Lecture 2 - Atomic properties: orbital energies, Aufbau principal, ionisation enthalpy, electron affinities

Question 1

Learning outcomes
CHECK GRAPHS!!!

Answer 1

2.1 Explain trends in orbital energies for all atoms and the origin of these changes, using shielding and penetration terms.
2.2 Understand (and mentally use) the long form of the periodic table
2.3 Discuss trends in Zeff using Slater’s Rules
2.4 Explain trends in ionisation energy and electron affinity
2.5 Explain how QM exchange energy influences ionisation energy

Question 2

What atoms can you use shrodingers equation for?

Answer 2

Only 1H
Wavefunctions for non hydrogen atoms can’t be determined exactly

Question 3

How do other elements orbital approximations compare to 1H?

Answer 3

1. They have the same angular functions
2. Similar radial functions but contracted (come closer to nucleus)
3. Orbital energies now depend on effective nuclear charge
4. Orbital energies now depend on l as well as n

Question 4

What equation for energy do you use for:
A) one electron
B) multi electrons

Answer 4

A) E = -Rₕ × (Z² / n²)
As for hydrogen Z=1 (atomic number) can simplify to
E = -Rₕ × (1 / n²)

B) E= -Rₕ × (Zeff ² / n²)

-Rₕ = Rydberg constant (should be capital H)

Question 5

What does Zeff equal?
Using this equation work out Li’s Zeff

Answer 5

Zeff = Z (atomic number) - S (shielding constant)

Lithium has 3 electrons, given S=1.7, Zeff = 1.3
The 1s electron shields the 2s electron from the full nuclear charge

Question 6

Zeff, shielding and Energy

Answer 6

• electrons are attracted by nucleus but repelled by other electrons (must consider both)
• assuming “other electrons” are spherically distributed, classical electrostatics tell us that field arising from them is eq. To a single charge at the Center of the spherical distribution
• this is Zeff and this number rises across a period
• other electrons in the same shell don’t fully shield (note repel is much stronger between different shells than the same shell)

Question 7

What are slaters rules (4 of them)

Answer 7

1. Electrons with a higher n contribute zero - no shielding
2. Electrons with the same value of n contribute 0.35 - not very much shielding ( a value of 0.30 is found to work better for 1s electrons )
3. Electrons with n-1 contribute 0.85 - better shielding
4. Electrons with n-2,n-3 etc contribute 1.00 - full shielding

Question 8

What is the Zeff for Li?

What is the Zeff for Al?

Answer 8

Li: 1s2 2s1
Therefore Z = 3

S = 2 x 0.85 = 1.7
So the outer electron experiences a Zeff of 3-1.7 =1.3 1.3

Al: 1s2 2s2 2p6 3s2 sp1
Therefore Z = 13

S = (2 × 0.35) + (8 × 0.85) + (2 × 1.00) = 9.50
Zeff = 13 - 9.50 = 3.50

Question 9

Effective nuclear charge (Zeff) for across a row

Answer 9

As one proton is added to the nucleus, one electron is added to the valence cell
This electron is not fully shielding (goes up by 0.65 at every point as S increases by 0.35 and Z increases by 1)
Test it!!!

1st and 2nd row is a straight line
3rd row is a staight line then a jump between Ca to Ga (big increase due to d block) then another straight line

Question 10

Effective nuclear charge (Zeff) for down a group

Answer 10

For groups 1 and 2, small increase between periods 2 and 3 (eg Li->Na) then constant after Na

When working out for example Na, K, Rb etc all have 1 electron in its n shell, 8 e- in its n-1 shell and then a proportional number of e- to its proton number in n-2 shell, giving the same number of Zeff =2.20

For all other groups, increase between periods 2 to 4 and then constant after
Again work it out to see pattern

Question 11

Penetration and energy notes - order of ns, np, nd, nf etc

Answer 11

• sketch radial distribution function (RDF)
  2s and 2p are both shielded from nuclear charge by 1s
• RDF shows that all the 2p most probable energy is closer the the nucleus then 2s, 2s has a small extra peak closer to the nucleus than 2p’s singular peak (ie 2s sounds some time close to the nucleus)
• therefore 2s electrons penetrate the shielding 1s electrons better

