George Mason University General Chemistry 211 Chapter 10

Name: George Mason University General Chemistry 211 Chapter 10
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Description: George Mason University General Chemistry 211 Chapter 10

George Mason University General Chemistry 211 Chapter 10
The Shapes (Geometry) of Molecules Acknowledgements Course Text: Chemistry: the Molecular Nature of Matter and Change, 7th edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material. Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor. 4/19/2017

Lewis Electron-Dot Symbols
A Lewis electron-dot symbol is a symbol in which the electrons in the valence shell of an atom or ion are represented by dots placed around the letter symbol of the element Note that the group (column) number indicates the number of valence electrons Group I Group II Group VII Group VIII Group VI Group IV Group V Group III Al . . Si : . S Cl : . Ar : . : P Na . Mg . 3s1 3s2 3s23p1 3s23p2 3s23p3 3s23p4 3s23p5 3s23p6 4/19/2017 2

Lewis Electron-Dot Formulas
A Lewis electron-dot formula is an illustration used to represent the transfer of electrons during the formation of an ionic bond The Magnesium has two electrons to give, whereas the Fluorines have only one “vacancy” each Consequently, Magnesium can accommodate two Fluorine atoms : F . Mg [ F ] - 2+ 4/19/2017 2

Lewis Structures The tendency of atoms in a molecule to have eight electrons (ns2np6) in their outer shell (two for hydrogen) is called the octet rule You can represent the formation of the covalent bond in H2 as follows: This uses the Lewis dot symbols for the hydrogen atom and represents the covalent bond by a pair of dots H  + 4/19/2017 2

The Electron Probability Distribution for the H2 Molecule
4/19/2017

Lewis Structures The shared electrons in H2 spend part of the time in the region around each atom In this sense, each atom in H2 has a helium (1s2) configuration : H 4/19/2017 2

Lewis Structures The formation of a bond between H and Cl to give an HCl molecule can be represented in a similar way Thus, hydrogen has two valence electrons about it (as in He) and Cl has eight valence electrons about it (as in Ar) . : Cl : H Cl . H + 4/19/2017 2

: H Cl Lewis Structures Hydrogen has no unbonded pairs
Formulas such as these are referred to as Lewis electron-dot formulas or Lewis structures An electron pair is either a: bonding pair (shared between two atoms) lone pair (electron pair that is not shared) Hydrogen has no unbonded pairs Chlorine has 3 unbonded pairs : H Cl 4/19/2017 2

Lewis Structures Rules for obtaining Lewis electron dot formulas
Calculate the number of valence electrons for the molecule from: group # for each atom (1-8) add the charge of Anion subtract the charge of a Cation Put atom with the lowest group number and lowest electronegativity as the central atom Arrange the other elements (ligands) around the central atom 4/19/2017 2

Lewis Structures Rules for Lewis Dot Formulas
Distribute electrons to atoms surrounding the central atom to satisfy the octet rule for each atom Distribute the remaining electrons as pairs to the central atom If the Central atom is deficient in electrons to complete the octet; move electron pairs from surrounding atoms to complete central atom valence electron needs, that is form one or more double bonds (possibly triple bonds) around the central atom 4/19/2017

Practice Problem Write a lewis structure for CCl2F2
Step 1: Arrange Atoms (Carbon is “Central Atom” because is has the lowest group number and lowest electronegativity Step 2: Determine total number of valence electrons 1 x C(4) x Cl(7) x F(7) = 32 Step 3: Draw single bonds to central atom and subtract e- for each single bond (4 x 2 = 8)  32 – 8 = 24 remaining Step 4: Distribute the 24 remaining electrons in pairs around surrounding atoms (3 electron pairs around each Fluoride atom) 4/19/2017

Writing Lewis Dot Formulas
The Lewis electron-dot formula of a covalent compound is a simple two-dimensional representation of the positions of electrons in a molecule Bonding electron pairs are indicated by either two dots or a dash In addition, these formulas show the positions of lone pairs of electrons 4/19/2017 2 12

The following rules allow you to write electron- dot formulas even when the central atom does not follow the octet rule To illustrate, draw the structure of: Phosphorus Trichloride Con’t on next slide 4/19/2017 2 13

