Theories of Covalent Bonding Molecular Geometry and Hybridization of Atomic Orbitals

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1 Chapter 10 Theories of Covalent Bonding and ybridization of Atomic Orbitals Drawing Lewis Structures 1) Place least electronegative element as the central atom. Recognize that C,S,P and N are often central atoms. and halogens are often bonded to central atoms. 2) Sum the total of valence electrons contributed by each atom in the molecule. Look at the Group number to help you. 3) Place bonds to central atoms using 2-electrons per bond. 4) Place an octet of electrons (octet rule) around bonded atoms remembering that only has 2 electrons---no octet. 5) Place remaining electrons around central atom which should have an octet if period 2 or less, but could be more than octet if period 3 or higher. 6) Some rules of thumb to have at your fingertips. forms 1-bond, C forms 4-bonds, N forms 3-bonds, O forms 2-bonds. Write the Lewis dot and skeletals structure of nitrogen trifluoride (N 3 ). Write the Lewis structure of nitrogen trifluoride (N 3 ). Step 1 N is less electronegative than --> N is central atom! Step 2 - Count valence electrons = A; Nitrogen = 5, luorine = 3 X 7 = 21 Write the Lewis dot and skeletal structures of the carbonate ion (CO 3 2- ). Write the Lewis dot and skeletal structures structure of the carbonate ion (BrO 3- ). Write the Lewis dot and skeletal structures structure of the carbonate ion CN? A = = 26 valence electrons Step 3 - Write structure with N central and three bonds and rest nonbonding octet electrons around the central atom. Step 4 - Write structure with N central and three bonds and rest nonbonding octet electrons. octet octet N octet octet Write the Lewis structure of the carbonate ion (CO 3 2- ). Step 1 C is less electronegative than O, put C in center Step 2 Count octet and valence electrons (N and A respectively) Valence electrons = = 24 valence electrons Step 3 - Arrange the atoms draw bonds between C and O atoms and complete octet on C and O atoms = 16 non-bonding electrons. [ O C ] O 2- O Write the Lewis structure of the carbonate ion (BrO 3- ). Look at the formula sometimes it gives clues to the central atom [BrO3] O Br O O Valence e- = 7 + 3(6) + 1 = 26 CN Valence e- = = 10 Carbon is central atom, watch for hydrogen--1 bond C N

2 A concept called resonance is used when more than one plausible Lewis structure can be drawn. Example: Ozone, O 3 Carbonate Ion- [CO3] 2- Resonance Structures 2 equally good Lewis structures O O O O O O Which structure is correct? Both are! Measured bond lengths show they are equal! Benzene, C 6 6 C - C single bond: 1.54 Å O O O O O O C = C double bond: 1.34 Å C - Bond in C 6 6 : 1.40 Å O O O a resonance hybrid structure Write resonance structures for the nitrate ion, NO 3-. Write resonance structures for the nitrate ion, NO 3-. PLAN: Valence e- = 5 + (3X6) + 1 = 24 e- A book-keeping method called formal charge is used to determine the best Lewis structure when multiple structures appear plausible. These are different two plausible structures...how do we decide? O C O O C O O C O To use the concept of formal charge, we draw the plausible Lewis structures and then for each atom determine it s formal charge. Atom ormal charge = # valence e - - Assigned e - to Atom Assigned Atoms = all from lone pair e! +! ( bonded e! ) O C O O C O Valence e # of Assinged e ormal Charge !1 This structure wins! 1. The best structure is one that minimizes total formal charge. Less or no charge is better--and it must equal net charge of ion or molecule. 2.! The best structure is one that places negative charge on the more electronegative atom.

3 What if more than one structure works? Example: Write 3 plausible Lewis structures for the thiocyanate ion [SCN] S C N Valence = 6 e- + 4 e- + 5 e- + 1 e- = 16 e- 3-plausible Lewis structures which one is best? [ S C N] [ S C N ] [ S C N ] ormal Charge or Multiple Structures [ S C N] [ S C N ] +1 0 [ S C CS = = 0 CC = = 0 CN = = -1 CS = = -1 CC = = 0 CN = = 0-2 N ] CS = = 1 CC = = 0 CN = = ormal charge must sum to charge of ion or molecule. 2. N is more electronegative than C or S, it should have a the most negative charge in the best structure. 3. The most plausible structure has the least amount of formal charge [ S C N] Structure on the left is best structure! There are three major exceptions to the octet rule. 1) Incomplete Octet - rare situation that occurs with Be, B and Al as central atoms. 2) Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 non-metals like P, S and halogens. 3) Odd-number electrons highly reactive species called radicals that have an odd number of electrons (uneven). Incomplete Octet: Occurs With Group 2A (Be) and 3A (B and Al) Draw Lewis structures for the following Be 2e - 2 2x1e - Be 2 Be Al3 4e - Al 3e - 3 3x7e - 24e - Al B 3e B x7e - B 24e - Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 nonmetals like P, S and halogens. S 6 S 6e e - P5 48e - P 5e e - 40e - S P Phosphorous pentachloride P Phosphorous trichloride P3 [I4] -1 Odd-Electron Molecules: Radicals are highly reactive species that have an odd number of electrons (uneven). C N 5e - NO O 6e - N O 11e - An Odd Number of Valence e - = No octet and radical N=O O Methyl radical Nitrosyl radical ydroxide radical

