Exam III (Chapter 7-10) Wednesday, October 3, 2012 Time: 6:00PM - 7:30PM SEC A 214A and 215A Announcements Chem 7 inal Exam Wednesday, Oct 10 1:30-3:30AM Chapter 1-12 70 or 75 multiple choice questions Chapter 10 The Shapes of Molecules Please note the following topics will be excluded from assessment. The page numbers refer to the 2nd Edition of the textbook. 1. Numerical problems involving the Rydberg equation (Chapter 7) 2. Spectral analysis in the laboratory (Chapter 7 p. 226-227) 3. Trends among the transition elements (Chapter 8 p. 261) 4. Trends in electron affinity (Chapter 8 pp. 265-266) 5. Pseudo-noble gas configuration (Chapter 8 p. 269) 6. Lattice energy (Chapter 9 pp. 283-285) 7. IR spectroscopy (Chapter 9 p. 292) 8. Numerical problems involving electronegativity (Chapter 9 p. 296) 9. Electronegativity and oxidation number (Chapter 9 p. 297) 10. Section 11.3: MO theory and electron delocalization (Chapter 11 p.343) 11. All sections in chapter 12 except 12.3 (types of intermolecular forces). Chemical bonds and the chemistry of an element is related to the number of valence electrons are in the outer shell (highest value of n quantum number) of the atom. Group e - configuration # of valence e - 1A ns 1 1 2A ns 2 2 3A ns 2 np 1 3 4A ns 2 np 2 4 5A ns 2 np 3 5 6A ns 2 np 4 6 7A ns 2 np 5 7 LOOK OR Group # or B group elements, the valence electrons are in the highest value ns orbital and the (n-1)d. Lewis dot structures are used to depict valence electrons and bonding between atoms. +1 +2 +3-3 -2-1 A chemical symbol represents the nucleus and all core e -. A single dot around the symbol represents one valence e -. GOAL: Draw and connect Lewis structures with geometry of a molecule (VSEPRT) and in Chapter 11 connect the geometry with how bond (VBT). Valence Bond theory explains how bonding occurs between atoms using hybridized. 1 2 3 4 formula Lewis structure VSEPRT formula Lewis structure VSEPRT 1 2 3 4 VB Theory Pyramidal Bipyramidal sp sp 2 sp 3 sp 3 d Valance Bond Theory sp 3 d 2
Must Learn How To Draw Lewis Structures 1) Put least electronegative element as the central atom. C,S,P and N are often central atoms. H and halogens are often bonded to central atoms. 2) Sum all valence e - from each atom in the molecule (careful with ions add or subtract e - ). Use the Group numbers! 3) Place bonds to central atoms using 2 e - per bond. 4) Place an octet of electrons (octet rule) around bonded atoms remembering that H--bonds have no octet. Know the incomplete and expanded octet exceptions. 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) Helpful Rules of Thumb: H forms 1-bond, C forms 4- bonds, N forms 3-bonds, O forms 2-bonds. Electronegativity is an element s inherent ability to draw electrons to itself when chemically bonded to another atom in a molecule (relative to Li)., O, N,, Br, C are highly electronegative with being the most electronegative 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 Step 3 - Write Lewis structrue with N central and three bonds and rest non-bonding octet electrons around the central atom. octet octet N octet octet N 5e - 7e - x 3 = 21e! Total 26e! 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. We count and code the bonding/non-bonding information into shorthand called AXnEm classification. A = Central Atom X = # of Bonded Domains N AX2E0 = AX2 E = # Non-Bonded Domains 4 electron groups 3 bonding 1 non-bonding shorthand AX3E1 There are three major exceptions to the octet rule. 1) Incomplete Octet - 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). 1. Incomplete Octet - no octet around central atom. Occurs with Be, B and Al as central atoms. Draw Lewis structures for the following BeH 2 Al3 B 3
