Download here

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Oxidation state wikipedia, lookup

Coordination complex wikipedia, lookup

Metalloprotein wikipedia, lookup

Bond valence method wikipedia, lookup

Transcript
Approaches to Solve Problems for Exam 2

Review How to Draw Lewis Structures
For molecules and polyatomic ions composed of nonmetals
1. Write the skeletal structure (the arrangement of atoms within the molecule):
Central atom = the atom with the lowest electronegativity (usually).
Hydrogen is always a terminal atom (on the end).
2. Count the total number of valence electrons:
Group number for each element = # valence electrons.
Add electrons for negatively charged ions
Subtract electrons for positively charged ions.
3. Draw a bond between the central atom and each surrounding atom.
Single bond = 1 pair of electrons
4. Place lone pairs of electrons about each terminal atom to complete their octets.
Octet = 4 electron pairs around an atom (eight electrons)
Hydrogen can only have 2 electrons.
5. If there are more electrons left, place them as lone pairs on the central atom.
This will sometimes lead to an "expanded octet" around the central atom.
Expanded octet = five or six electron pairs around an atom.
Only central atoms from the third period and above can have expanded octets.
6. If the central atom has an incomplete octet, use the electrons from surrounding atoms to make double or triple bonds.
Do not add electrons. "Borrow" them from surrounding atoms.
Double bond = 2 pairs of electrons
Triple bond = 3 pairs of electrons
C, N, O, P, and S commonly form multiple bonds.
Terminal F and Cl do not form multiple bonds.
7. Sometimes you just can't complete the octet for a central atom.
If there is an odd number of electrons, give the central atom 7 electrons instead of 8.
Boron and beryllium just don't have enough electrons to go around and often have incomplete octets.

Bond Properties
1.
Bond length
i)
A covalent bond length is the average distance between the nucleii of two covalently bonded atoms. Such
distances are based on experimental observation. It is observed that bonds become shorter as the bond order
increases. Such distances are based on experimental observation, and reflect the sizes of the bonded atoms.
ii) For the same pair of bonded atoms, a single bond is expected to be longer than a double bond, and a double
bond is expected to be longer than a triple bond. Bond lengths: single > double > triple
iii) The bond length can be viewed as the sum of the covalent radii of the bonded atoms.
Thus it is possible to predict relative bond lengths using periodic trends in atomic size.
Use periodic trends in atomic radii to identify the smaller atoms. The bond with the smaller atoms, will have a shorter
bond. In cases where bonds between the same atoms are compared, triple bonds are shorter than double bonds that are shorter than single bonds.
2.
Bond strength
i)
For the same pair of bonded atoms, a single bond is expected to be longer than a double bond, and a double
bond is expected to be longer than a triple bond.
ii) Consequently, the bond energies go in the reverse order. It takes more energy to break a double bond than it
does a single bond, and it takes more energy to break a triple bond than it does a double bond. Bond
energies: single < double < triple
Ionic bonds are stronger than polar covalent bonds. Polar covalent bonds are stronger than purely covalent bonds. The larger the difference in
electronegativity, the stronger the bond will be.
3.
Electronegativity
i)
Electronegativity is the ability of an atom in a molecule to attract electrons to itself.
ii) In general, electronegativity decreases as the atomic radius increases. Larger atoms have lower
electronegativities. Smaller atoms have higher electronegativities.
Within a group of the periodic table, atomic radius increases with atomic number. The electronegativity, therefore, decreases. Across a row of the
periodic table, atomic radius decreases with increasing atomic number. The electronegativity, therefore, increases.
4.

