Unit 1 Packet - dlongchem.wikispaces.comdlongchem.wikispaces.com/file/view/Unit+1+Packet.pdf · Unit 1 Packet: Background, Lewis ... VSEPR, Hybridization, Isomerism, Acids and Bases

Name: _______________________________

Unit 1 Packet: Background, Lewis Structures, Resonance,

Formal Charge, VSEPR, Hybridization,

Isomerism, Acids and Bases

Honors Organic Chemistry

2

Key Terms For Unit 1 General

Organic Chemistry

Orbital

Orbital Notation

Isomerism

Structural Isomerism

Geometric Isomerism

Bonding

Anion

Aufbau Principle

Bond Angle

Bond Length

Cation

Covalent Bond

Double Bond

Electronegativity

Formal Charge

Full Octet

Hindered Rotation

Hund’s Rule

Hybridization

Ionic Bond

Line Angle

Non-polar covalent bond

Octet Rule

Pauli exclusion Principle

Polar covalent bond

Pi bond (π)

Resonance hybrid

Sigma bond (σ)

sp hybrid orbital

sp2 hybrid orbital

sp3 hybrid orbital

Tetrahedron

Trigonal Planar

Valence Electrons

Valence Shell

Valence Shell Electron-

Pair Repulsion

Nomenclature

Alcohol

Aldehyde

Carbonyl group

Carboxyl group

Carboxylic Acid

Ether

Functional Group

Ketone

Acids and Bases

Ka

pKa

pH

Strong Acid

Weak Acid

Conjugate Acid/Base Pair

Bronsted-Lowry Acid

Bronsted-Lowry Base

Lewis Acid

Lewis Base

Nucleophile

Electrophile


3

What is Organic Chemistry?

Organic Chemistry:

What makes carbon so special?

What other elements are commonly involved?

Important Dicsovery: Friedrich Wohler (1828) – Accidental synthesis of Urea ( ____________ )

from ammonium cyanate (___________ ).

Atomic Structure Review

Any atom is made up of three subatomic particles: __________, __________ and __________.

Two atoms of the same element with different numbers of ___________ are called __________.

An atom of an element with different numbers of __________ and __________ are called _____.

Anions are _______________ charged ions. Ex: ________

Cations are _______________ charged ions. Ex: ________

In Chemistry, the subatomic particle that we are most concerned with is the ___________, and

particularly, the ____________ in the outermost, or _____________ shell. These are the

particles that allow different atoms to ________ with one another.

Electrons

Electrons are located in…

Principle Energy Levels:

Sublevels:

Orbitals:

Electrons can be represented visually using

orbital notation


4

Ground State Orbital Notation

Orbital Notation is a visual method to represent the ________ ________, ___________,

_________ and ____ of each electron in an atom or ion. The ground state of an atom represents

its lowest energy, preferred configuration.

The placing of electrons into an orbital notation is governed by three rules:

Aufbau Principle:

Hund’s Rule:

Pauli Exclusion Principle:

To practice using orbital notation, consider the ground state configurations of carbon and

bromine:

Carbon, Z = 6 Bromine, Z = 35

Lewis Dot Diagrams

Since we are primarily concerned with the __________ _________ in an atom, a scientist named

G. N. Lewis devised another method of representing an atom that is commonly used called a

Lewis Dot Diagram. Let’s consider how these are drawn for several elements.

H C P Xe I

**We can piece these diagrams together to make

Lewis structures for molecules. Ex: H2O, water**


5

Lewis Structures

History:

Purpose:

Process:

CH4 COCl2 NO3-

1. Predict the Arrangement of

the atoms:

2. Count up the valence

electrons

3. Connect the surrounding

atoms to the central atom with

single bond (one shared pair)

4. Determine how many

electron pairs you have left

5. Place lone pairs around

terminal atoms to satisfy octet

(stick any leftovers on central

atom)

6. If the central atom has not

achieved an octet, form

multiple bonds to do so


6

Lewis Structures – Additional Practice

CO

NH3

CH2O

CO32-

SOCl2

C2H4

Resonance Structures

Often, it is possible to draw more that one legal Lewis structure for a molecule.

e.g. NO3-

Each of these structures fulfills the requirements of a __________ ____________, therefore, they

are each valid resonance structures. No single one of these resonance structures gives a

_________ picture of what the ion looks like. A resonance structure is a Lewis structure that

contributes to the overall resonance _________ of a molecule or polyatomic ion.

