Chapter 10 The Shapes of Molecules
Molecules are visualized using Lewis Structures Molecular formula Step 1 Atom placement Step 2 Add A-group numbers ctet Rule Sum of valence e - Step 3 Remaining valence e - Draw single bonds. Subtract 2e - for each bond. Step 4 Lewis structure Give each atom 8e - (2e - for )
Molecules are visualized using Lewis Structures Molecular formula or N 3 Atom placement Sum of valence e - Remaining valence e - N N 5e - 7e - X 3 = 21e - Total 26e - Lewis structure
Molecules are visualized using Lewis Structures PRBLEM Write a Lewis structure for CCl 2 2, one of the compounds responsible for the depletion of stratospheric ozone. PLAN ollow the steps outlined in igure 10.1. SLUTIN Step 1 Carbon has the lowest EN and is the central atom. The other atoms are placed around it. Steps 2-4 C has 4 valence e -, Cl and each have 7. The sum is 4 + 4(7) = 32 valence e -. Make bonds and fill in remaining valence electrons placing 8e - around each atom. Cl Cl C Cl Cl C a. 2 S b. 2 c. SCl 2
Some molecules have two or more central atom or multiple bonds PRBLEM Write the Lewis structure for methanol (molecular formula C 4 ), an important industrial alcohol that is being used as a gasoline alternative in car engines. SLUTIN ydrogen can have only one bond so C and must be next to each other with filling in the bonds. There are 4(1) + 4 + 6 = 14 valence e -. C has 4 bonds and has 2. has 2 pair of nonbonding e -. C a. N 3 b. C 2 6
Some molecules have two or more central atom or multiple bonds PRBLEM Write Lewis structures for the following (a) Ethylene (C 2 4 ), the most important reactant in the manufacture of polymers (b) Nitrogen (N 2 ), the most abundant atmospheric gas PLAN or molecules with multiple bonds, there is a Step 5 which follows the other steps in Lewis structure construction. If a central atom does not have 8e -, an octet, then two e - (either single or nonbonded pair)can be moved in to form a multiple bond. SLUTIN (a) There are 2(4) + 4(1) = 12 valence e -. can have only one bond per atom. C C C C (b) N 2 has 2(5) = 10 valence e -. Therefore a triple bond is required to make the octet around each N... N N.. N N.. a. C b. CN c. C 2 N N
Some molecules that have multiple bonds may have different possible placement for these bonds. 3 can be drawn in 2 ways - Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure. B B A C A C Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs. is used to indicate that resonance occurs.
Some molecules that have multiple bonds may have different possible placement for these bonds. PRBLEM Write resonance structures for the nitrate ion, N 3-. PLAN After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms. SLUTIN N Nitrate has 1(5) + 3(6) + 1 = 24 valence e - N N N does not have an octet; a pair of e - will move in to form a double bond. N N N
If there is more than one possible structure, the use of formal charges indicate the most possible/stable structure. An atom owns all of its nonbonding electrons and half of its bonding electrons. ormal charge is the charge an atom would have if the bonding electrons were shared equally. ormal charge of atom = # valence e - - (# unshared electrons + 1/2 # shared electrons) B or A # valence e - = 6 # nonbonding e - = 4 # bonding e - = 4 X 1/2 = 2 ormal charge = 0 A C or B # valence e - = 6 # nonbonding e - = 2 or C # valence e - = 6 # nonbonding e - = 6 # bonding e - = 2 X 1/2 = 1 ormal charge = -1 # bonding e - = 6 X 1/2 = 3 ormal charge = +1
If there is more than one possible structure, the use of formal charges indicate the most possible/stable structure. Three criteria for choosing the more important resonance structure Smaller formal charges (either positive or negative) are preferable to larger charges. Avoid like charges (+ + or - - ) on adjacent atoms. A more negative formal charge should exist on an atom with a larger EN value.
If there is more than one possible structure, the use of formal charges indicate the most possible/stable structure. EXAMPLE NC - has 3 possible resonance forms - N C N C N C A B C formal charges -2 0 +1-1 0 0 0 0-1 N C N C N C orms B and C have negative formal charges on N and ; this makes them more preferred than form A. orm C has a negative charge on which is the more electronegative element, therefore C contributes the most to the resonance hybrid.
Some molecules contain atoms that do not follow octet rule 0 PRBLEM PLAN SLUTIN -1 0 0 Write Lewis structures for (a) 3 P 4 (pick the most likely structure); (b) BCl 2. Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. Note that (a) contains P which is a Period-3 element and can have an expanded valence shell. 0 0 (a) 3 P 4 has two resonance forms and formal charges indicate the more important form. +1 0 0 0 0 0 0 more stable 0 lower formal charges 0 0 (b) BCl 2 will have only 1 Lewis structure.
