Why do atoms bond?

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Why do atoms bond?
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Introduction to Bonding

• Atoms are generally found in nature in
  combination held together by chemical bonds.
• A chemical bond is a mutual electrical attraction
  between the nuclei and valence electrons of
  different atoms that binds the atoms together.
• There are two types of chemical bonds ionic,
  and covalent.

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Introduction to Bonding

• What determines the type of bond that forms?
• The valence electrons of the two atoms involved
  are redistributed to the most stable arrangement.
• The interaction and rearrangement of the valence
  electrons determines which type of bond that
  forms.
• Before bonding the atoms are at their highest
  possible potential energy

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Introduction to Bonding

• There are 2 philosophies of atom to atom
  interaction
• One understanding of the formation of a chemical
  bond deals with balancing the opposing forces of
  repulsion and attraction
• Repulsion occurs between the negative e- clouds
  of each atom
• Attraction occurs between the positive nuclei and
  the negative electron clouds

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Introduction to Bonding

• When two atoms approach each other closely enough
  for their electron clouds to begin to overlap
• The electrons of one atom begin to repel the
  electrons of the other atom
• And repulsion occurs between the nuclei of the
  two atoms

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Introduction to Bonding

• As the optimum distance is achieved that balances
  these forces, there is a release of potential
  energy
• The atoms vibrate within the window of maximum
  attraction/minimum repulsion
• The more energy
  released the stronger
  the connecting bond
  between the atoms

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Introduction to Bonding

• Another understanding of the form-ation of a
  chemical bond between two atoms centers on
  achieving the most stable arrangement of the
  atoms valence electrons
• By rearranging the electrons so that each atom
  achieves a noble gas-like arrangement of its
  electrons creates a pair of stable atoms (only
  occurs when bonded)

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Introduction to Bonding

• Sometimes to establish this arrange-ment one or
  more valence electrons are transferred between
  two atoms
• Basis for ionic
  bonding
• Sometimes valence
  electrons are shared between
  two atoms
• Basis for covalent
  bonding

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Introduction to Bonding

• A good predictor for which type of bonding will
  develop between a set of atoms is the difference
  in their electronegativities.
• Remember, electronegativity is a measure of the
  attraction an atom has for e-s after developing a
  bond
• The more extreme the difference between the two
  atoms, the less equal the exchange of electrons

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Introduction to Bonding

• Lets consider the compound Cesium Fluoride, CsF.
• The electronegativity value (EV) for Cs is .70
  the EV for F is 4.00.
• The difference between the two is 3.30, which
  falls within the scale of ionic character.
• When the electronegativity difference between two
  atoms is greater than 2.1 the bond is mostly
  ionic.

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Introduction to Bonding

• The take home lesson on electro-negativity and
  bonding is this
• The closer together the atoms are on the P.T.,
  the more evenly their e- interact, and are
  therefore more likely to form a covalent bond
• The farther apart they are on the P.T., the less
  evenly their e- interact, and are therefore more
  likely to form an ionic bond.

metal w/nonmetal ionic
nonmetal w/nonmetal covalent
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Introduction to Covalent Bonding

• In a co-valent bond
• The electronegativity difference between the
  atoms involved is not extreme
• So the interaction between the involved electrons
  is more like a sharing relationship
• It may not be an equal sharing relationship, but
  at least the electrons are being shared.

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Covalent Bonds
Lets look at the molecule Cl2

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each atom must have 8 valence e's
Cl
Cl
Notice 8 e- in each valence shell!!!
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Covalent Bonds
How about the molecule HCl?

(Polar Covalent) shared, but not evenly
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So whats the bottom line?
To be stable the two atoms involved in the
covalent bond share their electrons in order to
achieve the arrangement of a noble gas.
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Introduction to Ionic Bonding

• In an ion - ic bond
• The electronegativity difference is extreme,
• So the atom with the stronger pull doesnt really
  share the electron
• Instead the electron is essentially transferred
  from the atom with the least attraction to the
  atom with the most attraction

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An electron is transferred from the sodium atom
to the chlorine atom

Na
Cl
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Both atoms are happy, they both achieve the
electron arrangement of a noble gas.
Notice 8 e- in each valence shell!!!
-1
1
Cl
Na
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Very Strong Electrostatic attraction established
IONIC BONDS
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So whats the bottom line?
To be stable the two atoms involved in the ionic
bond will either lose or gain their valence
electrons in order to achieve a stable
arrangement of electrons.
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Introduction to Metallic Bonding

• In a metallic bond
• The resulting bond is a cross between covalent
  and ionic bonding
• Valence electrons are transferred from one metal
  atom to the surrounding metal atoms
• But none of the involved metal atoms want the
  electrons from the original atom, nor their own
  so they pass them on

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Introduction to Metallic Bonding

• What results is a sharing/transfer of valence
  electrons that none of the atoms in the
  collection own the valence electrons
• It resembles a collection of positive ions
  floating around in a sea of electrons
• Its this interaction that leads to the
  properties unique to metals
• Conductive
• Malleable and Ductile

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Sea of Electrons
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Bond Energies and Bonding

• As weve learned so far ionic com-pounds are
  formed by the transfer of electrons from a metal
  to a nonmetal
• The ionic compound is held together by the strong
  electrostatic attraction between oppositely
  charged ions.
• There is a tremendous
  amount of energy stored in
  the bonds formed in
  an ionic compound.