5.
Polarity
i)
When two atoms are joined by a covalent bond, the bond will have polar character when the
electronegativities of the two atoms are different. The atom with the highest electronegativity will have a
partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.
ii) The most polar bond is expected to occur between atoms with the greatest electronegativity difference.
Formal Charge
Calculate formal charge
i)
Draw the Lewis dot structure.
ii) Count the no. of lone electron pairs on the atom. The no. of lone pair electrons=2*the no. of lone pairs
iii) Count the no. of bonding electrons to the atom. This is 2*the no. of bonds.
iv) Calculate formal charge: Formal charge=Group no.−(no. of lone pair e-s+1/2no. of bonding e-s)
The formal charge on an atom in a molecule is the charge that it would have if all of the bonding electrons were shared equally. In calculating a
formal charge, lone pair electrons belong entirely to the atom on which they reside, and bonding electrons are divided equally between the
bonded atoms. If the no. of electrons assigned to an atom in this way is equal to the no. of valence electrons in the isolated atom, then the atom
has a formal charge of zero. If the no. of assigned electrons is larger than the no. of valence electrons, the formal charge is negative by this
amount, while if the no. of assigned electrons is smaller than the no. of valence electrons, the formal charge is positive by this amount.
To calculate the formal charge for an atom then, one subtracts from its no. of valence electrons all of its lone pair electrons and half of its
bonding electrons: Formal charge = (# valence electrons) - (# lone pair electrons) - 1/2(# bonding electrons)
For main group elements, the number of valence electrons is equal to the group number (1A - 8A).
6.
7.
Choose Lewis structure
We choose the best structure based upon two ideas:
i)
Electroneutrality Principle: Electrons are distributed in such a way that the charges on the atoms are as
close to zero as possible.
ii) Electronegativity: If there is a negative charge, it should be placed on the most electronegative atom.
Draw resonance structure
Write the Lewis structure. Each resonance structure differs only by where the electrons for a multiple bond
come from. When 2 or more atoms with lone pairs are bonded to an atom that needs more electrons, any
one of them can supply the electron pair for a multiple bond. To write all resonance structures,
systematically & sequentially supply the central atom with needed electrons from each of the possible
sources. Keep in mind that Period 2 elements must have an octet, but Period 3 elements can have 8 or more
electrons. Separate these different structures with a double-headed arrow to show that they are resonance
structures.
i.
Smaller formal charges are more favorable than larger ones.
ii.
Negative formal charges should reside on the more electronegative atoms. Conversely, positive formal
charges should reside on the least electronegative atoms.
iii.
Like charges should not be on adjacent atoms.

Predicting Molecular Shapes: VSEPR
8.
Electron Pair Geometry
Ideal Repulsion Shapes
i)
Draw the Lewis structure.
ii) Determine the electron-pair geometry from the no. of structural electron pairs.
Molecular Geometry
Determining Molecular Shapes (as well as Organic)
i)
Draw the Lewis structure.
ii) Determine the electron-pair geometry from the no. of structural electron pairs.
iii) Ignore the positions of lone pairs on the central atom & describe the shape the atoms take.
9.
Recall that for predicting geometry, double and triple bonds count as only one electron pair.
10. Chirality
A stereocenter is a carbon atom that is bonded to 4 different groups.
11. Hybridized orbitals (do more exercises & get familiar with the situations)
To determine the type of hybrid orbitals used, add the no. of atoms bonded to the atom in question & the no. of lone electron
pairs on that atom. This no. is the no. in the left-hand column of the chart, which indicates the type of hybrid orbitals used.
12. Hybridization & Bonding
Sigma bond: bonding electrons exist primarily directly between the 2 bonding atoms. Use hybrid
orbitals
i)
s, px, dx2-y2: px, s , dx2-y2
ii) py, pz, dxy:cannot form(not point toward to the other atom)
Pi bond: bonding electrons are found “above” & “below” the line directly between the 2 bonding atoms,
occupies different regions od space with sigma bonds.(2 p orbitals must be oriented in the same direction). Use
unhybridized p orbitals. Pi bonding leads to an overall planarity between the atoms bonded to the 2 sp2
hybridized central atoms.
13. Molecular Polarity
i)
Drawing the molecular structure.
ii) Assigning the polarity of each of the bonds.
iii) Adding the bond dipoles to determine the net molecular dipole.
iv) Deciding if the bond dipoles cancel. If they do cancel, the molecule is nonpolar.
If the molecule has sufficient symmetry, the bond dipole moments will cancel out (symmetric charge distribution) and the molecule will be
non-polar.If the molecule does not have sufficient symmetry, the bond dipole moments will not cancel out (unsymmetric charge distribution) and
the molecule will have a net dipole moment (polar).
14. Determine Intermolecular Forces
All chemical entities have attractions to other chemicals entities that allow them to form liquids or solids. In each case attractive forces arise due
to attraction of opposite charges on neighboring particles These forces are called intermolecular forces & categorized by how the charges arise.
Ion: ion forces.
Polar molecules: dipole forces. (dipole-induced dipole forces: partially charged atoms induce a dipole in the normally nonpolar molecule.
All molecules have induced dipole(London, dispersion) forces: nonpolar entities no preferred directionality to the induced dipoles, because of
this the dipole is often termed “temporary dipole” . Hold nonpolar molecule as liquid/solid.
Molecules with N-H, O-H or F-H bonds have strong dipole forces called hydrogen bonds.
i)
ii)
Determine if the molecule is polar or nonpolar. If nonpolar, the molecule exhibits only dispersion forces. If
polar, the molecule has both induced dipole & dipole based forces.
If the molecule has any N-H, O-H or F-H bonds, it also exhibits hydrogen bonds.
The properties of most compounds are controlled by dispersion forces with the exception of those compounds that can form hydrogen bonds.

