COMPOUND PROPERTIES
CHEMICAL BONDING
Chemical compounds are formed by the joining of two or more
atoms. A stable compound occurs when the total energy of the
combination has lower energy than the separated atoms = more
stable. The bonded state implies a net attractive force between
the atoms ... a bond. The common cases of bonds are classified
into two main groups: Intramolecular and Intermolecular.
INTRAMOLECULAR FORCES
Intramolecular forces involve sharing or transferring e– resulting in a
chemical bond. Creating intramolecular attractions that form a more
stable (read Noble Gas) e–configuration for component atoms.
Covalent bond: bond in which one or more pairs of e– are
shared (equally =non-polar, unequally = polar) between two
atoms.
Ionic bond: bond in which one or more e– from one atom (a
metal) are removed and attracted to another atom (a
nonmetal). Resulting in positive and negative ions which
attract each other due to their opposite charges.
Metallic bond: bonding that occurs between metals in which an
array of positive ions in a sea of e–are held together by strong
forces of attraction between the positive nuclei and the
delocalized e– of the metal atoms.
Properties
Composition
Physical State
Melting Point
Solubility in H2O
Solubility in Ethanol
Conductivity in H2O
Representative Unit
Ionic
Ions of metal and nonmetal
atoms
Crystalline (s)
High (>300˚C)
Usually High
Usually Low
Yes
Formula unit – simplest
neutral charge ratio of
cations and anions.
Compound Type
Molecular
Nonmetal atoms only
Amorphous (s), (l), or (g)
Low (<300˚C)
Usually Low
Usually High
No
Molecule – a neutral group
of atoms that act together
as a unit.
BASIC RULES OF LEWIS DIAGRAMS
Metallic
Metal atoms only
Crystalline (s)
High (>800˚C)
None
None
No (But yes as a solid)
Formula unit – simplest ratio of
repeating atomic structure.
HONC = 1234 RULE
RULE 1. Bonds are pairs of e– shared between two atoms.
RULE 2. Hydrogen can form only one covalent bond (Duplet
Rule).
RULE 3. (Octet Rule) Most covalently bonded atoms (except
for hydrogen) have a filled octet of valence e–.
INTERMOLECUAR FORCES
Intermolecular are the columbic attractions between dipoles, both
permanent and temporary that are physical bond. Columbic
attractions are the attraction between oppositely charged particles.
For example, the p+ in the nucleus of an atom have attraction for the
e– surrounding the nucleus as the positive p+ attracting the negative
e–. These forces are “weak” in comparison to the intramolecular
forces that hold molecules together.
Dispersion Forces (weak) : As e– may be on one side of the
atom or the other, a temporary dipole may form due to
unequal distribution of e–. As e– move, the dipole fluctuates.
This fluctuation in an atom produces a fluctuating electric field
that is felt by the e– of an adjacent atom. This atom then
polarizes so that its outer e– are on the side of the atom closest
to the + side (or opposite to the – side) of the dipole. This shift
is a temporary (read nearly instantaneous but reoccurring)
Dipole-Dipole Attractions (strong): The columbic attraction of
two dipoles (aka a polar molecule). Note: In a dipole, the
attractive force of the nucleus is NOT divided up among the e–
in the atom. Rather each e– gets approximately the full
attractive force of the nucleus (minus the repulsive effects of
other e–) but their tendency in location results in different
densities in the e– cloud that result in a “separation” of the
atoms charge.
Hydrogen bonding (VERY strong): a special type of dipoledipole attraction in which the columbic attraction between
polar molecules (in which hydrogen (H) is bound to a highly
electronegative atom) results in a permanent dipole. However,
they are still weaker than a covalent bond as e– are not
actually shared or transferred.
LEWIS DIAGRAMS for COVALENT BONDING
Looking at the section of the periodic table left, notice how each element,
as we move to the right, adds one more dot (valence e–). For example,
carbon with four valence electrons (group 4A) can become
isoelectronic with Ne (8 valence e–, group 8A) by sharing 4 electrons
from other non-metallic elements.