4 Chemists use Valence Shell Electron Pair Repulsion Theory to predict the shapes of molecules using these five electron group geometries. VSEPRT explains the geometry of molecules but NOT how covalent bonds are formed with that geometry. 1. Draw Lewis Structure from chemical formula. 2. Count all electron domains to get AXE code. formula Lewis structure VSEPRT ybrid 3. Group domains into bonding and non-bonding pairs of electrons. 4. Match the number of bonding and non-bonding domains to the proper VSEPRT geometry. Lewis Structure VSEPRT Valence Bond Theory The goal is to understand geometry (via VSEPRT) and to relate it to a picture of covalent bonding in molecules. formula Lewis structure VSEPRT sp sp 2 sp 3 sp 3 d ybrid Valence BondTheory sp 3 d 2 The 3-D geometry of a molecule is one of five basic arrangements of electron groups (domains). Valence Shell Electron Pair Repulsion Theory: the optimum arrangements of a given number of electron domains is the one that minimizes repulsion among them. Linear Pyramidal Bipyramidal Note each shape has a specific bond angle The total number of electron groups (domains) defines one of the five basic geometries. 2 EG 3 EG 4 EG The electron geometry is the geometry of all electron domains, whereas the molecular geometry describes the geometry of only the atoms bonded to the central atom. AX3E1 = electron geometery with bond angles. 5 EG 6 EG is trigonal pyramidal bond angles <109.5

5 An electron group (domain) is either a pair of bonding electrons or a pair of non-bonding electrons surrounding a central atom. Multiple bonds only count as 1-group or domain. Be N P 4 electron groups 3 bonding 1 non-bonding 2 electron groups bonding 5 electron groups 5 bonding 0 non-bonding 4 electron groups 3 bonding 1 non-bonding [ O C ] O 2- O 3 electron groups 3 bonding 0 non-bonding P We count and code the bonding/non-bonding information into shorthand called AXE classification. A = Central Atom N Be AX2E0 = AX2 X = # of Bonded Domains 4 electron groups 3 bonding 1 non-bonding 2 electron groups bonding E = # Non-Bonded Domains It s implied that E = 0 AX3E1 AX2 ow Predict Using VSEPRT 1. Draw a plausible Lewis structure for the molecule. 2. Determine the total number of electron domains and identify them as bonding or lone pairs. 3. Use the total number of electron domains to establish the electron geometry from one of the five possible geometric shapes. 4. Establish the AXnEm designation to establish the molecular geometry (or do both electron and molecular geometry together simultaneously) 5. Remember bond angles in molecules are altered by lone pairs of electrons (repulsion forces reduce angles). 6. Molecules with more than one central atom can be handled individually. 2 Electron Groups = Linear Electron and 1-Possible Be S C N O C O A = Central Atom Bond Angle AX 2 E 0 = AX 2 X = # of Bonded Domains Other Examples: CS 2, CN, Be 2 E = # Non-Bonded Domains 3 Electron Groups = Planar Electron and 2-Possible Geometries 4 Electron Groups = Electron and 3-Possible Geometries Examples: SO 3, B 3, NO 3-, CO 3 2- A Bond Angle AX 3 Bond Angle AX 4 Examples: C 4, Si 4, SO 4 2-, O 4-3-Electron Domain Examples: SO 2, O 3, Pb 2, SnBr 2 A AX2E1 AX 3 E 1 N 3 P 3 AX 2 E 2 O 3 2 O O 2 S 2 3 O +

