1. Chapter 7 Molecular Structure: Solids and Liquids
Electron Configuration of Ionic Compounds
Review

2. Naming Polyatomic Ions
Nick the Camel Craved and ate a Clam Supper in Phoenix.

3. Review
Formation of cations and anions.

4. Metals Form Positive Ions
Octets by losing all of their valence electrons.
Positive ions with the electron configuration of the nearest noble gas.
Positive ions with fewer electrons than protons.
Group 1A(1) metals  ion 1+
Group 2A(2) metals  ion 2+
Group 3A(13) metals  ion 3+
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5. Formation of Mg2+
Magnesium achieves an octet by losing its two valence electrons.
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6. Formation of Negative Ions
In ionic compounds, nonmetals
Achieve an octet arrangement.
Gain electrons.
Form negatively charged ions with 3-, 2-, or 1- charges.

7. Formation of a Chloride, Cl-
Chlorine achieves an octet by adding an electron to its valence electrons.
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8. Ions of Transition Metals
Form positively charged ions.
Lose ____ valence electrons to form 2+ ions.
Lose ____ electrons to form ions with higher positive charges.
Do not form octets like representative metals.

9. Ions of Transition Metals
Example: Fe forms Fe3+ and Fe2+
Fe s22s22p63s23p64s23d6
Loss of two 4s electrons
Fe2+ 1s22s22p63s23p6
Loss of two 4s electrons and one d electron
Fe3+ 1s22s22p63s23p6
Note: The 3d subshell is half-filled and stable.

10. Electron-Dot Formulas
Why look at Electron-Dot Formulas?
Ionic Compounds: Helps to determine formulas
Covalent Compounds: Help us to understand molecular
structures or molecular geometries.
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11. Molecular Geometries or Shapes

12. Multiple Covalent Bonds
In a single bond
______ pair of electrons is shared.
In a double bond,
______ pairs of electrons are shared.
In a triple bond.

13. Electron-Dot Formulas
Electron-dot formulas show
The order of bonded atoms in a covalent compound.
The bonding pairs of electrons between atoms.
The unshared (lone, non-bonding) valence electrons.
A central atom with an octet.
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14. Number of Covalent Bonds
The number of covalent bonds can be determined
from the number of electrons needed to complete
an octet.
Table 7.1
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15. Guide to Writing Electron-Dot Formulas
STEP 1 Determine the arrangement of atoms.
STEP 2 Add the valence electrons from all the atoms.
STEP 3 Attach the central atom to each bonded atom
using one pair of electrons.
STEP 4 Add remaining electrons as lone pairs to
complete octets (2 for H atoms).
STEP 5 If octets are not complete, form one or more
multiple bonds.

16. Electron-Dot Formula of SF2
Write the electron-dot formula for SF2.
STEP 1 Determine the atom arrangement.
S is the central atom.
F S F
STEP 2 Total the valence electrons for 1S and 2F.
1S(6e-) + 2F(7e-) = 20e-

17. Electron-Dot Formula SF2
STEP 3 Attach F atoms to S with one electron pair.
F : S : F
Calculate the remaining electrons.
20e e- = 16e- left
STEP 4 Complete the octets of all atoms by placing
remaining e- as 8 lone pairs to complete octets.
           
: F : S : F : or : F─S─F :
           

18. Electron-Dot Formula ClO3-
Write the electron-dot formula for ClO3− .
STEP 1 Determine atom arrangement.
Cl is the central atom O −
O Cl O
STEP 2 Add all the valence electrons for 1Cl and 3O
plus 1e- for negative charge on the ion.
1Cl(7e-) + 3 O(6e-) + 1e− = 26e-

19. Electron-Dot Formula ClO3-
STEP 3 Attach each O atom to Cl with one electron pair.
O −
 
O : Cl : O
Calculate the remaining electrons.
26e e- = 20e- left

20. Electron-Dot Formula ClO3-
STEP 4 Complete the octets of all atoms by placing the
remaining 20 e- as 10 lone pairs to complete
octets.
  −   −
:O: :O:
        │  
: O : Cl : O : or : O─Cl─O :
           
Note: Bonding electrons can be shown by a

21. Multiple Bonds In a single bond One pair of electrons is shared.
In a double bond,
Two pairs of electrons are shared.
In a triple bond.
Three pairs of electrons are shared.