The octet rule provides a way to find out how many covalent bonds are formed by the representative nonmetallic elements.
So far we have only considered molecules which have single bonds between pairs of atoms. Molecules can have multiple
bonds, e.g., double bonds O=O and triple bonds N≡N.
See how in this case each oxygen atom shares two of its e– with the
other oxygen atom. So altogether there are four e– shared between
the two oxygen atoms, forming TWO covalent bonds. This is called
a double bond.
HOW TO DRAW LEWIS STRUCTURES
ELECTRONEGATIVITY DIFFERENCE (ΔEN)
1)
2)
ΔEN, determines the bond type.
Ionic Bond
Covalent Bond
-Polar Covalent
-Non-Polar Covalent
Metallic Bond
M / NM + NM
NM / M / ML + NM
NM / M / ML + NM
NM + NM
M+M
Valence eFilled e- shell core
Separated Atoms
Ionic Bond
ΔEN ≥ 1.7
ΔEN < 1.7
1.7 < ΔEN ≥ 0.5
ΔEN < 0.5
ΔEN < 0.5
Polar Covalent Bond
Covalent Bond
Draw Electron Dot Diagrams for each atom in molecular formula.
Make the “Facts of your Case Match” your “Evidence”:
1) All atoms bond to reach their octet through SHARING eo H this is 2
o All other atoms this is 8.
2) All atoms in the molecular formula must be bonded together and bonded (represented by circling the shared
pair) so that all atoms participate in the structure. Some handy rules to remember are these:
o H - Hydrogen and the halogens bond once. (1A and 7A)
o O - The GROUP oxygen is in (6A) bonds twice.
o N - The GROUP nitrogen is in bonds (5A) three times.
o C - The GROUP carbon is in bonds (4A) four times.
A good thing to do is to bond all the atoms together by single bonds, and then add the multiple
bonds until the rules above are followed.
3) You cannot break up a “lone pair” of e- (as a lone pair represents a full orbital. NEVER break up roommates)
3) Redraw your structure to represent each shared pair as a — rather than ▪ ▪ and denote lone pairs.
4) Each atoms bonding and lone pairs should be drawn so that they are as far away from each other as possible
as they are charged the same and repel each other.
5) IF IONIC Place whole molecule in brackets and denote charge on upper right hand side.
NOTE THIS IS A TRIAL AND ERROR PROCESS, YOU NEED TO KEEP ANALYZING YOUR EVIDENCE UNTIL IT
MATCHES THE “FACTS” OF YOUR CASE.
ORBITALS AND SHAPES
V.S.E.P.R. Theory
Lewis diagrams are electron dot pictures which give
an excellent account of the number of valence
electrons in a covalent molecule. The dot diagrams do
NOT explain the reasons for covalent bonding - the
‘glue’- nor do they explain the shapes of molecules. A
good example is water. The Lewis diagram is:
VSEPR Theory
1. The shape of a molecule is determined by the
number of e- pairs (bonding pairs and lone
pairs) associated with the central atom of
the molecule.
VSEPR and Unshared Electron Pairs
1. Unshared pairs take up positions in the
geometry of molecules just as atoms do.
2. Unshared pairs have a relatively greater effect
on geometry than do atoms
3. Lone (unshared) electron pairs require more
room than bonding pairs (they have greater
repulsive forces) and tend to compress the
angles between bonding pairs
4. Lone pairs do not cause distortion when bond
angles are 120° or greater
A = Central Atom ; X = Ligands; E =Lone Pairs
Electron
Molecular
Name
Regions
Formula
2
AX2
Linear
3
AX3
Trigonal Planar
3
AX2E1
Bent
4
AX4
Tetrahedron
4
AX3E1
Trigonal Pyramidal
4
AX2E2
Bent
5
AX5
Trigonal Bipyramidal
5
AX4E1
See-saw
5
AX3E2
T-shaped
5
AX2E3
Linear
6
AX6
Octahedron
6
AX5E1
Square Pyramidal
6
AX4E2
Square Octahedron
The overwhelming experimental evidence shows that
water is in fact a bent polar covalent molecule shaped
something like:
or
or
or
or
Explanations of shape, etc. require quantum theory
and combinations of orbital that result in Valence Shell
Electron Pair Repulsion Models. AKA AP/College
Chemistry.
