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– A CHEMICAL BOND occurs between atoms Li+ F Ionic bond Electrostatic force
UA GenChem + CHEMICAL BONDING
+ FF Covalent bond The atoms in a molecular compound are connected by covalent bonds 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), there are exceptions
UA GenChem Electron Electron Distribution in Molecules Molecules • Electron distribution is Electron depicted with Lewis electron dot structures dot • Electrons are Electrons distributed as shared or BOND PAIRS and BOND unshared or LONE PAIRS. PAIRS. G. N. Lewis G. 1875 - 1946 1875 UA GenChem Bond and Lone Pairs
• Electrons are distributed as Electrons • shared or BOND PAIRS and unshared or shared BOND . H .. Cl : .. LONE PAIRS. LONE This is called a LEWIS ELECTRON DOT STRUCTURE
UA GenChem Building a Lewis Structure
1. Decide on the central atom; never H. 1. never Central atom tends to be the atom with the lowest Electronegativity. the Therefore, N is central (H cannot) 2. Count valence electrons 2. valence H = 1 and N = 5 Total = (3 x 1) + 5 (neutral) charge = 0 (neutral) = 8 electrons or 4 pairs electrons pairs
UA GenChem Building a Lewis Structure
3. Form a single bond between the central atom and surrounding atoms. H NH H Each bond line represents two electrons outside atoms, but not if H; place any leftover electrons on the central atom) 4. Remaining electrons form LONE PAIRS to complete octet as needed. (Start with terminal or H NH H
UA GenChem ..
•• 3 BOND PAIRS and 1 LONE PAIR. Lewis Structure: ClF21+
Step 1. Central atom Step 2. Count valence electrons Step Cl = 2xF = (Positive) charge = ? (Positive) TOTAL = Step 3. Form single bonds F Cl F 8 pairs of electrons pairs are now left. are
UA GenChem Lewis Structure: ClF2+
Remaining pairs become lone pairs, first on Remaining outside atoms and then on central atom outside .. :F .. .. Cl F : .. .. .. + Io ns ne e d Bra c ke ts And th e C o rre c t C h a rg e . Ea c h a to m no w h a s 8 e le c tro ns .
UA GenChem Hint: since H is always on the3 outside 2 making one bond try CH NH making
Step 1. Central atom (s) Step 2. Count valence electrons Step C= N= N= 5xH= charge = TOTAL = TOTAL Step 3. Form single bonds Lewis Structure: CH5N H HCNH HH 1 p a ir o f e le c tro ns a re no w le ft.
UA GenChem Lewis Structure: CH3NH2
R e m a ining p a irs b e c o m e lo ne p a irs , firs t o n o u ts id e a to m s a nd th e n o n c e ntra l a to m H HCNH HH
Ea c h a to m no w h a s 8 e le c tro ns .
UA GenChem .. Carbon Dioxide, CO2
1. Central atom = 1. 1. Valence electrons = __ or 1. Valence __ pairs __ Charge = Charge 3. Form single bonds. 4. Place lone pairs on outer 4. atoms; place any leftover electrons on the central atom. atom. O C O .. :O .. C .. O: ..
UA GenChem Carbon Dioxide , CO2
5. So that C has an 5. octet, we shall form DOUBLE BONDS DOUBLE between C and O. .. :O .. C .. O: .. Ea c h a to m h a s 8 e le c tro ns .. :O .. .. O .. C C .. O: .. .. O ..
UA GenChem Nitrogen, N2
1. Central atom = 1. 1. Valence electrons = __ 1. Valence or __ pairs or Charge = Charge Total electrons= Total 3. Form single bonds. 4. Place lone pairs on outer 4. atoms; place any leftover electrons on the central atom. atom. NN : N N: ..
UA GenChem .. Nitrogen, N2
3. Form single bonds. 1. Place lone pairs on outer 1. Place atoms; place any leftover electrons on the central atom. central 2. Arrange electrons to make an octet around each atom. each : N N: .. .. : N N: .. .. :N N:
UA GenChem Ne e d T R IP LE BO ND Double and even triple bonds Double are commonly observed for C, N, P, O, and S and H HH H
HC H CCCCCCH H :N
.. O ..
C N: C 2 F4 .. O ..