Step 1: Total all valence electrons in the molecular formula. That is, total the group numbers of all the atoms in the formula For a polyatomic anion, add the number of negative charges to this total For a polyatomic cation, subtract the number of positive charges from this total P 3s23p3 Cl 3s23p5 (2+3) + 3x(2+5) = 5+21 26 total electrons 5 e- (7 e-) x 3 Con’t on next slide 4/19/2017 2 14

Step 2: Arrange the atoms radially, with the least electronegative atom in the center Place one pair of electrons between the central atom and each peripheral atom P Cl 26 – 6 = 20 remaining Con’t on next slide 4/19/2017 2 15

Step 3: Distribute the remaining electrons to the peripheral atoms to satisfy the octet rule : Cl : Cl P : Cl : : 26 – (3 x 6 + 6) = 2 remaining Con’t on next slide 4/19/2017 2 16

Step 4: Distribute any remaining electrons (2) to the central atom. If the number of electrons on the central atom is less than the number of electrons required to complete the octet for that atom, use one or more electrons pairs from other atoms to form double or triple bonds : : P Cl : Phosphorus has an octet of electrons No double bonds required 4/19/2017 2 17

Exceptions to the Octet Rule
Although many molecules obey the octet rule, there are exceptions where the central atom has more than eight electrons Generally, if a nonmetal is in the third period or greater it can accommodate as many as twelve electrons, if it is the central atom These elements have unfilled “d” subshells that can be used for bonding 4/19/2017 2 18

For example, the bonding in phosphorus pentafluoride, PF5, shows ten electrons surrounding the phosphorus Total valence electrons 5 x 7 (F) + 5 (P) = 40 Distribute electrons to F atoms 5 x 6 = 30 Establish bonding pairs 5 x 2 = 10 Remaining electrons 40 – 30 – 10 = 0 Phosphorus has “0” non-bonding pairs : F : : F P Since Phosphorus is in Period 3, PF5 is a “hypervalent” molecule The Phosphorus utilizes electrons from other shells (vacant orbitals) to create a valence shell with more than 8 electrons 4/19/2017 2 19

In Xenon Tetrafluoride, XeF4, the Xenon atom must accommodate two extra lone pairs Total valence electrons 4 x = 36 Distribute electrons to F atoms 4 x 6 = 24 Establish bonding pairs 4 x 2 = 8 Remaining electrons 36 – 24 – 8 = 4 Add 2 non-bonding pairs to Xe Xe violates “octet” rule XeF4 is a “hypervalent” molecule and utilizes vacant “d” orbitals to create a valence shell with more than 8 electrons F : : : F Xe 4/19/2017 2 20

Delocalized Bonding: Resonance
The structure of Ozone, O3, can be represented by two different Lewis electron-dot formulas Experiments show, however, that both bonds are identical Ozone (O3) O : O : or 4/19/2017 2 21

Delocalized Bonding: Resonance
According to Resonance Theory, these two equal bonds are represented as one pair of bonding electrons spread over the region of all three atoms This is called delocalized bonding, in which a bonding pair of electrons is spread over a number of atoms Ozone (O3) O 4/19/2017 2 22

Resonance & Bond Order Recall (Chap 9) – Bond Order
The number of electron pairs being shared by any pair of “Bonded Atoms” or The number of electron pairs divided by the number of bonded-atom pairs Ex. Ozone 4/19/2017

Practice Problem In the following compounds, the Carbon atoms form a “single ring.” Draw a Lewis structure for each compound, identify cases for which “resonance” exists, and determine the C-C bond order(s). C3H4 C3H6 4/19/2017

Practice Problem C4H6 C4H4 4/19/2017

Practice Problem C6H6 4/19/2017

Formal Charge & Lewis Structures
In certain instances, more than one feasible Lewis structure can be illustrated for a molecule For example, H, C and N The concept of “formal charge” can help discern which structure is the most likely Formal Charge: An atom’s formal charge is: Total number of valence electrons Minus all unshared electrons Minus ½ of its shared electrons Formal Charges must sum to actual charge of species: Zero Charge for a Molecule Ionic Charge for an Ion H C N or : 4/19/2017 2 27

