Chemistry140 Lecture 22 Chapter 9 Bonding Molecular Structure
Chemistry-140 Lecture 22 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts v Chapter Highlights ð bonding types ð Lewis symbols & octets ð ionic bonding & ionic lattices l covalent bonding & Lewis dot structures l resonance structures l breaking the octet rule & formal charge l bond order & bond energy l VSEPR molecular shape & polarity
Chemistry-140 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-140 Lecture 22 Ionic, Covalent & Metallic Bonding ¶ Covalent bond: The sharing of electrons between two atoms that acts to hold the atoms together H : H
Chemistry-140 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 : : : M
Chemistry-140 Lecture 22 Lewis Symbols v Lewis symbols = electron-dot symbols ¶ Lewis symbols place one dot for each valence electron around the symbol of the element. For Example: S: [Ne]3 s 23 p 4 S
Chemistry-140 Lecture 22 Lewis Octets v 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]2 s 22 p 6 Ne
Chemistry-140 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) Na + Cl(g) Na+ + Cl -
Chemistry-140 Lecture 22 Energetics of Forming a Na-Cl Ionic Bond Na(g) Cl(g) + e- Na+(g) + e. Cl-(g) DE = +496 k. J/mol DE = -349 k. J/mol Therefore electron transfer costs 147 k. Jmol-1 !!!!!
Chemistry-140 Lecture 22 Energetics of Forming a Na-Cl Ionic Bond But, Coulomb’s law. . . E=k ¶ where: Q 1 and Q 2 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/C 2 E = (6. 022 x 1023)(8. 99 x 109) E = -411 k. J/mol
Chemistry-140 Lecture 22 Energetics of Forming a Na-Cl Ionic Bond Therefore: the overall DE for the reaction Na(g) + Cl(g) Na. Cl(g) DE = (-411 + 147) k. J/mol = -264 k. J/mol
Chemistry-140 Lecture 22 Ionic Lattices ¶ In order to maximize the attractions among ions, ionic solids exist in lattices, which are regularly repeating threedimensional arrays of ions
Chemistry-140 Lecture 22 Coordination Number ¶ Coordination number: the number of close contacts in the lattice array (equals 6 for this Na+ ion below)
Chemistry-140 Lecture 22 Lattice Energies ¶ Lattice energy: the energy required to separate the crystalline solid into the constituent gaseous ions. Na. Cl(s) Na+(g) + Cl-(g) It is a measure of the stability of the crystalline state Note: The lattice energy for Na. Cl(s) is -786 k. J/mol as compared to -264 k. J/mol that we calculated!!
Chemistry-140 Lecture 22 Lattice Energies In summary lattice energies: Œ increase as charges of the ions increase as sizes of the ions decrease Coulomb’s Law!!! E=k Ž increase with increasing coordination number
Chemistry-140 Lecture 22 Born-Haber Cycle Na+(g) + Step 4 Step 3 Na(g) + Step 1 Na(s) Cl-(g) Step 5 Cl(g) Step 2 + 1/2 Cl 2(g) D H fo Na. Cl(s)
Chemistry-140 Lecture 23 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts v Chapter Highlights l bonding types l Lewis symbols & octets l ionic bonding & ionic lattices ð covalent bonding & Lewis dot structures l resonance structures l breaking the octet rule & formal charge l bond order & bond energy l VSEPR molecular shape & polarity
Chemistry-140 Lecture 23 Covalent Bonding v Covalent bond: a bond formed between two atoms by sharing of electrons. ¶ Lewis Structures for H 2 and Cl 2 H + H Cl + Cl H H = H-H Cl Cl = Cl-Cl
Chemistry-140 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 Cl or Cl-Cl
Chemistry-140 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 + =N N= 1. 10 A N N N or =N N= N=N _ N N 1. 24 A 1. 47 A
Chemistry-140 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). v When sharing is not equal, the bond is called a polar bond. v Equal sharing is sometimes referred to as a nonpolar bond.