Another example of resonance, the sulfate ion (SO42-

), involves a concept called expanded octet.

e.g. SO42-


7

Resonance Hybrids – What does a molecule really look like?

Similar to restaurant Diet Coke mixtures, not all of the resonance structures that can be drawn for

a sulfate ion are created equal. How do we decide which ones contribute most to the actual

picture of what a molecule looks like? To do this, we have to minimize the formal charge of

each atom.

e. g. CO2

By minimizing the formal charge on each atom, we can isolate the resonance structure that best

represents what a molecule or ion looks like.

Try some others:

C2O42-

SO32-

Evidence for Resonance – Benzene

General Guidelines for Resonance Structures

1. Try to draw structures that are as low in energy as possible

2. The best structures tend to have the maximum number of bonds and the most octets

3. When structures are equivalent in terms of bonds and octets, minimize formal charge

to find the more stable structure

4. All structures must be valid. Only electrons may be moved to change between

structures, bonding sequence of atoms must remain the same.

5. Use curved arrows to show the movement of electrons. Only move lone pairs and

multiple bonds.

6. Separate resonance structures by a double headed arrow.

7. Resonance stabilization is very important when it delocalizes or spreads a charge over

two or more atoms

8. Negative formal charges are more stable on atoms with higher electronegativities.

A commonly cited example of evidence for the

resonance theory is the molecule benzene (C6H6). In a

Lewis diagram of benzene, the carbons can be

connected by either a single or double covalent bond.

The known length of a C=C double bond is 133 pm, the

known length of a C-C single bond is 154 pm, but the

bond length of every carbon-carbon bond in benzene is

139 pm, a value in between that of a single and double.


8

Worksheet – Resonance and Formal Charge

1. For each of the following compounds, draw the important resonance structures.

Indicate which structures are major and minor contributors or whether they have the

same energy

.. (-)

H :O: H

| | .. (-) |

a. H – C = N – O:

(+) ..

:O:

|| ..

b. H – C – NH2

.. :O – H

| c. H – C – H

(+)

H

| .. .. d. H – C – N = N:

(+) (-)

e. H – C – C ≡ N:

..

(-)

:O:

.. || .. f. H2C – C – O – CH3

..

(-)

:O: :O:

|| .. || g. H – C – C – C – H

| H

(-)

2. Draw the important resonance structures for the following ions

(+) (-)

a. H2C = CH – CH2 b. H2C = CH – CH2


9

Summary of Bonding Patterns

Atom Valence

Electrons

Positively

Charged Neutral

Negatively

Charged

B

C

N

O

Halogen

Important Note: The bonds shown do not always have to be single bonds

Ex:


10

Line-Angle Structures

Often times when we are drawing organic molecules, a large number of atoms are involved and

drawing standard Lewis structures can become somewhat hairy.

e.g. 2-methylpropane (aka isobutane)

A remedy for this is to represent molecules in what is called line-angle format.

There are several important rules to remember when drawing line-angle structures:

1. Every angle and every end of a line that does not have an element symbol represents

a carbon atom.

2. All non-carbon atoms (except hydrogens attached to a carbon) are shown with their

chemical symbol

3. Hydrogen atoms attached to carbon atoms are disregarded.

4. When four lines are drawn from an intersection, an attempt should be made to show

3-D perspective using dashes and wedges.

e.g. 5,5-dimethylhex-3-en-1-ol

becomes

2,4,6-trinitrotoluene

becomes

4-bromo-3-ethylpent-1-yne

becomes


11

Worksheet – Line-Angle Structures

For the following structures, if the standard structure is shown, draw the line-angle equivalent. If

the line angle structure is shown, draw the standard equivalent.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


12

VSEPR Theory

VSEPR:

VSEPR theory is used to predict the 3-D geometry of the terminal atoms around the central atom

of a molecule. VSEPR theory tells us the shape of the molecule and the bond angles to expect.

How to determine the 3-D geometry of a molecule from a Lewis Structure

1. Draw a Lewis structure for a molecule or polyatomic ion

2. Count the regions of electron density around the central atom

a. Any bond (single, double, or triple) counts as one region

b. A lone pair counts as a region

c. The number or regions tells you the “parent shape” or electron arrangement of the

structure

3. If there are any lone pairs around the central atom, the actual shape will be based on the

parent shape, but differ from it in practice.