When you can write several Lewis structures, choose the one having the least formal charge Form I Form II H C N or : 1 e- 4 e- 5 e- 4e- “domain” electrons group number I IV V -1 +1 FC: Total Valence e- – unshared e- – ½ shared e- FCH: [ ½(2)] = 0 FCC: [ ½(8)] = 0 FCN: [ ½(6)] = 0 FCH: [ ½(2)] = 0 FCC: [ ½(6)] = -1 FCN: [ ½(8)] = +1 Preferred Form - Form I (Least Formal Charge) Note: HCN is a neutral molecule Sum of Formal Charges in the preferred form (0) equals molecular charge (0) 4/19/2017 2 28

Ozone FCOA: [ ½(4)] = 0 FCOB: [ ½(6)] = +1 FCOC: [ ½(2)] = -1 FCOA: [ ½(2)] = -1 FCOB: [ ½(6)] = +1 FCOC: [ ½(4)] = 0 Both “Resonance” forms have the same formal charge and thus, are identical Note: Ozone (O3) is a neutral molecule Sum of Formal Charges (0) equals molecular charge (0) 4/19/2017 2 29

Boron Trifuoride BF3 B F  B F FC B = 3 – 0 -(1/2 * 6) = 0 Even though B violates “Octet Rule”, this is the preferred form because it has “less” formal charge FC B = 3 – 0 -(1/2 * 8) = -1 FC F = 7 – (1/2 * 4) = +1 Sulfur Dioxide SO2 S O  S O  FC S = 6 – 2 – (1/2 * 8) = 0 Preferred Form (Less Formal Charge) FC S = 6 – 2 – (1/2 * 6) = 1 4/19/2017

Resonance/Formal Charge – Nitrate Ion
Total Valence electrons - 3 x 6 (O) x 5 (N) + 1 (ion charge) = 24 Add 1 pair electrons between central atom and each other atom – 3 x 2 = 6 Add electrons to oxygen atoms to complete octet Nitrogen still missing 2 electrons to complete octet Borrow 2 electrons from one oxygen to form double bond Formal Charge – Nitrogen: 5 – (0 + ½*8) = 5 – 4 = +1 Formal Charge – Single bonded Oxygen: 6 – (6 + ½*2) = 6 – 7 = -1 x 2 = -2 Formal Charge – Double bonded Oxygen: 6 – (4 + ½*4) = 6 – 6 = 0 Net Charge of the ION is: (-2) + 0 = -1 4/19/2017

Resonance/Formal Charge – Cyanate Ion
FCN = 5 – (6 + ½*2) = -2 FCC = 4 – (0 + ½*8) = 0 FCO = 6 – (2 + ½*6) = +1 FCN = 5 – (4 + ½*4) = -1 FCC = 4 – (0 + ½*8) = 0 FCO = 6 – (4 + ½*4) = 0 FCN = 5 – (2 + ½*6) = 0 FCC = 4 – (0 + ½*8) = 0 FCO = 6 – (6 + ½*2) = -1 Preferred Form: Eliminate I – Higher formal charge on Nitrogen than Carbon & Oxygen Positive formal charge on Oxygen, which is more electronegative than Nitrogen Eliminate II – Forms II & III have the same magnitude of formal charges, but form III has a -1 charge on the more electronegative Oxygen atom Forms II & III both contribute to the resonant hybrid of the Cyanate Ion, but form III is the more important Note: Net formal charge in form III is same as ionic charge (-1) 4/19/2017

Formal Charge vs Oxidation No
“Formal Charge” is used to examine resonance hybrid structures , whereas “Oxidation Number” is used to monitor “REDOX” reactions Formal Charge - Bonding electrons are assigned equally to the atoms as if the bonding were “Nonpolar” covalent, i.e., each atom has half the electrons making up the bond Formal Charge = valence e- – (unbonded e ½ bonding e-) Oxidation Number - Bonding electrons are transferred completely to the more electronegative atom, as if the bonding were “Ionic” Ox No. = valence e- – (unbonded e- + bonding e-) 4/19/2017

