Bonding & Molecular Structure:

Bonding & Molecular Structure:
Chemistry Lecture 22 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts Chapter Highlights bonding types Lewis symbols & octets ionic bonding & ionic lattices covalent bonding & Lewis dot structures resonance structures breaking the octet rule & formal charge bond order & bond energy VSEPR molecular shape & polarity

Ionic, Covalent & Metallic Bonding
Chemistry Lecture 22 Ionic, Covalent & Metallic Bonding Ionic bond: Term given to the electrostatic (charge-based) attractive forces which hold oppositely charged ions together Li+ F-

Chemistry Lecture 22 Ionic, Covalent & Metallic Bonding Covalent bond: The sharing of electrons between two atoms that acts to hold the atoms together H :

Chemistry Lecture 22 Ionic, Covalent & Metallic Bonding Metallic bond: Is found in metals. Atoms of the metal are bound to several neighbors, holding the atoms together but allowing electrons to move freely. M :::

S Lewis Symbols S: [Ne]3s23p4 Lewis symbols = electron-dot symbols
Chemistry Lecture 22 Lewis Symbols Lewis symbols = electron-dot symbols Lewis symbols place one dot for each valence electron around the symbol of the element. For Example: S: [Ne]3s23p4 S

Attain the closed shell configuration of a Group 18 inert gas
Chemistry Lecture 22 Lewis Octets Octet rule: Atoms tend to gain, lose, or share enough electrons to become surrounded by eight valence electrons Attain the closed shell configuration of a Group 18 inert gas Ne: [He]2s22p6 Ne

Na + Cl Na+ + Cl - Ionic Bonding
Chemistry Lecture 22 Ionic Bonding The ionic bond is formed when ions of opposite charge (anions and cations) are attracted and held to one another by electrostatic attractions Na(g) Cl(g) NaCl(g) Na Cl Na Cl -

Energetics of Forming a Na-Cl Ionic Bond
Chemistry Lecture 22 Energetics of Forming a Na-Cl Ionic Bond Na(g) Na+(g) e DE = kJ/mol Cl(g) e Cl-(g) DE = kJ/mol Therefore electron transfer costs 147 kJmol-1 !!!!!

Chemistry Lecture 22 Energetics of Forming a Na-Cl Ionic Bond But, Coulomb’s law... where: Q1 and Q2 are the charges on the cation and anion, d is the distance between the nuclei (sum of the ionic radii) and k is a constant = 8.99 x 109 J-m/C2 E = (6.022 x 1023)(8.99 x 109) E = E = k -411 kJ/mol

Chemistry Lecture 22 Energetics of Forming a Na-Cl Ionic Bond Therefore: the overall DE for the reaction Na(g) Cl(g) NaCl(g) DE = ( ) kJ/mol = -264 kJ/mol

Chemistry Lecture 22 Ionic Lattices In order to maximize the attractions among ions, ionic solids exist in lattices, which are regularly repeating three-dimensional arrays of ions

Chemistry Lecture 22 Coordination Number Coordination number: the number of close contacts in the lattice array (equals 6 for this Na+ ion below)

Chemistry Lecture 22 Lattice Energies Lattice energy: the energy required to separate the crystalline solid into the constituent gaseous ions. NaCl(s) Na+(g) Cl-(g) It is a measure of the stability of the crystalline state Note: The lattice energy for NaCl(s) is kJ/mol as compared to -264 kJ/mol that we calculated!!