“Parent

Shape”

Regions of

Electron

Density

Example

Molecule w/ Lewis

Structure

Actual

Shape

Bond

Angles

Drawing Hybridization

of Central

Atom

CO2

BF3

CH4

NH3

H2O


13

Parent

Shape

Linear

Trigonal

Planar

Tetrahedron

Trigonal

Bipyramid

Octahedron

Pentagonal

Bipyramid


14

Hybridization

Consider the molecule methane, CH4. One carbon atom bonded to four different hydrogen

atoms. Their orbital notations are shown below.

C: ___ ___ ___ ___ ___ H: ___

1s 2s 2p 1s

Hund’s Rule:

Pauli Exclusion Principle:

We know that single bonds are formed by the overlap of two orbitals that each have one unpaired

electron – to be able to form 4 bonds to carbon, one electron will have to be promoted from the

2s to the empty 2p orbital.

C: ___ ___ ___ ___ ___

1s 2s 2p

At present, of the four bonds that carbon is making to hydrogen, there are three “1s overlapping

2p” bonds and one “1s overlapping 2s” bond. From your knowledge of s and p orbitals, draw a

diagram of what this molecule would look like.

There are a couple of problems with the diagram above.

First, experimental evidence shows us that the bond angles for the above molecule are

incorrect – all 4 hydrogen atoms should be equivalently spaced 109.5° from one

another.

Second, all experimental evidence tells us that the four bonds in methane are equal in

energy.

To take into consideration these two problems, a bonding theory called hybridization was

devised

Hybridization:


15

Hybridization Continued

sp hybrids: produced from one “s” and one “p” orbital. Two “p” orbitals remain unchanged and

available for π bonding.

Ex: Beryllium in BeCl2 – two sp hybrid orbitals form two σ bonds. Drawing

___ ___ ___ ___ ___ ___ ___ ___

2p Electron 2p Hybridization 2p

___ Promotion ___ ___ ___

2s 2s sp

Ex: Carbon in C2H2 – two sp hybrid orbitals form two σ bonds. Two “p” orbitals remain available for π

bonding. Drawing

___ ___ ___ ___ ___ ___ ___ ___


___ Promotion ___ ___ ___

2s 2s sp

sp2 hybrids: produced from one “s” and two “p” orbitals. One “p” orbital remains unchanged and

available for π bonding.

Ex: Boron in BF3 – three sp2 hybrid orbitals form three σ bonds. Drawing

___ ___ ___ ___ ___ ___ ___


___ Promotion ___ ___ ___ ___

2s 2s sp2

Ex: Carbon in C2H4 – three sp2 hybrid orbitals form two σ bonds. One “p” orbital remains available for π

bonding. Drawing

___ ___ ___ ___ ___ ___ ___


___ Promotion ___ ___ ___ ___

2s 2s sp2

sp3 hybrids: produced from one “s” and three “p” orbitals.

Ex: Carbon in CH4 – four sp3 hybrid orbitals form four σ bonds. Drawing

___ ___ ___ ___ ___ ___

2p Electron 2p Hybridization

___ Promotion ___ ___ ___ ___ ___

2s 2s sp3

Ex: Oxygen in H2O – two sp3 hybrid orbitals with unpaired electrons will form two σ bonds. The two sets

of paired electrons exist as lone pairs. Drawing

___ ___ ___ There is no electron

2p Electron promotion because Hybridization

___ Promotion all orbitals are occupied ___ ___ ___ ___

2s sp3


16

Orbital Diagrams for Unhybridized s, p and d Orbitals


17

Representing Molecules with 3-D Perspective: The Orbital Diagram

When we first learned to draw Lewis structures, we typically drew bond angles that were evenly

spaced in 2-D.

e.g. CH4

Upon the introduction of VSEPR theory, it became clear that the original manner of

representation did not give an accurate reflection of the 3-D shape of most molecules, so we

began drawing molecules with perspective.

e.g. CH4

As we continue to draw more and more complex molecules in 3-D, such as those that involve pi

bonds, their drawings become more and more intricate. Let’s consider a few.

Ethane, C2H6

Ethene, C2H4

Ethyne, C2H2

**Challenge** Allene, H2C=C=CH2


18

Worksheet: Hybridization and Lewis Structures

1. For each of the following, predict the hybridization and bond angles for the indicated

atoms.

a) :O:

.. || ..

H – O – C – O - H

˙˙ ˙˙

b) H H

| | ..