Formal Charge vs Oxidation No
FC (-2) (0) (+1) (-1) (0) (0) (0) (0) (-1) N 5 – (6 + ½ (1)) = -2 C 4 – (0 + ½ (8)) = 0 O 6 – (2 + ½ (6)) = +1 N 5 – (4 + ½ (4)) = -1 C 4 – (0 + ½ (8)) = 0 O 6 – (4 + ½ (4)) = 0 N 5 – (2 + ½ (6)) = 0 C 4 – (0 + ½ (8)) = 0 O 6 – (6 + ½ (2)) = -1 ON (-3) (+4) (-2) (-3) (+4) (-2) (-3) (+4) (-2) N 5 – (6 + 2) = -3 C 4 – (0 + 0) = +4 O 6 – (2 + 6) = -2 N 5 – (4 + 4) = -3 C 4 – (0 + 0) = +4 O 6 – (4 + 4) = -2 N 5 – (2 + 6) = -3 C 4 – (0 + 0) = +4 O 6 – (6 + 2) = -2 Note: Both Nitrogen (N) & Oxygen (O) are more electronegative than Carbon (C); thus, in the computation of Oxidation Number all the electrons are transferred to the N & O leaving C with no lone pairs and no bonded pairs Note: Oxidation Nos do not change from one resonance form to another (electronegativities remain same) 4/19/2017

The Valence-Shell Electron Pair Repulsion Model (VSEPR)
The Valence-Shell Electron Pair Repulsion (VSEPR) model predicts the shapes of molecules and ions by assuming that the valence shell electron pairs are arranged as far from one another as possible Molecular geometry – The shape of a molecule is determined by the positions of atomic nuclei relative to each other, i.e., angular arrangement Central Atom Place atom with “Lower Group Number” in center (N in NF3 needs more electrons to complete octet) If atoms have same group number (SO3 or ClF3), place the atom with the “Higher Period Number” in the center (Sulfur & Chlorine) 4/19/2017 2

VSEPR Model of Molecular Shapes
The following rules and figures will help discern electron pair arrangements Select the Central Atom (Least Electronegative Atom) Draw the Lewis structure Determine how many bonding electron pairs are around the central atom Determine the number of non-bonding electron pairs Count a multiple bond as “one pair” Arrange the electron pairs as far apart as possible to minimize electron repulsions Note the number of bonding and lone pairs 4/19/2017 2

To predict the relative positions of atoms around a given atom using the VSEPR model, you first note the arrangement of the electron pairs around that central atom Molecular Notation: A – The Central Atom (Least Electronegative atom) X – The Ligands (Bonding Pairs) a – The Number of Ligands E – Non-Bonding Electron Pairs b – The Number of Non-Bonding Electron Pairs Double & Triple Bonds count as a “single” electron pair The Geometric arrangement is determined by: sum (a + b) AXaEb 4/19/2017

Molecule Lewis Structure ALxNy Notation Geometric e- Pair Arrangement 3-D 3-D View & Isomers BeH2 AX2E0 a = 2 b = 0 a + b = 2 Linear BH3 AX3E0 a = 3 a + b = 3 Trigonal Planar CH2Li2 AX4E0 or AX2X2E0 a = 4 a + b = 4 Tetrahedral OH2 AX2E2 b = 2 BiF5 AX5E0 a = 5 a + b = 5 Bipyramidal 4/19/2017

Molecule Lewis Structure ALxNy Notation Geometric e- Pair Arrangement 3-D 3-D View & Selected Isomers SF5Cl AX6E0 or AX5X1E0 a = 6 b = 0 a + b = 6 Octahedral PCL4Br AX5N0 AX4X1E0 a = 5 a + b = 5 Trigonal Bipyramidal TeCl3Br AX4E1 AX3X1E1 b = 4 a = 1 SF4Cl2 AX4X2E0 XeF2 AX2E3 a = 2 b = 3 4/19/2017

Arrangement of Electron Pairs About an Atom: Basic Shapes
CS2 HCN BeF2 NO2+ 4/19/2017