Incomplete Octet: Occurs With Group 2A (Be) and 3A (Boron and Aluminum) Draw Lewis structures for the following 2. Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 nonmetals like P, S and halogens. Be 2e - 2H 2x1e - BeH 2 H Be H Al3 4e - Al 3e - 3 3x7e - 24e - Al B 3e B - 3 3 3x7e - B 24e - AX2 AX3 AX3 Draw Lewis structures for the following S 6 P5 Phosphorous trichloride [I4] -1 Expanded Octet (the largest class of octet exceptions)-occurs mostly with Period 3 nonmetals like P, S and halogens. Draw Lewis structures for P3 P5 and the carbonate anion. Determine the AXE designation for each. S 6 S 6e - 6 42e - P5 48e - P 5e - 5 35e - 40e - S P Phosphorous pentachloride P Phosphorous trichloride P3 [I4] -1 Draw Lewis structues and determine the AXE designation for each. P P [ O C ] O O 4 electron groups 3 bonding 1 non-bonding 5 electron groups 5 bonding 0 non-bonding AX3E1 2-3 electron groups 3 bonding 0 non-bonding AX5E0 AX3E0 Write the Lewis structure of the carbonate ion (BrO 3- ) and hydrogen cyanide, give AXE designation. MUST look to see if its an ion and add the necessary electron! [BrO3] O Br O O HCN Valence e- = 1 + 4 + 5 = 10 H C N Valence e- = 7 + 3(6) + 1 = 26 Electron Domains 4 electron groups 3 bonding 1 non-bonding Carbon is central atom, watch for hydrogen--1 bond AX3E1 2 electron groups 2 bonding 0 non-bonding AX2
Write resonance structures for the carbonate ion, CO 3-. 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 valence electrons (C and O) Valence electrons = 4 + 6 + 6 + 6 + 2 = 24 valence electrons Step 3 - Arrange the atoms draw bonds between C and O atoms and complete octet on C and O atoms. Which structure is correct? Electron Domains 3 electron groups 3 bonding 0 non-bonding All equally good and plausible resonance structures AX3 A concept called resonance is used when more than one plausible Lewis structure can be drawn. Example: Ozone, O 3 Write resonance structures for the nitrate ion, NO 3-. 2 equally good Lewis structures O O O O O O O O O Which structure is correct? Both are! O O O Measured bond lengths show they are equal! O O O a resonance hybrid structure Write resonance structures for the nitrate ion, NO 3-. PLAN: Count valence e- of atoms = 5 + (3X6) + 1 = 24 e- Most electronegative atom in center Surround and get an octet around N When more than one Lewis structure is plausible, we apply the concept of ORMAL CHARGE to figure out the best Lewis structure! Both are two plausible structures for CO2 O C O VS O C O Which is the best one? Electron Domains 3 electron groups 3 bonding 0 non-bonding AX3 USE ORMAL CHARGE
To use the concept of formal charge, we determine the formal charge for each atom. Example: Write 3 plausible Lewis structures for the thiocyanate ion [SCN] 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 - 6 4 6 6 4 6 # of Assinged e - 6 4 6 5 4 7 ormal Charge 0 0 0 +1 0!1 This structure wins! 1. The best structure is one that minimizes total formal charge. Net charge of ion or molecule must equal total formal charge. 2.! Also, the best Lewis structure places negative charge on the most electronegative atom. Example: Write 3 plausible Lewis structures for the thiocyanate ion [SCN] it s a -1 ion 1. Count the Valence e - S C N # of Valence = 6 e- + 4 e- + 5 e- + 1 e- = 16 e- 2. Draw the Lewis Structures With Least Electronegative Atom as central atom. [ S C N] S C N 1 2 3 [ ] [ S C N ] 3-plausible Lewis structures which one is best? Example: Write 3 plausible Lewis structures for the thiocyanate ion [SCN] 0 0-1 [ S C N] -1 0 0 [ S C N ] +1 0 [ S C CS = 6-4 -2 = 0 CC = 4-0 - 4 = 0 CN = 5-6 - 2 = -1 CS = 6-6 -1 = -1 CC = 4-0 - 4 = 0 CN = 5-2 - 3 = 0-2 N ] CS = 6-2 -3 = 1 CC = 4-0 - 4 = 0 CN = 5-6 - 1 = -2 1. 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. 0 0-1 [ S C N] Structure on the left is best structure! Write resonance structures for the nitrate ion, NCO - and determine the most plausible Lewis structure. Write resonance structures for the nitrate ion, NCO - and determine the most plausible Lewis structure. EXAMPLE: NCO - has 3 possible resonance forms - formal charges A B C -2 0 +1-1 0 0 0 0-1 orms B and C have negative formal charges on N and O; this makes them more preferred than form A. orm C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.
