Final Exam Review - PowerPoint Presentation

Slide1

Final Exam ReviewSlide2

Conceptual Reminders: Waves/Blackbody Radiation EnergySlide3

Example

A lamp rated at 40.0 J/sec emits violet light of wavelength 422 nm. What is the energy of the photons? How many photons of violet light can the lamp generate in 2.0 sec?

Applicable concepts/equations

E

total

= Ephoton* # photons

=

= 4.03x10

26

photons

 Slide4

Wave Particle DualitySlide5

Things light does:

Reflection/Refraction

Diffraction

Polarization

Interference****Photoelectric effect***

Can a wave do this?Can a Particle do this?

yes

yes

yes

yes

no

no

****Things we discussed in this course.

yes

no

no

yesSlide6

Wave/Particle Duality Review

https

://

www.youtube.com/watch?v=DfPeprQ7oGc

Double Split Experiment

Does this show that light has wavelike or particle like properties? Why?What would the results look like if they had particle like properties?

Wave-like: It creates an interference pattern.

Two bright lines behind the splits no interference pattern. Slide7

Wave/Particle Duality Review

Photoelectric Effect

Does this show that light has wavelike or particle like properties?

Why?

What

would the results look like if they had only wave-like properties? What is the effect of increasing the intensity of a laser of a frequency less

than the threshold frequency?  What is the effect of increasing the intensity of a laser of a frequency

greater the threshold frequency?

http://phet.colorado.edu/en/simulation/photoelectric

Particle: 1 photon=1 electron, which must be of high enough energy.

High total energy still won’t eject an electron if each isn’t of sufficient energy.

None: it must be OVER the threshold frequency to eject electrons.

The

rate of

the ejected electrons is increased. Slide8

Example Problem:

Calculate the

frequency of light that ejects electron from

the surface of a copper sheet that has a work function of 7.53 × 10

-19 J if the electron has a kinetic energy of 1.3*10-19 J. Draw the graph of what an experiment would look like if the frequency of the incident light started below the threshold frequency and was slowly increased. Write the equation of the resulting graph.

E

k

E

k

y

= m x + b

Threshold

Frequency

Y-intercept

= work function

Slope=h

frequency

KeSlide9

Wavefunction

and probability density.

Explain the difference between a

wavefunction

and a probability density. Give the symbol for each in your explanation and also state what sign (positive or negative) each is allowed to be. Draw a picture of the first and second

wavefunction and probability densities for the particle in a box. Probability density is the wavefunction squared. This has physical meaning that the wavefunction doesn’t and is how we can see the shape of orbitals.

How do

wavefunctions and probability densities relate to orbitals in hydrogen atoms? In other atoms? ......Lets look at the path we took to get to atoms

Ψ

Ψ

2Slide10

How did we get to the pictures of orbitals we have now….

Particle in a box (1d)

Particle in a box (2/3D)

Hydrogen Atom

Multi electron atoms= many types of approximation, no exact solutions

Schrodinger equation solutions

Didn’t cover

Subject of

Current research:

We saw result, of

appoximationsSlide11

Electron ConfigurationsSlide12

Electron Configurations: Ground states

Give the ground electron configuration for Sb.

[Kr]5s

2

4d105p3Give the ground electron configuration for neutral Copper.[

Ar]4s23d9[Ar]4s

13d10Give ground electron configuration for +1 copper.

[Ar]3d10Slide13

Electron Configurations: Excited states

Ground state: Sb.

[Kr]5s

2

4d105p3Ground state for neutral Copper.[

Ar]4s23d9[Ar]4s

13d10

For each of the previous neutral electron configurations, give an excited state. Keep same number of electrons, move at least one up in energy: Many

many

many

correct answers.

Sb: [Kr]5s

2

4d105p3  [Kr]5s1

4d105p4 or [Kr]4d105p5 ect…..Cu: [Ar]4s13d10  [Ar]4s23d9 or [Ar]3d104p1 ect…..Slide14

Quantum numbers

What is the quantum numbers for the highest energy electrons in phosphorous?

n=3, l=> p orbital so l=1

How many orbitals and electrons are in the n=3 and n=2 energy levels.

n=3

l= 0, 1, 2

m

l

=0;

-1,0,1

; -2, -1, 0, 1, 2=9 orbitals

n=2

l= 0, 1 ml=0; -1,0,1 =4 orbitals9+4=13 orbitals 2 electrons= 13*2=26 electronsSlide15

Periodic TrendsSlide16

Effective Nuclear Charge

Electronegativity

Electronegativity

Ionization Energy

Electron Affinity

Atomic Radius

Ionization Energy

Electron Affinity

Atomic Radius

SummarySlide17

Questions: Periodic Trends

Why is atomic radius the reverse of the other trends?

This is not the

answer and is not worth any points:

“Because all the other ones go up and to the right and atomic radius goes down and to the left”. The horizontal trends are based on effective nuclear charge which increases as you go across the periodic table. As effective nuclear charge is increased, electrons are held closer to the nucleus making the atom smaller. However this tighter control also causes higher ionization energy, higher electron affinity and higher electronegativity because the nuclear charge is felt more by its own, and other atoms’ electrons.

