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Lecture 6
Chemical Bonding I
Tutorial
1) Which compound from each set will have the higher melting temperature? Justify
your answer.
a. CaS or KCl
KCl and CaS are both ionic solids, thus, they both have very high melting temperatures
due to the strong forces of attraction between ions. According to Coulombs Law these
forces of attraction (lattice energy) increase as the distance between ionic centres
decrease and as the charges on the ions increase. An increase in ionic charge is, however,
the most significant factor in determining the strength of the bonds between ions. As the
ions in CaS carry charges of 2+ and 2-, while the ions in KCl carry charges of 1+ and 1-;
the forces of attraction between the ions in CaS are greater. This means that CaS must
absorb more energy to overcome these forces of attraction, and thus, it has a higher
melting temperature.
b. MgBr2 or CaI2
MgBr2 and CaI2 are both ionic solids, thus, they both have very high melting
temperatures due to the strong forces of attraction between ions. According to Coulombs
Law these forces of attraction (lattice energy) increase as the distance between ionic
centres decrease and as the charges on the ions increase. The two sets of ions carry the
same charges; however, Mg2+ has a smaller ionic radius than Ca2+, and Br- has a smaller
ionic radius than I-. Thus, the distance between the ionic centres of Mg2+ and Br- is less
than the distance between the ionic centres of Ca2+ and I-. This means that MgBr2 must
absorb more energy to overcome these forces of attraction, and thus, it has a higher
melting temperature.
c. MgCl2 or NaCl
MgCl2 and NaCl are both ionic solids, thus, they both have very high melting
temperatures due to the strong forces of attraction between ions. According to Coulombs
Law these forces of attraction (lattice energy) increase as the distance between ionic
centres decrease and as the charges on the ions increase. An increase in ionic charge is,
however, the most significant factor in determining the strength of the bonds between
ions. As the ions in MgCl2 carry charges of 2+ and 1-, while the ions in NaCl carry
charges of 1+ and 1-; the forces of attraction between the ions in MgCl2 are greater. This
means that MgCl2 must absorb more energy to overcome these forces of attraction, and
thus, it has a higher melting temperature.
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2) Which compound from each set is ionic? Explain.
a. H2O or CaO
CaO is an ionic compound as it is composed of a metal (Ca2+) and a non-metal (O2-).
H2O is a covalent compound as it is composed of two non-metals.
b. BCl3 or CaCO3
CaCO3 is an ionic compound as it contains a polyatomic ion.
BCl3 is a covalent compound as it is composed of non-metals.
3) Which bond from each set is most covalent? Explain.
a. O – O or N – O
The O – O bond is most covalent. The electronegativity difference between like elements
is zero. The smaller the electronegativity difference between two atoms, the more
covalent the bond.
b. B – F or N – F
The N – F bond is most covalent. The smaller the electronegativity difference between
two atoms, the more covalent the bond. The electronegativity difference between
Nitrogen and Fluorine is less than the electronegativity difference between Boron and
Fluorine. You can figure this out by looking at the periodic trends for electronegativity.
Electronegativity increases while moving to the right along a period or up a group. This
means that elements that are closer to one another in the periodic table have closer
electronegativity values. Fluorine is closer to Nitrogen than it is to Boron, so the
electronegativity difference between Fluorine and Nitrogen will be less.
4) Which is the most polar bond in each set? Explain.
a. B – C or B – F
The B – F bond is most polar. Bond polarity increases as the electronegativity difference
between the two bonding atoms increases.
b. S – O or S – F
The S – F bond is most polar. Bond polarity increases as the electronegativity difference
between the two bonding atoms increases. Fluorine is the most electronegative element
on the periodic table, so we know that it is more electronegative than oxygen.
5) Draw Lewis Structures for the following compounds:
a. PO43- phosphate
Step 1) Count the total number of valence electrons in the structure.
Phosphorus = 5 e- (Group 5A)
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Oxygen = 6 e- x 4 atoms = 24 e- (Group 6A)
Overall charge on the ion = -3 = 3 eTotal = 5 e- + 24 e- + 3 e- = 32 eStep 2) Put the least electronegative atom in the center and connect terminal atoms
with single bonds.
Step 3) Complete the octets of all terminal atoms
This structure has used all 32 electrons. All of the atoms in this structure have
complete octets, so we are done.
b. NH3
Step 1) Count the total number of valence electrons in the structure.
Nitrogen = 5 e- (Group 5A)
Hydrogen = 1 e- x 3 atoms = 3 e- (Group 1A)
Overall charge on species = 0 = 0 eTotal = 5 e- + 3 e- = 8 eStep 2) Put the least electronegative atom in the center and connect terminal atoms
with single bonds.
In this case we must put nitrogen in the center as hydrogen atoms can only act as
terminal atoms.
Step 3) Complete the octets of all terminal atoms.
A full octet for hydrogen contains 2 electrons. Thus, the octets of the terminal
atoms are already full.
Step 4) Add up the electrons you have used, and attach the unused electrons to the
central atom.
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We have used 6 electrons so far. In step 1 we determined that we had 8 valence
electrons, so we must attach the other 2 electrons to the central atom.
All of the atoms in this structure have complete octets, so we are done.
c. SO2
Step 1) Count the total number of valence electrons in the structure.
Sulfur = 6 e- (Group 6A)
Oxygen = 6 e- x 2 atoms = 12 e- (Group 6A)
Overall charge on species = 0 = 0 eTotal = 6 e- + 12 e- = 18 eStep 2) Put the least electronegative atom in the center and connect terminal atoms
with single bonds.
Step 3) Complete the octets of all terminal atoms.
Step 4) Add up the electrons you have used, and attach the unused electrons to the
central atom.
We have used 16 electrons so far. In step 1 we determined that we had 18 valence
electrons, so we must attach the other 2 electrons to the central atom.
Step 5) Make multiple bonds to complete the octet of the central atom.
We need one double bond to complete the octet around sulfur. It does not matter
which side receives the double bond as the structure is identical when flipped 1800.
All of the atoms in this structure have complete octets, so we are done.
d. O2
There is no central atom in a diatomic molecule. Fill the octets of each atom by making
multiple bonds. Diatomic molecules from Group 7A do not require multiple bonds to fill
their octets.
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