Crystal Solid Structures
15. Calculating density
If the edge length of the unit cell is 420.3 pm, what is the density of MgO in g/cm3. Solid magnesium oxide
has the same kind of crystal structure as NaCl
i)
Density = Mass / Volume
ii)
D = Mcell / Vcell
For a cubic cell the volume is the edge length cubed.
Vcell = (420.3 pm)3 = 7.425E+7 pm3
1 pm3 = 10-30cm3
Vcell in cm3 = 7.425E+7 pm3 x (10-30cm3 / pm3) = 7.425E-23 cm3
formula weight of MgO = 40.30 g/mol
mass per formula unit = formula weight / Avogadro's number
Mformula unit = (40.30 g/mol) / (6.022e23 units/mol) = 6.692E-23 g/unit
Z = number of MgO units per cell = 4 units
mass per unit cell = Z x Mformula unit
Mcell = (4 units) x (6.692E-23 g/unit) = 2.677E-22 g
D = (2.677E-22 g) / (7.425E-23 cm3) = 3.605 g/cm3
16. Calculating Avogadro’s Number from Ion Radii & Structure
i)
Experimental measurements of macroscopic samples of Al can be used to determine the volume of a mole
of the metal, in units of cm3/mol
ii) The crystal data can be used to create a conversion factor to convert volume of Al, in cm3, to atoms of Al.
17. The relationships for atomic substances that crystallize in cubic lattices with one atom per lattice point,
where a is the edge length of the unit cell and r is the radius of the atom
Lattice
atoms per cell
r to a relationship
r to a factor
Simple Cubic
1
2r=a
r = 0.5 a
Body-centered Cubic
2
4 r = (3)1/2 a
r = 0.43301 a
Face-centered Cubic
4
4 r = (2)1/2 a
r = 0.35355 a

Boiling Point & Solubility of Organic compounds
18. Boiling Point
i)
Boiling points increase as intermolecular attractive forces increase. For the alkanes, which are non-polar hydrocarbons, the
ii)
iii)
intermolecular forces are induced dipole forces. In the case of the alcohols, the intermolecular forces include induced dipole forces,
dipole forces, and hydrogen bonding.
Induced dipole forces increase with increasing molecular weight, so the boiling points of the straight chain alkanes increase as the
number of carbon atoms in the chain increases.
For the same reason, the boiling points of the straight chain alcohols also increase as the number of carbon atoms in the chain
increases.
Because of hydrogen bonding, the boiling point of an alcohol is higher than that of the alkane with the closest molecular weight,
which is the one with one more carbon atom.
19. Solubility
i)
Substances dissolve best when solute-solvent interactions are similar to solute-solute interactions. As a
result of this, like things tend to dissolve like things: non-polar solutes dissolve best in non-polar solvents,
and polar or ionic solutes dissolve best in polar solvents
ii) Since water is a polar solvent that is capable of hydrogen bonding, it will be a better solvent for polar
solutes, especially those that can enter into hydrogen bonding, than it will be for nonpolar solutes.
iii) The alcohols given each have a polar -OH end that can hydrogen bond, and a nonpolar hydrocarbon end.
When the hydrocarbon end is small, the -OH end dominates and solubility in water is high. Alcohols with
fewer than 4 carbon atoms are miscible with water. After that, as the length of the carbon chain increases
the solubility in water decreases.