The vertical trend is based on the number of energy shells, and therefore the shielding that occurs. More energy shells and more shielding cause larger electron clouds, but the lowered effective nuclear charge for the valence shells make all the other trends decrease. TL;DR version: effective nuclear charge and shielding are the reasons for all the trends, they are higher and lower respectively cause tighter control over the nucleus for small atoms.Slide18

Effective Nuclear Charge

Electronegativity

Electronegativity

Ionization Energy

Electron Affinity

Atomic Radius

Ionization Energy

Electron Affinity

Atomic Radius

Exceptions?Slide19

Periodic trends exceptions:

Write

the electron configurations for C, N, and O. Place in order of increasing ionization energy, increasing electron affinity and increasing electronegativity. For each characteristic, do these follow the trend? Why or why not?

Wrong Answer 1:

All: C, N, O

Yes

Because the trend goes up and to the right.

Wrong Answer 2: Variety of wrong/right answers

No

Because the trend goes up and to the right. But hydrogen is half filled.

Correct:

I.E. C, O, N : No, while the effective nuclear charge increases as you go to the right, nitrogen’s half filled shell has increased stability making it unlikely to ionize.

E.A. N, C, O: No

, while the effective nuclear charge increases as you go to the right, nitrogen’s half filled shell has increased stability making it unlikely to ionize.

E.N. C, N, O: Yes. Because it is speaking of atoms in a stable bond, electronegativity is only based on effective nuclear charge which increases as you go to the right. Slide20

You are on cash cab and you are asked where the d-block exceptions to electron affinity are. First you curse me for not having that be required exam knowledge, but then you think back on the p block exceptions and your logic used above. What do you guess for the d-block exceptions in order to claim your prize?

Hint: what is the p block exceptions to electron affinity?

Hint: What would be the equivalent of that with the

dblock

?

Half filled subshells are more stable and therefore have a lower electron affinity.

Half filled subshells are more stable and therefore have a lower electron affinity.

ns2np

5

and also ns

1

np

5Slide21

Ionization Energy:

I1

I2

I3

I4

Element 10.605

1.1101.455.10

Element 20.203

3.215

3.89

4.42

Which group do these elements belong to?

First is always smallest,

The “jump” indicates stable configuration

1st, 2nd , 3rd come off, then big jump: so group 3Big jump after 1st electron, so group 1Slide22

Other topics in

Quantum MechanicsSlide23

Why do we see the sun as different colors?

Which lights depicted in this diagram are emitted, which are scattered.

What causes the events depicted by the blue arrows in the atmosphere. Slide24

MO TheorySlide25

2p

Draw the MO diagram for C2. Write the electron configuration. Is it para or diamagnetic and state the bond order.

Describe the difference between the bonding and

antibonding

orbitals (energy levels, energy levels in relation to starting orbitals, and shapes).

E

C

N

CN

[

He

]s

2s

2

s

2s

*2

p

2p

4

diamagnetic

Bond order= ½ (6-2)=2

Bonding orbitals are a lower energy than

antibonding

and the original atomic orbitals. Most of the electron density in bonding orbitals are between the nuclei rather than outside of it. This adds to the bond order, rather than subtracts. Slide26

Bonding, shape and structure

ExamplesSlide27

Draw the

lewis

and line structure of CH

3

COO-: Tell me everything about everything about it!

Steric coordination number of each element

Hybridization of all applicable elements

All bond angles

Is it polar?

What are all the formal charges

What overlapping orbitals form each bond.

(next slide)

sp

3

sp

3Tetrahedral109.5Trig. Planar120Yes!+0 ExceptSlide28

sp

3

sp

3

sp

3

sp

3

sp

2

sp

2

sp

2

p

ppppp

s

ssSlide29

Line Structure Example

Draw the line structure of

HOCH

2CHCHCH2COOH

. What is the hybridization of each atom.

HO

OH

OSlide30

Resonance

If given two resonance structures of equal stability, how is the sigma and pi bonding electrons shared amongst the atoms in the molecule? How does this affect bond order? How does this affect the location of formal charges?

Draw an example of a molecule with resonance to help explain. Slide31

Gas LawSlide32

Conceptual questions.

In which of the two conditions shown is the gas more ideal?

(assume same number of particles/temperature in each container).

In which of the two conditions shown is the pressure higher?

(assume same number of particles in each container).Slide33

CH3OH can be synthesized by the reaction shown. What volume of H

2

gas (in L), at 748 mmHg and 86

o

C, is required to synthesize 25.8 g CH3OH?

Grams product

Volume Reactant

moles product

moles reactants

= 1.6

1

0

mol

H

2

conversion

conversion

Ideal gas law

=48.2 L H2

Part 2: If the reaction is known to only be 70.0% efficient (aka has a 70% yield) what volume is required (at the same conditions)?