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Bond Energies and Bonding

• It takes a lot of energy (A.K.A. bond energy) to
  pull the two ions apart once they have
  established their stable arrangement through
  bonding
• Energy can be released or absorbed when ions form
• Removing electrons from atoms requires an input
  of energy
• Remember from last chapter this energy is called
  ionization energy

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Energy and Ionic Bonding

• On the other hand adding electrons to atoms
  releases energy into the environment
• Remember this has to do with the atoms affinity
  for electrons
• Sometimes this energy is used to help remove the
  electron from another atom
• The ionization energy to remove 1 e- from each
  atom in a mole of Na atoms is 495.8 kJ

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Energy and Ionic Bonding

• A mol of Cl atoms releases 348.6 kJ when an e- is
  added to the atom
• Notice that it takes more energy to remove Nas
  e- than the amount released from the Cl atoms.
• Forming an ionic bond is a multi-step process
• The final step releases
  a substantial amount of energy
  (a.k.a. the driving force)

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Crystal Formation
At the beginning there is solid sodium and
chlorine gas. Na(s) Cl2(g)
A mol of sodium is converted from a solid to a
gas Na(s) energy ? Na(g)
Step 1
ENERGY IN
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One electron is then removed from each sodium
atom of form a sodium cation Na(g) energy ?
Na(g) e-
Step 2
ENERGY IN
Energy is required to break the bond holding
0.5mol of Cl2 molecules together to form a mole
of chlorine atoms Cl2(g) energy ? 2Cl(g)
Step 3
ENERGY IN
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The next step involves adding an electron to each
chlorine atom to form a chloride anion Cl(g)
e- ? Cl-(g) energy
Step 4
348.6 kJ/mol
ENERGY OUT
The final step provides the driving force for the
reaction. Na(g) Cl-(g) ? NaCl(s) energy
Step 5
787.5 kJ/mol
ENERGY OUT
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Crystal Formation

• Energy released in the final step is called the
  lattice energy
• Energy released when the crystal lattice of an
  ionic solid is formed
• For NaCl, the lattice energy is 787.5 kJ/mol,
  which is greater than the input of energy in the
  previous steps
• The lattice energy provides enough energy to
  allow for the formation of the sodium ion

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Crystal Formation

• We can use the lattice energy as a method for
  measuring the strength of the bond in ionic
  compounds.
• The amount of energy necessary to break a bond is
  called bond energy.
• This energy is equal to the lattice energy, but
• Bond energy moves into the system
• Lattice energy moves out of the system

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Bond energy
Lattice energy
Compound
kJ/mol (in)
kJ/mol (out)
861.3
-861.3
LiCl
817.9
-817.9
LiBr
759.0
-759.0
LiI
787.5
-787.5
NaCl
751.4
-751.4
NaBr
700.1
-700.1
NaI
2634.7
-2634.7
CaF2
3760.2
-3760.2
MgO
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Hydrate Formation

• In the construction of a crystal lattice,
  depending on the ions involved there can be small
  pores develop between ions in the ionic
  crystal.
• Some ionic compnds have enough space between the
  ions that water molecules can get trapped in
  between the ions
• Ionic compounds that absorb water into their
  pores form a special type of ionic compound
  called a hydrate.

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Trapped Water Molecules
Hydrated Crystal
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Hydrate Formation

• Hydrates typically have different properties than
  their dry versions - A.K.A. anhydrides
• Anhydrous CuSO4 is nearly colorless
• CuSO45 H2O is a bright blue color
• When Copper (II) Sulfate is fully hydrated there
  are 5 water molecules present for every Copper
  ion.
• The hydrated name would be Copper (II) Sulfate
  Pentahydrate

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Hydrate Formation

• Have you ever bought a new purse or camera and
  found a small packet of crystals labeled do not
  eat?
• These crystals are there to absorb water that
  might lead to mildew or mold
• The formula of a hydrate is XAYB Z
  H2O (Z is a coefficient indicating how many
  waters are present per formula
  unit)

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Percent Composition

• An important quantitative measurement that can be
  made for any chemical substance is a Percent
  Composition.
• The percent composition of a compound is a
  relative measure of the mass of each different
  element present in the compound.
• It gives you a rough comparison of the masses of
  the each component in the total sample

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Percent Composition

• percent composition in a compnd can be determined
  in 2 ways
• The 1st is by calculating the percent
  composition by mass from a chemical formula.
• The 2nd is a lab scenario where an unknown
  compound is chemically broken up into its
  individual components and percent compo-sition is
  determined by analyzing the results.