SO3 BF3 NO3− NO2– CO32− SO2 O3 PbCl2 SnBr2 4/19/2017

CH4 SiCl4 SO42- ClO4- NH3 PF3 ClO3 H3O+ H2O OF2 SCl2 4/19/2017

PF5 AsF5 SOF4 SF4, XeO2F2, IF4+, IO2F2- ClF3 BrF3 XeF2 I3- IF2- 4/19/2017

SF6 IOF5 BrF5 TeF5- XeOF4 XeF4 ICl4- 4/19/2017

Electron Pair Arrangement
4/19/2017

: Linear Geometry Two electron pairs (linear arrangement)
Double bonds count as a “single electron pair”  2 bonding pairs 0 non-bonding pairs AXaEb = a + b = = 2 (Linear) Thus, according to the VSEPR model, the bonds are arranged linearly (bond angle = 180o) Molecular shape of carbon dioxide is linear : Carbon is central atom because it has lower group number 4/19/2017 2

Trigonal Planar Geometry
Three electron pairs on Central atom The three groups of electron pairs are arranged in a trigonal plane. Thus, the molecular shape of COCl2 is trigonal planar. The Bond angle is 120o Central Atom - Carbon 3 bonding electron pairs (double bond counts as 1 pair) 0 non-bonding electron pairs a + b = = 3  Trigonal Planar Cl C : O 4/19/2017 2

Effect of Double Bonds Bond angles deviate from ideal angles when surrounding atoms and electron groups are not identical A double bond has greater electron density and repels two single bonds more strongly than they repel each other C H O    120o Ideal C H O    122o 116o Actual 4/19/2017

Effect of Lone Pairs The molecular shape is defined only by the positions of the nuclei When one of the three electron pairs in a trigonal planar molecule is a lone (non-bonding) pair, it is held by only one nucleus It is less confined and exerts a stronger repulsive force than a bonding pair This results in a decrease in the angle between the bonding pairs The normal Trigonal Planar angle between the bonding pairs is 120o 4/19/2017

Three electron pairs (Effect of ‘Lone’ pairs) (trigonal planar arrangement) Ozone has two bonding electron pairs about the central oxygen (double bond counts as 1 pair) There is one non-bonding lone pair These groups have a: Trigonal Planar arrangement AXaEb (a + b) = = 3 Since one of the groups is a lone pair, the molecular geometry is described as bent or angular SO3 BF3 NO3- CO32- O : <120o 4/19/2017 2

(Tetrahedral Arrangement)
Tetrahedral Geometry Four electron pairs (Tetrahedral Arrangement) Four electron pairs about the central atom lead to three different molecular geometries a + b = a + b = a + b = 2 + 2 = = = 4 :Cl: : :Cl C Cl: H N : : O H 4/19/2017 2

H H Tetrahedral Geometry :Cl: : :Cl C Cl: Cl N : : O
Molecular Geometries produced by variable non- bonding electron pairs :Cl: : :Cl C Cl: Cl Note impact of non-bonding electron pairs on bond angle H N : : O H AX4 109o AX4E 107o AX2E2 105o 107o 107o 105o CH4, SiCl4, SO42-, ClO4- PF3, ClO3-, H3O+ OF2, SCl2 4/19/2017 2

(trigonal bipyramidal arrangement)
Five electron pairs (trigonal bipyramidal arrangement) This structure results in both 90o and 120o bond angles : F : : F : : F P 90o axial 120o equatorial ASF5 SOF4 4/19/2017 2

Trigonal Bipyramidal : F F F : : Cl S Xe
Other molecular geometries are possible when one or more of the electron pairs is a lone pair Cl F : S F : Xe F : <90o (ax) <120o (eq) 180o <90o (ax) XeO2F2 IF4+ IOF2- ClF3 BrF3 XeF2 I3- IF2- 4/19/2017 2

Octahedral Geometry :F: : S :F F: Six electron pairs
(Octahedral arrangement) This octahedral arrangement results in: 90o bond angles F: : :F S :F: 90o SF6 IOF5 4/19/2017 2

(octahedral arrangement)
Other Geometries Six electron pairs (octahedral arrangement) I F : Xe F : Noble gases not always inert Xenon forms 6 electron domains Iodine violates octet rule Iodine is sp3d2 hybridized Iodine uses d orbitals square pyramidal square planar <90o 90o BrF5 TeF5- XeOF4 XeF4 ICl4- 4/19/2017 2

