Midterm TestTuesday, October 10th (This Tuesday) at 7:30pm

Practice is test on website.

Section A, M.W.F. 8:00 – 8:50 am, Room D634.

90 minute test

Chapters: 1, 2, 7, 8

Review Chapter 1

Units

Significant figures

SI System

Dimensional Analysis

Scientific notation

Prefixes and suffixes

Propagation through addition and multiplication

Precision/Acuracy

Chemical & physical properties

Dalton’s atomic theory of matter

Models of the atom – Rutherford's model

Subatomic particles - protons, neutrons, and electrons

Elemental Forms

Periodic table (groups and periods) – Common properties in each group

Elements - names and symbols

Atomic number and mass number - # of n’s, p’s, & e’s

Isotopes - calculating average atomic mass and percent abundance

Avogadro’s number and the mole

Chapter 2

Chapter 7Properties of waves - wavelength, frequency, amplitude, speed

Electromagnetic spectrum - speed of light

Planck’s equation and Planck’s constant

Wave-particle duality for light , electrons, etc.)

Atomic line spectra: Balmer and Rydberg Series

Ground vs. excited states

Heisenberg uncertainty principle

Bohr and Schrödinger models of the atom

Quantum numbers (n, l, ml)

Shells (n), subshells (s,p,d,f)

Shapes and properties of atomic orbitals –#nodes, # lobes

Chapter 8

Diamagnetic vs. paramagnetic vs. ferromagnetic substances

Electron spin

Pauli exclusion principle

Orbital box diagrams

Electron configurations

Aufbau order and its exceptions.

Predicting ions using electron configurations

Core vs. valence electrons

Effective nuclear charge

Periodic trends (atomic radius, ionization energy, electron affinity, ionic radius)

Chapter 9Bonding and Molecular Structure:

Fundamental ConceptsCountless arrangements of atoms are possible!!

Bonding behaviour:What does it depend on?

How do atoms form bonds?

What kind of bonds exist?

How do we predict bonding behaviour?

Valence ElectronsGiven an electron configuraton one can group the electrons into the core and valence e’s

Ex) C 6 e’s 1s2

core

2s22p2

valence

Atoms can form bonds by sharing or exchanging valence e’s.

The core e’s don’t participate, just like the noble gas is unreactive.

Then main objective for each atom is to achieve a noble gas configuration for its valence electrons

Ex) Ne 1s2

Has a full valence shell with 8 e’s C needs 4 more e’s to achieve a full shell. How?

2s22p6

Electron Dot Structures (G. N. Lewis)

The first row elements will either want to achieve He or Ne configurations.

Those who attain Ne configurations need 8 e’s in their valence shell

The number of valence electrons can neatly be depicted by arranging them around the atomic symbol

Ex)

•

C •••

C has 4 valence e’s Ex) N has 5 valence e’s

N •••

• •

Electron Dot Structures

Electrons placed around the four sides of the atom symbol individually or in pairs depending on the # of valence e’s.

Initially on e’s is placed on each side and are paired up when all four sides are occupied, after which they are paired until the valence shell is full.

Octet RuleNotice that all noble gas configurations have an outer shell with ns2.. . np6.

Ne = 2s22p6

Ar = 3s23p6

Kr = 4s23d104p6

Xe = 5s24d105p6

Ignoring the d and f e’s the valence shell contains 8 e’s.

Rn = 6s25d104f146p6

The d and f subshell contributions can be ignored if they are full

Therefore the valence e’s in groups 13(3A) to 18(8A) can be described in terms of ns and np e’s

In each case atoms want to have 8 electrons hence the

Octet rule.

An element can achieve octet status in one of three ways:

2. It can lose valence electrons to make an cation:

3. It can share valence electrons with another atom to make a covalent bond. This will generally involve two atoms of similar electronegativity (ability to attract electrons)

1. It can gain valence electrons to make an anion:

Ex) Cl [Ne]3s23p5Cl- [Ne]3s23p6=[Ar]

Octet Rule

Ex) K [Ar]4s1 K+ [Ar]

Ex) Cl2

•• •

Cl••

• •

•••

••Cl• •

+ Cl •••

• •

• •

•••Cl

• • ••

Nonmetals normally gain electrons to obtain a complete octet.

