Post on 20-Jan-2016
Students who haven’t passed the safety quiz
708411871401
Understanding “periodicity”
• Trends within a period
• what is changing?• what is constant?
• Trends within a group• what is changing?• what is constant?
11
Na12
Mg
11
Na
19
K
• First Ionization Energy (first ionization potential)
• The minimum energy needed to remove the highest-energy (outermost) electron from a neutral atom in the gaseous state, thereby forming a positive ion
Periodicity of First IonizationEnergy (IE1)
Like Figure 8-18
• Trends• Going down a group, first ionization energy
decreases.
• This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.
Fig. 8.15
• Generally, ionization energy increases with atomic number.
• Ionization energy is proportional to the effective nuclear charge divided by the average distance between the electron and the nucleus. Because the distance between the electron and the nucleus is inversely proportional to the effective nuclear charge, ionization energy is inversely proportional to the square of the effective nuclear charge.
• Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge.
• A dramatic increase occurs when the first electron from the noble-gas core is removed.
• Trends• Going down a group, first ionization energy
decreases.
• This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.
• Small deviations occur between Groups IIA and IIIA and between Groups VA and VIA.
• Examining the valence configurations for these groups helps us to understand these deviations:
• IIA ns2 • IIIA ns2np1
• VA ns2np3
• VIA ns2np4
It takes less energy to remove the np1 electron than the ns2 electron.
It takes less energy to remove the np4 electron than the np3 electron.
• Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb.
Sb is larger than As.As is larger than Br.
Ionization energies: Sb < As < Br
35Br
33As
51Sb
Ranking Elements by First Ionization Energy
Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasing IE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, ArPlan: Find their relative positions in the periodic table and apply trends!Solution:
a. Rn>Ar>Ne
b. Bi<Po<At
c. Na<Mg<Be
d. K<Cl<Ar
• Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge.
• A dramatic increase occurs when the first electron from the noble-gas core is removed.
Fig. 8.16
What are the units here????
• Left of the line, valence shell electrons are being removed. Right of the line, noble-gas core electrons are being removed.
Identifying Elements by Its Successive Ionization Energies
Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4
580 1,815 2,740 11,600Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table!Solution:
Identifying Elements by Its Successive Ionization Energies
Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4
580 1,815 2,740 11,600Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table!Solution:
The largest jump in IE occurs after IE3 so the element has 3 valence electrons thus it is Aluminum ( Al, Z=13), its electron configuration is :
1s2 2s2 2p6 3s2 3p1
• Electron affinity (E.A.) • The energy change for the process of adding an electron to a
neutral atom in the gaseous state to form a negative ion
• A negative energy change (exothermic) indicates a stable anion is formed. The larger the negative number, the more stable the anion. Small negative energies indicate a less stable anion.
• A positive energy change (endothermic) indicates the anion is unstable.
The Electron affinity of a molecule or atom is the energy change when an electron is added to the neutral atom to form a negative ion.
Overall periodic trends
Note: Electronegativity has similar trend as electron affinity
23
Reactivity of the Alkali Metals
Potassium video
Sodium video
Lithium video2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
2K(s) + 2H2O(l) 2KOH(aq) + H2(g)
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
Trend?
24
More Sodium Reaction Videos
Prepping Na
150 g Na in small pieces
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
http://www.theodoregray.com/PeriodicTable/
100 g Na in one piece
Electronic Configuration Ions
• Na 1s 2 2s 2 2p 6 3s 1 Na+ 1s 2 2s 2 2p 6
• Mg 1s 2 2s 2 2p 6 3s 2 Mg+2 1s 2 2s 2 2p6
• Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al+3 1s 2 2s 2 2p 6
• O 1s 2 2s 2 2p 4 O- 2 1s 2 2s 2 2p 6
• F 1s 2 2s 2 2p 5 F- 1 1s 2 2s 2 2p 6
• N 1s 2 2s 2 2p 3 N- 3 1s 2 2s 2 2p 6
Isoelectronic Atoms and Ions
• H- 1 { He } Li+ Be+2
• N- 3 O- 2 F- { Ne } Na+ Mg+2 Al+3
• P- 3 S- 2 Cl- { Ar } K+ Ca+2 Sc+3 Ti+4
• As- 3 Se- 2 Br- { Kr } Rb+ Sr+2 Y+3 Zr+4
• Sb- 3 Te- 2 I- { Xe } Cs+ Ba+2 La+3 Hf+4
Trends when atoms form chemical bonds
Empirical Observation
“when forming ionic compounds, elements tend to lose or gain electrons to be more like the nearest noble gas”
Metals tend to lose e-’s
Nonmetals tend to gain e-’s
Are ions bigger or smaller than atoms?
