Draw Lewis structures for each of the following structures and assign formal charges to each atom:
a)
20 electrons
b) ( N and O are bonded to one another)
14 electrons
c)
32 electrons
d)
22 electrons
2. Draw two possible resonance structures for the isocyanate ion ( ) and using formal charges determine which structure has greater contribution to the resonance hybrid.
16 electrons
Both structures possess the same magnitude of formal charges. However, structure B has greater contribution to the resonance hybrid since the negative charge is carried by oxygen which is more electronegative than nitrogen.
3. Arrange the bonds in each of the following sets in order if increasing polarityCl:
B-F
b)
S-Br
DEN 0.28
0.38
0.36
4. Classify each of the following bonds as ionic, polar covalent or non-polar covalent:
a)
polar covalent
b)
polar covalent
c)
non-polar covalent
d)
ionic
5. Use bond energies listed in Table 9.5 in your textbook to find for the reactions shown below:
a)
Bond Breaking (BB)
Bond Forming (BF)
b)
Bond Breaking (BB)
Bond Forming (BF)
Use the data provided below to calculate the lattice energy of RbCl . Is this value greater or less than the lattice energy of NaCl ? Explain.
Electron affinity of ionization energy of
Bond energy of
Sublimation energy of
This equation can be written as the sum of the following:
sublimation
ionization energy
bond energy of
electron affinity of Cl
lattice energy
This value would be expected to be smaller than NaCl . This is because is a larger ion than Na and would be further apart from the anion. Lattice energy is inversely proportional to the distance between the ions.
7. Arrange the following compounds in order of increasing lattice energy:
Calcium has a +2 ion and oxygen has -2 ion, while both NaF and CsI possess +1 and -1 charges. Since lattice energy is directly proportional to the charges, CaO would have the largest value.
Sodium ion and fluoride ions are smaller than cesium and iodide ions. Since lattice energy is inversely proportional to the size of the ions, CsI would have the lowest value.
8. Oxalic acid is a weak acid that can lose two hydrogens to form the following anions:
Draw Lewis structures for the two anions above, and comment on the relative strength and length of their bonds.
In , since the resonance structures are not equivalent, the C -O bonds are of different length ( bond shorter than bond) and different strengths ( bond stronger than bond).
In , since the resonance structures are equivalent, all bonds are of the same length and strength, and have values intermediate to those in .
9. Bond energies can be combined with values for other atomic properties to obtain values that cannot be measured directly. Use bond energy and other data found in your textbook to calculate for the ionic dissociation of chlorine gas:
Bond dissociation
ionization energy
Electron affinity
Tetrazene is a thermally unstable nitrogen hydride with the atom sequence shown below. It decomposes above to form hydrazine and nitrogen gas. Draw a Lewis structure for tetrazene and calculate the for its decomposition.
Bond Breaking (BB)
Bond Forming (BF)
Thionyl chloride ( ) can have the 3 skeletal structures shown below. Complete the Lewis structure for each, assign formal charges and determine which structure is the most plausible for this compound.
Most
plausible structure
(lowest formal charges)
Not acceptable FC's
Not acceptable FC's
(S has negative FC)
(very large values)
Rank the length of the bond length in the following ions:
(2 resonance structures)
(3 resonance structures)
Bond order:
3
1.5 (3/2)
1.33 (4/3)
Bond length:
shortest
intermediate
longest
Two compounds are isomers if they have the same chemical formula but a different arrangement of atoms. Use bond energies available in Table 9.3 in your test to estimate for each of the following isomerization reactions and indicate which isomer is more stable.
a)
Ethanol
Dimethyl ether
Bond Breaking (BB)
Bond Forming (BF)
1
360 kJ
1
414 kJ
Ethanol is the more stable isomer since it has stronger bonds (larger bond energy values)
b)
Ethylene oxide
Acetaldehyde
Bond Breaking (BB)
2 C-O 360 kJ
Bond Forming (BF) 736 kJ
Acetaldehyde is the more stable isomer since it has stronger bonds (larger bond energy values)
