CH 07 (PowerPoint)

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					    Chemical Bonding,
Molecular Shape & Structure
 Chemical Bonding
Problems and questions —
• How is a molecule or polyatomic ion
  held together?
• Why are atoms distributed at strange
  angles?
• Why are molecules not flat?
• Can we predict the structure?
• How is structure related to chemical and
  physical properties?
    Chemical Bonds—it’s all about
             potential energy!
   An atom has relatively high potential energy in the
    form of valence electrons.
   Nature minimizes the potential energy by shifting
    valence electrons to form chemical bonds.
   By bonding with other atoms, potential energy is
    decreased creating stable compounds.


                         Dot Diagram
              Cl         Practice Sheet
                  The Octet Rule
    The Noble Gases, group 8A, do not react with other
    elements.
•   Described as “stable”, “inert”, and “inactive”.
•   The outer most “s” and “p” sublevels are completely
    filled with 8 electrons, satisfying the octet rule.




                                             Ne
    Ionic Bond
• essentially complete electron transfer from an element of low
  IE (metal) to an element of high electron affinity (EA)
  (nonmetal)

    Na(s) + 1/2 Cl2(g)  Na+ + Cl-        NaCl (s)

• primarily between metals (Grps 1A, 2A and transition metals)
 - and nonmetals (esp O and halogens)
   NON-DIRECTIONAL bonding

                  -
                   Covalent Bond
A covalent bond is a balance of attractive and repulsive forces.
When bringing together two atoms that are initially very far apart.
Three types of interaction occur:




(1) The nucleus-electron attractions (blue arrows) are greater than
the (2) nucleus-nucleus and (3) electron-electron repulsions (red
arrows), resulting in a net attractive force that holds the atoms
together to form an H2 molecule.
Covalent Bonding
         Energy and Stability
         As P.E. decreases when atoms bond, energy is released i.e.,
         atoms lose P.E. when they bond loss of P.E. implies higher
         stability

A graph of potential energy versus internuclear distance for the H2 molecule.
   Molecules & Molecular Compounds
There are seven elements that are always found as
 diatomic molecules
Iodine, Hydrogen, Nitrogen, Bromine, Oxygen, Chlorine,
 Fluorine
          I Have No Bright Or Clever Friends
                Covalent Bonds
• The distance between the two bonded atoms at their minimum
  potential energy is the bond length




• When forming a covalent bond atoms release energy, the same
  amount of energy must be added to separate the bonded atoms
• Bond Energy is the energy required to break a chemical bond and
  form neutral isolated atoms
         II. Covalent Bonds      A. Introduction

-   Covalent bond = sharing of 2 electrons.

-   2 shared electrons with                   (Single Bond).
-   4 shared electrons with                   (Double Bond).
-   6 shared electrons with                   (Triple Bond).

-   We frequently show the structure as a Lewis Structure -
    covalent bonds with lines and nonbonding valence
    electrons as dots.

- Note: Group IVA usually forms 4 bonds; VA three bonds;
   VIA two bonds; and VIIA (along with H) one bond.
              Chemical Bonds

Bond Type      Single       Double        Triple
# of e’s        2            4             6
Notation       —             =             
Bond order      1            2             3
Bond
               Increases from Single to Triple
strength
Bond length    Decreases from Single to Triple
Strengths of Covalent Bonds
Electronegativity is defined as the ability of an atom in a
molecule to attract electrons to itself
II. Bonding and Molecular Structure
   A. Ionic, covalent, and polar bonds: electronegativity

           H—H
      C = 2.1 2.1
                       DC = 0  equal sharing of electrons
          Cl—Cl                 = nonpolar covalent bond
      C = 3.0 3.0                                                 nonmetal
                                                                     +
           d+ d–                                                  nonmetal
           H—Cl        DC = 0.9  unequal sharing of electrons
      C = 2.1 3.0                 = polar covalent bond


                                                             metal
           Na+Cl–      DC = 2.1  transfer of electrons
                                                               +
      C = 0.9 3.0                = ionic bond
                                                            nonmetal
          generally:    when DC < 1.9  covalent
                                > 1.9  ionic
Molecular Shape and Molecular Polarity
           Bond and Lone Pairs

• Valence electrons are distributed as
  shared or BOND PAIRS and unshared or
  LONE PAIRS.
             ••
                  •
       H     Cl   •
             ••
                      lone pair (LP)
       shared or
       bond pair

  This is called a LEWIS structure.
                                      8A
1A
            Valence
            Electrons
      2A    3A   4A   5A    6A        7A




     Number of valence electrons is
     equal to the Group number.
        Lewis Symbols
        Represent the number of valence electrons as dots
        Valence number is the same as the Periodic Table Group Number
n= 1

        H                                                          He
n= 2
        Li        Be          B    C       N       O       F       Ne
 Groups 1          2          3    4        5       6       7       8


  For example,

       Na; Is2, 2s2, 2p6, 3s1 = [Ne] 3s1

       Lewis Structure = Na
Rules for Drawing Lewis Structures
First sum the number of valence electrons from each atom
1) The central atom is usually written first in the formula
2) Complete the octets of atoms bonded to the central atom (remember
   that H can only have two electrons)
3) Place any left over electrons on the central atom, even if doing so it
   results in more than an octet
4) If there are not enough electrons to give the central atom an octet , try
   multiple bonds
Question 1-3. Draw Lewis structures of the following.