22. Electron-Dot Formula of CS2
Write the electron-dot formula for CS2.
STEP 1 Determine the atom arrangement. The C
atom is the central atom. S C S
STEP 2 Total the valence electrons for 1C and 2S.
1C(4e-) + 2S(6e-) = 16e-

23. Electron-Dot Formula CS2
STEP 3 Attach each S atom to C with electron pairs.
S : C : S
Calculate the remaining electrons.
16e e- = 12e- left

24. Electron-Dot Formula CS2
STEP 4 Attach 12 remaining electrons as 6 lone pairs
to complete octets.
: S : C : S :
STEP 5 To complete octets, move two lone pairs
between C and S atoms to give two double bonds.
: S : : C : : S : or : S = C = S :

25. Multiple Bonds in N2 In nitrogen N2,
Octets are achieved by sharing three pairs of electrons, which is a triple bond.
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26. Some Electron-Dot Formulas
Table 7.2
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27. Electronegativity and Polarity
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28. Electronegativity Electronegativity values
Indicate the attraction of an atom for shared electrons.
Increases from left to right going across a period on the periodic table.
Is high for the nonmetals with fluorine as the highest.
Is low for the metals.

29. Some Electronegativity Values for Group A Elements
Electronegativity increases
High values
Low values
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30. Nonpolar Covalent Bonds
A nonpolar covalent bond
Occurs between nonmetals.
Is an equal or almost equal sharing of electrons.
Has almost no electronegativity difference (0.0 to 0.4).
Examples:
Atoms Electronegativity Type of Bond Difference
N-N = Nonpolar covalent
Cl-Br = Nonpolar covalent
H-Si = Nonpolar covalent

31. Polar Covalent Bonds A polar covalent bond
Occurs between nonmetal atoms.
Is an unequal sharing of electrons.
Has a moderate electronegativity difference (0.5 to 1.7).
Examples:
Atoms Electronegativity Type of Bond Difference
O-Cl = 0.5 Polar covalent
Cl-C = 0.5 Polar covalent
O-S = 1.0 Polar covalent

32. Comparing Nonpolar and Polar Covalent Bonds
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33. Ionic Bonds An ionic bond Occurs between metal and nonmetal ions.
Is a result of complete electron transfer.
Has a large electronegativity difference (1.8 or more).
Examples:
Atoms Electronegativity Type of Bond Difference
Cl-K – 0.8 = 2.2 Ionic
N-Na 3.0 – = 2.1 Ionic
S-Cs 2.5 – 0.7 = 1.8 Ionic

34. Predicting Bond Types Table 7.4
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35. Range of Bond Types Table 7.5
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36. Learning Check nonpolar covalent (NP), polar covalent (P), or
Use the electronegativity difference to identify the type of bond between the following as:
nonpolar covalent (NP), polar covalent (P), or
ionic (I).
A. K-N
B. N-O
C. Cl-Cl
D. H-Cl

37. Polar Molecules A polar molecule Contains _________ bonds.
Has a separation of positive and negative charge called a dipole indicated with + and -.
Has dipoles that do not cancel.
+  • •
H–Cl H—N—H
dipole │ H
dipoles do not cancel

38. Nonpolar Molecules A nonpolar molecule Contains _____________ bonds.
Cl–Cl H–H
Or has a symmetrical arrangement of polar bonds.
O=C=O Cl │
Cl –C–Cl Cl
dipoles cancel

39. Polar Bonds and Nonpolar Molecules
For example, the bond dipoles in CO2 cancel each other because CO2 is linear.

40. Determining Molecular Polarity
STEP Write the electron-dot formula.
STEP Determine the polarity of the bonds.
STEP Determine if any dipoles cancel or not.
Example: H2O
. .
H─O: H2O is polar │ H
dipoles do not cancel

41. Polar Bonds and Polar Molecules
In water, the molecule is not linear and the bond dipoles do not cancel each other.
Therefore, water is a polar molecule.

42. Attractive Forces Between
Particles
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43. Ionic Bonds In ionic compounds, ionic bonds
Are strong attractive forces.
Hold positive and negative ions together.
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44. Dipole-Dipole Attractions
In covalent compounds, polar molecules
Exert attractive forces called dipole-dipole attractions.
Form ________ dipole attractions called hydrogen bonds between hydrogen atoms bonded to F, O, or N and typically a lone pair or unshared pair of electrons.

45. Dipole-Dipole Attractions
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46. Dispersion Forces Dispersion forces are
Weak attractions between nonpolar molecules.
Caused by temporary dipoles that develop when electrons are not distributed equally.
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47. Comparison of Bonding and Attractive Forces
Table7.8
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48. Melting Points and Attractive Forces
Ionic compounds require large amounts of energy to break apart ionic bonds. Thus, they have high melting points.
Hydrogen bonds are the strongest type of dipole-dipole attractions. They require more energy to break than other dipole attractions.
Dispersion forces are weak interactions and very little energy is needed to change state.