Chapter 4 Review Answers
15. Atoms of which combinations of elements tend to form ionic bonds? Why?
Ionic bonds are formed between metals and non-metals
Metals donate electrons and non-metals accept electrons, this results in the formation of stable ions (valence shell
is either filled or empty)
16. Atoms of which types of elements tend to form covalent bonds? Why
Covalent bonds are formed between non-metals. The valence shells of non-metals are more than half full and they
have a high E.N. (due to a low ionization energy and high electron affinity) so they are more likely to gain electrons.
By sharing electrons they can complete their octet and become stable.
18. Electrons are said to be “delocalized” in metallic bonds
a. In your own words, describe what this means, and how it is different from the locations of valence electrons
in ionic and covalent bonds
Delocalized electrons do not remain in one location or in association with one specific rather they become a part
of a larger group of shared electrons (the “sea” or “pool” of electrons surrounding a cluster of metal cations). In
ionic bonds, the electrons are transferred specifically from one to another atom and held there to form discrete
ions. In covalent bonds, an electron from one atom pairs up with an electron from another atom and is held in a
bonding orbital that is shared between the two nuclei.
b. How does the “delocalization” of electrons in metallic solids explain “malleability” the property unique to all
metals?
When a force is applied to a metal the ions in the metal crystals can slide past each other without breaking the
metallic bonding because the pool of delocalized electrons helps to keep the metal together by continuing to
exert a uniform attraction on the positive ions
19. Explain why ionic solids are brittle.
Ionic solids are brittle because the oppositely charged ions are aligned in a regular pattern. When the crystals are
stressed, the ions along a plane move in a way that causes like charges to be aligned beside each other. The
repulsive electrostatic force then causes the crystal to break apart.
21. What experimentally observed property of methane makes it necessary to invoke the concept of hybridization to
explain the structure of methane?
The tetrahedral shape of methane made it necessary to invoke the concept of hybridization as this shape doesn’t
match the shape that would be expected if the s and p orbitals were used to form the C-H bonds in methane
22. List the five possible shapes of hybrid orbitals.
sp orbitals – linear
sp2 orbitals – trigonal planar
sp3 orbitals – tetrahedral
sp3d orbitals – trigonal bipyramidal
sp3d2 orbitals – octahedral
23. Describe two ways in which a non-polar molecule can temporarily become a dipole.
A dipole can be induced in a non-polar molecule by an ion or a polar molecule
Dipoles can also be spontaneously formed due to the random motion of the electrons which can result in
momentary, uneven distributions of charge.
24. Explain why symmetrical molecules are non-polar and asymmetrical molecules can be polar.
In symmetrical molecules, any polar bonds are aligned so that the polarities of the bonds cancel, leaving the
molecule non-polar. If an asymmetrical molecule has any polar bonds, the asymmetry will prevent them from
cancelling each other.
26. List two properties of metallic and ionic solids that are
a. the same - They both have a crystalline structure, ionic solids have a high melting point, many metals have high
melting points
b. different – metals are malleable while ionic solids are brittle, metallic substances conduct electric current while
ionic solids do not (ionic solutions do conduct)
27. List two properties of ionic solids and polar molecular solids that are
a. the same – they are both brittle and they are often soluble in water
b. different – ionic solids have much higher melting points then polar covalent solids, and ionic solids tend to be
harder than polar covalent solids.
29. The melting point of rubidium chloride is 718°C and its solubility in water is 91 g/100mL. The melting point of
rubidium bromide is 693°C and that for rubidium iodide is 646°C. Would you predict their solubility in water to
increase or decrease as the melting point decreases? Why?
The higher melting point of RbCl, is likely due to a stronger electrostatic attraction (ionic bond) between Rb and Cl
compared to Rb and Br or Rb and I. This stronger ionic bond will make RbCl less soluble in water as the ions will be
much more attracted to each other then they will be to the water molecules. As the electrostatic attraction
between the ions decreases the ionic solid will become more soluble in water. (Check RbBr – 98 g/100mL and RbI –
152 g/mL)
30. Use the periodic table to help you draw the orbital diagrams (for the valence shell only) for the following elements
and their most likely ions (if there is one):
a. calcium
b.
c.
d.
e.
nitrogen
aluminum
neon
beryllium
31. Earlier in this chapter, ozone is pictured as a resonance structure as shown below:
a. Analyze the Lewis structures and apply VSEPR theory to predict the electron group arrangement and the
molecular shape of the ozone molecule, Use Lewis resonance structures to roughly sketch the structure
The electron group arrangement is trigonal planar, and the molecular shape is bent
b. Experimental evidence shows that the bond angle in the molecules is 116.8°. With this fact in mind, how
likely is your answer to part (a) to represent the actual molecular structure?
It is very likely as the bond angle for a V-shaped (or bent) molecule is a little less than 120°.