Chemistry-140 Lecture 23 Electronegativity v 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
Chemistry-140 Lecture 23 Electronegativity Values
Chemistry-140 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
Chemistry-140 Lecture 23 Bond Polarity Examples: F 2 HF Li. F 4. 0 - 4. 0 = 0 4. 0 - 2. 1 = 1. 9 4. 0 - 1. 0 = 3. 0 nonpolar ionic d+ d- _ H F d+ represents a partial positive charge d- represents a partial negative charge
Chemistry-140 Lecture 23 Drawing Lewis Structures Œ 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 Ž Complete the octets of the atoms bonded to the central atom (i. e. the peripheral atoms)
Chemistry-140 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
Chemistry-140 Lecture 23 Drawing Lewis Structures Question: Draw the Lewis structure for PCl 3.
Chemistry-140 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-140 Lecture 23 Drawing Lewis Structures Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. NOTE: In a binary (twoelement) compound, the first element listed is usually the central one with the others surrounding it _ _ _ Cl P Cl Cl
Chemistry-140 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-140 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
Chemistry-140 Lecture 23 Drawing Lewis Structures Question: Draw the Lewis structure for HCN.
Chemistry-140 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 [1 + 4 + 5)] = 10 valence electrons
Chemistry-140 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-140 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-140 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 _ =N H C=
Chemistry-140 Lecture 24 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts v Chapter Highlights l bonding types l Lewis symbols & octets l ionic bonding & ionic lattices l covalent bonding & Lewis dot structures ð resonance structures ð breaking the octet rule & formal charge l bond order & bond energy l VSEPR molecular shape & polarity
Chemistry-140 Lecture 24 Formal Charge & Lewis Structures v 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.
Chemistry-140 Lecture 24 Assigning Formal Charge ΠAll bonding electrons are divided equally between the atoms that form bonds All nonbonding electrons are assigned to the atom on which they reside
Chemistry-140 Lecture 24 Assigning Formal Charge v 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 + 1/2 BE)
Chemistry-140 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-140 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 [6 + 4 + 5 + 1] = 16 valence electrons
Chemistry-140 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]-
Chemistry-140 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-140 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]-
Chemistry-140 Lecture 24 Drawing Lewis Structures Answer: Step 5: Calculate formal charge for each atom. [S = C = N]0 0 -1 [C = S = N]- [S = N = C]-2 +2 -1 0 +1 Note: the total formal charge on each molecule is equal to the charge on the molecule -2
Chemistry-140 Lecture 24 Drawing Lewis Structures Answer: Step 6: Decide on the most probable structure. [S = C = N]0 0 -1 [C = S = N]- [S = N = C]-2 +2 -1 0 +1 The structure that results in the least amount of formal charge separation throughout the molecule -2
Chemistry-140 Lecture 24 Resonance Structures ¶ There are times when more than one Lewis structure involving multiple bonds seems equally stable _ O O =O ozone _ O=O O v Resonance structures: structures that differ only in the placement of electrons.
Chemistry-140 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
Chemistry-140 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
Chemistry-140 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. ƒ Molecules in which an atom has more than an octet.
Chemistry-140 Lecture 24 Odd Numbers of Electrons ¶ Some examples are Cl. O 2, NO and NO 2. They tend to be very rare and very reactive. React to pair unpaired electrons. + N = O O O _ = N N = O O _ N = O _ _ O
Chemistry-140 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-140 Lecture 24 Less Than an Octet ¶ Formal charge assignments generally support these deviations from the octet rule. _ F _F _B F
Chemistry-140 Lecture 24 Consequences For example, BF 3 acts as a Lewis acid to form an adduct with NH 3 which is a Lewis base _ _ H F _ _ _ H N B F H F _ _ H F _F + _ _B _N H F H
Chemistry-140 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-140 Lecture 24 More Than an Octet ¶ For example: After satisfying the octet rule for the peripheral F-atoms in PF 5, we have 10 valence electrons at P-atom. The extra electrons can reside in a 3 d-orbital. _ _ F F _F _P F F _
Chemistry-140 Lecture 24 Lewis Structures Question: Draw the Lewis structure for IF 4 -.
Chemistry-140 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-140 Lecture 24 Drawing Lewis Structures Answer: Step 2: Arrange atoms showing connectivity and draw a single bond between atoms. Recall: In a binary (twoelement) compound, the first element listed and the heavier is usually the central one with the others surrounding it. _ F _ _ F I F _ F
Chemistry-140 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 _ _ F I F _ F
Chemistry-140 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 _ _ F I F _ F Remember: No double bonds to F
Chemistry-140 Lecture 24 Term Test #2 Friday November 9 th, 2001 6: 30 P. M.