H – C – C – O : (-)

| | ˙˙

H H

c) CH3

| (+)

CH3 – N – CH3

|

CH3

d) H H

| |

H – C – C (+)

| |

H H

e) H

.. | ..

:Cl – C = C – C = N – H

˙˙ | |

H H

f)

H – C ≡ C – H

a. Hybridization: _____________ Bond Angle: _______

b. Hybridization: _____________ Bond Angle: _______

c. Hybridization: _____________ Bond Angle: _______

d. Hybridization: _____________ Bond Angle: _______

e.

i. Carbon - Hybridization: _____________ Bond Angle: _______

ii. Nitrogen - Hybridization: _____________ Bond Angle: _______

f. Hybridization: _____________ Bond Angle: _______

2. From question #1, which structures are ions?

3. What type of orbitals are overlapping between the atoms in the following?

a)

H

| ..

H – C = O:

b)

H H

| |

H – C = C = C – H

c)

H

| ..

H – C – O:

| |

H H


19

4. Draw Lewis Structures for the following. Remember, hydrogen and terminal

halogens do not make double bonds.

a. CO2

b. CCl4

c. NH4+

d. N2H4

e. NH2-

f. CO3-2

g. NO2+

h. BF3

5. State the hybridization of each central atom from Question #4

a. ______

b. ______

c. ______

d. ______

e. ______

f. ______

g. ______

h. ______

..

6. Use wedges and dashes for σ bonds to draw an orbital diagram of CH3 – C – O:

Draw any “p” orbitals, label bond angles. || |

:O: H


20

Worksheet – Bonding and Hybridization

1. For each of the following compounds:

a. Give the hybridization for each atom except hydrogen

b. Give the approximate bond angles for each atom except hydrogen

c. Draw an orbital diagram using lines, wedges and dashed lines for sigma bonds.

Draw the “p” orbital interaction for pi bonds

..I. H3O

+

II. (CH3)4N+

III. CH3 – C = N – H

| H

IV. CH2O

2. For each of the following:

a. Draw the Lewis structure

b. Indicate what type of orbitals are overlapping to form each bond

c. Give approximate bond angles for each atom except hydrogen

..I. CH3 – C ≡ C – C = O

| ˙˙ H

.. II. H2N – CH2 – CN:

III. (CH3)2NH

˙˙

IV. CH3 – CH = C(CH3)2


21

3. Predict the hybridization and geometry of each carbon atom in the following anion:

:O: (-)

|| .. CH3 – C – CH2

4. Draw orbital diagrams of the pi bonding in the following compounds. Use lines,

dashes and wedges to show sigma bonds

a. CH3 – C – CH3

|| :O:

.. b. CH3 – C ≡ C – C = O

| ˙˙ H

c. CH3 – CH = CH – CH2CH3 (you may realize there are two ways to do this one,

circle the six coplanar atoms in this molecule)


22

Isomerism

In the simplest terms, isomers are compounds with the same ______________ ____________

but a different arrangement of atoms. We will investigate this topic in greater detail in the

future, but for now, it will suffice to introduce a few categories of isomerism and give examples

of each.

1. Structural Isomers – isomers that have different ______-___-_______ bonding

Example: C5H12

2. Cis-Trans Isomers (aka Geometric) – involves either a double bond or a ring;

isomers have the same ______-___-______ bonding, but a different spatial

arrangement about the double bond or ring (because of ___________ rotation)

Example: but-2-ene, CH3-CH=CH-CH3

Extra Practice: Draw all of the isomers of C6H14


23

Worksheet – Bonding Part 2

1. 2-pentyne has the formula CH3 – C ≡ C – CH2CH3. Use dashed lines and wedges to

draw a 3-D diagram of this molecule. Draw p orbitals as clouds. Circle the four

atoms that are in a straight line.

2. Which of the following show cis-trans isomerism? Draw the cis and trans isomers of

the ones that do.

a. CH2=C(CH3)2

b. CH3CH=CHCH3

c. CH3C≡CCH3

d.

e. CH3CH = C – CH2CH3

| CH2CH3


24

Worksheet – Bonding Part 2 (cont.)