Metals normally lose electrons to obtain a complete octet.

Atomic charge = # protons – # electrons

Elements in groups 1, 2 and 3, ionic charge is group #:

Elements in groups 15, 16 and 17, ionic charge is group # -18:

Ex) Cl to Cl- O to O-2 S to S2-

Octet Rule

Ex) Na to Na+ Ca to Ca2+ Al to Al3+

P to P3-

Ex) Al3+ has 13 protons and 10 electrons q = 13 – 10 = +3

Ex) Al: 3(A) → q = +3 Ca: 2(A) → q = +2 Li: 1(A) → q = +1

Ex) As: 15 → q = -3 S: 16 → q = -2 F: 17 → q = -1

Ionic CompoundsRecall that opposite charges attract each other.

A cation and an anion, experience an electrostatic force, given by:

This force pulls them together to make an ionic bond.

1 22

q eq eF k

r

Na+ Cl-

rq1 q2

Cl

r-

Na+

e = unit of charge

k = 8.988 × 109 N m2/C2

Coulonb’s Constant

The energy released to form an ionic bond:

1 2q eq eE k

r

Exercise

Compute the force that a sodium cation and chloride anion experience when 10.00 nm apart

F = ke2q1q2/r2 q1 = +1 q2 = -1

r = 10 nm = 1.000*10-8 m

F = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1) (1.000*10-8 m)2

F = -2.307*10-12 N

Compute the energy of formation of the ionic bond in NaCl, where the bond length is 279 pm

E = ke2q1q2/r q1 = +1 q2 = -1

r = 279 pm = 2.79*10-10 m

E = (8.988 × 109 N m2/C2)(1.6022 *10-19 C)2(1)(-1) (2.79*10-10 m)

E = -8.27*10-19 N m = J

How energy for one mole of NaCl bonds?

E(total) = E(bond) *(# of bonds)

= (-8.27*10-19 J/bonds)(6.022*1023 bonds/mol) = 498,000 J/mol = 498 kJ/mol

LiF formationLi loses all e’s in its valence shellLarge reduction in atomic radius:Li – 152 pm Li+ - 78 pm Vol = 1/7th

F gains electron its valence shellExpansion in the radius because effective Z* is decreased due to extra electron.F – 71 pm F- - 133 pm Vol = 6x

Ionization Energy consumedAs an atomic gas.

Electron Affinity Energy releasedAs an atomic gas.

LiF formation

Energy is released to form the ionic bond

NaFNa

FNaF

FNaF

NaFNa

NaFNa

FNaF

Ionic MaterialsLattice - 3-D pattern of ions

- Minimize repulsive forces-Maximize attractive forces

- Large lattice energy released

- Brittle

- No net charge

- High melting point

Formation of ionic materials

Whether element from ionic compounds depends on the balance between:

1) Ionization energy2) Electron affinity 3) Lattice energy4) Phase transition energies5) Bond energies

Dissociation energy

Electron Affinity

Ionizationenergy Lattice

energy

Formation energy

vaporization

Covalent CompoundsCovalent bond - valence electrons are shared between two or more atoms. i.e. bonding electrons are “co-valent”.

Electrons are shared to complete the valence shell of each atom

The combination of electrons forms one entity.

In Ionic systems - the cation and anion are separate entities

The wavefunctions for each electron constructively interfere and form a combined wavefunction, called a molecular orbital

1s(1) 1s(2) 1

Atom 1 Atom 2

Molecule

Nucleus 1 Nucleus 2

The lowest energy point is at the average bond length.

Bond dissociation energy -Energy is required to break the bond.

H H H + H Edissociation = +435 kJ/mol. .