Lattice Energies In summary lattice energies:
Chemistry Lecture 22 Lattice Energies In summary lattice energies: increase as charges of the ions increase increase as sizes of the ions decrease Coulomb’s Law!!! increase with increasing coordination number E = k

Born-Haber Cycle Na+(g) + Cl-(g) Na(g) + Cl(g)
Chemistry Lecture 22 Born-Haber Cycle Na+(g) Cl-(g) Step 3 Step 4 Step 5 Na(g) Cl(g) Step 1 Step 2 Na(s) /2 Cl2(g) NaCl(s) DHfo

H + H H = H-H Cl + Cl Cl = Cl-Cl
Chemistry Lecture 23 Covalent Bonding Covalent bond: a bond formed between two atoms by sharing of electrons. Lewis Structures for H2 and Cl2 H + H H = H-H Cl + Cl Cl = Cl-Cl

Cl Covalent Single Bonds
Chemistry Lecture 23 Covalent Single Bonds Different multiplicities of covalent bonds are possible. Single bonds are covalent bonds in which one pair of electrons is shared by the two atoms Cl or Cl-Cl

Multiple Covalent Bonds
Chemistry Lecture 23 Multiple Covalent Bonds Different multiplicities of covalent bonds are possible. Double bonds are covalent bonds in which two pairs of electrons are shared by the two atoms. Triple bonds are covalent bonds in which three pairs of electrons are shared by the two atoms N + = or N N N N _ = N N N N = 1.10 A A A

Chemistry Lecture 23 Bond Polarity Recall: for a covalent bond the bonding electrons are equally shared between two atoms. Recall: for an ionic bond the bonding electrons are separated between the ions (electrostatic attraction). When sharing is not equal, the bond is called a polar bond. Equal sharing is sometimes referred to as a nonpolar bond.

Chemistry Lecture 23 Electronegativity Electronegativity: the ability of an atom in a molecule to attract electrons to itself. The higher an element's electronegativity, the better it competes for electrons. Electronegativity is related to ionization energy and electron affinity. The scale (Pauling scale) has no units

Electronegativity Values
Chemistry Lecture 23 Electronegativity Values

Polar vs. Nonpolar Bonds
Chemistry Lecture 23 Polar vs. Nonpolar Bonds Electronegativity difference between two atoms of a bond is related to the polarity of the bond. The greater the electronegativity difference, the more polar the bond. > 2.0 = ionic, < 0.5 = nonpolar between 0.5 and 2.0 = polar

_ H F Bond Polarity Examples: F2 HF LiF 4.0 - 4.0 = 0 4.0 - 2.1 = 1.9
Chemistry Lecture 23 Bond Polarity Examples: d+ represents a partial positive charge d- represents a partial negative charge F2 = 0 nonpolar HF = 1.9 polar LiF = 3.0 ionic H F _ d+ d-

Drawing Lewis Structures
Chemistry Lecture 23 Drawing Lewis Structures Sum the valence electrons from all atoms in the species. Write the atomic symbols for the atoms involved so as to show which atoms are connected to which. Draw a single bond between each pair of bonded atoms Complete the octets of the atoms bonded to the central atom (i.e. the peripheral atoms)

Chemistry Lecture 23 Drawing Lewis Structures Place leftover electrons on the central atom, even if it results in the central atom having more than an octet If there are not enough electrons to give the central atom an octet, form multiple bonds by pulling terminal electrons from a peripheral atom and placing them into the bond with the central atom

Drawing Lewis Structures Draw the Lewis structure for PCl3.
Chemistry Lecture 23 Drawing Lewis Structures Question: Draw the Lewis structure for PCl3.

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 1: Sum the valence electrons. P has 5 and each Cl has 7 for a total of [5 + (3 x 7)] = 26 valence electrons

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. NOTE: In a binary (two-element) compound, the first element listed is usually the central one with the others surrounding it Cl P Cl _ Cl

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 3: Complete the octets on the atoms bonded to the central atom. NOTE: This accounts for 24 of the 26 valence electrons Cl P Cl _ Cl

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 4: Place the remaining electrons on the central atom to complete the octet. Since this gives an octet to each atom we are finished Cl P Cl _ Cl

Drawing Lewis Structures Draw the Lewis structure for HCN.
Chemistry Lecture 23 Drawing Lewis Structures Question: Draw the Lewis structure for HCN.

