Electron Configuration 2 & 12

1. ElectronConfiguration2 & 12

2. 2.3 Electron Arrangement 2.3.1 Describe the electromagnetic spectrum 2.3.2 Distinguish between a continuous spectrum and a line spectrum 2.3.3 Explain how the lines in the emission spectrum of hydrogen are related to electron energy levels 2.3.4 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. High energy Low energy Low Frequency High Frequency X-Rays Radiowaves Microwaves Ultra-violet GammaRays Infrared . 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. • http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/linesp16.swf

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. 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. Ehigh-Elow= hv = hc/λ h=3.983 x 10-13 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 -1312 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 = -1312 / 42 = - 82 kJ E2 = -1312 / 22 = - 328 kJ E4 – E2 = - 82 kJ – (- 328 kJ)= 246 kJ

16. 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 10-13 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 ConfigurationHL only 12.1.3 State the relative energies of s, p, d, and f orbitals in a single energy level 12.1.4 State the maximum number of orbitals in a given energy level. 12.1.5 Draw the shape of an s orbital and the shapes of px, py and pz orbitals 12.1.6 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. 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p Increasing energy 3s 2p 2s 1s

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

29. 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p Increasing energy 3s 2p 2s 1s • The next electrons go into the 2s orbital • only 11 more

30. 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p Increasing energy 3s 2p 2s 1s • The next electrons go into the 2p orbital • only 5 more

31. 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p Increasing energy 3s 2p 2s 1s • The next electrons go into the 3s orbital • only 3 more

32. 7p 6d 5f 7s 6p 5d 6s 4f 5p 4d 5s 4p 3d 4s 3p Increasing energy 3s 2p 2s 1s • 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.