Representative cation
Na → Na+ + e
Representative anion
F + e → F
Cations are always smaller than parent atom- decreased e- repulsion (clouds contract)- if emptying valence shell, “n” decreases
Anions are always larger than parent atoms- increased e- repulsion (clouds expand)
Cations
Contract
Anions
Add
Trends in ion size
Trends in atom & ion size
Trends in ion size
Ranking Ions According to Size
Problem: Rank each set of Ions in order of increasing size. a) K+, Rb+, Na+ b) Na+, O2-, F - c) Fe+2, Fe+3
Plan: We find the position of each element in the periodic table and apply the ideas of size: i) size increases down a group, ii) size decreases across a period butincreases from cation to anion. iii) size decreases with increasing positive (or decreasing negative) charge in an isoelectronic series.iv) cations of the same element decreases in size as the charge increases.Solution: a) since K+, Rb+, and Na+ are from the same group (1A), they increase in size down the group: Na+ < K+ < Rb+
b) the ions Na+, O2-, and F- are isoelectronic. O2- has lower Zeff than F-, so it is larger. Na+ is a cation, and has the highest Zeff, so it is smaller: Na+ < F- < O2-
c) Fe+2 has a lower charge than Fe+3, so it is larger: Fe+3 < Fe+2
Chapter #9 - Models of Chemical Bonding
9.1) Atomic Properties and Chemical Bonds
9.2) The Ionic Bonding Model
9.3) The Covalent Bonding Model
9.4) Between the Extremes: Electronegativity and Bond Polarity
9.5) An Introduction to Metallic Bonding
Sodium Chloride
Depicting Ion Formation with OrbitalDiagrams and Electron Dot Symbols - I
Problem: Use orbital diagrams and Lewis structures to show the formation of magnesium and chloride ions from the atoms, and determine the formula of the compound.Plan: Draw the orbital diagrams for Mg and Cl. To reach filled outer levels Mg loses 2 electrons, and Cl will gain 1 electron. Therefore we need two Cl atoms for every Mg atom.Solution:
2 Cl
Mg+2 + 2 Cl-
Mg
+
Mg + ..Cl
Cl
....
......
....
Mg+2 + 2 Cl.. ....
..
Depicting Ion Formation from OrbitalDiagrams and Electron Dot Symbols - II
Problem: Use Lewis structures and orbital diagrams to show the formation of potassium and sulfide ions from the atoms, and determinethe formula of the compound.Plan: Draw orbital diagrams for K and S. To reach filled outer orbitals,sulfur must gain two electrons, and potassium must lose one electron.Solution:
2 K
S2 K+ + S - 2
+
K
K..
+ S
........ ..
.. 2 K+ + S
..
2 -
Three Ways of Showing the Formation ofLi+ and F - through Electron Transfer
Lewis Electron-Dot Symbols for Elements in Periods 2 & 3
The Reaction between Na and Br to Form NaBr
The ElementsThe Reaction!
Melting and Boiling Points of Some Ionic Compounds
Compound mp( oC) bp( oC)
CsBr 636 1300NaI 661 1304MgCl2 714 1412KBr 734 1435CaCl2 782 >1600NaCl 801 1413LiF 845 1676KF 858 1505MgO 2852 3600
Figure 9.11: Potential-energy curve for H2.
Covalent Bonding in Hydrogen, H2
Figure 9.10: The electron probability distribution for the H2 molecule.
Covalent bonds
animation
http://www.chem1.com/acad/webtext/chembond/cb03.html
This is a good overview.
http://www.chem.ox.ac.uk/vrchemistry/electronsandbonds/intro1.htm
For elements larger than Boron, atoms usually react todevelop octets by sharing electrons. H, Li and Be striveto “look” like He. B is an exception to the noble gas paradigm.It’s happy surrounded by 6 electrons so the compound BH3 is stable.
Try drawing a Lewis structure for methane.
Draw Lewis dot structures for the halogens.
Try oxygen and nitrogen.
Notice that these all follow the octet rule!
These also follow the octet rule!
Bond Lengths and Covalent Radius
Figure 9.14: The HCl molecule.
Figure 9.12: Molecular
model of nitro-glycerin.
What is the formula for thiscompound?
Rules for drawing Lewis structures
1. Count up all the valence electrons
2. Arrange the atoms in a skeleton
3. Have all atoms develop octets (except those around He)
Make some Lewis Dot Structures with other elements:
SiH4 NH3H2O
C2H6 C2H6OCH2O
Figure 9.9: Model of CHI3
Courtesy of Frank Cox.
Make some Lewis Dot Structures with other elements:
CH4 NH3H2O
C2H6 C2H6OCH2O
Look at all these structures and make some bonding rules:
The normal number of bonds that common elements make in covalent structures.