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 1: Sum the valence electrons. H has 1, C has 4 and N has 5 for a total of [ )] = 10 valence electrons

Drawing Lewis Structures This accounts for 4 valence electrons
Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. NOTE: Since H can only form one covalent bond it can never be the central atom. The choices are HCN or HNC. Formula is written HCN!!! H C N _ This accounts for 4 valence electrons

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 3: Complete the octets on the atoms bonded to the central atom. H C N _ BUT: There are only 6 valence electrons left. If we put them on N we do not achieve an octet at C !!

Chemistry Lecture 23 Drawing Lewis Structures Answer: Step 4: Try using multiple bonding to share the electrons between C and N. A triple bond is required to give an octet to each atom = H C N _

Formal Charge & Lewis Structures
Chemistry Lecture 24 Formal Charge & Lewis Structures Formal charges: a way of assigning a relative charge to each atom in the molecule When several different Lewis structures seem plausible, the one in which the formal charges are minimized is generally the preferred one.

Assigning Formal Charge
Chemistry Lecture 24 Assigning Formal Charge All bonding electrons are divided equally between the atoms that form bonds All nonbonding electrons are assigned to the atom on which they reside

Assigning Formal Charge FC = VE - (NBE + 1/2BE)
Chemistry Lecture 24 Assigning Formal Charge Formal charge: the number of valence electrons for the element minus the number of electrons assigned by rules 1 and 2. formal charge on an atom in a molecule = {# valence electrons normally found for that atom - [(# non-bonding electrons) + 1/2(# bonding electrons)]} FC = VE - (NBE + 1/2BE)

Applying Formal Charge to Lewis Structures
Chemistry Lecture 24 Applying Formal Charge to Lewis Structures Question: There are three possible structures for SCN-. Use formal charge to decide the most likely structure.

Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 1: Sum the valence electrons. S has 6, C has 4 and N has 5 and there is an extra electron represented by the single negative charge of the ion. Total of [ ] = 16 valence electrons

Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. [S C N]- _ [C S N]- _ [S N C]- _

[ S C N ]- _ [ C S N ]- _ [ S N C ]- _
Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 3: Complete the octets on the atoms bonded to the central atom. [ S C N ]- _ [ C S N ]- _ [ S N C ]- _ BUT: Each of these leaves us with only four electrons at the central atom!

Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 4: Use multiple bonding to share the electrons between peripheral atoms and the central atom until octets are achieved. [S C N]- = [C S N]- = [S N C]- =

[S C N]- = [C S N]- [S N C]- 0 0 -1 -2 +2 -1 0 +1 -2
Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 5: Calculate formal charge for each atom. [S C N]- = [C S N]- [S N C]- Note: the total formal charge on each molecule is equal to the charge on the molecule

[S C N]- = [C S N]- [S N C]- 0 0 -1 -2 +2 -1 0 +1 -2
Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 6: Decide on the most probable structure. [S C N]- = [C S N]- [S N C]- The structure that results in the least amount of formal charge separation throughout the molecule

_ _ O O O = O O O = Resonance Structures ozone
Chemistry Lecture 24 Resonance Structures There are times when more than one Lewis structure involving multiple bonds seems equally stable O O O = _ O O O = _ ozone Resonance structures: structures that differ only in the placement of electrons.

_ O O O = Resonance Structures
Chemistry Lecture 24 Resonance Structures Resonance forms rapidly interconvert so that the structure appears to be a blend of all the forms. O O O = _

_ O O O Resonance Structures
Chemistry Lecture 24 Resonance Structures The molecule does not oscillate rapidly between two or more different forms. There is only one form of the molecule. Ozone has two equivalent O-O bonds whose length is intermediate between single and double bonds O O O _

Exceptions to the Octet Rule
Chemistry Lecture 24 Exceptions to the Octet Rule Most second-period elements (n = 2), notably C, N, O and F are always observed with octets, but.… Molecules with an odd number of electrons. Molecules in which an atom has less than an octet. Molecules in which an atom has more than an octet.