Alcohol
20. Classification 1°, 2°, 3°
i)
The molecule is an alcohol. An alcohol contains an -OH (hydroxyl) group that is bonded to a tetrahedral
carbon atom. If this tetrahedral carbon atom is bonded to only one C atom, the alcohol is a primary alcohol;
if it is bonded to two C atoms, it is a secondary alcohol; & if it is bonded to three C atoms it is a tertiary
alcohol
21. Oxidation
i)
A primary alcohol can be oxidized to an aldehyde, which in turn can be oxidized to a carboxylic acid . The oxidation of a secondary
alcohol (C-OH carbon atom bonded to two other C atoms) gives a ketone, while tertiary alcohols (C-OH carbon atom bonded to three
other C atoms) are not readily oxidized

Carboxylic Acids
22. Intermolecular Forces in Carboxylic Acids
i)
In water
Water molecules are able to function as both donors and acceptors.
The -OH group in alcohols and the -COOH group in carboxylic acids can enter into both donor and
acceptor H-bonding interactions with water.
The O atom in the carbonyl group of the ketone is an acceptor atom, and can form hydrogen bonds with
water acting as the donor.
Both of the O atoms in the -COO- group of the ester are acceptor atoms, and can form hydrogen bonds with
water acting as the donor.
ii) Pure Substance
In order to have hydrogen bonding in a pure compound, the molecules must contain both a donor and an
acceptor atom. The -COOH group in carboxylic acids, and the -OH group in alcohols provide both of these.
Therefore, the carboxylic acid and the alcohol form intermolecular hydrogen bonds in the pure state.
Hydrogen bonding is not expected to be important in pure compounds in which all of the H atoms are
covalently bonded to C atoms, even when an acceptor atom is present in the molecule. The O atom in the
C=O group of the ketone is an acceptor atom, but there is no donor present in the molecule. Similarly, the
-COO- group of the ester provides acceptor atoms but there are no donors.
Because oxygen is more electronegative than carbon or hydrogen, carboxylic acids are polar molecular compounds. Therefore, the intermolecular
forces dipole forces. Induced dipole forces are always present. Induced dipole forces are also called dispersion forces, or London forces. There is
also hydrogen bonding. (dipole forces, induced dipole, hydrogen bonding)
23. Boiling point of Carboxylic Acids
i)
Pure compounds that form intermolecular hydrogen bonds are expected to have higher boiling points than
those of roughly the same molecular weight and shape that do not form hydrogen bonds. Because
carboxylic acids form intermolecular hydrogen bonds and aldehydes do not, a carboxylic acid is expected
to have a higher boiling point than an aldehyde that has roughly the same molecular weight and shape.
While aldehydes do not form intermolecular hydrogen bonds, they are polar molecules and are expected to
have stonger intermolecular attractive forces due to dipole interactions than non-polar molecules do.
Alkanes are not capable of forming intermolecular hydrogen bonds and are non-polar. Therefore, they have
lower boiling points than comparable carboxylic acids or aldehydes.
24. Solubility of Carboxylic acid
i)
Because the C=O group is a hydrogen bond acceptor and the OH group is both a hydrogen bond donor and
acceptor, carboxylic acids form extensive hydrogen bonds with water. Alkanes are nonpolar and do not
form hydrogen bonds with water. Therefore, carboxylic acids are more soluble in water than hydrocarbons
of similar molecular weight.
25. Ester formation
i)
A carboxylic acid, RCOOH, will react with an alcohol, R'OH, to give an ester, RCOOR'

Functional Groups
alcohol -OH. carboxylic acid -COOH. ester -COOC- aldehyde
-CHO. ketone C-CO-C amine-NH2 amide C–CONX2(X=H,R)

Amine
26. Classification
i)
Amines are classified as primary (1°), secondary (2°)
or tertiary (3°) depending on the number of carbon
atoms bonded to nitrogen.
27. Amide Formation
i)
A carboxylic acid, RCOOH, will react with an amine, R'NH2, to
give an amide, RCONHR'