Practice Problem In the ICl4– ion, the electron pairs are arranged around the central iodine atom in the shape of a. a tetrahedron b. a trigonal bipyramid c. a square plane d. an octahedron e. a trigonal pyramid Ans: a I Cl AX4 AXaEb a + b = = 4 (AX4 – Tetrahedral) 4/19/2017

Dipole Moment The dipole moment () is a measure of the degree of charge separation (molecular polarity) in a molecule The product of the magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges  = Q  r Example: What is the dipole moment in Debyes of a molecule with a: Bond Length = 127 pm Electron Charge (e) 1.6x10-19 C 1 Debye = 3.34 x C  m 4/19/2017 2

Dipole Moment In the previous example the Dipole Moment was calculated on the assumption that each of the atoms bore a full charge of 1 electronic unit , e, (+1 & -1). That is: Q = 1 e = x Coulombs (C) Such a molecule would be nearly ionic Atoms in asymmetric covalent molecules would not exhibit full ionic charges, thus, the value of Q would be less than 1 e, depending on the relative electronegativity differences and the bond length 4/19/2017

Dipole Moment and Molecular Geometry
The polarity of individual bonds within a molecule can be viewed as vector quantities Thus, molecules that are perfectly symmetric have a zero dipole moment. These molecules are considered nonpolar d- d+ 4/19/2017

However, molecules that exhibit any asymmetry in the arrangement of electron pairs would have a nonzero dipole moment. These molecules are considered polar d- d+ H N : NH3 PF3 ClO3 H3O+ 4/19/2017 2

Formula Molecular Geometry Dipole Moment AX Linear Can Be nonzero AX2E0 Zero AX3E0 Trigonal Planar AX2E1 Trigonal Planar Bent AX4E0 Tetrahedral AX3E1 Tetrahedral Trigonal Pyramidal AX2E2 Tetrahedral Bent AX5E0 Trigonal Bipyramidal AX4E1 Trigonal Bipyramidal SeeSaw AX3E2 Trigonal Bipyramidal T-Shaped AX2E3 Trigonal Bipyramidal Linear AX6E0 Octahedral AX5E1 Octahedral Square Pyramidal AX4E2 Octahedral Square Planar 4/19/2017

Practice Problem N is more Electronegative than H N is more EN than Ca
The Nitrogen atom would be expected to have the positive end of the dipole in the species a. NH4+ b. Ca3N2 c. HCN d. AlN e. NO+ Ans: e N is more Electronegative than H N is more EN than Ca N is more EN than C N is more EN than Al O is more EN than Nitrogen 4/19/2017

Practice Problem  N O  -  N O  - N O 
Which of the following molecules is polar? a. BF3 b. CBr4 c. CO2 d. NO2 e. SF6 Ans: d  N O  ● ● ● + -  N O  ● ● ● + - N O  ● ● ● The Lewis structures for BF3, CBr4, CO2, and SF6 do not have any non-bonding electrons on the central atom The Lewis structure for NO2 shows one double bond and a lone non-bonding electron on the Nitrogen The VSEPR Molecular Geometry for NO2 is AX2E (a + b = = 3) - Trigonal Planar Formal Charge on N is 5 – 1 – ½ (6) = +1 NO2 molecule is polar 4/19/2017

Practice Problem Which of the following compounds is nonpolar?
a. XeF2 b. HCl c. SO2 d. H2S e. N20 Ans: a HCL is ionic and very polar SO2 has AX2E1 Trigonal Planar Bent geometry with a dipole moment (polar) H2S has AX2E2 Tetrahedral Bent geometry and with a dipole moment (polar) N2O has AX2E0 linear with asymmetric geometry. Since oxygen is more EN than N, the molecule is polar XeF2 has AX2E3 Trigonal Bypyramidal Geometry, but linear molecular geometry (nonpolar) 4/19/2017

Geometric Configuration Determined by the sum (a + b)
Equation Summary VSEPR Model - AXaEb Geometric Configuration Determined by the sum (a + b) Dipole Moment =  = Q x r 4/19/2017

George Mason University General Chemistry 211 Chapter 10

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