Odd Numbers of Electrons
Chemistry Lecture 24 Odd Numbers of Electrons Some examples are ClO2, NO and NO2. They tend to be very rare and very reactive. React to pair unpaired electrons. O N _ = O N _ = O N _ = +

Chemistry Lecture 24 Less Than an Octet Some lighter elements (H, Be, B) tend to be surrounded by less than an octet of electrons. For example: Hydrogen has a valence-shell capacity of only 2 electrons. Beryllium often is surrounded by four electrons and boron is often surrounded by six electrons, due mostly to their small size

Chemistry Lecture 24 Less Than an Octet Formal charge assignments generally support these deviations from the octet rule. F B _

_ _ _ B N N B F H H F Consequences
Chemistry Lecture 24 Consequences For example, BF3 acts as a Lewis acid to form an adduct with NH3 which is a Lewis base F B _ H N F B _ H N _ +

Chemistry Lecture 24 More Than an Octet A larger group of compounds are those in which the central atom is surrounded by more than an octet of electrons. Elements of the third period or lower are capable of expanding their octets, due to the availability of d orbitals.

Chemistry Lecture 24 More Than an Octet For example: After satisfying the octet rule for the peripheral F-atoms in PF5, we have 10 valence electrons at P-atom. The extra electrons can reside in a 3d-orbital. F P _

Lewis Structures Question: Draw the Lewis structure for IF4-.
Chemistry Lecture 24 Lewis Structures Question: Draw the Lewis structure for IF4-.

Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 1: Sum the valence electrons. I has 7 and each F has 7 and we count 1 for the anion for a total of [7 + (4 x 7)] +1 = 36 valence electrons

Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. Recall: In a binary (two-element) compound, the first element listed and the heavier is usually the central one with the others surrounding it. F F I _ F F

Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 3: Complete the octets on the atoms bonded to the central atom. NOTE: This accounts for 32 of the 36 valence electrons F I _

Drawing Lewis Structures Remember: No double bonds to F
Chemistry Lecture 24 Drawing Lewis Structures Answer: Step 4: We must place the remaining electrons on the central atom even though this gives the I-atom more than an octet. F I _ Remember: No double bonds to F

Term Test #2 Friday November 9th, 2001 6:30 P.M.
Chemistry Lecture 24 Term Test #2 Friday November 9th, 2001 6:30 P.M.

Contents: SIX “Problems”!!
Chemistry Lecture 24 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 6-9 A Periodic Table & ALL Required Constants will be Supplied

Chemistry Lecture 25 Molecular Shape The geometry of a molecule, along with its size determines in large part its chemical behavior C Cl _

(R. J. Gillespie & R. S. Nyholm)
Chemistry Lecture 25 VSEPR Model The geometry of molecules, can be predicted using VSEPR rules (valence-shell electron-pair repulsion) (R. J. Gillespie & R. S. Nyholm) The best arrangement of electron pairs is the one that maximizes the separation among them, since this minimizes the repulsions among the electron pairs..

Chemistry Lecture 25 Electron-pairs Electron pairs are differentiated as being bonding electron pairs or nonbonding electron pairs (lone pairs). (For the purposes of VSEPR, a multiple bond counts as a single bonding pair or as a single region of electrons.) 1 lone pair H _ N 3 bonding pairs

electron pairs in total
Chemistry Lecture 25 Electron-pairs The electron-pair geometry is the geometry described by the regions of electrons around the central atom. H _ N NH3 (ammonia) has FOUR electron pairs in total tetrahedral geometry

bonding electron pairs & ONE lone pair
Chemistry Lecture 25 Electron-pairs The molecular geometry is the geometry described by the atoms only!! (central atom and the peripheral atoms) NH3 (ammonia) has THREE bonding electron pairs & ONE lone pair H _ N Triangular pyramidal N

Chemistry Lecture 25 Building VSEPR Models Sketch the Lewis dot structure of the molecule or ion Count the total number of electron pairs around the central atom (multiple bonds only count as one pair) and arrange them in the way that minimizes electron-pair repulsions Describe the molecular geometry in terms of the angular arrangement of bonding-pairs