Polymer formation
28. Additional Polymers
i)
The polymer is formed in an addition reaction in
which the electrons in the double bonds of the
monomers become the bonds which hold the units
together in the polymer, as shown below for two
ethylene units: The reaction continues with
additional monomer units to form long polymer
chains.
29. Condensation Polymers
i)
Polyamides
A polyamide is formed in a step-growth reaction, in
which water molecules are split out as the bonds in
the polymer form.
ii) Polyesters
A polymer formed by splitting out water, which
forms according to the following scheme.
In this type of polymerization reaction, water
molecules are split out as the bonds in the polymer
form.
31. Average Rate from Conc-Time Data
i)
The average rate of disappearance is calculated as a
change in concentration with a change in time for the
reactant over the time period in question.
For a reactant, the change in concentration is negative and a minus sign is introduced so that the rate
will be positive.
32. Rate Law
i)
Rate= K[reactant1]order1[reactant2]order2
ii) The order for a reactant tells the effect of changing the concentration of that reactant.
If a reaction is 1st order in a reactant, any change in that reactant’s concentration will have the same
effect on the rate. Doubling the concentration will double the rate
If the reaction is 2nd order, the change in rate will equal the change in concentration squared.
1.
2.
3.
4.
Three types of chemical bonds: Ionic bond, covalent bond, and metallic bond. Characterize for each what
types of elements typically combine to form the bond.

An ionic bond occurs when electrostatic forces exist between ions of opposite charges. Ions occur when
valence electrons are transferred from one atom to another. Metals gain electrons, whereas nonmetals tend
to lose electrons if electron transfer occurs

A covalent bond occurs when valence electrons are shared between two atoms. This commonly occurs
when the two atoms are nonmetals.

A metallic bond occurs when a metal atom bonds to several neighboring metal atoms. Some of the valence
electrons are relatively free to move throughout the 3-dimenstional structure of the metallic substance.
Arrange the oxides of magnesium, calcium & barium in order of increasing magnitude of lattice energies.

The trend in magnitude of the lattice energies: E∝Q(+)Q(-)/d, where Q(+) is the charge of the cation, Q(-)
is the charge of the anion, and d is the closest distance between a cation & anion.

For the oxides of magnesium, calcium & barium, the product of charges is a constant because the cations
are all 2+, and the oxide ion is 2-. Thus, the factor that causes the lattice energies to differ is the distance
term, d. The trend in sizes is Mg2+<Ca2+<Ba2+ which means that the trend in Metal—O distances is
Mg—O<Ca—O<Ba—O. The trend in lattice energies should be BaO<CaO<MgO because lattice energy
increases in magnitude with decreasing distance between the ions.
Which structure has a shorter C—O covalent bond distance? CO or CO2

The greater no. of covalent bonds between 2 atoms the stronger bond strength & the shorter bond distance.

The Lewis structure for carbon monoxide is :C≡O: & carbon dioxide is

CO contains a triple bond & CO2 contains a double bond. Therefore,
CO has a shorter
C—O covalent bond distance.
The sulfate ion is tetrahedral with four equal S O distances of 149 pm. Draw a reasonable structural
formula consistent with these facts.

Each of the five atoms involved belong to Group VIA, which
tells us there are 30 electrons (6e− for each atom). Further, there are two additional electrons for the net ion
charge of −2. It is possible to place the 32 valence electrons in an octet structure having only single bonds.
There are two objections to this structure (Fig. 9-28). The first is that the predicted bond distance, rS +rO =
104+66 = 170 pm, is much too high. Secondly, the calculated formal charge on the sulfur, +2, is rather high.
Regardless, this structure appears widely in textbooks. The reasoning is that the short bond length is the
result of the strong attraction between the +2 sulfur and −1 oxygens (formal charges).
A structure like that in Fig. 9-29 places zero formal charge on the sulfur and −1 on each of the singly
bonded oxygens. The shrinkage in bond length due to double-bond formation helps to account for the low
observed bond distance (149 instead of 170 pm, calculated). Other resonance structures (making a total of
six) with alternate locations of the double bonds are, of course, implied. Structures like this, with an
expanded valence level beyond the octet, are generally considered to involve d orbitals of the central atom.
This is the reason that second-period elements (C, N, O, F) do not form compounds requiring more than
eight valence electrons per atom (the 2d subshell simply does not exist).
5. Explain the observations that the bond length in N+2 is 2 pm greater than in N2, while the bond length in
NO+ is 9 pm less than in NO.

