Atomic Structure Part 3

Dr. Chris UP Aug.2016

SLATER RULES TO CALCULATE EFFECTIVE CHARGE

“Shielding” of outer electrons

Radial part squared of the Schroedinger eq.

An electron here is shielded from the

nucleus by 2s and 1s

Effective nuclear charge Zeff vs. nuclear charge Z

Average nr of electrons

between the electron and the nucleus

Slater’s rule

Group el. in the same group

el. in same shell n and

orbital nr < l

el. in shell n-1

el. in shell < n-2

[ 1s] 0.30 - - -

[ns np ] 0.35 - 0.85 1

[nd] or [nf] 0.35 1 1 1

Examples

Fe atom which has nuclear charge 26 and electronic configuration 1s22s22p63s23p63d64s2

http://www.wikihow.com/Determine-Screening-Constant-and-Effective-Nuclear-Charge

Example V vs. V2+

Configuration for V: In V2+ : the 2 4s electrons or 2 3d electrons missing ? -> We estimate the effective nuclear charge that acts on the 4s vs. 3d electrons

Solution

Configuration V: (1s) (2s 2p) (3s 3p) (3d) (4s 4p) [Ar] 4s2 3d3

Shielding for one 4s electron: S = 10 + 11 * 0.85 + 0.35 Z* = 3.30 Shielding for one 3d electron: S = 18 + 2 * 0.35 Z* = 4.30 => The 4s electrons are less attracted than 3d

Effect on atomic radii

Irregularity: Zn -> Ga

Shielding of 3d10 in Zn: (1s)(2s 2p)(3s 3p)(3d)(4s 4p)(4d) 10 el 8el 10el 2el => S= 18+ 9*0.35 S = 21.15 => Z* = 8.85

Compare with 4p1 of Ga: (1s)(2s 2p)(3s 3p)(3d)(4s 4p)(4d) 10 el 18el 2el 1el => S= 10 + 18*0.85 + 2*0.35 => S = 26 => Z* = 5.0

Recalling: Uncertainty principle Assume an electron moves in an H-atom with diameter of 2 * Bohr Radius = 2 * 0.529 A = 2 * 0.529 10-10 m = 1.058 10-10 m = ∆x What is the uncertainty for the speed v ? (me = 9.1 10-31 kg, h=6.624 10-34 J s) ∆x ∆v m = h/2π

TERM SYMBOLS

Electronic configuration as “Terms” Angular momentum L Is the sum of all individual l of each electron The angular quantum number l = 0,1,.. (n-1)

Example: n = 2 => l = 0, 1 => s and p

ml = 0 ml = -1 ml = 0 ml = +1

L = -1 + 0 +1 = 0 => Term “S”

… and spin multiplicity 2S + 1

The magnetic quantum number can have the values of - l , - l+1, …. 0, …. l-1, l Therefore we have 2 l +1 spin states for one l

Example: S = 3 * ½ = 3/2 -> multiplicity 2S+1 = 4 (“Quartet”)

ml = 0 ml = -1 ml = 0 ml = +1

Examples

From Shriver/Atkins Chap.20

Microstates for 2 el. in a p orbital 5 states belong to L = 2 (“D”)

9 states belong to L = 1 (“P”)

1 state with L=0 (“S”)

Russell-Saunder Coupling

The angular orbital momentum L couples with spin S

Each state: 2S+1L J is 2*J + 1 degenerated (same energy)

Possible states in np2

L = 2 S= 0 -> J = 2 (2J+1 = 5 x degenerated) L = 1 S= 1 -> J = 2,1,0 (5x , 3x and 1x = 9 states) L = 0 S = 0 -> J = 0 (1 state) Therefore there are 5 electron configurations with slightly different energies.

Which is the ground state ?

Hund’s rules: 1. maximum spin => multiplicity 3 (S = 1)

2. highest L => L = 1 (“P”)

3. Shell < ½ filled -> lowest J

Shell > ½ filled -> highest J

=> 3P0

Exercises 1

(1) Order the terms for increasing energy (2) Find all J-values for each of these 5 states (3) And find the ground-state

Answers

http://chemistry.illinoisstate.edu/standard/che362/homework/ 362ps11solns.pdf

Exercises 2

For a electron configuration 2s1 2p1: • Determine all possible term symbols • their degeneracy • and their energy order

Answers 2 The orbital angular momentum is l1=0 and l2=1 The total momentum ranges from |l1-l2| …. l1+l2

-> L = 1 (“P”) The spin S can be |s1-s2| … s1+s2 -> ½ - ½ = 0 and ½ + ½ = 1 2 states are possible: 1P and 3P The total angular momentum J = |L-S| … |L+S| -> for 1P: (S=0, L=1) J = 1 -> for 3P: (S=1, L=1) J = 2, 1, 0