Element # BondsCNO
H, Halogens
Element # BondsC 4N 3O 2
H, Halogens 1
Rules for drawing Lewis structures
1. Count up all the valence electrons
2. Arrange the atoms in a skeleton
3. Have all atoms develop octets (except those around He)
4. Satisfy bonding preferences!
A model of ethylene.
A model of acetylene.
A model of COCl2.
The Relation of Bond Order,Bond Length and Bond Energy
Bond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol)
C O 1 143 358C O 2 123 745C O 3 113 1070
C C 1 154 347C C 2 134 614C C 3 121 839
N N 1 146 160N N 2 122 418N N 3 110 945
Table 9.4
Conceptual Problem 9.103
Fig. 9.14
Figure 9.15: Electronegatives of the elements.
The Periodic Table of the Elements2.1
0.9 1.5
0.9 1.2
0.8 1.0 1.3
0.8
0.7
0.7
1.0
0.9
1.5 1.6 1.61.5 1.8
1.2
1.1
1.8 1.8 1.9 1.6
1.4 1.6
1.5
1.8
1.7
1.9
1.9
2.2 2.2
2.2
2.2
2.2
1.9
2.4
1.7
1.9
2.0 2.5 3.0 3.54.0
He
Ne
Ar1.5 1.8 2.1 2.5 3.0
1.6 1.8 2.0 2.4 2.8 Kr
Xe
Rn
2.52.1
2.2
1.9
2.01.9
1.81.7
1.81.8
1.1 1.1 1.1 1.1
1.3
1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.21.3
1.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.31.3 1.5
0.9
1.3 2.2
Electronegativity
1.1
Th Pa U Np No Lr
1.3
Ce Pr Nd Pm Yb Lu
Fig. 9.16
Fig. 9.17
Determining Bond Polarity from Electronegativity Values
Problem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity.Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values.Solution: a) the EN of O = 3.5 and of H = 2.1: O - H
the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O
0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5
Fig. 9.18
Percent Ionic Character as a Function ofElectronegativity Difference (En)
Fig. 9.19
Lewis Structures of Simple Molecules
C
H
H H
H
Cl
O
O O
K+
KClO3
CF4
..
..H C O H
H
H
H
H
C
Ethyl Alcohol (Ethanol)
Potassium Chlorate Carbon Tetrafluoride
......
..
..
..
.. ...... ..
.... C
F
FF
F
......
..
.. ..
..
....
CH4
Methane
Resonance: Delocalized Electron-Pair Bonding - I
Ozone : O3 ......
..
O O
O ..
........ ..OOO ....
I II
O
O
O
..
........
Resonance Hybrid Structure
One pair of electron’s resonances between the two locations!!
Resonance:Delocalized Electron-Pair Bonding - II
C
CC
CC
C
CC
C
C C
C
C
CC
CC
C
H
HHH
H H
H
H
H
H
H
H
H
HHH
H
HResonance Structure Benzene
Lewis Structures of Simple Molecules Resonance Structures -III Nitrate
N
O
O O
N
O
O O
..
..
..
..
..
..
....
.... ..
.... ..
....
.... ..
N
O
O O......
.. ..
Lewis Structures for Octet Rule Exceptions
Cl
F
F
F ....
..
..
....
..
.... ..
..
..
..
BCl
Cl
Cl......
..
......
Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons!
Each chlorine atom has8 electrons associated. Boron has only 6!
Cl ClBe....
..
.. ....
Each chlorine atom has8 electrons associated. The beryllium has only 4 electrons.
NO O
... .... ..
..NO2 is an odd electron atom.The nitrogen has 7 electrons.
Resonance Structures - Expanded Valence Shells
.... S
F
F
F
FF
F......
.. ......
.. .. .. ....
.. .. ....
Sulfur hexafluoride
....
.. PF
F
F
FF......
......
..
...... ..
..
Phosphorous pentafluoride
O
S
O
O OH H
..
....
.... ..
.. ......
O
S
O
O OH H.. .... ..
........
Sulfuric acid
S = 12e- p = 10e-
S = 12e-
Resonance Structures
Lewis Structures of Simple Molecules
Resonance Structures-VSO O
O
O
SO O
O
O
. .
. . . .
. .. .
. .. .. .
. .
. .
. .
. .
. .
-2
. .
. .
. .. .. .
-2 Sulfate
S
O
O
O Oxx
x = Sulfur electrons o = Oxygen electrons
o o
o o
o o
o o
o o
x o
x x
x o
o o
o o
o o
o o
o *
o *-2
o o
Plus 4 othersfor a total of 6
. .
. .
VSEPR: Valence Shell Electron Pair Repulsion:
A way to predict the shapes of molecules
Pairs of valence electrons want to get as far away from each other as possible in 3-dimensional space.