The electron configurations are written for the four molecules. The computed
bond orders are 3 for N2 and 5/2 for N2+. N2, therefore, has a stronger bond and
should have the shorter bond length. The computed bond orders are 5/2 for NO
and 3 for NO+. The NO+ cation has the stronger bond
6. Some ligands are multifunctional, that is, they have two or more atoms that can
bind to the central metal atom or ion. Each binding site occupies a different
corner on the coordination surface. Ethylenediamine (abbreviated en) is such a ligand. The two binding
atoms are nitrogens and the two binding sites must be cis to each other because of the shape and size of the
en. How many geometrical isomers of [Cr(en)2Cl2]+ should exist and which isomer(s) might display
optical activity?

Two geometrical isomers exist, cis and trans (Fig. 9-48). Each
en can be represented by an arc terminating at the two binding
sites. By drawing other arrangements of arcs while preserving
positions of the chlorines, one can see that (b) is a distinct
mirror image of (a); however, the mirror image of (c) is the
exact same structure as (c). In other words, only the cis isomers can be optically active
7. Explain why metals are usually lustrous (mirror-like)

In the band model there is a continuum of empty energy levels, rather than discrete energy levels. This
situation allows light quanta of all energies within a wide range of wavelengths to be absorbed equally, then
the energized electrons will re-emit the light when they fall back into their ground-state orbitals. This is the
mechanism for reflection of light of all frequencies, which we call “luster.”
8. BaTiO3 crystallizes in the perovskite structure. This structure may be described as a barium-oxygen
face-centered cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying
the face-centers, and titanium ions occupying the centers of the unit cells. (a) If titanium is described as
occupying holes in the Ba-O lattice, what type of hole does it occupy? (b) What fraction of the holes of this
type does it occupy? (c) Suggest a reason why it occupies those holes of this type but not the other holes of
the same type?

(a) These are octahedral holes.

(b) The octahedral holes at the centers of the unit cells constitute just one-fourth of all the octahedral holes
in a face-centered cubic lattice.

(c) An octahedral hole at the center of a unit cell has six nearest-neighbor oxide ions and is occupied by a
titanium ion. The other octahedral holes are located at the centers of the edges of the unit cell and have six
nearestneighbors each, as is the case with any octahedral hole. However, two of the six neighbors are
barium ions (at the unit-cell corners terminating in a given edge) and four are oxide ions. The proximity of
two cations, Ba2+and Ti4+, would be electrostatically unfavorable.
9. The melting point of quartz, a crystalline form of SiO2, is 1610◦C, and the sublimation point of CO2 is
−79◦C. How similar do you expect the crystal structures of these two substances to be?

The big difference in melting points suggests a difference in type of crystal binding. The intermolecular
forces in solid CO2 must be very low to be overcome by a low-temperature sublimation. CO2 is actually a
molecular lattice held together only by the weak van der Waals forces between discrete CO2 molecules.
SiO2 is a covalent lattice with a three-dimensional network of bonds; each silicon atom is bonded
tetrahedrally to four oxygen atoms and each oxygen is bonded to two silicon atoms.
10. Which would have the higher melting point, (a) V or Ca, (b) MgO or KCl? Explain.

(a) Vanadium is expected to have a much greater charge density in the electron sea since it has three 3d
electrons in addition to the two 4s electrons to contribute; calcium only has two electrons in the outside
11.
12.
13.
14.
15.
orbit. Further, the ionic cores, being smaller, will be closer together. As a result of these factors, the crystal
forces will be greater in vanadium than calcium. The actual melting points are 1890◦C for vanadium and
845◦C for calcium. The relative closeness of the cores is indicated by the densities, which are 6.11 g/cm3
for V and 1.55 g/cm3 for Ca.

(b) MgO will have a higher melting point. The first consideration is that the smaller cation and anion in
MgO allows for closer approach. Further, both the cation and the anion carry twice as much charge as K+
and Cl−. This becomes clear when we consider that the electrostatic force is proportional to the product of
the charges and inversely proportional to the square of the distance between them. The actual melting
points are 2800◦C for MgO and 776◦C for KCl.
Account for the differences in melting point between (a) and (b), between (c) and (d), and between these
two differences as shown

