1. Electron Configuration 2 & 12

2. 2.3 Electron Arrangement
Describe the electromagnetic spectrum
Distinguish between a continuous spectrum and a line spectrum
Explain how the lines in the emission spectrum of hydrogen are related to electron energy levels
Deduce the electron arrangement for atoms and ions up to Z=20

3. Bohr’s Model Why don’t the electrons fall into the nucleus?
Move like planets around the sun.
In circular orbits at different levels.
Amounts of energy separate one level from another.

4. Bohr postulated that: Fixed energy related to the orbit
Electrons cannot exist between orbits
The higher the energy level, the further it is away from the nucleus
An atom with maximum number of electrons in the outermost orbital energy level is stable (unreactive)
Think of Noble gases

5. Low energy
High energy
Radiowaves
Microwaves
Infrared .
Ultra-violet
X-Rays
GammaRays
Low Frequency
High Frequency
Long Wavelength
Short Wavelength
Visible Light

6. Wavelength and frequency

7. How did he develop his theory?
He used mathematics to explain the visible spectrum of hydrogen gas
Lines are associated with the fall of an excited electron back down to its ground state energy level.

8. The line spectrum
electricity passed through a gaseous element emits light at a certain wavelength
Can be seen when passed through a prism
Every gas has a unique pattern (color)

9. Continuous line spectrum
Carbon
Helium
Continuous line spectrum

10. "Those who are not shocked when they first come across quantum theory cannot possibly have understood it. “ (Niels Bohr on Quantum Physics)

11. Wavelengths and energy
Understand that different wavelengths of electromagnetic radiation have different energies.
c=vλ
c=velocity of wave (2.998 x 108 m/s)
v=(nu) frequency of wave
λ=(lambda) wavelength

12. Bohr also postulated that an atom would not emit radiation while it was in one of its stable states but rather only when it made a transition between states.
The frequency of the radiation emitted would be equal to the difference in energy between those states divided by Planck's constant.

13. h=3.983 x 10-13 Jsmol-1= Plank’s constant
Ehigh-Elow= hv = hc/λ
h=3.983 x Jsmol-1= Plank’s constant
E= energy of the emitted light (photon)
v = frequency of the photon of light
λ = is usually stated in nm, but for calculations use m.
This results in a unique emission spectra for each element, like a fingerprint.
electron could "jump" from one allowed energy state to another by absorbing/emitting photons of radiant energy of certain specific frequencies.

14. Energy must then be absorbed in order to "jump" to another energy state, and similarly, energy must be emitted to "jump" to a lower state.
The frequency, v, of this radiant energy corresponds exactly to the energy difference between the two states.
In order for the emitted energy to be seen as light the wavelength of the energy must be in between 380 nm to 750 nm

15. For Hydrogen only!
En= -R/n2, where R is kJ/mol and n is principle quantum number (energy level)
Example: Calculate the energy required to ionize a mole of electrons from the 4th to the 2nd energy level in a hydrogen atom?
E4 = / 42 = - 82 kJ
E2 = / 22 = kJ
E4 – E2 = - 82 kJ – (- 328 kJ)= 246 kJ

16. E = hc/λ, therefore λ = hc/E
What is the wavelength of light emitted when electrons go from n=4 to n=2 ? Is it visible to our eyes?
E = hc/λ, therefore λ = hc/E
λ = [(3.983 x kJsmol-1)(2.998 x 108 ms-1)]/(246 kJmol-1)
= 4.85 x 10-7 m
Convert to nm and see if its visible! (1 nm = 1 x 10-9 m)
(4.85 x 10-7 m)( 1nm) = 485 nm (Its probably the green line)
1 x 10-9 m

17. Bohr’s Triumph
His theory helped to explain periodic law (the trends from the periodic table)
Halogens (gp.17) are so reactive because it has one e- less than a full outer orbital
Alkali metals (gp. 1) are also reactive because they have only one e- in outer orbital

18. Drawback
Bohr’s theory did not explain or show the shape or the path traveled by the electrons.
His theory could only explain hydrogen and not the more complex atoms

19. The Quantum Mechanical Model
Energy is quantized. It comes in chunks.
A quanta is the amount of energy needed to move from one energy level to another.
Since the energy of an atom is never “in between” there must be a quantum leap in energy.
Schrödinger derived an equation that described the energy and position of the electrons in an atom

20. Electron Configuration HL only
State the relative energies of s, p, d, and f orbitals in a single energy level
State the maximum number of orbitals in a given energy level.
Draw the shape of an s orbital and the shapes of px, py and pz orbitals
Apply the Aufbau principle, Hund’s rule and the Pauli exclusion principle to write electron configurations for atoms and ions up to Z=54.

21. S orbitals 1 s orbital for every energy level 1s 2s 3s
Spherical shaped
Each s orbital can hold 2 electrons
Called the 1s, 2s, 3s, etc.. orbitals

22. P orbitals Start at the second energy level 3 different directions
3 different shapes
Each orbital can hold 2 electrons

23. The D sublevel contains 5 D orbitals
The D sublevel starts in the 3rd energy level
5 different shapes (orbitals)
Each orbital can hold 2 electrons

24. The F sublevel has 7 F orbitals
The F sublevel starts in the fourth energy level
The F sublevel has seven different shapes (orbitals)
2 electrons per orbital

25. Summary
Starts at energy level

26. Electron Configurations
The way electrons are arranged in atoms(up to Z = 54)
Aufbau principle- electrons enter the lowest energy first.
This causes difficulties because of the overlap of orbitals of different energies.
Pauli Exclusion Principle- at most 2 electrons per orbital - different spins
Hund’s Rule- When electrons occupy orbitals of equal energy they don’t pair up until they have to .

27. Increasing energy 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p 3s 2p

28. Increasing energy 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p 3s 2p
Phosphorous, 15 e- to place
The first to electrons go into the 1s orbital
Notice the opposite spins
only 13 more

29. Increasing energy 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p 3s 2p
The next electrons go into the 2s orbital
only 11 more

30. Increasing energy 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p
The next electrons go into the 2p orbital
only 5 more

31. Increasing energy 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p
The next electrons go into the 3s orbital
only 3 more

32. Increasing energy 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s
The last three electrons go into the 3p orbitals.
They each go into separate shapes
3 unpaired electrons
1s22s22p63s23p3

33. Orbitals fill in order Lowest energy to higher energy.
Adding electrons can change the energy of the orbital.
Half filled orbitals have a lower energy.
Makes them more stable.
Changes the filling order

35. Write these electron configurations
Titanium - 22 electrons
Copper – 29 electrons
Chromium - 24 electrons

36. Titanium - 22 electrons
1s22s22p63s23p64s23d2
Vanadium - 23 electrons 1s22s22p63s23p64s23d3
Chromium - 24 electrons
1s22s22p63s23p64s23d4 is expected
But this is wrong!!

37. Copper’s electron configuration
Copper has 29 electrons so we expect
1s22s22p63s23p64s23d9
But the actual configuration is
1s22s22p63s23p64s13d10
This gives one filled orbital and one half filled orbital.
Remember this exceptions!!!!!!

38. Chromium is actually 1s22s22p63s23p64s13d5 Why?
This gives us two half filled orbitals.
Slightly lower in energy.
The same principal applies to copper.