PERCENTAGES
PERCENT COMPOSITION
It is a pet peeve of Mr. Butryn’s that students do not know what a percentage means. To calculate the percent composition of a component in a compound:
A percentage is ratio between an expected or wanted value and what actually was 1.Find the molar mass of the compound by adding up the masses of each atom in
measured. Percentages mean:
the compound using the periodic table or a molecular mass calculator.
2. Calculate the mass due to the component in the compound you are for which
Part X 100%
you are solving by adding up the mass of these atoms.
Note: Sig Figs still matter. Also when a part is taken
Whole
3. Divide the mass due to the component by the total molar mass of the compound
over a whole, the units of each much be the same
and multiply by 100.
and thus cancel out. When you multiply by that
100%, the units that remain is just %.
Percent Composition = Mass due to specific component x 100 %
Total molar mass of compound
EMPERICAL VS MOLECULAR FORMULAS
Thus, a chemical formula is a ratio of atoms i.e. X2Y3. That is, for every 2 X’s there are 3 Y's or for 3 Y's there are 2 X’s. This is an example of the simplest mole ratio of the
atoms/ions of the molecules/crystal’s structure. It cannot be simplified to any greater degree. This is thus an example of an empirical formula. The two most commonly
seen chemical formulas are the empirical formula and the molecular formula.
Both formulas, molecular and empirical, tell you what elements make up a specific chemical compound. For example, H2O (both a molecular and an empirical formula) has
hydrogen (H) atoms and oxygen (O) atoms. By looking at CaCl2 (calcium chloride) you will know that it has calcium (Ca) and chlorine (Cl) ions in it.
Molecular and empirical formulas differ by the numbers in their subscript. The subscript is the ’2′ in H2O. For a molecular formula, the subscript describes the total number of
atoms in a particular molecule. Hexane, a molecule with six atoms of carbon and fourteen atoms of hydrogen is expressed as C 6H14 in a molecular formula. For an
empirical formula, the subscript describes the ratio of atoms in a particular molecule. Again, hexane with its six carbon atoms and fourteen hydrogen atoms would now be
expressed C3H7 because that is the ratio of 6:14.
EMPERICAL FORMULA
MOLECULAR FORMULA
Empirical formulas can be calculated using experimental data: Given that a certain
compound is 69.58% Ba, 6.090% C and 24.32% O, calculate the empirical formula
of this compound.
Molecular formulas can be calculated using empirical formulas given the
molar mass of the molecular compound:
Step 1, % to g: Assume that you have 100.00 g of the compound, :
1. 69.58% Ba
= 69.58 g Ba
2. 6.090% C
= 6.090 g C
3. 24.32% O
= 24.32 g O
Step 2, g to mol: Convert the mass of each element to moles.
Ba = 69.58 g Ba ÷ 137.3 0 g/mol Ba
= 0.5068 mol Ba
C = 6.090 g C ÷ 12.01 g/mol C
= 0.5071 mol C
O = 24.32 g O ÷ 16.00 g/mol O
= 1.520 mol O
Step 2, mol ratio: Divide through each value by the smallest number of moles
0.5068 mol Ba ÷ 0.5068 mol = 1 Ba
= 1 Ba
0.5071 mol C ÷ 0.5068 mol = 1.001 C = 1 C
1.520 mol O
÷ 0.5068 mol = 2.999 O = 3 O
Given that the empirical formula of a compound is CH and the molar mass is
104 g/mol, calculate the molecular formula.
mass of C
= 12.00 g/mol C
mas of H
= 1.01 g/mol H
Empirical formula mass:
mass of CH
= 12.00 g/mol + 1.01 g/mol
mass of CH
= 13.01 g/mol CH
To find the number of CH units in the compound:
# CH units
# CH units
= 104 g/mol ÷ 13.0 g/mol
= 8.00 units
Molecular formula = 8(CH) or rather, C8H8