Final Exam CHEM 341

Question 1

Relationship to find formula for Ionic solids

Answer 1

Cation CN/Anion CN =# of anions/# of cations

Question 2

2 ways to change conductivity

Answer 2

1) Increase Temp (means that more Energy for e- to be promoted)
2) Doping (n-doping (add 1 e-)) and (p doping (1 less e-))

Question 3

Diodes

Answer 3

Diodes make n/p junctions where Fermi levels line up. This faciitates electron migration into holes.
This movement produces electricity/light

Question 4

Forward bias

Answer 4

PROMOTES e- and holes combing by negative potential on n-side and positive potential on p side

Question 5

Reverse bias

Answer 5

PREVENTS e- and holes combing by applying (+) potential to to n side and a (-) potential to the p side

Question 6

Superconductors def

Answer 6

Conductors of nearly infinate conductance

Question 7

Superconductors properties

Answer 7

Superconductors work better at low tempertures
More specifically, below Tc or critical temperature

Question 8

0D defects

Answer 8

Schottky: there is a vacancy in the lattice
Frenkel: the ions are moved to the interstitial site

Question 9

Factors that change 0D defects

Answer 9

High temp makes Schotttsky more likely because more energy causes atoms/molecules to shift
Frekel more common with small CN

Question 10

1D defects

Answer 10

Line defects (line of atoms ends prematurely/branches out)

Question 11

Meisner effect

Answer 11

Superconductors can expel a B field when below Tc

Question 12

Magnetic moment

Answer 12

mu = root N*(N+2)
Where n is unpaired electrons

Question 13

MO general rules

Answer 13

of AO=# of MO
Everytime you have a bonding orbital you also have an anti-bonding orbital

Question 14

Low symmetry groups (3)

Answer 14

C1= no symmetry other than identity
Cs= Only 1 mirror plane and identity
Ci= only an inversion center

Question 15

High Symmetry groups (5)

Answer 15

Cinfinityv=linear molecules w/o inversion
Dinfinityh= linear molecule w/inversion
Td=tetrahedral with 4 ligands that are identical
Oh=octahedral with identical ligands
In= 120 symmetry elements

Question 16

Point groups for propeller molecules

Answer 16

Dn=symmetric propeller
Cn=asymmetric propeller

Question 17

Eclipsed conformation pt group

Answer 17

Dnh

Question 18

Staggered conformation pt group

Answer 18

Dnd

Question 19

BCS Theory and why low temp

Answer 19

When e- are in the lattice formation, there are (+) areas around the e- beacause the (+) charges are attracted to the e-
Another e- is attracted to this (+) area and creates cooper pairs. The cooper pairs repulse and switch partners. This interchange is like the electrons moving together. This makes it difficult to impede e- flow
Low temps make it so that the lattice does not vibrate as much and slows e- movement

Question 20

MO orbitals 3 conditions for bonding

Answer 20

1) Symmetry of the orbitals so that same phase overlaps
2) Atomic orbital energy similar
3) distance between atoms short enough to overlap orbitals not too short that e-/nuclei create repulsion

Question 21

LCAO steps

Answer 21

1) Find pt group
2)Determine symmetry related fragments of the molecule (ex. bridging, terminal, central atoms)
3) For each fragement, determine the Mulliken labels assoc with AOs of the fragment
a)unmovable atoms
b) sum of characters associated e/relevant atomic orbitals
c) multiply each column for each mulliken symbol
d) REDUCE
e)make matches, matches are bonding and get antibonding
non-bonding left over after bonding

Question 22

Coordinate system for MO

Answer 22

prinicpal axis of rotation is the z axis

Question 23

Principal axis of rotation

Answer 23

Axis with the highest n

Question 24

Degeneracy
how to tell from Char table and Mulliken symbols

Answer 24

2/3 in the first column (identity column) signals and degeneracy
If there is a degeneracy, objects have the same energy level or frequency
The Muliken label E always represents double degeneracy and T represents triple degeneracy

Question 25

Bond Order

Answer 25

BO=0.5(e- in bonding MO- e- in the antibonding MO)

Question 26

Ordering d orbitals in MO

Answer 26

On AO side, d orbital most stable for transition metal complexes (lowest in energy)

Question 27

Predicting IR spectra steps

Answer 27

STEP 1) Draw molecule and assign a coordinate sys
STEP 2) Find pt group + character table
STEP 3) Find reducible rep
3a) uma for each mulliken symbol operation
3b)add up the total x/y/z (ex. z,x,Ry, Rx, and x)
3c) muliply both rows to get reducible rep
STEP 4) reduce (Jacob’s) to determine irriducible rep
STEP 5) Subtract translational and rotational irriducible reps translational (x,y,z) and rotational (Rx, Ry, Rz) to get vibrational modes
STEP 7) Determine if modes are IR or Raman active
IR ACTIVE: has a function of x,y,z
RAMAN ACTIVE: quadratic function of x,y,z

Question 28

What does reducible rep describe

Answer 28

All possible motions of the molecule

Question 29

Madelung Constant

Answer 29

M= # of neighbors/distance

Question 30

Cp ligands are

Answer 30

Weak pi acceptors

Question 31

How to check for optical isomers

Answer 31

If the structure has no mirror plane-> check optical isomer
If mirror image of structure with no mirror image non-superimposible- optical isomer!

Question 32

Mixing criteria and results of mixing on energy

Answer 32

Non-bonding and anti/bonding pair of same mulliken symbol (result non-bonding lowered, bonding lowered, anti-bonding raised in energy)
Bonding and Anti-bonding pairs, the more stable pair increases in energy and less stable pair increases in energy

Question 33

Determining high/low spin for TM complexes (3 factors)

Answer 33

1) 2+ row periodic table- ALWAYS STRONG FIELD= large Deltanaught
2) Ligands via spectrochemical series
top- CO/CN =complex will be strong field
middle- H2O/NH3/en (look @ 3)
bottom-Halgoens/OH- =weak field, small delta naught
or (H(halogens)ONK)
3) Look at the oxidation state of the metal=larger oxidation state= larger deltanaught

Question 34

Trans effect

Answer 34

Trans influence: The PtiX bond is influenced by the PtiT bond, because both use the Pt px and dx2-y2
orbitals. When the PtiT s bond is strong, it uses a larger contribution of these orbitals and
leaves less for the PtiX bond . As a result, the PtiX bond is weaker, and its
ground state (sigma-bonding orbital) is higher in energy, as in Figure 12.14b . This groundstate, thermodynamic effect is called the trans influence . It contributes to the reaction rate
by lowering the activation barrier for PtiX bond breaking. The ranking below predicts the
order for the trans influence on the basis of the relative s@donor properties of the ligands

trans effect: pi acceptors make metal more positive and weaken bond and so complex wants the incoming ligand more

both stabilize the transition state