3. State the relationships between the following pairs of structures. Your choices are:

identical compound, cis-trans isomers, structural isomers, totally different molecules.

a. CH3CH2CH2CH3 and (CH3)3CH

b. CH2 = CH – CH2Cl and CHCl = CHCH3

c. CH3 CH3 CH3

\ / and \

CH = CH CH = CH

\

CH3

d. CH3 CH3 CH2

\ / and ||

CH = CH CH3 – C – CH3

e. and

f. and

g. and


25

Acids and Bases – Bronsted-Lowry Definition

Bronsted-Lowry Acid:

Bronsted-Lowry Base:

Example reaction:

H :O: H H :O:

| || .. .. | | || .. (-)

H – C – C – O – H + H – N – H H – N – H + H – C – C – O:

| ˙˙ | | (+) | ˙˙

H H H H

Bronsted-Lowry Bronsted-Lowry Conjugate Conjugate

Acid Base Acid Base

The reaction of Bronsted-Lowry Acid and a Bronsted-Lowry Base produces a new acid and a

new base. To distinguish the products from the reactants, they are referred to as the conjugate

acid and conjugate base.

Acids and Bases – Lewis Definition

Lewis Acid:

Lewis Base:

Example reaction:

F H F H

| | (-) | | (+)

F – B + :N – H F – B – N – H

| | | |

F H F H

Lewis Acid Lewis Base

The electron pair involved in a Lewis Acid-Base reaction is shared to form a new covalent bond.

Ex: Using curved arrows, show how acetaldehyde, CH3CHO, can act as a Lewis base.

:O:

|| + H – A

CH3 – C – H


26

Equilibrium

Any acid and base reaction is represented by an equilibrium. An equilibrium is a representation

of a reaction that takes into account both the forward and reverse reactions present in a system.

Both forward and backward arrows are written to show this ebb and flow.

Let’s take a look at the reaction of HCl with water:

H2O + HCl H3O+ + Cl

-

An important question to ask here is _________________________________________? That

answer hangs on the following considerations:

1. Which of the two acids (HCl and H3O+) are stronger?

2. Which of the two bases (H2O and Cl-) are stronger?

Having a clear answer to any one of these questions can lead you to know which direction the

equilibrium favors. Equilibrium always drives in the direction of the ________. Since Cl- is

the weaker base and H3O+ is the weaker acid, the equilibrium will move in the direction of the

products.

H2O + HCl H3O+ + Cl

-

Acid Strength – Ka and pKa

Acids vary in terms of their strength, or acidity. Strong acids (such as HCl) will react almost

completely with water, while weak acids (such as acetic acid) react only slightly. The exact

measure of the strength of an acid is given by its acid _________________ constant, Ka.

For the reaction of any acid (shown as HA) with water, the acidity constant, Ka, is

HA + H2O A- + H3O

+

Ka = [H3O+][ A

-] [ ] Concentration in

[HA] mol/L

The stronger the acid in the above equation, the more the reaction favors the ____________.

This results in a large value (greater than 1) for the Ka.

Ka values for acids can vary anywhere from the order of 1015

for the strongest acids to 10-60

for

the weakest. These numbers are unwieldy, so we use the concept of pKa to make the numbers

more workable.

pKa = - log(Ka)

Ka = 10-pKa


27

Acid Strength – Ka and pKa - Continued

For the convenience of the numbers, pKa values are usually listed for us to determine the relative

strength of an acid or base.

The lower the pKa value, the stronger the acid. The lower the pKa value of the conjugate acid,

the weaker its conjugate base.

Thus, a very strong acid will have a very ______ conjugate base. This makes logical sense with

what we learned above.

The Relationship Between [H30+], pH Ka and pKa

Chances are that you have heard of the concept of pH before. pH is a measure of the

__________ or _____________ of a water solution, with a low pH corresponding to an acidic

solution and a high pH corresponding to a basic solution.

Although the values for pH and pKa of a strong acid would both tend to be small, and the values

for pH and pKa of a strong base would both be large, the concepts of pH and pKa are inherently

different.

pH = p[H+] = p[H3O

+] = - log[H3O

+]

While the pKa takes into account all of the components of an acid base reaction, pH only focuses

on the concentration of the __________________ (H3O+).

Thus, the pH of a solution __________ change with depending on how much acid or base is

added to the solution (molarity), while the pKa _________.

Example Problem:

A sample of 6.00 g of acetic acid (CH3COOH) is diluted with enough water to make 1 L of

solution. After waiting for equilibrium to become established, the pH is measured to be 2.87.

What is the pKa of acetic acid?

Additional Relationship:

(Henderson-Hasselbalch Equation)

][

][log

cidconjugatea

aseconjugatebpHpKa


28

Acid Strength – What makes one stronger than the other?