So… 2s will experience a higher Zeff
And so is lower in energy (more -ve)
Therefore in order of energy ns<np<nd<nf

Question 12

Understanding ionisation energy - theory
Between vacuum (ie distance is infinity and energy = 0) and orbital (with energy E)

Answer 12

As this distance between these to energy levels increase, the orbital becomes more stable
So orbital energy becomes more negative
The orbital is at a lower energy
So ionisation energy will be larger

Question 13

Orbital energies (using E eq for multiple e-)

Answer 13

1. As we go down a group energy becomes less negative (where n increases) for eg 1s (H), 2s (Li), 3s (Na)
2. Across a row orbital energies become more negative because Zeff increases
3. 2s energy is lower than 2p and falls more rapidly, due to better penetration

LOOK AT GRAPHS

Question 14

Orbital energy key points

Answer 14

1. H energy only depends on n whereas all other depending on n and l due to shielding and penetration
2. Orbital energies show largest difference in valence region
   Eg for Cl Z=17 energy 3s, 3p and 3d differ most
3. Orbital energies become more similar as you get further
   away from valence region
   Eg for iodine Z=50 energy 3s, 3p and 3d are very similar
4. Orbitals can cross in energy
   Eg 4s and 3d near Ca/Sc (why K’s 4s fills before 3d)
5. Order only holds for neutral atoms

Question 15

Why is 4s lower in energy in 3d

Answer 15

Due to shielding by core electrons and better penetration by 4s in RDF 4s is lower in energy than 3d
CHECK GRAPH

Question 16

Paulis exclusion principle
Aufbau principle

Answer 16

No two electrons in the same atom can have identical values for all 4 of their quantum numbers
Electrons fill from lowest energy level subshells to highest

Question 17

Ionisation energy - across a row and exception 1

Answer 17

M(g) -> M+(g) + e- under standard conditions

General rule: As Zeff increases, energy becomes more negative, therefore IE increases

Exception 1:
B < Be
Be: 1s2 2s2
B: 1s2 2s2 2p1
Easier to ionise 2p1 electron as it is higher in energy (less-ve)

2p is less penetrating
Well shielding from z by 1s and 2s
Zeff lower, energy less -ve, IE decreases
GRAPH

Question 18

Ionisation energy - across a row and exception 2

Answer 18

Exchange energy:
Each pair of electrons with parallel spins in an atom leads to a lowering of energy (ie for 1 pair (̶ (̶ E=-K and (̶ )̶ is E=0K)
Energy = -K stabilisation for every pair
It explains p^n configurations found for atoms and is a quantum mechanical effect

N>O
N: 1s2, 2s2, 2p3
So N = (̶ (̶ (̶ E = -3K
N+ = (̶ (̶ - E = -K
Leading to an increase by 2K

O: 1s2 2s2, 2p4
So O = )̶(̶ (̶ (̶ and O+ = (̶ (̶ (̶
Both have E=-3K

Question 19

Ionisation energy down the group and the exception

Answer 19

Zeff has a small rise and then is constant (as seen before)
N value increases
E becomes less negative
IE decreases

Exception: Ga>Al
Ga is after d-block and so Zeff is high
E is more negative so IE increases

Tl>In
Tl is just after f block so same reasoning as above
Also discretions for 6s due to relativistic effects (6s is particularly stable)

Question 20

Equation linking ionisation energy, energy, and zeff

Answer 20

-IE ≈ E ∝ -(Zeff ² / n²)

Question 21

Electron attachment enthalpy equation and general trend

Answer 21

Standard eq: X(g) + e- -> X-(g) ie opposite to IE

Consider F(g) + e- -> F-(g)
highly exothermic as complete octet formed so lots of E given out

General trend becomes more exothermic as we go from left to right (Li to F) as Zeff increases (ie less favourable to add an e-)

Question 22

What are the two exceptions for the electron attachment enthalpy

Answer 22

C > N in terms of exothermicness
This is due to exchange energies from C-> C- and N -> N-
C = (̶ (̶ -, C- = (̶ (̶ (̶ goes from -K to -3K therefore 2K more -ve in energy

N = (̶ (̶ (̶ N- = (̶)̶ (̶ (̶ exchange energy is -3K for both (no change)

Li>Be
This is due to electron adding to 2s orbital (rather than 2p for Be to F)