Unit 11 - MOT

Question 1

Failures of LEM

Answer 1

• needs the use of resonance to describe the delocalization of electrons (not all electrons reside between 2 atoms)
• LEM describes only ground state bonding. Molecules are known to exist in excited states
• Fails to interpret certain molecular properties (like BO)

Question 2

Molecular Orbital Theory (MOT)

Answer 2

• Bonding e-‘s may be delocalized over the entire molecule, not localized between specific pairs of atoms
• atomic orbitals (a.o.s) of different atoms are mixed to make molecular orbitals (MO’s) that may extend over the entire molecule
• the mixing of x a.0’’s results in x MO’s (3 atomic orbitals = 3 MO’s)

Question 3

MOT: 1st Row Homonuclear Diatomics

Answer 3

• H and He
• Each atom uses its 1s orbital
• 2 atoms means two 1s a.o.’s mix so two MO’s
• a plus combination and a minus combination (both have positive signs and one has positive one has negative)
• antibonding node forms in the minus combination
• • means antibonding

Question 4

Antibonding Node

Answer 4

• destructive interference
• happens when one lobe has positive sign and one has negative sign

Question 5

Valance MO Correlation Diagram - 1s

Answer 5

• shows how atomic orbitals (AOs) of two atoms “correlate” (connect) to the molecular orbitals (MOs) they form as the atoms come together
• energy: no star n MO < n a.o’s < n MO with *
• dashed lines from a.o.’s to MO’s
• Same rules as with atomic orbitals: lowest energy first, maximize spin multiplicity in degenerate orbitals (Hund), opposite spins pair

Question 6

MOT Bond Order

Answer 6

• 1/2(# of bonding e-‘s - # of antibonding e-‘s)
• lower bond order means weaker
• bond order of 0 is unstable

Question 7

Molecular Orbital Electron Configuration

Answer 7

• lowest to highest enegry
• Ex: He2 is (σ1s)^2 (σ*1s)^2

Question 8

MOT: 2nd Row Homonuclear Diatomics - 2s

Answer 8

• 2s, 2px, 2pz valence a.o.’s used in making MO’s for each atom
• has a radial node (bc just s) because #RN=n - l - 1 so 2 - 0 - 1 = 1
• one of each node for each atom, so if l=s then 1 RN then 2 on the whole drawing (bc two atoms)
• two ways of drawing them (one with constructive and one with destructive interference)

if antibonding:
- has a antibonding node (destructive interference)

• define coordinates

Question 9

MOT: 2nd Row Homonuclear Diatomics - 2pz

Answer 9

• for 2pz and 2pz, the minus combination is 2pz - 2pz where theres 2 angular nodes and the middle two lobes have + + or - - combination with constructive interference. not antibonding nobe. σ2pz. vertical nodes
• for 2pz and 2pz, the plus combination is 2pz + 2pz where theres 2 angular nodes 1 antibonding node and the middle two lobes have + - or - + combination with descrutive interference. σ*2pz. vertical nodes

Question 10

MOT: 2nd Row Homonuclear Diatomics - 2px

Answer 10

• for 2px and 2px, the plus combination is 2px + 2px where theres 1 angular node the top and bottom two lobes each have - - or + + combination with constructive interference. pi bond because a one contains the bond axis. π2px. horizontal node
• for 2px and 2px, the minus combination is 2px - 2px where theres 1 angular node 1 antibonding node the top and bottom two lobes each have - - or + + combination with destructive interference. π*2px. horizontal node AN and vertical antibonding node

Question 11

MOT: 2nd Row Homonuclear Diatomics - 2py

Answer 11

• for 2py and 2py, the plus combination is 2py + 2py where theres 1 angular node behind the front lobes and the front lobes each have - - or + + combination with constructive interference. pi bond because a one contains the bond axis. π2py. horizontal node
• for 2py and 2py, the minus combination is 2py - 2py where theres 1 antibonding node vertically and the front lobes each have - - or + + combination with destructive interference. π*2py. vertical antibonding node

Question 12

Valence MO Correlation Diagram - 2nd Row Homonuclear Diatomics: First Order (without atomic orbitals)

Answer 12

/σ2s < σ2s< σ2p < π2p < π2p < σ*2p

• depends on coordinates so subscript (2px, 2pz) may change on diff coordinates
• this is where z is bond axis
• applies to O - Ne
• when you draw the correlation diagram, IE1 is the energy from the a.o.’s to 0 (NOT MO’s)

Question 13

Valence MO Correlation Diagram - 2nd Row Homonuclear Diatomics: First Order (with atomic orbitals)

Answer 13

2s < σ2s < σ2s* < 2p < σ2p < π2p < π2p* < σ2p*

Question 14

Valence MO Correlation Diagram - 2nd Row Homonuclear Diatomics: Second Order (without atomic orbitals)