   CCl4

   CH2O

   C2H2

   CH3OH

   CH3CHCH2

   HCN                                           Check Answer
Answer 1-3. Draw Lewis structures of the following.



            Cl
   Cl       C Cl             O
            Cl         H C        H        H C    C     H

        CCl4            CH2O                     C2H2


        H
                         H H H
  H C            O H   H C       C C   H   H C N
        H                H
  CH3OH                 CH3CHCH2            HCN
         Exceptions to the Octet Rule

Central Atoms Having Less than an Octet
• Relatively rare.
• Molecules with less than an octet are typical for
  compounds of Groups 1A, 2A, and 3A.
• Most typical example is BF3.
• Formal charges indicate that the Lewis structure with an
  incomplete octet is more important than the ones with
  double bonds.
         Exceptions to the Octet Rule

Central Atoms Having More than an Octet
• This is the largest class of exceptions.
• Atoms from the 3rd period onwards can accommodate
  more than an octet.
• Beyond the third period, the d-orbitals are low enough in
  energy to participate in bonding and accept the extra
  electron density.
           Sulfite ion, SO3              2-

Step 1. Central atom = S
Step 2. Count valence electrons
             S= 6
             3 x O = 3 x 6 = 18
             Negative charge = 2
             TOTAL = 6 + 18 + 2 = 26 e-
                                   or 13 pairs


  Step 3. Form sigma bonds                    O
   10 pairs of electrons are left.
                                     O        S   O
        Sulfite ion, SO3                    2-   (2)
Remaining pairs become lone pairs,
     first on outside atoms
     then on central atom.
                              ••
                          •
                          •   O ••
                     ••               ••
                 •
                 •   O        S      O ••
                                     ••
                     ••       ••

  Each atom is surrounded by an octet of
   electrons.
NOTE - must add formal charges (O-, S+) for complete dot diagram
      Carbon Dioxide, CO2
1. Central atom = __C____
2. Valence electrons = _16_ or _8_ pairs
3. Form sigma bonds.
                          O         C      O
         This leaves __6__ pairs.

4. Place lone pairs on outer atoms.

                 ••            ••
             •                      •
             •   O
                 ••
                      C       O••
                                    •
C. Formal charges

(use this easier method.)

1. Divide the electrons in each bond equally between the two atoms sharing
them.
2. Count the number of electrons each atom now has and compare this
number to its normal valence.
    -more electrons than normal valence  negative formal charge
    -fewer electrons than normal valence  positive formal charge

Question 1-4. Draw the Lewis structures, then determine the formal charge
on each atom in the following molecules or ions. Check your answer by
clicking on the arrow.

H3O+          CH3O–         CH3+           CO            N3–
                                                                    Check Answer
C. Formal charges


  Answer 1-4. Compare your answers.
                              H                        H
          H
                         H C O                    H C
      H O H                   H                        H
    O: 6 - 5 = +1        C: 4 - 4 = 0
                                               C: 4 - 3 = +1
                         O: 6 - 7 = -1

                          1   2   3           1    2   3
       C O                N N N          or   N N N

   C: 4 - 5 = -1        N1: 5 - 5 = 0         N1: 5 - 6 = -1
   O: 6 - 5 = +1        N2: 5 - 4 = +1        N2: 5 - 4 = +1
                        N3: 5 - 7 = -2        N3: 5 - 6 = -1
C. Formal charges


 When two or more nonequivalent Lewis structures are possible, the better
 (more stable) one is the one with:
     1. fewer formal charges
                                                      In decreasing
     2. more octets                                   order of
     3. a – charge on a more electronegative atom,    importance
     or a + charge on a more electropositive atom


 Question 1-5. Draw the most stable Lewis structure for each of the
 following compounds. Check your answers by clicking on the arrow.


 COCl2              BF3            (CH3)2SO           HOCN
                                                                Check Answer
C. Formal charges


  Answer 1-4. The preferred Lewis structures are shown here. Note that
  formal charges for all atoms in the preferred Lewis structure are 0.


       O                   F                  F +1

  Cl   C Cl            F   B   F    not   F    B   F
                                              -1



       H O H
  H C      S   C   H           H O C N
       H       H
Hybridization
— Assume bonding involves only valence orbitals
— Methane, CH4:           H

                           H   C    H

                               H

          Isolated atoms                Valence orbitals
                H                             1s1

                C                  2s22p2 (2p: 2px, 2py, 2pz)

    H atoms in CH4 will use 1s orbitals
    Of the two types of orbitals (2s and 2p)
    Which will C atoms use for bonding in CH4?

