151F09_W9_LV_3

151F09_W9_LV_3 - General Chemistry CHEM 151 Week 9 UA...

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Unformatted text preview: General Chemistry CHEM 151 Week 9 UA GenChem REVIEWING CHEMICAL BONDING In a chemical reaction between two atoms, their valence electrons are reorganized. Atoms lose or share electrons because that leads to a more stable state (full electron shells). During the process, a net attractive force occurs between atoms Li+ F- Ionic bond Electrostatic force UA GenChem + CHEMICAL BONDING + F-F Covalent bond The atoms in a molecular compound are connected by covalent bonds (sharing) The number of covalent bonds that each atom forms is determined by the number of electrons that the atom must share to achieve a noble gas configuration (full shell) The tendency in molecular compounds is to have a structure in which each atom has eight valence electrons (OCTET RULE) UA GenChem Electron Distribution in Molecules Electron distribution is depicted with Lewis electron dot structures Electrons are distributed as: shared = BOND PAIRS or unshared = LONE PAIRS. G. N. Lewis 1875 - 1946 UA GenChem A Lewis Electron Dot Structure LONE PAIRS H BOND PAIRS .. Cl : .. UA GenChem Building a Lewis Structure Ammonia, NH3 1. Count and total valence electrons H = 1 and N = 5 Total = (3 x 1) + 5 (neutral) charge = 0 = 8 electrons or 4 e- pairs 2. Decide on the central atom; never H. Central atom tends to be the atom with the lowest electronegativity Therefore, N is central (H cannot) UA GenChem Building a Lewis Structure 3. Form a single bond between the central atom and surrounding atoms. H N H H Each bond line represents two electrons 4. Count electrons, see how many are missing 8 - 6 = 2 electrons 5. Remaining electrons form LONE PAIRS to complete octet as needed. 6. Count your electrons, make sure you end up with your original total H N .. H H UA GenChem 3 BOND PAIRS and 1 LONE PAIR (3x2) + 2 = 8 electrons Building Lewis Structures H H C H H H .. N H H H O H .. .. H .. F .. CH4 NH3 H2O HF UA GenChem Is there a pattern for C, N, O, F and H? UA GenChem Is there a pattern for C, N, O, F and H? H H C H H Element H C N O F H .. N H H H O H .. .. H .. F .. Bonds (pairs) 1 4 3 2 1 Lone pairs 0 0 1 2 3 UA GenChem Lewis Structure: CH5N Hint: H is always on the outside and only has one bond try CH3NH2 Step 1. Count valence electrons C= 4 N= 5 5xH= 5 charge = 0 TOTAL = 14 2. Central atom (s) 3. Form single bonds 14 12 = 2 H H C N H H H 1 pair of electrons are now left. 4. Subtract existing electrons UA GenChem Lewis Structure: CH3NH2 5. Add lone pairs H H C N H H H 6. Count electrons to match total (6 x 2) + 2 = 14 electrons .. UA GenChem Carbon Dioxide, CO2 1. Valence electrons = 16 or __ 8 __ pairs 2. Central atom = C 3. Form single bonds 4. Subtract existing electrons = 16 4 = 12 5. Add lone pairs (outer atoms first) place any leftover electrons on the central atom (4 x 2) + 8 = 16 Does Each Atom Have an OCTET? O C O .. :O .. C .. O: .. UA GenChem 6. Count total electrons Carbon Dioxide , CO2 .. :O .. So that C has an octet, we form DOUBLE BONDS between the C and the O atoms C .. O: .. Each atom has 8 electrons Total electrons (4 x 2) + 8 = 16 .. :O .. .. O .. C C .. O: .. .. O .. UA GenChem Nitrogen, N2 1. Valence electrons = 10 or _ pairs __, 5 Charge = 0 Total electrons= 10 2. Central atom = N 3. Form single bonds 4. Subtract existing electrons 10 2 = 8 N N 5. Place lone pairs 6. Count total electrons (3 x 2) + 4 = 10 : N N: .. UA GenChem .. Does Each Atom Have an OCTET? Nitrogen, N2 So that each N has an octet, we form a TRIPLE BOND between the two N atoms : N N: .. .. : N N: .. .. :N Each atom has 8 electrons Total electrons (3 x 2) + 4 = 10 N: UA GenChem Double and triple bonds are commonly observed for C, N, P, O, and S H H C H H H H H C C C C C C H :N .. O .. C N: C2F4 .. O .. UA GenChem Types of Bond Names The first bond between atoms is a sigma () bond .. :O .. C .. O: .. .. :O .. C .. O: .. .. O .. C .. O .. The second/ third bond between two atoms is a pi () bond UA GenChem Identify the total number of and bonds :C H N: H 1 . = ____ . = ____ 2 : F: H H 6 . = ____ . = ____ 0 .. C C C C H 9 . = ____ 2 . = ____ CH3NH2 UA GenChem Ozone, O3 1. Valence electrons = 18 or __ __ pairs 9 Charge = 0 Total electrons= 18 2. Central atom = 3. Form single bonds. 