Covalent bonds form if energy is released when atoms bond:

H + H H H E = -435 kJ/mol. .The negative energy of reaction means that the product (H2) is more stable than the reactants (2 × H).

2 electrons are shared completing the 1s orbital for each H

Covalent Compounds

Lewis structure diagramsThe exact solutions of molecular quantum mechanics are highly complex

These Lewis diagrams reflect the underlying quantum mechanics and serve primarily as a bookkeeping device for the valence electrons in the moleculeThe guiding principle behinds Lewis diagrams is that each atom in molecules

achieve noble gas electron configuration by sharing electrons

This is known as the octet rule because the majority of noble gases have 8 valence e-, except for H which only requires two e’s to complete its valence.

The electrons of a chemical bond are represented by a dash

There can also be non-bonding electrons, which are written as dots ●●

The Lewis dot diagrams of atoms can be combined to depict bonding in molecules

•• •

Cl••• •

•••

••Cl• •

+ Cl •••

• •

• •

•••Cl

• • ••

Cl••

• •

• •

••

••Cl• • •

•

• • ••Cl••

• •

••Cl• •-

Steps for Drawing Lewis Electron Dot Structures

1. Determine the central atom. The rest are terminal atoms.

2. Determine the total number of valence electrons.

3. Use one pair of electrons to make a single bond between each pair of bonded atoms.

4. Use any remaining electrons as lone pairs around each terminal atom (except H) so that each terminal atom has a complete octet, if possible.

5. Allow for any deficit or excess of e’s only for the central atom.

6. Check the central atom for too few or too many electrons.

7. If there are too few electrons, increase the bond order of one or more bonds by sharing non-bonded electrons. (i.e. make double or triple bonds, as necessary)

8. Calculate formal charge for all atoms, and indicate any which are not zero.

Ex) CF4 •

C •••

•• •F••

• •

Central Terminal4 v.e’s 7 v.e’s

•C •

• ••

• •

F••

• •

• •F••

•

••

• •F••

•

••

• •F •••

• •C

• •F•• ••

• •F•• ••

• •F ••

• •

••

F••

••

When finished review it to verify that correct number of atoms and electrons were used and that the octet rule is obeyed

Remember that Lewis dot structures do not show a molecule’s shape.

If there is more than one acceptable solution, the true electron distribution is a hybrid of the possible distributions. This is called resonance.

If it is impossible to avoid having one atom with too few or too many electrons, make sure it is the central atom. Elements in the 1st or 2nd period can never have more than eight electrons under any circumstance.

Molecules with odd numbers of electrons form free radicals and cannot fully obey the rules.

Tips

The central atom can be generally be chosen using a few rules:

1. The central atom is never H or F.

2. If you have many atoms of one element and one atom of another, the lone atom is the central one.

3. Given a choice, the central atom is not O.

4. Given a choice, C is central.

Larger molecules are treated as a sequence of central atom problems.

Tips

Ex) Choose the central atom for the following molecules: (a) BBr3 (b) CH3Cl (c) CH2O (d) POCl3

Draw Lewis dot structures for the following molecules

(a) NH3 (b) C2H6 (c) C2H4 (d) C2H2

N •••

• •

H •

H•

H•

H

H

HN -• •

-

•C •

••

•C •

••

H•

H•

H•

H•

H•

H •

C CHH

H H

HH

a) b)

Draw Lewis dot structures for the following molecules

(a) NH3 (b) C2H6 (c) C2H4 (d) C2H2

•C •

••

•C •

••

H•H •c) d)•

C ••

•

•C •

••

H•

H•

H • H•

H•

H•

H • H•C •• •

•C• •

•

•

C CH

H H

HC CH H

C •

• •

•

C• • •

•

H•H •

Ex) ClO-

•• •Cl••

• •]q[

#p = 17

#e = 18

qCl = #p - #e = 17 – 18 = -1

10 core 8 valence

#p = 8

#e = 9

qO = #p -#e = 8 – 10 = -2

2 core8 valence

Total charge = -3 ????

We have miscounted the e’s

The bonding e’s have been double counted!!!

