ANSWER SHEET for Chemistry C4 Exam View Study Guide
1. b
c = ; speed of light = (wavelength)(frequency) = 3 x 108 m/s
2. c
All forms of electromagnetic radiation has the same speed. See page 98 of your text.
3. a
Because the product of  equals a constant, they are inversely proportional. As one increases, the
other must decrease. The example in class was pairs of numbers equally 36.
[36 x 1; 18 x 2; 12 x 3; 9 x4; 6 x 6]
4. d
The photoelectric effect does not occur with red light regardless of the light’s intensity. This is due to the
fact that red light has the longest wavelength, and therefore, the lowest frequency. According to Planck’s
equation, Ephoton = h , where h = Planck’s constant &  = frequency, there is a minimum frequency for
electron excitation. {h = 6.626 x 10-34 Joule second}
5. b
Photoelectric effect’s definition: emission of electrons from metal when light shines on the metal
6. b
Energy must be absorbed for an electron in ground state to rise to a higher excited state.
7. c
Bohr’s study of H gas revealed emission spectra where differences in of visible light = different colors
(Balmer series). Others worked to reveal the Lyman series (UV light) & Paschen series (IR light) for H.
8. c
Electrons exist only at discrete energy levels. This was demonstrated by H’s emission-line spectrum
when H atoms emitted only specific frequencies of light, indicating H’s electron existed only in very
specific energy states. See pg 101 of your text.
9. b
de Broglie proposed that all matter exhibits wave-like properties. Experiments confirmed that electrons
had wave-like properties of diffraction & interference in addition to their particle-like properties. The
combination of Heisenberg’s uncertainty principle & Schrodinger’s wave equations is the foundation of
the quantum model of the atom.
10. d
Heisenberg’s uncertainty principle states that it is impossible to know simultaneously both the position &
velocity of electrons or other particles. This is now one of the fundamental principles of our current
understanding of light & matter.
n = principle quantum number = main energy level = 1, 2, 3, 4, …
l = angular momentum quantum number = shape of orbital {s, spherical, p, dumbbell, d, clover leaf, f,?}= 0, < n-1
m = magnetic quantum number = orientation within orbital (x, y, z for p orbitals) = -l to +l
s = spin quantum number = clockwise (= + ½) or counterclockwise (= - ½)
11. c
12. d
13. b
14. a
(p 107) “For a specific main energy level, the # or orbital shapes possible is equal to n.”
15. b
16. b
17. a
18. c
All s orbitals are spherical in shape with a maximum of 2 electrons.
19. d
s orbitals begin at 1s, p begin at 2p, d orbitals begin at 3d, & f begin at 4f.
20. c
One can count the # of elements in each block to determine the maximum # of electrons per orbital.
s=2
d =10
p=6
f = 14
21. c
Although the # of electrons per orbital equals the # of elements in that block, there are only 2 electrons
allowed in each suborbital, which means the # of suborbitals per block is half that number. s orbital has
only 2/2, or 1, orientation. p orbitals have 6/2, or 3, possible orientations: px, py, & pz. d orbitals have
10/2, or 5, suborbitals: dx2-y2, dxy, dyz, dxz, & dz2. f orbitals have 14/2, or 7, different suborbitals. (p108)
22. b
See above.
23. c
2s, 2px, 2py, & 2pz
24. a
According to Pauli’s exclusion principle, no 2 electrons can have the same set of 4 quantum #s, so the
spins must be different for electrons in the s orbital.
25. d
p 110. The number of electrons per main energy level is 2n2.
26. b
2n2 = 18  n2 = 18/2 = 9  √n2 = √9  n = 3
27. c
2n2 = 8  n2 = 8/2 = 4  √n2 = √4  n = 2
28. c
2n2  2(3)2 = 2 x 9 = 18  18 – 15 = 3
29. b
Pauli exclusion principle limits each suborbital to 2 electrons.
30. b
p 111:Aufbau principle states that electrons must enter their lowest-energy available orbital first.
31.d
p112: “Orbitals of equal energy are each occupied by 1 electron before any is occupied by a 2nd electron
and all electrons in singly occupied orbitals must have the same spin” is Hund’s rule.
32. a
p 112: see # 24 above
33. c
Pauli exclusion principle, again
34. d
p 111: If one has to enter the lowest available energy level first, one is following Aufbau’s principle.
35. b
p 111.
36. a
Sc-45 has the electron configuration of 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d1
37. b
p 111, Figure 16
38. d
p 118, Table 5 & p 119, second paragraph
39. d
One can count the # of electrons (those #s in the superscripts) & this equals the # protons = atomic #.
40. c
One follows the sequence outlined in Figure 16, p 111, or note Al’s configuration in Table 4, p 117.
Aufbau’s order: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d10, 5p6, 6s2, 4f14, 5d10, 6p6, 7s2, 5f14, 6d10, 7p6