32. Predict which bond in the following groups is the most ionic in character. Calculate ΔEN for each to check your
predictions
a. H-Cl (ΔEN = 0.96 ) , H-Br(ΔEN = 0.76), H-F (ΔEN = 1.78 ), the H-F bond is the most ionic in character
b. Na-O(ΔEN = 2.51 ), Li-O(ΔEN = 2.46 ), K-O(ΔEN = 2.62 ), the K-O bond is the most ionic in character
33. Classify the following bonds as mostly ionic, polar covalent, or mostly covalent by looking at the location of the
elements on the periodic table . Check your classifications by calculating ΔEN for each
a. Li-Cl – metal and non-metal – ionic bond (ΔEN = 2.18 )
b. S-S – same atom – pure covalent (ΔEN = 0 )
c. C-N – two non-metals - polar covalent (ΔEN = 0.49 )
d. Na-O – metal and non-metal – ionic (ΔEN = 2.51)
34. Use the periodic table to help your write the condensed electron configuration for the following elements and
their most likely ion (if there is one)
a. lithium
Li[He]2s1,
Li+[He]2s0
2
6
b. argon
Ar[Ne]3s 3p
no ion
2
5
c. chlorine
Cl[Ne]3s 3p
Cl-[Ne]3s23p6
d. phosphorus
P[Ne] 3s23p3
P3-[Ne]3s23p6
35. Use VSEPR theory to identify the electron group arrangement (VSEPR shape), the molecular shape, and the bond
angle of the following molecules shoes central atoms have.
VSEPR shape
Molecular shape
Bond angle
a. 4 bonding pairs and 1 lone pair
trigonal bipyramidal
see-saw
<90°, 120°
b. 6 bonding pairs and 0 lone pairs
octahedral
octahedral
90°
c. 3 bonding pairs and 2 lone pairs
trigonal bipyramidal
T-shaped
<90°
d. 3 bonding pairs and 0 lone pairs
trigonal planar
trigonal planar
120°
e. 2 bonding pairs and 2 lone pairs
tetrahedral
V-shaped (or bent)
<109.5°
f. 4 bonding pairs and 2 lone pairs
octahedral
Square planar
90°
36. Determine whether each of the following compounds will be polar or non-polar
a. CO2 – non-polar (although the bonds are polar)
b. H2S - polar
c. SiO2 – not applicable it is network covalent
d. PCl3 – polar
40. In the 1950’s, the reaction of hydrazine, N2H4, with chlorine trifluoride, ClF3, was used as a rocket fuel
a. Draw the Lewis structures of hydrazine and chlorine trifluoride
Hydrazine
chlorine trifluoride
b. Identify the hybrid orbitals used in each one
Hydrazine – sp3 orbitals (there are 4 electron groups around the nitrogen atoms)
Chlorine trifluoride – sp3d orbitals (there are 5 electron groups around chlorine)
c. Identify the molecular shape and polarity of chlorine trifluoride
Chlorine trifluoride would be T-shaped, this is an asymmetrical molecule so it would be polar
42. Use VSEPR theory and Lewis structures to predict the number of bonding pair and lone pair around the central
atom so that you can identify the VSEPR shape, molecular shape, and bond angle for the following molecules and
ions
Lewis structure
Electron pairs
VSEPR shape
Molecular shape
Bond angle
XeF2
2 BP, 2LP
Trigonal
bipyramidal
linear
180°
BCl4-
4BP, 0LP
tetrahedral
Tetrahederal
109.5°
5BP, 1LP
octahedral
Square pyramidal
<90°
(chlorine octets should be full)
SF5-
43. Which compound in each of the following pair has the higher boiling point? Explain your choice in each case
a. NH3 or PH3 – NH3 would have the higher boiling point because it would have hydrogen bonding between
molecules, while PH3 would only have dipole-dipole forces. (NH3 b.p.= -33.34°C , PH3 b.p. = -87.7°C)
b. C2H6 or C4H10 – These are both non-polar molecules but C4H10 is larger so it would have stronger London
Dispersion forces making it boiling point higher (C2H6 b.p.= -88.5°C , C4H10 b.p. = 1°C)
c. SeCl4 or SiCl4 – has a see-saw shape so it is a polar molecule, while SiCl4 has a tetrahedral shape so it is nonpolar. Therefor SeCl4 will have the higher boiling point because it has dipole-dipole forces between molecules
while SiCl4 only had London dispersion forces. (SeCl4 b.p. = 191.4° , SiCl4 b.p. = 57.65° )
This is a bad example as SeCl4 is not actually see-saw shape and is actually part of a cluster – google it!
44. For each of the following elements or compounds, predict which would have the higher boiling point and explain
how you made your choice
a. O2 or N2 – These are both non-polar molecules so the only force attracting molecules to each other are London
dispersion forces, because oxygen is slightly bigger then Nitrogen it will have more London dispersion forces and
therefore a slightly higher boiling point (O2 b.p. = -183° , N2 b.p. = -195.8° )
b. Ethanol (CH3CH3OH) or methoxy methane (CH3OCH3) – methoxy methane is a slightly polar molecule so it will
have dipole-dipole forces attracting molecules to each other, ethanol has a O-H bond so it will have hydrogen
bonding between molecules giving it a higher boiling point (CH3CH3OH b.p. = 78.37°C , CH3OCH3 b.p. = -24°C )
45. Benzene, C6H6(l), is a clear, colourless, and flammable liquid with a pleasant smell, It is now known to be a potent
carcinogen, Before its carcinogenic properties were known, it was used for many application that exposed it to
the general public, such as for decaffeinating coffee, in high-school experiments, and even as an aftershave lotion.
Today, it remains an important industrial chemical, but safety regulations govern its use
a. A benzene ring has two resonance structures; one of them is shown here labelled “A” Draw the other
resonance structure.
b. Chemists sometimes draw benzene as a shown in the image labelled “b” what does the circular dashed line
represent? How does this representation compare with the use of resonance structures? Which form do you
think best portrays the structure of benzene
The circular dashed line represents the 6 delocalized electrons from the three double bonds,
shows that they are able to travel throughout the entire carbon ring.