1. Chemistry EM spectrum notes

2. Light is a small part of all the radiation (something that spreads from a source) called electromagnetic radiation. Electromagnetic radiation is energy in the form of waves (of electric and magnetic fields). Electromagnetic radiation includes radio waves, microwaves, infrared, visible light, X-rays, and Gamma rays. All these together are considered the Electromagnetic Spectrum.

4. As all the forms of electromagnetic radiation are waves, they all have similar properties.
All electromagnetic radiation travels at the speed of light (c), 299,792,458 m/s (3 x 108) in a vacuum

5. •The crest is the top of the waves, the trough is the bottom of the waves, and the amplitude is a measurement from the rest or zero line to a crest or trough

7. •The wavelength (λ – lambda) is the distance between successive crests/troughs and is measured in meters (often nm = 10-9 m)

8. in (per second – can be written as s-1) or Hz (Hertz)
•The wavelength (λ – lambda) is the distance between successive crests/troughs and is measured in meters (often nm = 10-9 m)
•The frequency (ν – nu) is the number of waves that pass a point in one second and is measured
in (per second – can be written as s-1) or
Hz (Hertz) 1 s

9. How many hertz is the first wave?
How many hertz is the second wave?

10. How many hertz is the first wave?
1 wave per second = 1 Hz
How many hertz is the second wave?
2 waves per second = 2 Hz

11. The speed of a wave is directly proportional to the wavelength and the frequency; c = λν is the formula c ν λ

12. Example. A certain violet light has a wavelength of 413 nm
Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?

13. Example. A certain violet light has a wavelength of 413 nm
Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?
ν = c λ

14. Example. A certain violet light has a wavelength of 413 nm
Example. A certain violet light has a wavelength of 413 nm. What is the frequency of the light?
ν = c λ
ν =
3.00 × 108 m/s
413 × 10-9 m

15. Example. A certain violet light has a wavelength of 413 x 10-9 m
Example. A certain violet light has a wavelength of 413 x 10-9 m. What is the frequency of the light?
ν = c λ
ν =
3.00 × 108 m/s
413 × 10-9 m
ν = 7.26 × 1014 Hz

16. This lead Max Planck to theorize that light must carry energy in basic minimum amounts that he called quanta. Like a delivery person cannot correctly deliver half a box, the electrons in atoms cannot gain a fraction of a quantum of energy (it has to be in whole numbers).

17. h = Planck’s constant = 6.626 × 10-34 Js ν = frequency in Hz or 1/s
He proposed that this energy was directly proportional to the frequency of the electromagnetic radiation and a constant, now called Planck’s constant. E = h ν
E = energy in Joules (J)
h = Planck’s constant = × Js
ν = frequency in Hz or 1/s
You do not need to
memorize this number E h ν

18. Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?

19. Example. What is the energy content of one quantum of the light with a wavelength of 413 x 10-9 nm?
Note: wavelength is not in the energy equation, but frequency is. So first, you must solve for the frequency. As seen in the earlier example, a wavelength of 413 x 10-9 m gives a ν = 7.26 × 1014 Hz.

20. Example. What is the energy content of one quantum of the light with a wavelength of 413 x10-9 m?
ν = 7.26 × 1014 Hz
E = h × ν
E = × Js × 7.26 × /s

21. Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?
ν = 7.26 × 1014 Hz
E = h × ν
E = × Js × 7.26 × /s

22. Example. What is the energy content of one quantum of the light with a wavelength of 413 nm?
ν = 7.26 × 1014 Hz
E = h × ν
E = × Js × 7.26 × /s
E = 4.81 × J

23. In 1905 Einstein used Plancks work to propose that electromagnetic radiation had a dual wave-particle nature. As a particle, electromagnetic radiation carries a quantum of energy of energy, has no mass, and is called a photon.

24. So to get an electron to emit from a metal, it must be struck with a photon having quantum energy big enough, or nothing will happen. Each metal requires a different quantum energy, thus each metal can be identified by the frequency of light needed to emit electron.

25. This idea was expanded upon to develop an idea of where the electrons were in an atom. It was found that low pressure gases could be trapped in a tube and electrified, and would then glow a color particular to the gas inside.

26. Furthermore this light could be passed into a prism, and instead of getting the entire spectrum (rainbow) of colors, only certain wavelengths of light would be seen as small bars of color, called a line-emission spectrum.

27. This would indicated that the electrons in an atom were only absorbing specific amounts of energy from the electricity, causing the electrons to move from their ground state (normal position close to the nucleus) to an excited state (higher energy position further away from the nucleus). The electrons do not stay in the excited state for long and fall back to their ground state, losing the energy equal to what they gained.