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 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 electron geometry is the geometry of all electron groups, whereas the molecular geometry describes the geometry of only the atoms bonded to the central atom. AX3E1 = electron geometery with 109.5 bond angles. The goal is to understand geometry (via VSEPRT) and to relate it to a picture of covalent bonding in molecules. formula Lewis structure VSEPRT Electron Group is trigonal pyramidal bond angles <109.5 VB Theory Pyramidal Bipyramidal sp sp 2 sp 3 sp 3 d sp 3 d 2 The 3-D geometry of a molecule is one of five basic arrangements of electron groups (domains). The total number of electron groups (domains) defines one of the five basic geometries. 2 EG 3 EG 4 EG Planar 5 EG 6 EG Bipyramidal
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 109.5 bond angles. is trigonal pyramidal bond angles <109.5 How 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 = Electron and 1-Possible 3 Electron Groups = Planar Electron and 2-Possible Geometries Be S C N Bond Angle Examples: SO 3, B 3, NO 3-, CO 3 2- A AX 3 O C O A = Central Atom AX 2 E 0 = AX 2 X = # of Bonded Domains Other Examples: CS 2, HCN, Be 2 E = # Non-Bonded Domains Examples: SO 2, O 3, Pb 2, SnBr 2 3-Electron Domain A AX 2 E 1 4 Electron Groups = Electron and 3-Possible Geometries 5 Electron Groups = Bipyramial Electron and 4-Possible Geometries Bond Angle AX 4 Examples: CH 4, Si 4, SO 4 2-, O 4 - P 5 AX 5 AX 4 E 1 As 5 SO 4 Equatorial Position S 4 XeO 2 2 I 4 + IO 2 2 - AX 3 E 1 NH 3 P 3 AX 2 E 2 O 3 H 3 O + H 2 O O 2 S 2 3 Br 3 AX 3 E 2 Axial Position AX 2 E 3 Xe 2 I 3 - I 2 -
6 Electron Groups = Electron and 3-Possible Geometries AX 6 S 6 IO 5 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. Br 5 Te 5 - XeO 4 AX5E1 AX 4 E 2 Xe 4 I 4 - 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. 1. 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 <109.5 0 due to the repulsion of the nonbonding electron pair. <109.5 0 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). 3 5. The -C- bond angle will be less than 120 0 due to the electron density of the C=O. 124.5 0 111 0 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. (b) Br 5-42 valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal.
More Than One Central Atom More Than One Central Atom In acetic acid, CH 3 COOH, there are three central atoms. We assign the geometry about each central atom separately. What is the geometry around these atoms? Take one atom at a time and apply the rules of electron domains. ethane CH 3 CH 3 ethanol CH 3 CH 2 OH Predicting the Shape With Multiple Central Atoms Determine the shape around each of the central atoms in acetone, (CH 3 ) 2 C=O. ind the shape of one atom at a time after writing the Lewis structure. 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). tetrahedral trigonal planar tetrahedral Rank O N Br >120 0 <120 0 Differences in elements electronegativity between bonding atoms result in the formation of polarcovalent bonds and net dipole moments in molecules. Draw a Lewis structure, show the AXE designation, determine electron and molecular geometry and whether polar or non-polar of: Polar Bond Polar Bond Polar Bond Polar Bond C3H No Net Dipole Moment Net Dipole Moment C4 Think of the dipole moment as a molecule with separated charges + and -. CH4
Draw a Lewis structure, show the AXE designation, determine electron and molecular geometry and whether polar or non-polar of: AX4 EG Tetrahedra MG CH3 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 C4 C4 Polar bonds Not Polar Molecule No Dipole Moment CH3 Polar bond Polar Molecule Has Dipole Moment Dipole Moment Dipole Moment No Net Dipole Moment No Net Dipole Moment Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. rom electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: (a) Ammonia, NH3 Predicting the Polarity of Molecules PROBLEM: rom electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: (a) Ammonia, NH3 (b) Boron trifluoride, B3 (c) Carbonyl sulfide, COS (atom sequence SCO) PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity. (b) Boron trifluoride, B3 SOLUTION: The dipoles reinforce each other, so the overall molecule is definitely polar. (a) NH3 ENN = 3.0 (c) Carbonyl sulfide, COS (atom sequence SCO) ENH = 2.1 bond dipoles molecular dipole 10- Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 10.9 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Predicting the Polarity of Molecules (b) B3 has 24 valence e! and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar. 1200 Chapter 11 (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to. Because all are at the same angle and of the same magnitude, the molecule is nonpolar. Theories of Covalent Bonding (c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar ("EN), so the molecule is polar overall. 10-11-