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What is the percent composition of Hydrogen
Oxygen in Water (H2O)?
1st Assume you have a mole of the compound in
question, and calculate its molar mass
(21) (116)
18 g H2O
2nd Use the MM of each component and the MM
of the compound to calculate the percent by
mass of each component
(21) 2 g/mol
H
x 100
11.1
O
100 11.1 88.9
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Calculating PC Using Analysis Data

• In this method, the mass of the sample is
  measured, then the sample is decomposed or
  separated into the component elements
• The masses of the component ele-ments are then
  determined and the percent composition is
  calculated as before
• divide the mass of each element by the total mass
  of the sample and multiply by 100.

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Find the percent composition of a compound that
contains 1.94g of carbon, 0.48g of Hydrogen, and
2.58g of Sulfur in a 5.0g sample of the compound.

• Calculate the percents for each component by the
  equation (Component Mass/Total Sample Mass) x
  100

C 1.94g/5.0g x 100 38.8
H 0.48g/5.0g x 100 9.6
S 2.58g/5.0g x 100 51.6
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Classroom Practice 1
Calculate the percent composition of Mg(NO3)2.
Mg 16.2 N 18.9 O 64.0
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Empirical Formulas

• Percent compositions can be used to calculate
  the a simple chemical formula of a compound,
  called an empirical formula
• Empirical formula is the simplest ratio of the
  atoms in a compound
• Ionic compounds are always written as empirical
  formulas

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Empirical Formulas

• Procedure for calculating Empirical Formula
• convert the percent compositions into moles
• compare the mols of each compo-nent to calculate
  the simplest whole number ratio
• divide each amount in moles by the smallest of
  the mole amounts
• This sets up a simple ratio

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Calculate the empirical formula of a
compound that is 80.0 Carbon and 20.0 Hydrogen
by mass
Since we dont know the original mass of the
sample, we can assume a 100 g sample

• We have 80 grams of Carbon and 20 grams of
  Hydrogen
• We need to calculate the number of moles of each
  element that we have.

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Calculating Empirical Formulas
1 mole C
80.0g C
12.01 g C
1
CH3
1 mole H
20.0g H
2.97
1.008 g H

• Now we need to calculate the smallest whole
  number ratio in order to find the empirical
  formula.
• Divide each component by the smallest number in
  moles

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Calculating Empirical Formulas
Determine the empirical formula of a compound
containing 2.644g of Au and 0.476g of Cl.
1 mol Au
2.664g Au
.01352mol Au
197 g Au
1
1 mol Cl
.476g Cl
.01345mol Cl
1
35.4 g Cl
AuCl
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Classroom Practice 2
Determine the empirical formula for a compound
which is 54.09 Ca, 43.18 O, and 2.73 H.
CaO2H2 or Ca(OH)2
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Molecular Formulas

• The empirical formula for a compound provides the
  simplest ratio of the atoms in the compound
• However, it does not tell you the actual numbers
  of atoms in each molecule of the compound
• For instance the empirical formula for glucose is
  CH2O (121)
• While the molecular formula for glucose is C6H12O6

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Molecular Formulas

• A molecular formula indicates the numbers of each
  atom involved in the the compound
• The molecular formula is always a multiple of the
  empirical formula
• To calculate the molecular formula you must have
  2 pieces of info.
• Empirical formula
• Molar mass of the unknown compound (always given)

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Calculating Molecular Formula
Find the molecular formula of a compound that
contains 56.36 g of O and 54.6 g of P. The molar
mass of the compound is 189.5 g/mol.
1st find the Emp. Formula
1 mol O
56.36 g O
3.525mol O
15.99 g O
1.99
PO2
1 mol P
54.6g P
1.763mol P
1
30.97 g P
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Calculating Molecular Formula
Now determine the mass of the empirical formula
PO2 (130.97g P)(215.99g O) 62.95g
MM Given in the problem 189.5 g/mol
189.5 g/mol
3.01
P3O6
Molecular Formula 3(PO2)
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One More A Good 1
Methyl acetate is a solvent commonly used in some
paints, inks, and adhesives. Determine the
molecular formula for methyl acetate, which has
the following chemical analysis 48.64 C, 8.16
H, and 43.20 O. The Molar Mass of the compound
in question is reported as 74g/mol.
1st determine the empirical formula 2nd determine
the molecular formula
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1 mole C
48.64g C
4.050mol C
1.50
12.01 g C
2.702 mol
1 mole H
8.16g H
8.095 mol H
2.99
1.008 g H
2.702 mol
1 mole O
43.20g O
2.702 mol O
15.99 g O
1.00
2.702 mol
C1.5H3O1
2
C3H6O2
36632
74g
C3H6O2
74g
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Classroom Practice 3
A hydrocarbon is 84.25 carbon and 15.75
hydrogen and has a molecular weight of 114. What
is its molecular formula?
C8H18