The crystal forces in (b) and (d) are largely van der
Waals. Compounds (a) and (c), containing the polar
hydroxide group, are capable of hydrogen bonding.
In the case of (c), the hydrogen bonding is from the
hydroxide of one molecule to the doubly bonded
oxygen of the neighboring molecule; and the
resulting intermolecular (between molecules)
attraction leads to a very large increase in the melting point as compared with (d), the non-hydrogenbonded
control. In the case of (a), the molecular structure allows intramolecular (within a molecule) hydrogen
bonding from the hydroxide group of each molecule to the doubly bonded oxygen of the same molecule; in
the absence of strong intermolecular hydrogen bonding the difference in melting point as compared with
the reference substance (b) should be small, related perhaps to differences in crystal structure or to the van
der Waals forces, which should be slightly larger for (b) than for (a) because of the extra CH3 group.
Which of the following pairs of liquids is/are miscible (can be mixed)? Why, or why not? (a) butane
(C4H10) and pentane (C5H12); (b) butane and water; (c) 1-butanol (C4H9OH) and water.

(a) Miscible; forces of attraction between like and unlike molecules are about the same. (b) Not miscible;
mixing would disrupt the strong H-bonds in water; there is no especially strong attraction between unlike
molecules to compensate. (c) Miscible; both components have hydrogen bonding. Breaking of H-bonds in
water is compensated for by the formation of H-bonds between unlike molecules.
Explain what’s wrong with each of the following names: (a)3-methyl-2-propanol; (b)
3,3-dimethyl-2-pentene; (c) 1,4-dichlorocyclyobutane; (d) 2-propanal; (e) 2-methyl-1-butyne; (f)
pentanoicacid.
 (a) The longest chain is four carbons. The correct name is 2-butanol.
 (b) Such a compound is impossible because it would require five bonds on the third carbon.
 (c) Positions 1 and 4 are equivalent to 1 and 2. The correct name is 1,2-dichlorocyclobutane. (d) Such a
compound is not possible because the aldehyde carbon must be at the end of the chain.
 (e) Such a compound won’t work because it would require five bonds on the second carbon.
 (f ) The suffix “acid” should be a separate word. The correct name is pentanoic acid.
Which has more isomers, (a) dimethylbenzene or (b) dimethylcyclohexene? Explain.
 Because all corners of the benzene ring are equivalent, (a) has only three isomers (structural) with the
methyl groups in positions 1,2; 1,3; or 1,4. However, (b) can have many more structural isomers; for
example, 1,2 is different from 1,6 and 1,3 is different from 2,4. There are also many geometric isomers;
for example, the 3,4 structural isomer must have cis and trans forms. Optical isomerism also occurs; for
instance, both carbons 3 and 4 of the 3,4 isomer are chiral.
Adenosine triphosphate, ATP, is an extremely important molecule because it stores energy that is released
as needed for energy for living organisms. (a) If there is a chiral carbon in ATP, identify that carbon by the
number provided in the sketch (b) Considering the structure and composition of ATP, identify the
elements involved at the points of the strongest negative charges. (c) If there are any, identify carbons that
are substituted with an alcohol group. (d) This compound contains three rings. Identify which of the three
rings, if any, are heterocyclic.

(a) The chiral carbons are 2, 3, 4, and 5 (four different substituents). (b)
Both nitrogen and oxygen are expected to be negative points; however,
the oxygens that are doubly bound to the phosphorus in the three
phosphates are the most negative points. There are oxygens that are
negative poles in the alcohol groups (—OH). It is true that there are
nitrogens that service a negative pole, but the oxygens are more
electronegative, eliminating the nitrogens from consideration.

(c) #3 and #4;

(d) All three rings are heterocyclic because they contain at least one atom
other than carbon (N and O in these rings).
16. Compare the carbon-oxygen bond lengths in the formate ion, HCO2- , and on
the carbonate ion CO32- . In which ion is the bond longer?


17.
18.
19.
20.
21.
22.
23.
24.
Plan: Write the resonance structures for the two molecules. In cases where bonds
between the same two atoms are compared, shorter
& stronger: triple bonds > double bonds > single
bonds. If more than one resonance structure is
possible, average their contribution.
Two equivalent plausible Lewis structure exist
for HCO2- , with 18 valence electrons: One
resonance structure has a carbon-oxygen single
bond & a double bond. We predict that the
carbon-oxygen bond will be halfway between a
single & double bond

In carbonate, one resonance structure has 2
carbon-oxygen single bonds & a double bond.
We predict that the carbon-oxygen bond will be
closer to a single bond than a double bond.