Balloon Analogy for the MutualRepulsion of Electron Groups
Two Three Four Five Six
Number of Electron Groups
AX2 Geometry - Linear
Cl ClBe
..
.. ....
..
.. ..1800
BeCl2
Gaseous beryllium chloride is an example of a molecule in which the central atom - Be does not have an octet of electrons, and is electron deficient.Other alkaline earth elements also have the same valence electron configuration, and the same geometry for molecules of this type. Therefore this geometry is common to group II elements.
Molecular Geometry = Linear Arrangement
CO O..
..
..
1800
Carbon dioxide also has the same geometry, and is a linear molecule, but in this case, the bonds between the carbon and oxygens are double bonds.
CO2
The Two MolecularShapes of theTrigonal PlanarElectron-GroupArrangement
AX3 Geometry - Trigonal Planar
BF3B
F
F F..
..
......
.... ..
..N
O
O O
1200
1200
1200
NO3-
Boron Trifluoride
Nitrate Anion
All of the boron Family(IIIA)elements have the same geometry. Trigonal Planar !
AX2E SO2
....
.... ....
....
....
....
..S
O O
The AX2E molecules have a pair ofElectrons where the third atom would appear in the space around the central atom, in the trigonal planargeometry.
-
The Three Molecular Shapesof the TetrahedralElectron-GroupArrangement
AX4 Geometry - Tetrahedral
C
H
H H
H
CH4
Methane
C
H
H
H
H
109.50
All molecules or ions with four electron groups around a central atomadopt the tetrahedral arrangement
H
H
N
H
H
H
HH++
109.50
109.50
N
H
..107.30
all angles arethe same!
Ammonia is in a tetrahedral shape, but it has only an electron pair in one location, so the smaller angle! Ammonium Ion
The Four Molecular Shapes of the TrigonalBipyramidal Electron-Group Arrangement
AX5 Geometry - Trigonal Bipyramidal
Br
F
F
F
..
..
..
..
..
..
..
..
..
....
86.20
AX3E2 - BrF3
I
I
I
..
..
.... ..
....
..
.. 1800
AX2E3 - I3-
P
Cl
Cl
Cl
Cl
Cl
..
.
.
..
.. ..
..
..
..
..
..
....
.. ....
AX5 - PCl5
The Three MolecularShapes of the Octahedral Electron-GroupArrangement
AX6 Geometry - Octahedral
S
F
F
F
F
F
F
......
....
..
..
..
........
......
.... ..
AX6
Sulfur Hexafluoride
I
F
F F
FF
.. ..
.. ............
..
....
..
.... ..
AX5EIodine Pentafluoride
Xe
F
FF
F..
..
..
..
..
..
..
........
......
Xenon Tetrafluoride Square planar shape
Using VSEPR Theory to Determine Molecular Shape
1) Write the Lewis structure from the molecular formula to see the relative placement of atoms and the number of electron groups.
2) Assign an electron-group arrangement by counting all electron groups around the central atom, bonding plus nonbonding.
3) Predict the ideal bond angle from the electron-group arrangement and the direction of any deviation caused by the lone pairs or double bonds.
4) Draw and name the molecular shape by counting bonding groups and non-bonding groups separately.
Hybrid Orbital Model
The sp Hybrid Orbitals in Gaseous BeCl2
The sp3 Hybrid Orbitals in NH3 and H2O
The sp3d Hybrid Orbitals in PCl5
The sp3d2 Hybrid Orbitals in SF6
Sulfur Hexafluoride -- SF6
Figure 10.26: Sigma and pi bonds.
Figure 10.27: Bonding in ethylene.
Figure 10.28: Bonding in acetylene.
Restricted Rotation of -Bonded Molecules
A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene
Postulating the Hybrid Orbitals in a Molecule
Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following:
a) Methyl amine, CH3NH2 b) Xenon tetrafluoride, XeF4
Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms,from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.
Postulating the Hybrid Orbitals in a Molecule
Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2 b) Xenon tetrafluoride, XeF4
Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms,from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.Solution:a) For CH3NH2: The shape is tetrahedral around the C and N atoms.Therefore, each central atom is sp3 hybridized. The carbon atom has four half-filled sp3 orbitals:
Isolated Carbon Atom2s 2p sp3
Hybridized Carbon Atom
The N atom has three half-filled sp3 orbitals and one filled with a
lone pair.
2s 2p sp3
C
H
H
H
H
H
N
..
b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty.
5 s 5 p 5 d
Hybridized Xe atom:
5 d
Isolated Xe atom
sp3d2
b) continued:For XeF4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 900 to each other, whereas if the two polar positions are chosen, the two electron groups will be 1800 from each other. Thereby minimizing the repulsion between the two electron groups.
Xe
F
F
F
F..
..
Xe
F
F F
F
Square planar
1800