    — If both are used: 2 different types of C-H bonds
     (Contrary to experimental facts)
    — Neither of the ―native‖ atomic orbitals
      of C atoms are used; instead, new hybrid orbitals are used.
         Hybridization of atomic orbitals
The mixing of the ―native‖ atomic orbitals to form
special orbitals for bonding is called hybridization.

The 4 new equivalent orbitals formed by mixing the
one 2s and three 2p orbitals are called sp3 orbitals.

The carbon atom is said to undergo sp3 hybridization,
i.e. is sp3 hybridized.

Energy-level diagram showing the sp3 hybridization

The hybrids will consist of 75% p character and 25% s
character so they should look much more like p orbitals
than s orbitals.
Energy-level diagram showing the formation of sp3
                    orbitals
The formation of four sp3 hybrid orbitals by combination of an atomic
s orbital with three atomic p orbitals. Each sp3 hybrid orbital has two
lobes, one of which is larger than the other. The four large lobes are
oriented toward the corners of a tetrahedron at angles of 109.5°.
The bonding in methane. Each of the four C-H bonds results from
head-on (s) overlap of a singly occupied carbon sp3 hybrid orbital
with a singly occupied hydrogen 1s orbital. Sigma bonds are formed
by head-to-head overlap between the hydrogen s orbital and a singly
occupied sp3 hybrid orbital of carbon.
                       Sp2 Hybridization
Consider ethylene C2H4 molecule
               H             H
                   C     C                 Lewis structure
               H             H


— 12 valence e-s in the molecule

— What orbitals do the carbon atoms use to bond in ethylene?

— 3 effective electron pairs around each carbon

      • VSEPR model predicts a trigonal planar
        geometry    120o angles

      • sp3 orbitals with tetrahedral geometry and
        109.5o angles will not work here.
                         sp2 hybridization
E.g. the molecular geometry is trigonal planar with bond angle =
120°. To explain its geometry, we can use the following rational.
sp2 signifies one s and two p orbitals are combined.
The sigma bonds in ethylene.
A carbon-carbon double bond consists of
      a sigma bond and a pi bond.
(a) The orbitals used to form the bonds in ethylene.
       (b) The Lewis structure for ethylene.
            Other sp2 hybridized carbon atoms

An atom surrounded by 3 effective electron pairs
uses sp2 hybridized orbitals for bonding.

 Example

 H2CO formaldehyde
        H
                 ..
            C    O..       Lewis Structure
        H
 — 12 valence electrons
 — 3 effective pairs around C

Sp2 hybridized orbitals are used to form the C-H bonds
and the C-O σ bond, the un-hybridized 2pz orbital is used
to form the C=O π bond.
                           sp Hybridization

Carbon in carbon dioxide, CO2 uses another type of hybridization
(rather than sp2 or sp3)
                                O=C=O
2 hybrid orbitals required to meet the 180° (linear) geometry
requirement are sp orbitals.
2 effective pairs around C atom         sp hybrid orbitals
3 effective pairs around O atom         sp2 orbitals

                           2p                                 2p
                                                              sp
  Energy




                                Hybridization
                           1s
                                                Orbitals in sp hybridized
           Orbitals in a
                                                orbitals in CO2
           free C atom
                         sp hybridization

 Now consider BeCl2 which has linear molecular geometry
 determined experimentally.
 In hybridization scheme that best describes this compound is that




The combination of one s and one p orbital gives two sp hybrid
orbitals oriented 180° apart. Two unhybridized p orbitals remain and
are oriented at 90° angles to the sp hybrids.
The hybrid orbitals in the CO2 molecule
(a) Orbitals predicted by the LE model to describe (b) The
             Lewis structure for carbon dioxide
              Other Examples of sp Hybridization

Example
N2 molecule               :N        N:        Lewis Structure
N atom: 2s22p3
   2 effective pairs around each N atom in the Lewis structure
   – Linear (180°) geometry
   – 2 sp orbitals for each N atom:
      .1 sp orbital for forming the σ bond
      .1 sp orbital for holding the lone pair

    See Fig. 14.20
    The remaining un-changed 2p orbitals are used to form
    the 2 π bonds.
    Each triple bond consists of one σ and two π bonds.
          (a) An sp hybridized nitrogen atom
(b) The s bond in the N2 molecule (c) the two p bonds
                in N2 are formed when
                Other Examples of sp Hybridization


1. Acetylene gas C2H2                    H–C   C–H

     What is the hybridization of each carbon atom?

     How many π bonds?

     How many σ bonds?
           5    4   3       2
2.     H   C    C   C       C        H

                    H       C1
                        H        H
                            H
                For Practice
•   SiF4
•   BCl3
•   CCl4
•   CO2



Predict hybridization and show orbitals of them.
   Give it a try

				
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