That means the shortest C-O bond is the bond in
HCO2These have two or more resonance structures.
Write all the resonance structures for each. (a)
Nitric acid (b) Nitrate ion, NO3
There are 3 plausible Lewis structures for both
nitric acid, HNO3 and nitrate ion NO3-, with 24
valence electrons for both. They are formed
using one lone pair from a different one of the
outer O atoms to make the second bond in the
double bond to complete the octet of the N
atom.
Several Lewis structures can be written for
perbromate ion, BrO4-the central Br with all
single Br-O bonds, or with one, two, or three Br=O double bonds. Draw the Lewis structures of these
possible resonance forms, and use formal charges to predict the most plausible one.

BrO4- has 32 electrons. Formal charges for single bonded O atoms are always -1. Formal charges for double
bonded O are always 0. Each tie electrons are moved from lone pairs into bonding pairs into bonding pairs
the positive formal charge on Br goes down. The Lewis structures that follows the octet rule for all the
atoms is the 1st one. Since smaller formal charges are more favorable than larger ones, we predict that this
fourth resonance structure, with three double bonds & most 0 formal charges, is the most plausible.
Several Lewis structures can be written for thiosulfate ion, S2O32-. Write the Lewis structures of these
possible resonance forms.

The molecule has 32 valence electrons.
Could CaCl2 possibly have the NaCl structure?

No, the ration of the ions in the unit cell must reflect the empirical formula of the compound.
What is the monomer in natural rubber? Which isomer is present in
natural rubber, cis or trans?

Isoprene, 2-methyl-1,3-butadiene; cis-isomer
Explain why ethanol CH3CH2OH, is soluble in water in all
proportions, but decanol, CH3(CH2)9OH, is almost insoluble in
water.

CH3(CH2)9OH is a large molecule than CH3CH2OH. The polar end
of the alcohol group will interact well with the water; however, the
nonpolar end of the molecule will not. The longer nonpolar end of
decanol will not be miscible in water, lowering the solubility compared to smaller, more polar ethanol.
How are rubber molecules modified by vulcanization?

Vulcanized rubber has short chains of sulfur atoms that bond together(crosslink) the polymer chains of
natural rubber.
In terms of structural types & occupancy of holes, how are the fluorite & zinc blende unit cells similar &
how do they differ

Similarities: Unit cells are face centered cubic. Lattice points are located at the corners & in the centers of
the faces. Anions are located in tetrahedral holes formed by the cations.(alternate description of ZnS)
Difference: F- are located in all 8 of tetrahedral holes, S2-/Zn2+ are located in half of tetrahedral holes
formed by the other.
In the most common description of the cesium chloride unit cell, a cesium lies at the center of a cube
defined by chloride ions. Does this mean that CsCl has a body centered cubic structure type? Explain

No. In a unit cell, 8 chloride ions are located on the corners and a cesium ion at the center. Because the
atom at the center of the cell is different from the atoms at the corners, the unit cell is simple cubic, not
body-centered.
Why a copper wire doesn’t break when bending

It has a structure of face centered cubic. For this structure, it has good ductility & can endure tensile force.
Why is diamond the hardest known common substance whereas graphite is a soft, solid lubricant. Density:
diamond>graphite.

In diamond structure, here is a 3-D network of strong covalent bonds. This makes diamond extremely hard.
Because of its hardness, it can make cutting on glass. It has a closer packed arrangement of atoms, which
makes it have more atoms in the same volume, compared with graphite.

In graphite, there are flat layers of carbon atoms. These layers are held by weak Van der Wal’s forces, and
hence can easily slip one over the other. This makes graphite extremely soft & slippy
Many biological molecules, including steroids & carbohydrates, contain many –OH groups. What need
might biological systems have for this particular functional group?

–OH groups are a common site of hydrogen bonding intermolecular forces. Their presence would increase
the solubility of the biological molecule in water & create specific interactions with other biological
molecules.
What are some examples of thermoplastics? What are the properties of thermosetting plastics when
heated & cooled

Examples: milk jugs (polyethylene), cheap sun glasses & toys (polystyrene), CD audio discs
(polycarbonate). Thermoplastics soften & flow when heated.

25.
26.
27.
28.
29.