There are three factors that influence the strength of any particular acid. They all have to do

with an acid’s conjugate base.

1. Electronegativity – The more electronegative the atom, the more likely it is to be

willing to accommodate a ____________ charge. For the elements in the 2nd

period,

this trend is easy to see. Since F is the most _________________ element, its anion

F- should be the easiest to make.

HF > H2O > NH3 > CH4

Acid Strength

2. Size – The relative size of an atom or species, the easier it can accommodate a

negative charge. Atoms or species that are larger allow for charge to be more spread

out, which tends to make ions considerably happier. For the elements in column 7A,

this trend is clearly apparent. As you increase in principle ________ ________, the

diameter of the atoms increases, outer electrons are held less tightly and the atom

becomes more ______________.

HI > HBr > HCl > HF

Acid Strength

3. Resonance – A species’ ability to spread charge around between ___________ atoms

contributes highly to the stability of the _____________ _______. Consider HSO4-,

the conjugate base of sulfuric acid (pKa = -10).

Three additional resonance forms that are all _____________ in energy exist for the

bisulfate ion. This is the major contribution to the stability of the conjugate base, and

in turn, the strength of the acid.


29

Acids and Bases – Worksheet

1. Identify the conjugate acid and conjugate base of each of the following. Write

“none” if there is none.

Conjugate Acid Conjugate Base

CH4

NH3

CH3OCH3

CCl4

2. Using your pKa table, predict the products of the following reaction. If there is no

reaction, write NR.

a. CH3COOH + NH3

b. + NaNH2

c. + HCl

3. In which direction do the following equilibriums lie? Explain your choice.

+ -OH + H2O

(+)

CH3CH2NH2 + H2S CH3CH2NH3 + HS-

4. Use your pKa table to suggest an appropriate base to accomplish the following

reaction:

R – C ≡ C – H R – C ≡ C (-)


30

Acids and Bases – Worksheet (Cont.)

5. For the following list of compounds:

a. Circle the strongest acid

b. Put a box around the strongest base

c. Show the conjugate acid and base of each compound listed. Write none if there is

none.

d. Is the first compound a strong enough acid to protonate the third? Explain your

reasoning.

Conjugate Acid Conjugate Base

CH3OH

NH3

H2S

SH-

6. Explain why phenol (see below) is more acidic than methanol, CH3OH.

7. 425 mL of an aqueous solution containing 13.0 g of a mystery acid (FW = 142 g/mol)

is found to have a hydronium ion concentration of 0.000743 mol/L. What is the pKa

of this mystery acid?

8. Show the product of a reaction between NH3 and the following secondary carbocation

(see below). Show any formal charges that are required.

NH3 +


31

Material Covered on the Unit 1 Test

1. Be able to define and give examples of the all of the key terms on page 2 of this

packet

2. Know the structure and names of all functional groups mentioned in page 2 of this

packet

3. Be able to draw valid Lewis structures for polyatomic ions and organic molecules

4. Be able to assign formal charges to atoms in Lewis structures and also be able to

determine which structures are the major and minor contributors to the resonance

hybrid

5. Be able to predict bond angles and molecular geometry (shape) from Lewis structures

6. Be able to determine the hybridization that carbon, nitrogen, oxygen and sulfur atoms

have undergone in a molecule or ion

7. Memorize the relative order of electronegativities of the following elements:

F, O, Cl, N, Br, S, I, C, P, H.

8. Be able to assign polarity to bonds in Lewis structures

9. Be able to draw Lewis structures and line-angle diagrams for structural isomers if

given a molecular formula

10. Be proficient in the use of curved arrows to show electron movement when drawing

contributing resonance structures

11. Be able to write and interpret condensed structural formulas

12. Draw orbital diagrams of sigma and pi bonding

13. Be able to draw representations of 3-D molecules by using wedges and dashed lines

14. Be able to do calculations between pH, pKa and Ka.

15. Be able to describe the effects of electronegativity, size and resonance on acid

strength – be able to illustrate these with examples.

16. Use curved arrows in acid base reactions

17. Use pKa table to predict products of acid-base reactions and label conjugates in

Bronsted-Lowry reactions

18. Predict directions of equilibria using pKa tables

Documents

Unit 1 Packet - dlongchem.wikispaces.comdlongchem.wikispaces.com/file/view/Unit+1+Packet.pdf · Unit 1 Packet: Background, Lewis ... VSEPR, Hybridization, Isomerism, Acids and Bases