The bonding e’s are shared and therefore count only as half an electron for each atom

#e = core e’s + ½(bond e’s) + non-bond e’s

# e’s Cl = 10 + ½ (2) + 6 = 17

qCl = 17 – 17 = 0

# e’s O = 2+ ½ (2) + 6 = 9

qCl = 8 – 9 = -1

• •O•• •

••

Therefore the –ve charge is on O

It needs it to complete its valence

Extra e added to complete Valence on O.

Formal ChargeFormal charge (Qf) is the charge on an atom assuming that every bond is completely covalent.

As a general rule, we want to keep the formal charge on each atom as close to 0 as possible (without giving any atom more than a complete octet).

All bonding electrons are shared equally

Qf = #p’s – [ (# core e’s) + ½ (# bonded e’s) + (# non-bonded e’s)]

Ex) BF3 B •••

•• •F••

• •

• •F••

•

••

• •F •••

• • Qf(B) = 3 –[½ (6) + 0] = 0

Qf(F) = 7 –[½ (2) + 6] = 0

Note: valence of B is incomplete!!

Qf = (# valence e’s) – [ ½ (# bonded e’s) + (# non-bonded e’s)]

•• •F••

• • •• •F•• ••

• •F •••

• • Qf(B) = 3 –[ ½ (8) + 0] = -1

Qf(F) = 7 –[ ½ (2) + 6] = 0

Let’s try another arrangement taking an extra set of bonding electrons from one of the F’s

or Qf(F) = 7 –[ ½ (4) + 4] = +1

Total q = -2 + 0 + 1 + 1 = 0

Large charge separation!!!

Therefore the previous arrangement is preferred

B --• •F ••• •

• •F•• ••

• •F••

• •

0

0

00

+1

00

]0

B•

••

-1

Consider the anion NO3-.

N •••

• • • •O•• •

•

•

O•

•• ••

• •O• •

•

• •0

0 -1

0Incomplete valence on two O’s

Lets try to fill them using the remaining 2 e’s on N

•• N •

• •

• •O•

• •

•

• •

O

•• ••

• •O• •

•• •

0 -1

0

0

Charge separation is the sameAs aboveTotal charge is -1

All valences are complete

N

O

•• ••

• •O• •

• •O• •

••

]-

• •O• •

•

N

O

•• ••

• •O• •

•

]-

This arrangement is also correct? This one too!!!

• •N

O

•• ••

• •O• •

O• •

• •]-

Resonance

When there are several permutations that are equivalently correct, the actual situation is an average between them.

This phenomenon is known are resonance, which can be depicted using bidirectional arrows

O = N = O

O

-1

O = N _ O

O

-1O _ N = O

O

-1

C

C

C

C

C

C

Ex) O3

Ex) Benzene, C6H6

+1 -1

O _ O = O:..

...... .... ..

O = O _ O:.. ....+1-1

:O _ O _ O:.. .. .... .. ..-1 -1

X

+2X O = O =O.. .... ..

C

C

C

C

C

C

.

..

.

.

. C

C

C

C

C

C

?

Bond polarity and electronegativity

F-Li+ F F

In a symmetric molecule such as H2, the concept of covalence is unambiguous: the electrons are evenly shared by the two atoms.

But in LiF, we have complete electron transfer: Li+ and :F-, i.e. ionic bonding

There is a continuum of behaviour between these two extremes

Ionic bonds are completely polarized towards the opposite charged ions.

As the electronegativity difference decreases, e’s are more likely to be shared, but unequally: polar covalent bonds

In polar covalent bonds, there is a bond dipole, which is indicated by a vector

There are also partial positive and negative charges, indicated by δ+ and δ-

d

Pauling Electronegativity

General trend in element electronegativity

We use Pauling electronegativity values to determine bond polarity.

Electronegativity is the ability of an atom in a molecule to attract electrons to itself.

A difference in Pauling electronegativity between elements of more than 2 units is enough to cause ionic bonding

Evaluating Lewis diagrams - Revisited

How do you chose between several possible Lewis diagrams?