6 5 Electron Groups = Bipyramial Electron and 4-Possible Geometries 6 Electron Groups = Electron and 3-Possible Geometries P 5 AX 5 AX 4 E 1 As 5 SO 4 S 4 XeO 2 2 I 4 + AX 6 S 6 IO 5 Equatorial Position IO Br 3 AX 3 E 2 Axial Position AX 2 E 3 Xe 2 I 3 - I 2 - Br 5 Te 5 - XeO 4 AX5E1 AX 4 E 2 Xe 4 I 4 - Non-bonding electrons repulse bonding electrons and alter the bond angles in molecules. Electron lone pairs render the normal 109 tetrahedral angle less than 109! Double-bonds and/or triple bonds in molecules also decrease bond angles in molecules (think repulsion by electron rich regions). 122 C C C C Predicted Bond Angles Actual Bond Angles bonding-pair vs. bonding pair repulsion lone-pair vs. bonding < pair repulsion < lone-pair vs. lone pair repulsion Double-bond vs. Single-bond repulsion > Single-bond vs, Single-bond repulsion The electron geometry is the geometry of all electron domains, whereas the molecular geometry describes the geometry of only the atoms bonded to the central atom. AX3E1 = electron geometery with bond angles. Predicting Shapes Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) P 3 and (b) CO 2. is trigonal pyramidal bond angles <109.5

7 Predicting Shapes Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) P 3 and (b) CO Count the valence electrons and draw Lewis structure for P3: VE = 5 + 3(7) = 26 e- 2. Count the electron domains and find electron geometry and molecular from core 5 electron domain shapes (using AXE designation and sub-shapes) 3. There are 4 electron domains so the electron geometry is tetrahedral 4. The designation is AX3E1 so the molecular geometry is trigonal pyramidal. 5. The -P- bond angles should be < due to the repulsion of the nonbonding electron pair. < Predicting Shapes with Two, Three, or our Electron Groups (b) or CO 2, C has the lowest EN and will be the center atom. Predicting Shapes with Two, Three, or our Electron Groups (b) or CO 2, C has the lowest EN and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. 1. Draw the Lewis structure 2. Count the electron domains and establish electron geometry from 5 shapes 3. There are 3 electron domains so the electron geometry is trigonal planar 4. The molecular geometry designation is AX3E0 so the molecular geometry is also trigonal planar (no lone Type AX pairs) The -C- bond angle will be less than due to the electron density of the C=O Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) Sb 5 and (b) Br 5. Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) Sb 5 and (b) Br 5. (a) Sb 5-40 valence e - ; all electrons around central atom will be in bonding pairs; shape is AX 5 - trigonal bipyramidal. More Than One Central Atom In acetic acid, C 3 COO, there are three central atoms. We assign the geometry about each central atom separately. (b) Br 5-42 valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal. What is the geometry around these atoms? Take one atom at a time and apply the rules of electron domains.

8 Predicting the Shape With Multiple Central Atoms More Than One Central Atom Determine the shape around each of the central atoms in acetone, (C3)2C=O. ind the shape of one atom at a time after writing the Lewis structure. tetrahedral ethane C3C3 tetrahedral electron domain and molecular geometry tetrahedral trigonal planar >1200 ethanol C3C2O <1200 Electronegativity is an element s inherent property to draw electrons to itself when chemically bonded to another atom in a molecule. The units are dimensionless (all relative measurements to Li). Rank O N Br Differences in elements electronegativity between bonding atoms result in the formation of polar-covalent bonds and net dipole moments in molecules. Polar Bond d Polar Bond P No Net Dipole Moment on rb ola Po lar B on d Net Dipole Moment Think of the dipole moment as a molecule with separated charges + and -. or a poly-atomic molecule we must consider the vector sum of polar bonds in the molecule to see if there is a net dipole moment. Dipole Moment Dipole Moment No Net Dipole Moment Valence Bond Theory explains covalent bonding by the spatial overlap of atomic on bonding atoms and the sharing of electron pairs. Bonding in 2 1s1 + 1s1 Electrons that must have opposite spins. Dipole Moment No Net Dipole Moment Bonding in 1s1 + 2p5 Bonding in 2 2p5 + 2p5