UA GenChem Even more
. H2CO C H▬O▬H O H▬C▬H Possible skeletons C▬H▬O▬H Formal Charge= group #– unshared e– # bonds= Rules for formal charge – 1) Keep as close to zero as possible 2) If greater than zero spread out as far apart as possible 3) Keep more positive element with pos. formal charge keep more negative elements with neg. formal charge
UA GenChem Is th e re a p a tte rn fo r C , N, O a nd H ?
H HC H HH H H CCCCCCH T y p ic a lly c a rb o n m a ke s ___ b o nd s with ___ uns h a re d p a irs T y p ic a lly h y d ro g e n m a ke s ___ b o nd s with ___ uns h a re d p a irs : O:
HC .. O .. H
UA GenChem T y p ic a lly o xy g e n m a ke s ___ b o nd s with ___ uns h a re d p a irs H N N N C N: H
Typically Nitrogen makes ___ bonds with ___ unshared pairs
Element H O N C Bonds (pairs) 1 2 3 4 Lone pairs 0 2 1 0 UA GenChem .. .. .. Is there a pattern for C, N, O and H? Types of Bond Names
The first bond between atoms is a sigma (σ) bond .. :O .. C .. O: .. .. :O .. C .. O: .. .. O .. C .. O .. The second/ third bond between two atoms is a pi (π) bond UA GenChem Identify the total number of σ and π bonds :C
H .σ = ____ .π = ____ : F:
.σ = ____ .π = ____ .. CCCC H .σ = ____ .π = ____ CH3NH2 UA GenChem Ozone, O3
1. Central atom = 1. 1. Valence electrons = __ or 1. Valence __ pairs __ Charge = Charge Total electrons= 3. Form single bonds. 4. Place lone pairs on outer atoms; place any leftover electrons on the central atom. atom. OOO .. .. .. : O O O: .. ..
UA GenChem Ozone, O3 Ozone, .. .. .. : O O O: .. ..
.. .. .. : O O O: .. ..
. .. .. .. : O O O: .. ..
OR .. .. .. : O O O: .. ..
.. .. .. O : .. O O :
UA GenChem .. .. .. : O O O: .. OR Experimental data
S inglebonds ( ― ) arelonge than doublebonds ( = ) which are r longe t han triplebonds (≡ ) r .. .. .. : O O O: ..
OR Experimental evidence .. .. .. O : .. O O :
UA GenChem Resonance
. .. .. .. .. .. .. O : O O O : ↔ : .. O O : .. O3 has two equivalent structures 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 Do NO3 1- 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 :
4 bonds/ 3 locations = 1.33 or 4/3 UA GenChem BOND ORDER UA GenChem Organic chemists’ shorthand for carbon and Hydrogen
• C2H6 • CH3CH2CH2OH UA GenChem When electrons become delocalized, they occupy a larger volume, which reduces electronelectron repulsion and thus stabilizes the molecule Experimental Electron Density Map Benzene C6H6
C C H H H Resonance H H H C C C C H C C H H
H H C C 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 Super Shorthand Super, super Shorthand Organic chemist’s shorthand UA GenChem . Draw the Lewis Structure of BeF2 Include formal charges. UA GenChem Violations of the Octet Rule
Usually occur with Boron, elements of higher periods, or compounds of noble gases. noble Too many electrons SF4 .. :F .. B .. F: .. Too few electrons ..
F F :F: ..
Odd # e BF3 F F
Free Radical S
NO2 O-N=O UA GenChem . Do Lewis for SF4 and SF6 UA GenChem BF3 . UA GenChem Do NO2 and N2O4 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 in two extreme cases (hypothetical): When electrons are equally shared (Formal Charge) When the most electronegative atoms takes all the a bond (Oxidation number). electrons in 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 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 Oxidation Number
[ N-C=O ]-1 [ N=C=O ]-1 [ N=C-O ]-1
Oxidation Number Compare the number of valence electrons the atom would have by itself to the number of electrons it would have if the most electronegative atom in a bond takes both electrons. 526=3 662=2 [ N=C-O ]-1
400=+4 The oxidation numbers are equal to the expected charges if the compound was ionic UA GenChem Oxidation Number of O in OF2
1. 2. 3. 4. -2 -1 0 +2 UA GenChem O2F2 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 oxidation numbers; SCN I3
UA GenChem end UA GenChem ...
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