Chapter 7 COVALENT BONDING

7.1 Lewis Structures; The Octet Rule 7.2 Molecular Geometry Valence-Shell Electron-Pair Repulsion (VSEPR) 7.3 Polarity of Molecules 7.4 Atomic Orbitals; Hybridization

The steps in converting a molecular formula into a Lewis structure. Molecular formula Atom placement Sum of valence e - Remaining valence e - Lewis structure Place atom with lowest EN in center Add A-group numbers Draw single bonds. Subtract 2e - for each bond. Give each atom 8e - (2e - for H) Step 1 Step 2 Step 3 Step 4

Molecular formula Atom placement Sum of valence e - Remaining valence e - Lewis structure For NF 3 N FF F N 5e - F 7e - X 3 = 21e - Total 26e - : :: :::: : : :

SAMPLE PROBLEMWriting Lewis Structures for Molecules with One Central Atom SOLUTION: PROBLEM:Write a Lewis structure for CCl 2 F 2, one of the compounds responsible for the depletion of stratospheric ozone. PLAN:Follow the steps outlined in Figure Step 1: Carbon has the lowest EN and is the central atom. The other atoms are placed around it. C Steps 2-4: C has 4 valence e -, Cl and F each have 7. The sum is 4 + 4(7) = 32 valence e -. Cl F F C F F Make bonds and fill in remaining valence electrons placing 8e - around each atom. :: : :: : : : : :: :

SAMPLE PROBLEMWriting Lewis Structure for Molecules with More than One Central Atom PROBLEM:Write the Lewis structure for methanol (molecular formula CH 4 O), an important industrial alcohol that is being used as a gasoline alternative in car engines. SOLUTION:Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds. There are 4(1) = 14 valence e -. C has 4 bonds and O has 2. O has 2 pair of nonbonding e -. COH H H H : :

SAMPLE PROBLEMWriting Lewis Structures for Molecules with Multiple Bonds. PLAN: SOLUTION: PROBLEM:Write Lewis structures for the following: (a) Ethylene (C 2 H 4 ), the most important reactant in the manufacture of polymers (b) Nitrogen (N 2 ), the most abundant atmospheric gas For 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. (a) There are 2(4) + 4(1) = 12 valence e -. H can have only one bond per atom. CC H HH H : CC H HH H (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 :.. N : N :

Resonance: Delocalized Electron-Pair Bonding Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs. O 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. is used to indicate that resonance occurs.

SAMPLE PROBLEMWriting Resonance Structures PLAN: SOLUTION: PROBLEM:Write resonance structures for the nitrate ion, NO 3 -. 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. Nitrate has 1(5) + 3(6) + 1 = 24 valence e - N does not have an octet; a pair of e - will move in to form a double bond.

Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e - - (# unshared electrons + 1/2 # shared electrons) For O A # valence e - = 6 # nonbonding e - = 4 # bonding e - = 4 X 1/2 = 2 Formal charge = 0 For O B # valence e - = 6 # nonbonding e - = 2 # bonding e - = 6 X 1/2 = 3 Formal charge = +1 For O C # valence e - = 6 # nonbonding e - = 6 # bonding e - = 2 X 1/2 = 1 Formal charge = -1 Formal charge is the charge an atom would have if the bonding electrons were shared equally.

Resonance (continued) 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. Three criteria for choosing the more important resonance structure:

EXAMPLE: NCO - has 3 possible resonance forms - Resonance (continued) ABC formal charges Forms B and C have negative formal charges on N and O; this makes them more preferred than form A. Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.

SAMPLE PROBLEMWriting Lewis Structures for Octet Rule Exceptions PLAN: SOLUTION: PROBLEM:Write Lewis structures for (a) H 3 PO 4 (pick the most likely structure); (b) BFCl 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. (a) H 3 PO 4 has two resonance forms and formal charges indicate the more important form lower formal charges more stable (b) BFCl 2 will have only 1 Lewis structure.