Answer 14

σ2s<σ2s<π2p<σ2p<π2p<σ*2p
- applies to Li2 ->N2

Question 15

Valence MO Correlation Diagram - 2nd Row Homonuclear Diatomics: Second Order (with atomic orbitals)

Answer 15

σ2s < σ2s< π2p < σ2p < π2p < σ*2p
- applies to B - N

Question 16

Diamagnetic

Answer 16

molecule is repelled by a magnetic field, # of unpaired e-‘s = 0

Question 17

Paramagnetic

Answer 17

molecule is attracted by a magnetic field, # of unpaired e-‘s > 0

Question 18

LUMO

Answer 18

lowest unoccupied moleuclar orbital

Question 19

HOMO

Answer 19

highest occupied molecular orbital

Question 20

Full MO Configuration

Answer 20

• lowest to highest energy
• just count number of valence e-‘s
• based on first or second order
• you can do [ He ][ He ] as shorthand for first 2
• Ex: B2 has 6 valnce e-‘s, second order
• [ He ][ He ] (σ2s)^2(σ2s)^2 (π2p)^2 or (σ1s)^2(σ1s)^2(σ2s)^2(σ*2s)^2 (π2p)^2

Question 21

why is antibonding higher in energy and why is bonding lower than a.o’s

Answer 21

Why Bonding MO’s Lower:
- Bonding MOs form by constructive interference of atomic orbitals (AO)
- When the phases match (same sign) between the two atoms along the bond:
- Electron density increases between the nuclei
- This attracts both nuclei toward the shared electron cloud
- Energy decreases → stabilizing
- therefore bonding MO’s energy is lower in energy
- sigma bonds are more stable

Why Antibonding MO’s Higher:
- Antibonding MOs form by destructive interference of atomic orbitals
- When the phases dont match (opposite signs) between the two atoms along the bond:
- A node forms between the nuclei
- electron density between nuclei decreases
- Nuclei repel each other because there’s no electron glue
- Energy increases → destabilizing
- because unstable , more energy

Question 22

Valence MO Correlation Diagram - 2nd Row Heteronuclear Diatomics

Answer 22

• same diagram except in heteronuclear diatomics, the more electronegative atom lowers the energy of its orbitals, so MOs are shifted toward that atom
• follows first and second order
• the more electronegative atom (like O or F) → orbitals lower in energy than the less electronegative atom (C, B, N)
• know total number of valence e-‘s

Question 23

MOT: 2nd Row Heteronuclear Diatomics

Answer 23

• you draw the same thing excpet you polarize the e-‘s toward the atom that has higher electronegativity for bonding MO’s and lower electronegativity for antibonding MO’s
• aka make cloud bigger for one that has higher or lower electronegativity

Question 24

Combined LEM and MOT

Answer 24

• for a molecule with delocalized e-‘s (ex: resonance structures or π e-‘s), a common approach is to use the LEM for σ framework (delocalized e-‘s)
• therefore combine LEM and MOT
• this is because LEM is featured to describe the localized electron bonding while MOT describes electron delocalization much better

Rules:
1. the σ and π systems are orthaganol (perpendicular)
2. LPs participiating in resonance are in the π system (delocalized) not the σ system
3. only LPs in the σ system (localized) contribute to the SN and participate in hybridization

Question 25

How to Draw Energy Level Diagram for π MO's

Answer 25

1. split up number of antibonding and number of bonding evenly 2. if odd number, its a nonbonding (nd) π MO 3. the number of electrons in the system is the number of delocalized e-'s

Question 26

How to Draw the π MO's (based on energy level diagram)

Answer 26

- you draw the nuclei as zigzag and alternate up and down. this is the number of nuclei - lowest π MO has 0 nodes then you add one as it goes up - detemine what p orbitals are being used (px, py, pz) so you can draw it correctly (like in front if needed) - # of a.o.'s = # of π MO's - draw circles according to whats being used (like if its px, you draw the circles in front) - combine lobes that have no node between them - bonding and antibonding MO's must have contribution on adjacent atoms (therefore when they're not adjacent, you get nonbonding) - you can do shaded having the same sign - remember that if heteronuclear, follow the rules for which one is bigger (more contributation)

Question 27

Total BO

Answer 27

1/2(# π bonding e-'s - antibonding e-'s)

Question 28

Specific BO

Answer 28

total BO/# of participating A-B units where A is an atom and B is an atom Ex: in NO2 - there's 2 NO units so the denominator is 2

Question 29

Conjugated System

Answer 29

Adjacent atoms have the same oritenation for p orbitals participiating in the π system

Question 30

Acyclic

Answer 30

Chain molecule (not ring/circle)