28. Niels Bohr used this to develop a model of the atom where the electrons could only be in certain, specific energy level (n) orbits around the nucleus. Just as you cannot go up half a rung on a ladder, the electron could not go up a partial energy level. The electrons gained or lost enough energy to move a whole number amount of energy levels (n) away from or closer to the nucleus, or it did not move.

29. He calculated the amount of energy needed for an electron of hydrogen to move between each energy level (n) (which was not constant) and his calculations agreed with experimental results.

31. The quantum mechanical model starts with a Principal Quantum Number (n), which is the basic energy level of an electron, and often matches the period number. Possible values (currently) are 1-7.

32. The quantum mechanical model starts with a Principal Quantum Number (n), which is the basic energy level of an electron, and often matches the period number. Possible values (currently) are 1-7.

33. Inside the sublevels are orbitals, specific regions with a 90% probability of finding electrons.
• s –orbitals are spherically shaped clouds around the nucleus
• p -orbitals are bar-bell shaped clouds with the nucleus between the lobes
• d and f are much more complex in shape

34. Each sublevel has room for a different amount of electrons, because an orbital can hold two electrons, then each sublevel has a different amount of orbitals

35. • s –sublevel can hold 2 electrons, so it has 1 orbital (shape)
• p –sublevel can hold 6 electrons, so it has 3 orbitals (shapes)
• d –sublevel can hold 10 electrons, so it has 5 orbitals (shapes)
• f –sublevel can hold 14 electrons, so it has 7 orbitals (shapes)

36. The s sublevel is simply a sphere centered on the nucleus.
                                            
The p sublevel has three orbitals.  These are often referred to a dumbbell shape.
                                                                                                                                                    
The d sublevel has five orbitals:
                                                                                                                                                                                                                                                            
The f sublevel has seven orbitals:
                                                                                                                                                                                                                                                                                                                        

37. To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2
if n=1, then

38. To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2
if n=1, then 2 electrons will fit
if n=4,

39. To know the maximum amount of electrons that could be in any principal quantum level (and the number of elements that could be represented) use the formula 2n2
if n=1, then 2 electrons will fit
if n=4, then 32 electrons will fit

40. In order to show on paper where electrons are likely to be located in an atom, orbital filling diagrams and electron configurations are drawn or written. When this is done, three rules must be followed:

41. 1. Aufbau principle – electrons fill lower energy levels first, thus 1 before 2 and s before p, etc.
a. orbitals within a sublevel are equal in energy (called degenerate)
b. the principal energy levels often overlap, making them seem a little out of order
c. boxes are used to represent orbitals

42. Another way of writing the aufbau principle diagram:1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p

43. 2. Pauli Exclusion principle – an orbital (box) can hold a maximum of two electrons (arrows)
a. for two electrons to fit, they have to have opposite spins
for one electron in the orbital
for two electrons in the orbital (opposite spins)

44. 3. Hund’s Rule – when electrons occupy degenerate orbitals, one electron is placed into each orbital with parallel spins before doubling up
Ex. _____ _____ _____ NOT _____ _____ _____
3p p

46. Orbital Notation shows the arrows in the boxes to represent the electrons in an atom. To shorten this process, an electron configuration can be written. It leaves out the information about the number of orbitals in each sublevel, so it will be expect you remember that information.

47. It has the general form nΔ°
n = principal quantum number (1-7…)
Δ = sublevel letter (s, p, d, or f)
° = number of e- in that orbital (1-14)

48. Ni = 28 e-

49. Sn = 50 e-

50. A few elements have electron configurations that do not follow the normal pattern. These two groups are represented by the first elements in their group:
By normal configuration:
Cr -- 1s2 2s2 2p6 3s2 3p6 3d4 4s2 …
Cu -- 1s2 2s2 2p6 3s2 3p6 3d9 4s2 …

51. By actual (seen in nature) configuration:
Cr -- 1s2 2s2 2p6 3s2 3p6 3d5 4s1 …
Cu -- 1s2 2s2 2p6 3s2 3p6 3d10 4s1 …

52. If writing out the entire electron configuration is too much, we can use the previous (in the periodic table) noble gas to take the place of part of the electron configuration:
Polonium:
1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p4
Xenon:
1s22s22p63s23p64s23d104p65s24d105p6
Polonium: [Xe] 6s24f145d106p4

53. When the electron configuration is written for an element using the noble gas configuration the electrons written after the noble gas are the ones that appear on the outside of the atom, called valence electrons..

54. Reviewing Flame Test Lab
When a substance is heated in a flame, the substance’s electrons absorb energy from the flame and move from a ground state electron configuration to an excited state. The electron takes in thermal energy and then emits the energy as EM radiation (light).