9 Major Points and Themes of Valence Bond Theory 1. Pauli Exclusion Principle olds: 2-electrons per overlapped bond with opposing spins. 2. Greater orbital overlap gives stronger bonds. Depends on orbital shapes and how they overlap. 3. Bonding is accounted by mixing or blending or hybridization of pure valance atomic. Connect the dots and it becomes easy to see and understand. formula Lewis structure VSEPRT ybrid Valence Bond Theory explains how bonds are made 4. The number of hybrid formed equals the number of atomic combined. sp sp 2 sp 3 sp 3 d sp 3 d 2 5. The types of hybrid combined varies with the types of mixed or blended together. USE VSEPRT to help! Linear planar Bipyramidal We use pure atomic (think ground state electronic structure and those ) to describe bonding in some molecules. Bonding in 1s 1 + 1s 2 2s 2 2p 5 1s 1 + 2p 5 Bonding in carbon presents a problem as combining atomics fails. Valance Bond Theory solves this by allowing the blending or mixing of pure atomic in a process called hybridization. Pure atomic (valence ) only two bond are possible in the ground state but we don t observe C2 Bonding in 2 hybridization 1s 2 2s 2 2p 5 + 1s 2 2s 2 2p 5 2p 5 + 2p 5 By hybridizing 4 bonds are possible. sp 3 hybridized By combing or mixing different numbers of pure atomic we make hybrids that match one of the VSEPRT geometries. or example 1 pure s orbital + 1 p-orbital combine to give and two sp hybrids that when superimposed form a linear geometry for bonding. The process of combining pure atomic to form hybrid on central bonding atoms in a molecule is called hybridization. sp 3 hybrid s-orbital + p-orbital --> 2 sp hybrid --> 2-superimposed sp hybrid s-orbital + Three p- -> our sp 3 hybrids = s-orbital + Two p-orbital --> 3 sp 2 hybrids = Trig Planar

10 Some generalized rules and comments on VBT and the formation of hybridized. The logic is connected all the way to Lewis and VSEPRT 1. The number of hybrid obtained equals the number of atomic mixed. formula Lewis structure VSEPRT ybrid 2. The name of and shape of a hybrid orbital varies with the types of atomic mixed. (s + p vs s + two p) 3. Each hybrid orbital has a specific geometry that matches one of five VSEPRT shapes (show below). sp sp 2 sp 3 sp 3 d sp 3 d 2 sp sp 2 sp 3 sp 3 d Linear Planar Bipyramidal sp 3 d 2 Linear Pyramidal Bipyramidal Electron AXnEm ybridization Linear Linear AX2 sp Electron AXnEm ybridization Linear Linear AX2 sp planar planar V-shaped bent AX3 AX2E1 sp 2 planar planar V-shaped bent AX3 AX2E1 sp 2 pyramidal V-shaped bent AX4 AX3E1 AX2E2 sp 3 pyramidal V-shaped bent AX4 AX3E1 AX2E2 sp 3 bipyramidal bipyramidal Seesaw T-shaped Linear AX5 AX4E1 AX3E2 AX2E3 sp 3 d bipyramidal bipyramidal Seesaw T-shaped Linear AX5 AX4E1 AX3E2 AX2E3 sp 3 d Square pyramidal Square planar AX6 AX5E1 AX4E2 sp 3 d 2 Square pyramidal Square planar AX6 AX5E1 AX4E2 sp 3 d 2 Determine the VSEPRT geometry, the bond angles and the hybridization of each indicated atom in the following molecule? ow many sigma and pi bonds are in the molecule? Determine the electron domain, molecular geometry, the bond angles and the hybridization of each indicated atom in the following molecule? ow many sigma and pi bonds are in the molecule? trig planar 120, sp 2 sp 2 sp 2 bent, <109.5, sp 3 sp 3 tetrahedral, 180, sp 3 linear 180, sp

11 Linking VSEPRT To Valence Bond Theory ybrids Atomic Orbitals Mixed Linear AX2 Trig Planar AX3 AX4 Trig Bypyr AX5 AX6 s + p s + 2 p s + 3 p s + 3 p + d s + 3 p + 2d An sp hybrid is formed from the combination of a one pure 1s orbital and a one 2p orbital from a central bonding atom producing two new called sp. 2s # ybrid Orbitals ormed ybrid Shape Orbitals Leftover for Pi bonds Two sp Three sp 2 our sp 3 ive sp 3 d Six sp 3 d 2 Two p one p none our d Three d s-orbital p-orbital ybridization Two sp hybrid sp hybrid superimposed --The number of hybrid formed is equal to the number of pure combined! --When superimposed the sp-hybrid give us bonding for a linear molecules. Show the bonding scheme and hybridized used in Be 2 Show the bonding scheme and hybridized in Be 2 2 unhybridized unoccupied p- 2 left-over p- hybridization ybridized Be Atom s + p ybridization = 2 sp Isolated Be Atom two sp hybrids on Be two lone p- After hybridization we have on the central atom, 2 pure p- and two sp hybrids. An sp 2 hybrid is formed from the combination of a one pure 1s orbital and a two 2p from a central bonding atom producing two new called sp 2. 3-atomic, s and two p s combine to form 3- sp 2 hybrid sp hybrid:ethylyne: C!C:Linear Sigma bonds (! bonds) and Pi bonds (" bonds)are two different types of covalent chemical bonds that form as a result of end to end spatial overlap of atomic or hybridized (! bonds) or side to side overlap on bonding atoms (" bonds) Lone p that are not hybridized Superimposed ybrid form a triginal planar geometry sp 2 = Triginal planar geometry, 120 bond angle sp hybrid