Total number of sub-states: 1P1 -> 2J+1 = 3 states with same energy 3P0 -> 1 state 3P1 -> 3 degenerate states 3P2 -> 5 - “ -

Exercises 3

Transitions between electronic states

1) Spin – Spin rule: during a transition the spin S of the system cannot change: S = 0 2) Laporte Rule: L = 1 the orbital momentum must change J = 0 the total momentum cannot change

Exercises 4: electronic transitions

Which transitions are allowed and why :

Answers 4

ELECTRONIC CONFIGURATION AND PERIODIC TABLE

Electron Configuration and Periodic Table

Hund’s Rules tell us: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin). Pauli Exclusion Principle: Each electron in an atom must at least have one different quantum number

Hund’s Rules / Pauli principle Which electron configuration obey both rules ?

1s 2s 2p

1s 2s 2p

1s 2s 2p

Periodic Table build up

Comprehension questions (1) For an electron in 4f: what quantum numbers (n,l,m,s) are possible for it ? (2) Give the correct notation for this set of quantum numbers: and which element has this configuration ?

n l m

2 1 1

3 2 -1

3 1 1

2 0 0

Lanthanoids and Actinoids (Period 6 and 7)

How many elements are in row 5, 6 and 7 ? What is the electronic configuration for Lead, Gold, Cerium, Tungsten, Silver, Mercury, Americum, Tin, Antimon, Osmium

Exceptions from Aufbau Principle

La (Z=57) should have configuration: but has: [Xe] 4f0 5d1 6s2

Ac (Z=89) should have configuration: but has: [Rn] 5f0 6d1 7s2

Exceptions from Hunds’ Rules Find the electron configuration for: Cr Cu Mo Pd Ag Au

Characteristics of Periods

(1) Number of valence electrons increases (how many VE in the period K -> Kr ?)

(2) Valency towards Hydrogen Increases from 1 to 4 and back 3 to 1 (write the hydrogen compounds for all elements in period 3)

(3) Size of atoms

(4) Metallic character (5) Number of shells is the same (example: how many shells do all elements in period 4 have ?)

Characteristics of Groups (1) Same number of valence electrons (example: how many VE do all Halogens have ?)

(2) Same valencies (example: oxygen compounds of group 1, 2 and 4)

(3) Properties of elements (4) Atomic size increases down (5) Metallic character increases down (example: compare Chlorine with Iodine)

(6) Number of electron shells (example: Li 2,1 (2) Na 2,8,1 (3) K 2,8,8,1 (4) Rb 2,8,18,8,1 (5)

Electron Configuration of ions Find the configuration for: Cl- 1s2 2s2 2p6

V3+

Fe2+ Br- S2- Mn2+ Mn3+ Mn7+

ELECTRO NEGATIVITY (EN)

Introduced by Pauling 1932

EN values for each element based on the comparison of bond energies Arbitrary scale with F as χ = 4 as highest possible Bond energies cannot always be measured accurately -> modifications by other scientists

http://www.webelements.com/chlorine/electronegativity.html

Mulliken EN Mulliken developed an easier system to calculate EN. EN is regarded to be the average of Ionization Potential (IP) and Electron Affinity (EA) of an element: χ (A) = IP (A) + EA (A) / 2 To make the values close to Pauling’s EN, a factor of 2.6 is introduced, so that: χ (A) = IP (A) + EA (A) / (2 * 2.6) = IP (A) + EA (B) / 5.2

Example:

Calculate the EN for Fluorine (IP : 17.4 eV/atom EA: 3.62 eV/atom)

Allred – Rochow (AR) EN

They rely on the effective nuclear charge of an element. The empirical formula is (related to Pauling): Χ (A) = (Z* e) e / r2 = Z* e2 / r2 => 0.359 Z*/r2 + 0.744

Charge felt by the VE

Coulomb attraction

Calculate the EN for Lead: Shielding S is 76.7, atomic radius 530 pm (Z = 82)

In Angstrom

Factors influencing EN

(1) Atomic size

(2) Number of inner shells (=> Z*)

(3) Charge on an ion

(4) Hybridization (the more s-character, the higher the EN)

Importance of EN (1) Ionic / covalent character of bonds

(see next page)

(2) Strength of bases and acids (cf. CH3-COOH vs. CH2F-COOH vs. CF3-COOH)

(3) Location of nucleophilic or electrophilic attack sites in organic and inorganic reactions cf.

Bond characteristics

Linus Pauling suggested an estimation of % ionic character for a bond:

http://wwwchem.uwimona.edu.jm/courses/CHEM1902/IC10K_MG_Fajans.html