O - MSU Chemistry

46 perfect papers Median score 120 1 CEM 141 Exam 2 FS13 300 13 perfect scores 250 Mean 111 Frequency 200 150 Frequency 100 50 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 Score 2 Excep=ons to the octet rule: Hydrogen : only two electrons are needed (duet) Too few electrons : elements leI of carbon Too many electrons : p block & period 3 or higher Odd number of electrons : radicals 3 Elements to the leI of C Valence electrons 16 : : Too few electrons ? = Be = Cl: : : : Cl
: : : Cl − Be − Cl : Why not? Bad charges on atoms 4 Elements to the leI of C : Valence electrons : : Too few electrons 16 : : Cl − Be − Cl : Forms a polymer 5 Too many electrons P block elements in period 3 or greater SF4
Valence electrons ? : 6 from S and 7 each from F atoms = 34 electrons 10 electrons around sulfur : : F : : : : : : F _ S _ F : _ Expanded Octet _ : F : 6 NO2
Valence electrons? 5 from N and 6 each from O for a total of 17 An odd number and therefore a radical ! . : : N = O : . Resonance hybrids : O = N _ O : : : : : : : _ : _ O N O
: O
: 4 electrons in single bonds : _ _ O N O
. _ 7 : Do the atoms in a molecule have a charge? Simplest way of assigning a charge is to use the concept of a formal charge Formal charge on an atom is the charge calculated by assuming all electrons are shared equally between bonded atoms FC = number of valence electrons on atom – number of lone pair electrons on atom -‐1/2 number of shared electron pairs 8 : = O - Cl : : : : N
: Nitrosyl chloride N atom FC = 5(valence electrons)-4(lp electrons)-1/2(4 shared electrons) = -1
O atom FC = 6(valence electrons)-2(lp electrons)-1/2(6 shared electrons) = +1
Cl atom FC = 7(valence electrons)-6(lp electrons)-1/2(2 shared electrons) = 0
9 : = O - Cl : : : : N
: Nitrosyl chloride N atom FC = 5(valence electrons)-4(lp electrons)-1/2(4 shared electrons) = -1
O atom FC = 6(valence electrons)-2(lp electrons)-1/2(6 shared electrons) = +1
Cl atom FC = 7(valence electrons)-6(lp electrons)-1/2(2 shared electrons) = 0
-‐1 +1 0 N = O - Cl
10 Lewis structures may not be unique 16 valence electrons : : N 2O
: : : N = N = O : : : : N ≡ N − O : : N − N ≡ O : -‐1 +1 0 0 +1 -‐1 -‐2 +1 +1 N = N =O
N ≡ N −O
N −N ≡O
11 When you have more than one valid Lewis structure the best one is that which has the smallest formal charges AddiJonally any negaJve charge should be on the most electronegaJve atoms 12 Lewis structures may not be unique 16 valence electrons : : N 2O
: : : N = N = O : : : : N ≡ N − O : : N − N ≡ O : -‐1 +1 0 0 +1 -‐1 -‐2 +1 +1 N = N =O
N ≡ N −O
N −N ≡O
13 : : : : Ozone 18 valence electrons 0 +1 -‐1 O = O−O
: :O − O = O :
-‐1 : : : : :O = O − O :
: : : : :O − O − O :
+1 0 O = O−O
-‐1/2 +1 -‐1/2 O
− O
−O
14 Lewis diagrams tell you something about the electron distribu=on and bonding in a molecule but nothing about the shape of the molecule. 15 Why? COCl2
NH 3
SO 4 2−
H 2O
Why? CO 2
CEM 141 Lecture 22, 3/11/2013 Valence shell electron pair repulsion theory VSEPR Electron pairs around the central atom in a molecule repel one another and get as far away from each other as possible. Both bonding and non-‐bonding electron pairs are included We will call the regions of space occupied by the electron pairs -‐ domains How many domains ? = O
C
O
O
3 electron domains Single and double and triple bonds are each ONE domain 18 : 2 domains : N ≡ N − O : 2 domains 2 domains : : N = N = O : : : : N 2O
: N − N ≡ O : 19 SF4
5 : How many domains? _ : F : _ : : F : : : : : : F _ S _ F : 20 Basic arrangements of electron domains around a central atom
electron pairs domain arrangement 2 linear 3 trigonal 4 tetrahedral 5 trigonal bipyramidal 6 octahedral 21 VSEPR Procedure 1. Determine central atom 2.Determine the number of domains around central atom 3. Determine shape associated with domains 4. Arrange terminal atoms and lone-‐pairs to minimize the interac=on energy 5. Determine shape of molecule Number of domains around central atom equals the sum of the number of bonding domains and the number of lone pair domains To ﬁnd the number of domains around a central atom when no H atoms are involved: divide the number of valence electrons by 8. The integer is then number of bonding domains and the remainder is the number of electrons in the lone pairs. Three Electron Domains : electrons are Trigonal planar 3 atoms –molecule has Trigonal planar Shape 1 lone pair 2 atoms – molecule has Bent Shape CO3 2-‐ 24 valence electrons 3 domains & 3 atoms no lone pairs Electron domains and molecule have a Trigonal Planar shape 24/8 = 3 + R(0) BF3 24 valence electrons 3 domains & 3 atoms no lone pairs Electron domains and molecule have a Trigonal Planar shape 24/8 = 3 + R(0) −
3
NO
24 valence electrons 24/8 = 3 +R(0) 3 domains and therefore 3 bonds 27 4 Electron Domains: Tetrahedral electron shape Tetrahedral shape (CH4) Trigonal pyramidal shape (NH3) 2 lone pairs 1 lone pair Bent shape (H2O)

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