12 sp hybrid:ethylyne: C!C:Linear Example 2: sp 2 hybridizaton scheme B 3. Lone p that were not hybridized on each carbon atom are able to form Pi bonds in a side to side overlap. A pair of electrons is shared in this region of space. " bonds overlap side to side Boron ybrid Box Diagram sp 2 hybrid on each carbon atom use end to end overlap to form a sigma bond. Boron Orbital Box Diagram Bonding of pure p-orbital in with sp 2 hybridized in B 3 geometry = sp 3 hybrid Example: sp 3 orbital hybridization: C 4. combine to generate four sp 3 Note the number of hybrids formed is the number of atomic combined! sp 3 hybridization mixes one 2s orbital with three 2p to produce four sp 3 on each carbon atom. End to end overlap with a 1s orbital from gives four sigma bond in C 4. which are represented collectively as: sp 3 sp 3 = geometry = bond angle This is the ground state configuration of valence atomic the four sp 3 hybrid form a tetrahedral shape C 4 Example 3: sp 3 hybrid in 2 O. sp 3 is tetrahedral shape. In water we have AX 2E 2 sp 3 hybridization mixes one 2s orbital with three 2p to produce four sp 3. The e- are distributed throughout the hybrids ready for bonding. End to end overlap with a 1s orbital from gives four sigma bond in C 4. Note the lone pairs occupy 2-of the sp 3 What is the electron geometry, the molecular geometry at each carbon atom? Use that information to determine the hybridization around each carbon atom in nicotinic acid? ow many sigma and pi bonds are in nicotinic acid? What is the electronic geometry? What is the molecular geometry? What contribute to bonding?

13 Example 2: sp 3 hybridization in N 3. sp 3 d hybridization in P 5. Bipyramidal Electron AX 5E 0 BiPyramidal Isolated P atom Electron AX 3E 1 Pyramidal The sp 3 d 2 hybrid in S 6 Electron AX 6E 0 Describe the types of bonds and in acetone, (C 3 ) 2 CO and in CO2 and in CN? Step 1 Step 2 Step 3 formula Lewis structure VSEPRT ybrid Describe the types of bonds and in acetone, (C 3 ) 2 CO. PLAN: Draw the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid taking note of geometries predicted from VSEPRT. Draw the and show overlap. SOLUTION: sp 3 hybridized Postulating ybrid Orbitals in a Molecule PROBLEM: PLAN: Use partial orbital diagrams to describe mixing of the atomic of the central atom leads to hybrid in each of the following: (a) Methanol, C 3 O (b) Sulfur tetrafluoride, S 4 Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid. Use partial orbital box diagrams to indicate the hybrid for the central atoms. sp 3 hybridized sp 2 hybridized " bonds # bond SOLUTION: (a) C 3 O The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e - pairs.

14 Postulating ybrid Orbitals in a Molecule (a) Methanol, C 3 O SOLUTION: (a) C 3 O The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e - pairs. Copyright The McGraw-ill Companies, Inc. Permission required for reproduction or display. Postulating ybrid Orbitals in a Molecule (b) S 4 has a seesaw shape with 4 bonding and 1 nonbonding e - pairs. S atom hybridized S atom single C atom hybridized C atom single O atom hybridized O atom 11- Bond order is the number of bonds between two bonded atoms. igher bond orders give shorter bond lengths and require more energy to break a bond. Single bond between 2 atoms = order = 1 Double bond between 2 atoms = order = 2 Triple bond between 2 atoms = order = 3 S C N C-N: Bond order = 2 S N N Bond order = 3 Note how bond energies (energy required to break a bond) goes up as bond order increases. S-C: Bond order = 2 S- Bond order = 1 Bond Lengths Triple bond < Double Bond < Single Bond Ethane (C 3 C 3 ) sp 3 hybrid sp hybridization showing " and # bonds in acetylene (C 2 2 ). both C are sp 3 hybrids sp 3 -sp 3 C overlap forms a " bond and C are s-sp 3 overlaps to " bonds overlap in one position - " p overlap - #