Electron-Pair Geometries (2)
Chemistry Lecture 25 Electron-Pair Geometries (2) AX2 Linear 180 o

Chemistry Lecture 25 Electron-Pair Geometries (3) AX3 Trigonal Planar 120 o

Chemistry Lecture 25 Electron-Pair Geometries (4) 109.5 o AX4 Tetrahedral

Chemistry Lecture 25 Electron-Pair Geometries (5) AX5 Trigonal Bipyramidal 90 o 120 o Axial Equatorial

Chemistry Lecture 25 Electron-Pair Geometries (6) AX6 Octahedral 90 o

Deviations From Electron-Pair Geometries
Chemistry Lecture 25 Deviations From Electron-Pair Geometries

Electron-Pair Repulsions
Chemistry Lecture 25 Electron-Pair Repulsions lone-pair//lone-pair > lone-pair//bond-pair bond-pair//bond-pair > Nonbonding electron pairs repel other electron pairs more than do single-bonding electron pairs. Electrons in multiple bonds repel other electron pairs more than do single-bonding electron pairs.

Molecular Shapes Trigonal No lone pairs Bipyramidal
Chemistry Lecture 25 Molecular Shapes Trigonal Bipyramidal No lone pairs

Chemistry Lecture 25 Molecular Shapes Seesaw One lone pair

Chemistry Lecture 25 Molecular Shapes T-Shaped Two lone pairs

Chemistry Lecture 25 Molecular Shapes Linear Three lone pairs

VSEPR Models Question:
Chemistry Lecture 25 VSEPR Models Question: Using the VSEPR model, predict the molecular geometry of SnCl3-.

_ Cl Sn VSEPR Models Answer:
Chemistry Lecture 25 VSEPR Models Answer: Step 1: The Lewis structure for the SnCl3- anion is: Sn Cl _

The molecular geometry is trigonal pyramidal
Chemistry Lecture 25 VSEPR Models Answer: Step 2: The central atom is surrounded by one nonbonding electron pair and three single bonds. Therefore, FOUR electron pairs; tetrahedral electron-pair geometry. The molecular geometry is trigonal pyramidal

VSEPR Models Question:
Chemistry Lecture 25 VSEPR Models Question: Using the VSEPR model, predict the molecular geometry of SF4.

S F _ VSEPR Models Answer:
Chemistry Lecture 25 VSEPR Models Answer: Step 1: The Lewis structure for the SF4 molecule is:. S F _

Chemistry Lecture 25 VSEPR Models Answer: Step 2: The central atom is surrounded by one nonbonding electron pair and four single bonds. Therefore, FIVE electron pairs; trigonal bipyramidal electron -pair geometry, but there are two possiblities

Chemistry Lecture 25 VSEPR Models Answer: Step 2: The central atom is surrounded by one nonbonding electron pair and four single bonds. Therefore, FIVE electron pairs; trigonal bipyramidal electron -pair geometry, only one possibility is correct!

Chemistry Lecture 25 Molecular Polarity

Chemistry Lecture 25 Molecular Polarity A polar molecule is one in which the centers of positive and negative charge do not coincide. CCl4 CHCl3

Textbook Questions From Chapter # 9
Chemistry Lecture 25 Textbook Questions From Chapter # 9 Valence electrons, octet rule: 28, 32 Lewis structures: 38, 40 Bond properties: 46, 48, 50, 56 Bond polarity & formal charge: 62, 66 Molecular geometry: 76, 80, 82, 86, 96

Term Test #2 Friday November 9th, 2001 6:30 P.M.
Chemistry Lecture 25 Term Test #2 Friday November 9th, 2001 6:30 P.M.

Contents: SIX “Problems”!!
Chemistry Lecture 25 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 6-9 A Periodic Table & ALL Required Constants will be Supplied

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