Valence Shell Electron Pair Repulsion (VSEPR) Theory
VSEPR - Valence Shell Electron Pair Repulsion Theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to maximize repulsions. These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-group arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-groups are defining the object arrangement,but the molecular shape is defined by the relative positions of the atomic nuclei. Because valence electrons can be bonding or nonbonding, the same electron-group arrangement can give rise to different molecular shapes. A - central atom X -surrounding atom E -nonbonding valence electron-group AX m E n integers
igure 10.2 Electron-group repulsions and the five basic molecular shapes. linear trigonal planar tetrahedral trigonal bipyramidal octahedral
igure 10.3 The single molecular shape of the linear electron-group arrangement. Examples CS 2, CN, Be 2
igure 10.4 The two molecular shapes of the trigonal planar electrongroup arrangement. Class Examples Shape S 2, 3, PbCl 2, SnBr 2 Examples S 3, B 3, N 3-, C 3 2-
actors Affecting Actual Bond Angles Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. Effect of Double Bonds ideal 120 0 C 120 0 greater electron density larger EN 116 0 122 0 C real Effect of Nonbonding(Lone) Pairs Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. Cl Sn Cl 95 0
igure 10.5 The three molecular shapes of the tetrahedral electrongroup arrangement. Examples C 4, SiCl 4, S 4 2-, Cl 4 - N 3 P 3 Cl 3 3 + 2 2 SCl 2
igure 10.6 Lewis structures and molecular shapes.
igure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement. P 5 As 5 S 4 S 4 Xe 2 2 I 4 + I 2 2 - Cl 3 Br 3 Xe 2 I 3 - I 2 -
igure 10.8The three molecular shapes of the octahedral electron-group arrangement. S 6 I 5 Br 5 Te 5 - Xe 4 Xe 4 ICl 4 -
igure 10.9 A summary of common molecular shapes with two to six electron groups.
igure 10.10 The steps in determining a molecular shape. Molecular formula Step 1 See igure 10.1 Lewis structure Step 2 Count all e - groups around central atom (A) Electron-group arrangement Step 3 Bond angles Note lone pairs and double bonds Step 4 Molecular shape (AX m E n ) Count bonding and nonbonding e - groups separately.
SAMPLE PRBLEM 10.6 Predicting Molecular Shapes with Two, Three, or our Electron Groups PRBLEM SLUTIN P Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) P 3 and (b) CCl 2. (a) or P 3 - there are 26 valence electrons, 1 nonbonding pair The shape is based upon the tetrahedral arrangement. P <109.5 0 The -P- bond angles should be <109.5 0 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. The type of shape is AX 3 E
SAMPLE PRBLEM 10.6 Predicting Molecular Shapes with Two, Three, or our Electron Groups continued (b) or CCl 2, C has the lowest EN and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. Cl C Cl C does not have an octet; a pair of nonbonding electrons will move in from the to make a double bond. Cl C Cl The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. The Cl-C-Cl bond angle will be less than 120 0 due to the electron density of the C=. 124.5 0 Cl C Cl Type AX 3 111 0
SAMPLE PRBLEM 10.7 Predicting Molecular Shapes with ive or Six Electron Groups PRBLEM Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) Sb 5 and (b) Br 5. SLUTIN (a) Sb 5-40 valence e - ; all electrons around central atom will be in bonding pairs; shape is AX 5 - trigonal bipyramidal. Sb Sb (b) Br 5-42 valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal. Br
SAMPLE PRBLEM 10.8 Predicting Molecular Shapes with More Than ne Central Atom PRBLEM Determine the shape around each of the central atoms in acetone, (C 3 ) 2 C=. PLAN ind the shape of one atom at a time after writing the Lewis structure. SLUTIN tetrahedral C C C tetrahedral trigonal planar C C C >120 0 <120 0
igure 10.11 The tetrahedral centers of ethane and ethanol. ethane C 3 C 3 ethanol C 3 C 2
igure 10.12 The orientation of polar molecules in an electric field. Electric field Electric field N
SAMPLE PRBLEM 10.9 Predicting the Polarity of Molecules PRBLEM 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, N 3 (b) Boron trifluoride, B 3 (c) Carbonyl sulfide, CS (atom sequence SC) PLAN Draw the shape, find the EN values and combine the concepts to determine the polarity. SLUTIN (a) N 3 EN N = 3.0 EN N = 2.1 N bond dipoles N molecular dipole The dipoles reinforce each other, so the overall molecule is definitely polar.
SAMPLE PRBLEM 10.10 Predicting the Polarity of Molecules continued (b) B 3 has 24 valence e - and all electrons around the B will be involved in bonds. The shape is AX 3, trigonal planar. B 120 0 (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. (c) CS is linear. C and S have the same EN (2.0) but the C= bond is quite polar(δen) so the molecule is polar overall. S C