4. Subtract existing electrons 18 4 = 14 O O O 5. Place lone pairs (outer atoms first) .. .. .. : O O O: .. .. UA GenChem 6. Count total electrons Does Each Atom Have an OCTET? Ozone, O3 .. .. .. : O O O: .. .. .. .. .. : O O O: .. .. . .. .. .. : O O O: .. .. OR .. .. .. : O O O: .. .. .. .. .. O : .. O O : UA GenChem .. .. .. : O O O: .. OR Experimental data Single bonds ( ) are longer than double bonds ( = ) which are longer than triple bonds ( ) .. .. .. : O O O: .. OR Experimental evidence .. .. .. O : .. O O : Identical Bond Lengths Shorter than single bonds but longer than double bonds! UA GenChem Resonance O3 . has two equivalent structures .. .. .. : O O O: .. .. .. .. O : .. O O : These equivalent structures are called RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two. Neither of the individual structures is correct, but these are the best we can do with dots and sticks. UA GenChem Resonance In resonance structures the placement of the atoms is the SAME, but the distribution of the electrons changes. 3 resonance structures NO3 1- .. :O .. N .. O: .. .. O .. N .. O: .. .. : .. O N O: .. : O: : O: .. : O: .. In the actual molecule, the electrons are "delocalized" over the entire molecule. One could say there are 4/3 bonds between each pair of atoms. How many resonance structure does the carbonate (CO32-) ion have? UA GenChem The Answer .. .. 2: O : : O: C :O : .. .. O .. C .. O: .. .. :O .. C .. O .. : O: : O: C .. .. :O : .. Average C to O bond order? :O : .. :O : UA GenChem 4 bonds/ 3 locations = 1.33 or 4/3 Resonance When electrons become delocalized, they occupy a larger volume, which reduces electron-electron repulsion and thus stabilizes the molecule Experimental Electron Density Map Benzene C6H6 H H H H C C C C H C C H H H H C C H C C H H C C H H H C C C C H C C H H Does the Electron Density map show single and double bonds? UA GenChem Resonance Organic chemist's shorthand UA GenChem Violations of the Octet Rule Usually occur with Boron, elements of higher periods, or compounds of noble gases. SF4 Too many electrons .. :F .. B .. F: .. .. Too few electrons :F: .. BF3 F F Free Radical S F F O-N=O Odd # e- NO2 UA GenChem Other Exceptions: ClF2+ Step 1. Count valence electrons 7 Cl = 14 2xF = (remove 1 e-) (Positive) charge = 20 TOTAL = 2. Central atom 3. Form single bonds 4. Subtract existing electrons 20 4 = 16 F Cl F UA GenChem Other Exceptions : ClF2+ 5. Place lone pairs (outer atoms first) 6. Count total electrons (2 x 2) + 16 = 20 .. .. .. : F Cl F : .. .. .. Each atom now has 8 electrons. + Ions need Brackets And the Correct Charge. UA GenChem Charge Distribution Once molecular structure is established, chemist are very interested in analyzing electron distribution in a molecule because it affects its stability and its reactivity. How do they do it? 1. Determine bond polarity and assign partial charges 2. Calculate charge distribution in a bond (hypothetical): Assuming electrons are equally shared (Formal Charge) UA GenChem Formal Charge Consider the Ozone molecule (O3): O-O=O Formal Charge Compare the number of valence electrons the atom would have by itself to the number of electrons it actually has (Assuming the bonding electrons are equally divided) 6-7=-1 Valence electrons from group number on periodic table 6-6=0 O-O=O 6-5=+1 Total formal charge -1 + 1 + 0 = 0 UA GenChem Formal Charge Consider the ion NCO-: 5-7=-2 4-4=0 6-5=+1 5-6=-1 4-4=0 6-6=0 5-5=0 4-4=0 6-7=-1 [ N-C=O ]-1 [ N=C=O ]-1 [ N=C-O ]-1 Assign formal charges to each atom in each resonance structure. Determine the most stable structure based on these criteria: Smaller formal charges (whether positive or negative) are preferable to larger ones; Like charges on adjacent atoms are not desirable; A more negative formal charge should reside on a more electronegative atom. UA GenChem Formal Charge Consider the ion NCO-: 5-7=-2 4-4=0 6-5=+1 5-6=-1 4-4=0 6-6=0 . 5-5=0 4-4=0 6-7=-1 [ N-C=O ]-1 [ N=C=O ]-1 [ N=C-O ]-1 Which structure is the most stable? Why? UA GenChem Putting it All Together Build the Lewis structures; Identify number of bonding and lone pairs; Identify possible resonance structures; Assign Bond Polarity; Assign formal charges and decide on most stable structure; SCNI3UA GenChem ...
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