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Theories of Covalent Bonding This only! 11.1 Valence Bond (VB) Theory and Orbital ization Valence Bond Theory explains covalent bonding by the spatial overlap of atomic on bonding atoms and the sharing of electron pairs. Bonding in H 2 1s 1 + 1s 1 Electrons that must have opposite spins. 11.2 The Mode of Orbital Overlap and the Types of Covalent Bonds Bonding in H 1s 1 + 2p 5 11.3 Orbital (MO)Theory and Electron Delocalization Bonding in 2 2p 5 + 2p 5 11- It s natural to think of using pure atomic to describe bonding in some molecules. It works well for some...but fails with carbon. Bonding in H 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 with use of atomic only...we don t observe CH2 Bonding in 2 1s 2 2s 2 2p 5 + 1s 2 2s 2 2p 5 2p 5 + 2p 5 hybridization fixes the problem! By hybridizing 4 bonds are possible. sp 3 hybridized ization combines or mixes different numbers of pure atomic that match one of the VSEPRT geometries. or example 1 pure s orbital + 1 p-orbital combine to give and 2 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
Some generalized rules and comments on VBT and the formation of hybridized. Connect the dots and it becomes easy to see and understand. 1. The number of hybrid obtained equals the number of atomic mixed. formula Lewis structure VSEPRT 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). Valence Bond Theory explains how bonds are made sp sp 2 sp 3 sp 3 d sp 3 d 2 sp sp 2 sp 3 sp 3 d sp 3 d 2 Planar Bipyramidal planar Bipyramidal Electron AXnEm ization AX2 sp planar planar V-shaped bent AX3 AX2E1 sp 2 Determine the VSEPRT geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How many sigma and pi bonds are in the molecule? pyramidal V-shaped bent AX4 AX3E1 AX2E2 sp 3 bipyramidal bipyramidal Seesaw T-shaped AX5 AX4E1 AX3E2 AX2E3 sp 3 d Square pyramidal Square planar AX6 AX5E1 AX4E2 sp 3 d 2 Determine the electron domain, molecular geometry, the bond angles and the hybridization of each indicated atom in the following molecule? How 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 Linking VSEPRT To Valence Bond Theory s Atomic Orbitals Mixed # Orbitals ormed 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 Two sp Three sp 2 our sp 3 ive sp 3 d Six sp 3 d 2 linear 180, sp Shape Orbitals Leftover for Pi bonds Two p one p none our d Three d
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 Example sp hybrid: Show the bonding scheme and hybridized used in Be 2 2 unhybridized unoccupied p- 2 left-over p- ization s-orbital p-orbital Two sp hybrid sp hybrid superimposed s + p ization = 2 sp --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. After hybridization we have on the central atom, 2 pure p- and two sp hybrids. Show the bonding scheme and hybridized in Be 2 Isolated Be Atom hybridization two sp hybrids on Be ized Be Atom two lone p- 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 Superimposed form a triginal planar geometry sp 2 = Triginal planar geometry, 120 bond angle Example 2: sp 2 hybridizaton scheme B 3. geometry = sp 3 hybrid Boron Box Diagram combine to generate four sp 3 Note the number of hybrids formed is the number of atomic combined! Boron Orbital Box Diagram Bonding of pure p-orbital in with sp 2 hybridized in B 3 which are represented collectively as: sp 3 sp 3 = geometry = 109.5 bond angle
Example: sp 3 orbital hybridization: CH 4. This is the ground state configuration of valence atomic the four sp 3 hybrid form a tetrahedral shape 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 H gives four sigma bond in CH 4. CH 4 Example 3: sp 3 hybrid in H 2 O. sp 3 is tetrahedral shape. In water we have AX 2E 2 What is the electronic geometry? What is the molecular geometry? What contribute to bonding? 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 H gives four sigma bond in CH 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? How many sigma and pi bonds are in nicotinic acid? Example 2: sp 3 hybridization in NH 3. Electron AX 3E 1 Pyramidal sp 3 d hybridization in P 5. Bipyramidal Electron AX 5E 0 BiPyramidal The sp 3 d 2 hybrid in S 6 Electron AX 6E 0 Isolated P atom
Electron AX ne m izaton AX 2 sp Describe the types of bonds and in acetone, (CH 3 ) 2 CO and in CO2 and in HCN? planar planar V-shaped bent AX 3 AX 2E 1 sp 2 Step 1 Step 2 Step 3 pyramidal V-shaped bent AX 4 AX 3E 1 sp 3 AX 2E 2 formula Lewis structure VSEPRT bipyramidal bipyramidal Seesaw T-shaped Square pyramidal Square planar AX 5 AX 4E 1 AX 3E 2 AX 2E 3 sp 3 d AX 6 AX 5E 1 sp 3 d 2 AX 4E 2 Describe the types of bonds and in acetone, (CH 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 sp hybrid:ethylyne: HC!CH: 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 sp 3 hybridized sp 2 hybridized # bonds $ bond sp hybrid sp hybrid:ethylyne: HC!CH: 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 sp 2 hybrid on each carbon atom use end to end overlap to form a sigma bond.