1. Most important is achieving the Octet rule: any structure that obeys the octet rule is better than any structure that does not

2. Any structure that minimizes the sum of the absolute values of the Formal Charge is better

3. The diagram that associates negative Formal Charge with more electronegative elements and positive Formal Charge with electronegative elements is to be preferred over others

Let us now apply this rule to some more complex cases: 1) OCN- (cyanate) 2) CO2 (carbon dioxide) 3) H2CSO (sulfine)

..O _ C N:.. : O = C = O

.. ..

.. ..-1

..

..H _C = S _ O:..H +1 -1

..

..H _C = S _ O:..H

..H _C _ S = O..H

.. ..

..

..O C _ O: :

..O _ C O..: ::: O _ C _ O

.. ..

.. ......

Let us now apply this rule to some more complex cases: 1) OCN- (cyanate) 2) CO2 (carbon dioxide) 3) H2CSO (sulfine)

..O _ C N:.. :

O = C = O.. .... ..

-1

+1 -1

....

O C _ N: :..O = C = N.. ..

.. -1

+1 -1-1-1 -2

-1

-1

+1

-1

ExerciseState whether the bond is ionic or covalent and draw the dipole vectors and partial charges for:

H - Br O - C

O - S

H - F

F - IC - Br

2.1 2.8

+ -

C 3.5 2.5

- +

C 2.1 4.0

+ -

C

K - F

0.8 4.0

+ -

I

O - Na

2.5 2.8

+ -

C 4.0 2.8

+

C 3.5 0.9

- +

I 3.5 2.5

- +

C

%-ionic character

Continuum in behaviour between pure covalent and ionic character

Molecules such as H2, F2 are at 0 on this scale

Common “ionic” compounds such as NaCl cover a wide range on this scale but are all over 2

The formal charge separation gives rise to ionic forces of attraction contributing to the bond energy.

Therefore all bonding must be considered as having ionic and covalent contributions.

Partial ChargeThe bond dipole can be quantified by calculating the partial charge (Q) on each atom.

Xvalence nonbonding bonding

X Y

Q (X) e e e

Partial charge, similar to formal charge, except that the electronegativities of the atoms, govern the distribution of the bonding electrons between the atoms.

valence nonbonding bonding

1

2Q (X) e e e

f

Recall the formula for formal charge

Therefore, pure covalency assume electronegativities are the same

Ex) H-Cl Q(Cl) = 7 – [6 + {3.0/(2.1+3.0)}*2] = -0.2

Q(H) = 1 – [0+ {2.1/(2.1+3.0)}*2] = 0.2

Bond order: length and energy

We have seen that chemical bonds between the same pairs of elements may be single, double or triple bonds

Quadrupole bonds also exist between some of the d-elements

How do these bonds differ?

The higher the bond order the larger the bond dissociation energies

The higher the bond order, the shorter will be the bond will be.

C—C C—N C—O N—O

C—Si C—P C—S

Si—Si Si—P Si—S P—S

Predicting Bond Lengths

154 147 143 136

194 187 181

234 227 221 214

Bond lengths shows the same trends as atoms size

Size decreases from left to right across the Periodic Table, and increases down any group

Increasing the bond order always shortens the bonds; however the %-shortening is not a very regular parameter, therefore not simply predicted.

Lewis diagrams and Molecular Shape

Four electron pairs define the tetrahedral shape family, as in SiCl4

There is a direct link between Lewis diagrams and molecular shape

The Valence Shell Electron Pair Repulsion theory states a molecule’s shape is determined by the electron pairs that surround central atoms

These electron pairs (EP) include both bond pairs (BP) and lone pairs (LP)

We define five shape families based on the need to accommodate mutually repelling electron pairs around a central atom

The balloons illustrate these “natural” shapes

The 5 Shape Families

The five shape families that electron pairs develop are illustrated above

Linear – Trigonal-planar – Tetrahedral – Trigonal-bipyramidal – Octahedral

In this case each molecule has all its electron pairs as bond pairs around the central atom