Chemistry-140 Lecture 24 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 6 -9 A Periodic Table & ALL Required Constants will be Supplied
Chemistry-140 Lecture 25 Chapter 9: Bonding & Molecular Structure: Fundamental Concepts v Chapter Highlights l bonding types l Lewis symbols & octets l ionic bonding & ionic lattices l covalent bonding & Lewis dot structures l resonance structures l breaking the octet rule & formal charge ð bond order & bond energy ð VSEPR molecular shape & polarity
Chemistry-140 Lecture 25 Molecular Shape v The geometry of a molecule, along with its size determines in large part its chemical behavior _ Cl _ _ Cl C Cl _ Cl
Chemistry-140 Lecture 25 VSEPR Model v 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-140 Lecture 25 Electron-pairs v 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 H H 3 bonding pairs
Chemistry-140 Lecture 25 Electron-pairs ¶ The electron-pair geometry is the geometry described by the regions of electrons around the central atom. _ _ _ H N H H NH 3 (ammonia) has FOUR electron pairs in total tetrahedral geometry
Chemistry-140 Lecture 25 Electron-pairs ¶ The molecular geometry is the geometry described by the atoms only!! (central atom and the peripheral atoms) _ _ _ H NH 3 (ammonia) has THREE bonding electron pairs & ONE lone pair H Triangular pyramidal N
Chemistry-140 Lecture 25 Building VSEPR Models Œ 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 Ž Describe the molecular geometry in terms of the angular arrangement of bonding-pairs
Chemistry-140 Lecture 25 Electron-Pair Geometries (2) AX 2 Linear 180 o
Chemistry-140 Lecture 25 Electron-Pair Geometries (3) AX 3 Trigonal Planar 120 o
Chemistry-140 Lecture 25 Electron-Pair Geometries (4) AX 4 Tetrahedral 109. 5 o
Chemistry-140 Lecture 25 Electron-Pair Geometries (5) AX 5 Trigonal Bipyramidal 90 o 120 o Equatorial Axial
Chemistry-140 Lecture 25 Electron-Pair Geometries (6) AX 6 Octahedral 90 o
Chemistry-140 Lecture 25 Deviations From Electron-Pair Geometries
Chemistry-140 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.
Chemistry-140 Lecture 25 Molecular Shapes No lone pairs Trigonal Bipyramidal
Chemistry-140 Lecture 25 Molecular Shapes One lone pair Seesaw
Chemistry-140 Lecture 25 Molecular Shapes Two lone pairs T-Shaped
Chemistry-140 Lecture 25 Molecular Shapes Three lone pairs Linear
Chemistry-140 Lecture 25 VSEPR Models Question: Using the VSEPR model, predict the molecular geometry of Sn. Cl 3 -.
Chemistry-140 Lecture 25 VSEPR Models Answer: Step 1: The Lewis structure for the Sn. Cl 3 - anion is: _ Cl _ _ Cl Sn Cl
Chemistry-140 Answer: Lecture 25 VSEPR Models Step 2: The central atom is surrounded by one nonbonding electron pair and three single bonds. Therefore, FOUR electron pairs; tetrahedral electronpair geometry. The molecular geometry is trigonal pyramidal
Chemistry-140 Lecture 25 VSEPR Models Question: Using the VSEPR model, predict the molecular geometry of SF 4.
Chemistry-140 Lecture 25 VSEPR Models Answer: Step 1: The Lewis structure for the SF 4 molecule is: . _ F _ _ _ F S F F
Chemistry-140 Answer: Lecture 25 VSEPR Models 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-140 Answer: Lecture 25 VSEPR Models 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-140 Lecture 25 Molecular Polarity
Chemistry-140 Lecture 25 Molecular Polarity A polar molecule is one in which the centers of positive and negative charge do not coincide. CCl 4 CHCl 3
Chemistry-140 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
Chemistry-140 Lecture 25 Term Test #2 Friday November 9 th, 2001 6: 30 P. M.
Chemistry-140 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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