization energy will always be higher than the first. True False

Answer

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22 _________ have the lowest first ionization energies

• f the groups listed.

A Alkali metals B Transition elements C Halogens D Alkaline eath metals E Noble gases

Answer

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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

Answer

Slide 54 / 102 Ionization Energy and PES

Ionization 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

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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 attractions

move for answer move for answer move for answer move for answer

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24 The following PES spectrum shows the valence "p"

• rbital 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

Answer

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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

Answer

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26 Below is an actual PES spectrum of palladium (Pd). Which of the following would be TRUE? (Note: the

• uter 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

Answer

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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 Intensity 10 10 100 28.6 1.09 1.72 Binding Energy Intensity 10 10 100 39.6 1.40 2.45

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

Answer

Slide 61 / 102 Ionization Energy and Metallic Character

Metals 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-metal 1st Ionization Energy (kJ/mol) Na metal 496 O non-metal 1314

Slide 62 / 102 Ionization Energy and Metallic Character

We can predict, based on ionization energies, where the metals and non-metals are on the periodic table. semi-metals

• r 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 Character

Let'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.

C Si Ge Sn Pb

move for answer

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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

Answer

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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

Answer

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30 Which of the following has the elements correctly

• rdered by increasing metallic character?

A Li < Be < B B Ca < K < Ga C Ga < Ca < K D Rb < Cs < As E Ga < As < Ba

Answer

Slide 69 / 102 Ionization Energy and Light

As 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 Light

Which 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

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Ionization Energy and Light

What would be the necessary wavelength required to remove

• ne 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

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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

Answer

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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 Answer

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33 What frequency of light would be required to ionize the first electron of cesium (1st IE = 376 kJ/mol)?

Answer

Slide 75 / 102 Periodic Law and Electronegativity

As 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. water O H H

Slide 76 / 102 Periodic Law and Electronegativity

Let's compare the electronegativities of H and O within the water molecule. O H 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. O H H

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Periodic Law and Electronegativity

Trends in electronegativity for periods 2-4. S O 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 Electronegativity

Trends 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 Electronegativity

The 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 4.0 S 2.6 Cl 3.2 Br 3.0

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34 Of the atoms below, __________ is the most electronegative. A Si B Cl C Rb D Ca E S

Answer

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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

• f 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

Answer

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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

Answer

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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

Answer

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 Metals

The 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 Metals

Some 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

• f 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”.

Slide 93 / 102 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

Answer

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

Answer

Slide 97 / 102

40 Which of the following elements would form colored complex ions? A F B Co C Ca D Al E Na

Answer

Slide 98 / 102

41 Which of the following elements would serve as a semiconductor? A Ge B C C F D Pb E Y

Answer

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

Answer

Slide 100 / 102

43 Which would be the halogen with the smallest atomic radii? A Ne B F C At D Pb E Fr

Answer

Slide 101 / 102 Now that we have a good understanding

• f 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