Not all the shape-determining electrons pairs need be bond pairs

General Shape Family Scheme (1)

0 LP 2 BP

1 LP 1 BP X X..

X....2 EP’s 3 EP’s

0 LP 3 BP 1 LP 2 BP 2 LP 1 BP

X

..X

Lone pairs and shape: 4 Electron Pairs

Linear

Hydrogen Fluoride, HF1 bond pair3 lone pairs

When there are four electron pairs, the shape family is tetrahedral

Four possibilities exist: with 0, 1, 2 or 3 lone pairs

0 LP – tetrahedral shape

1 LP – trigonal-pyramidal shape

2 LP – bent shape

3 LP – linear shape

General Shape Family Scheme

X X: X:......X:

4 BP 0 LP 3 BP 1 LP 2 BP 2 LP 1 BP 3 LP

General Shape Family Scheme

5 BP 0 LP 3 BP 2 LP 2 BP 3 LP 1 BP 4 LP

X X..

4 BP 1 LP

X.. .. ..

X .... X

..

..

....

Special features of 5 EP shapesIn the trigonal bipyramidal geometry, alone among the 5 shape families, the sites are not identical

There are three equivalent sites which are identical to those in trigonal-planar, with bond angles of 120 . These are called the equatorial sites

Axial Sites - above and below the plane in a mutually linear relationship.

Note - the angles are different

Note - if there are LP’s, these will always first occupy equatorial sites

General Shape Family Scheme

X..X X

.... ..

X.. ..

..X

........X

......

..

6 BP 0 LP 5 BP 1 LP 4 BP 2 LP 3 BP 3 LP 2 BP 4 LP 1 BP 5 LP

Some special features of VSEPRmultiple bonds, are found to occupy almost the same volume of space

Hence, multiple bonds can be considered as if they are single BP of electrons

Ex) NO2

The N has 4 EP’s - 1 LP and 2 BP’s

LP are considered to be larger than BP because constrained by only a single positively charged nucleus as opposed to two for a BP

For 5 EP since neighboring LP’s at 120 experience much less repulsion than at 90o they tend to cluster on the equatorial sites of the trigonal-bipyramidal geometry

bent shape

For 6 EP case LP prefer to be opposite since otherwise at 90o they would experience strong repulsive forces

Geometry of large molecules

VSEPR method extends to larger molecules by treating them as a chains central atoms

The only thing that remains is to twist the bits of the molecule about any single bonds to minimize congestion between the bonded atoms

The geometry at the S, O and N are similar to those in water and ammonia, i.e. bent and trigonal pyramidal

The geometry at the C1 and C2 atoms are tetrahedral, while at C3 it is trigonal planar

The complete molecule simply twists the bits and pieces about any given single bond

Shape and polarity

Recall that most covalent bonds are polar

Symmetry in a molecule can cause bond dipole vectors to cancel each other making the molecule non-polar even though its individual bonds are very polar

Therefore molecules must be polar Not always!!!

Permanent Dipole MomentsMolecules that have permanent dipole moment are called polar

The unit of dipole moment is the Debye

Note that for molecules like H2 and CCl4, there is no permanent dipole moment!

Such molecules are non-polar.

Measuring dipole momentsMolecules which have permanent dipole moments respond to an applied electric field

This is an arrangement similar to that of a capacitor

The dipole moments in the molecule tend to align with the applied electric field

By measuring the degree of alignment as a function of applied field strength, the size of the dipole moment in Debye units can be determined

Concepts from Chapter 9

DRAWING LEWIS ELECTRON DOT DIAGRAMS

Octet rule

Resonance structures

Bond polarity (ionic, polar covalent and covalent bonds)

Ionic vs. covalent compounds

Electronegativity

Dipole vectors

Calculating formal charges and partial charges

Bond order

Bond lengths

VSEPR and predicting shapes of molecules

Polarity of molecules