COVALENT BOND POLARITY AND MOLECULAR POLARITY Caiafa06

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COVALENT BOND POLARITY AND MOLECULAR POLARITY
Caiafa06

• OBJECTIVES
• TO RECOGNIZE AND MODEL POLAR COVALENT COMPOUNDS.
• TO APPLY VSEPR AND IONIC CHARACTER TO POLARITY.
• TO IDENTIFY POLAR BONDS AND ALSO TO IDENTIFY
  POLAR MOLECULES.
• To differentiate a polar BOND from a polar
  MOLECULE.
• SYMMETRY A molecule is symmetrical if it has two
  identical halves when cut through the central
  atom. If two different halves are obtained in ANY
  aspect, the molecule is ASYMMETRICAL and might be
  polar.

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1. A bond is considered POLAR if the electrons are
   not shared equally as indicated by an ionic
   character above about 0.4 and below 1.7.
2. In a bond between two atoms, the more
   electronegative atom is the NEGATIVE POLE and
   pulls the electrons closer to itself but does not
   ever gain them no ions are formed.
3. The less electronegative atom has the electrons
   pulled further away from it without the loss of
   electrons or the formation of positive ions, the
   less electronegative atom is the POSITIVE POLE of
   the bond.
4. A molecule can only be termed polar if
5. It has polar bonds
6. The bonds are ASYMMETRICAL around the central
   atom(VSEPR)

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• SKILL- PREDICTING POLARITY
• STEP ONE calculate ionic character or EACH type
  of bond in the molecule.
• STEP TWO render a VSEPR Lewis dot structure,
  assess symmetry.
• STEP THREE assign the least electronegative atom
  as the positive pole d.
• Assign the most electronegative atom as the
  negative pole d-.
• Using the VSEPR charts, determine the geometry
  and symmetry.
• IF the molecule is ASYMMETRICAL and the bonds are
  POLAR, the molecule is polar.

• EXAMPLE ONE, WATER H2O
• STEP ONE the ionic character of water is 3.5
  2.2 1.3 which is a means the O to H bond is
  POLAR COVALENT.
• STEP TWO
• The oxygen has 6 valence electrons and each H has
  one each for a total of 8 valence electrons.
• Two bonds are needed to connect the H atoms to
  the O, there are 4 bonded electrons which leaves
  4 remaining. Remember H cannot be central and
  never gets dots in any VSEPR structure.
• The remaining four electrons go on O as H is not
  an octet element. This gives two bonds and 2
  pairs on the central O for a BENT geometry, which
  is asymmetrical.

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The 2 pairs and 2 bonds around the central O
gives a BENT geometry which is asymmetrical. The
bonds are polar therefore water is POLAR. Caiafa06
d
H is the least electronegative atom so each is a
positive poles.
H
d
d-
IMP the d- of oxygen attracts to the d of a
hydrogen of an adjacent water molecule, this
attraction is called a permanent DIPOLE
attraction. When H is bonded to F,O,N this DIPOLE
is called H-BONDING.
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The 2 pairs and 2 bonds around the central O
gives a BENT geometry which is asymmetrical. The
bonds are polar therefore water is POLAR.
The little red arrows indicate the BOND polarity,
always point to the most electronegative atom of
the bond. Note this does not indicate a double
bond
H is the least electronegative atom so each is a
positive pole.
H
d
O
H
d
d-
The big red arrow is MOLECULAR polarity, points
to most electronegative atom and is the sum
(resultant) of the little red bond polarity
arrows.
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EXAMPLE TWO, NFH2

1. Both the N-F and N-H bonds are polar covalent.
2. H cannot be central, therefore N is central.
3. Each H adds 1 valence electron, F has 7 and N has
   5 for a total of 14 total valence electrons.
4. There are three bonds, accounting for 6 electrons
   leaving 8 electrons to account for.
5. Assign 6 of the eight remaining to complete the
   Fluorine octet and place the remaining 2 on the
   Nitrogen.
6. The Fluorine is the most electronegative atom and
   is the negative pole (d-).
7. Hydrogen is least electronegative and is the
   positive pole (d)

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8 The molecule is 3 bonds and one pair on the
central atom, It is therefore pyramidal and
asymmetrical. 9 This is an asymmetrical
molecule with polar bonds and is therefore polar.
N
F
H
d
d-
H
d