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 minimize repulsions. These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-pair (group) arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-pairs, N p, are defining the object arrangement or electron pair geometry, 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-pair arrangement can give rise to different molecular shapes. AX m E n A - central atomX -surrounding atomE -nonbonding valence electron-group integers VSEPR

Electron-group repulsions and the five basic molecular shapes. linear trigonal planar tetrahedral trigonal bipyramidal octahedral

The single molecular shape of the linear electron-group arrangement. Number of Electron pairs, N p =2. Examples: CS 2, HCN, BeF 2

The two molecular shapes of the trigonal planar electron group arrangement. N p =3. Class Shape Examples: SO 3, BF 3, NO 3 -, CO 3 2- Examples: SO 2, O 3, PbCl 2, SnBr 2

Factors 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. ideal larger EN greater electron density real Lone pairs repel bonding pairs more strongly than bonding pairs repel each other Effect of Double Bonds Effect of Nonbonding(Lone) Pairs

The three molecular shapes of the tetrahedral electron- group arrangement. N p =4. Examples: CH 4, SiCl 4, SO 4 2-, ClO 4 - NH 3 PF 3 ClO 3 H 3 O + H 2 O OF 2 SCl 2

Lewis structures and molecular shapes.

The four molecular shapes of the trigonal bipyramidal electron-group arrangement. N p =5. SF 4 XeO 2 F 2 I F 4 + I O 2 F 2 - ClF 3 BrF 3 XeF 2 I 3 - I F 2 - PF 5 AsF 5 SOF 4

The three molecular shapes of the octahedral electron- group arrangement. N p =6. SF 6 I OF 5 BrF 5 TeF 5 - XeOF 4 XeF 4 I Cl 4 -

A summary of common molecular shapes with two to six electron groups.

The steps in determining a molecular shape. Molecular formula Lewis structure Electron-group arrangement Bond angles Molecular shape (AX m E n ) Count all e - groups around central atom (A) Note lone pairs and double bonds Count bonding and nonbonding e - groups separately. Step 1 Step 2 Step 3 Step 4

SAMPLE PROBLEMPredicting Molecular Shapes with Two, Three, or Four Electron Groups PROBLEM:Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF 3 and (b) COCl 2. SOLUTION:(a) For PF 3 - there are 26 valence electrons, 1 nonbonding pair The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be < due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. < The type of shape is AX 3 E

SAMPLE PROBLEMPredicting Molecular Shapes with Two, Three, or Four Electron Groups continued (b) For COCl 2, C has the lowest EN and will be the center atom. There are 24 valence e -, 3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond. 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 due to the electron density of the C=O Type AX 3

SAMPLE PROBLEMPredicting Molecular Shapes with Five or Six Electron Groups PROBLEM:Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF 5 and (b) BrF 5. SOLUTION:(a) SbF valence e - ; all electrons around central atom will be in bonding pairs; shape is AX 5 - trigonal bipyramidal. (b) BrF valence e - ; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX 5 E, square pyramidal.

SAMPLE PROBLEMPredicting Molecular Shapes with More Than One Central Atom SOLUTION: PROBLEM:Determine the shape around each of the central atoms in acetone, (CH 3 ) 2 C=O. PLAN:Find the shape of one atom at a time after writing the Lewis structure. tetrahedral trigonal planar >120 0 <120 0

The tetrahedral centers of ethane and ethanol. ethane CH 3 ethanol CH 3 CH 2 OH

The orientation of polar molecules in an electric field. Electric field OFF Electric field ON

SAMPLE PROBLEMPredicting the Polarity of Molecules (a) Ammonia, NH 3 (b) Boron trifluoride, BF 3 (c) Carbonyl sulfide, COS (atom sequence SCO) PROBLEM:From 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: PLAN:Draw the shape, find the EN values and combine the concepts to determine the polarity. SOLUTION:(a) NH 3 EN N = 3.0 EN H = 2.1 bond dipoles molecular dipole The dipoles reinforce each other, so the overall molecule is definitely polar.

SAMPLE PROBLEMPredicting the Polarity of Molecules continued (b) BF 3 has 24 valence e - and all electrons around the B will be involved in bonds. The shape is AX 3, trigonal planar. F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar (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.