Tutorial β Quantum Theory and Atomic Structure 1
Practice 1. What are the quantum numbers of Cu? l = 0, s block
l = 1, p block
n=1 n=2
l = 2, d block
n=3 n=4
l = 3, f block
2
Practice 1. What are the quantum numbers of Cu? (continued) n = 4, l = 2 What are ml and ms?
-2
-1
0
+1
+2
Therefore the quantum numbers for this electron are: n = 4, l = 2, ml = +1, ms = -1/2
3
Practice 2. What are the principal quantum numbers of the final electron for each of the following: a) H
n=1
ml = -l to +l
l=0 ml = 0 1
ms = + 2 b) F
0
n=2 l=1 ml = 0
ml = -l to +l
1
m s = -2 c) Au
-1
+1
ml = -l to +l
n=6 l=2 ml = +1 1 m s = -2
0
-2
-1
0
+1
+2 4
Practice 3. Write the full and condensed electronic configuration of the following: a) Ni 1s22s22p63s23p64s23d8 [Ar]4s23d8 b) Nd
c) Ne
5
Practice 3. Write the full and condensed electronic configuration of the following: a) Ni 1s22s22p63s23p64s23d8 [Ar]4s23d8 b) Nd 104p665s224d10 105p666s2 4f4 1s222s222p663s223p664s223d10
[Xe]6s24f4 c) Ne
6
Practice 3. Write the full and condensed electronic configuration of the following: a) Ni 1s22s22p63s23p64s23d8 [Ar]4s23d8 b) Nd 1s22s22p63s23p64s23d104p65s24d105p66s24f4 [Xe]6s24f4 c) Ne 1s22s22p6 [He]2s22p6
7
Practice 4. Which of the following quantum numbers describes any electron in Na (not just the final valence electron)? a) n = 4, l = 1, ml = 0, ms = +1/2 b) n = 3, l = 0, ml = 0, ms = +1/2 c) n = 2, l = 1, ml = -2, ms = +1/2 d) n = 3, l = 1, ml = 0, ms = +3/2
1 +l. 1 Na is=all in row an electron m numbers from If lsniswill 3,either then3,lbe =so2, 1 and 0or β m spin +-l2to 2 in can So l =allowed 0,have thennm=l 3, = 02 or 1 arenifall If nl ==1,2,then thenml =l =1-1, 0 and +1 and 0 are allowed
8
Practice 5. Scientists at the Breakthrough Listen project have recently received radio frequency pulses from across the galaxy. If these radio frequency pulses had a measured wavelength of 300 m, what would be the energy of each wave?
What do we know: Ξ» = 300 m What do we want to learn: E=? How are they related? E = (h*c)/Ξ»
πΈ = πΈ =
ββπ Ξ»
3.00x108 π β ) π 300 π
(6.626x10β34 π½π
πΈ = 6.626x10β28 π½ πΈ = 6.63x10β28 π½
9
Practice 6. What is the wavelength associated with the following transition in Hydrogen? Is this absorbance or emission? What do we know: n1 = 2 n2 = 3 What do we want to learn: Ξ»=?
How are they related? 1 1 1 =π
2β 2 Ξ» π1 π2
1 1 1 =π
2β 2 Ξ» π1 π2 1 1 1 = π
( 2 β 2 ) Ξ» 2 3 1 1.096776x107 =( )(0.1388888) Ξ» π 1 1523300 = Ξ» π 1π =Ξ» 1523300 6.56469507x10β7 π = Ξ» 656 ππ = Ξ»
In this case, the sign indicates a gain of energy, which means the photon Is being absorbed
10
Practice Radial Probability Distribution (sum of Ο2)
7. Draw the radial probability distribution for px orbital.
11
Practice 8. Chester the dog saw a squirrel out in the yard. He went to chase the squirrel but his 40 kg frame was too slow to catch the squirrel. As Chester could only run 2 km/h, what was Chesterβs de Broglie wavelength?
What do we know: m = 40 kg v = 2 km/h What do we want to learn: Ξ»=? How are they related? β Ξ»= ππ£
Ξ»= 2 ππ 1β 1 πππ 1000 π π£= β β β β 60 πππ 60 π 1 ππ π£ = 0.56 π/π
β ππ£
6.626x10β34 π½ Ξ»= π 40 ππ β (0.56 π ) Ξ» = 2.9817x10β35 π Ξ» = 2.98x10β26 ππ
12
Practice 9. Which of the following forms of electromagnetic radiation has the highest energy? Lowest Energy?
a) Radio waves
1 m, very low energy, does not harm us
b) Ultraviolet Radiation
10 nm, enough energy to cause sunburns over time
C) Green Light
514 nm
D) X-Rays
0.01 nm, very high energy
13
Practice 10. How many d electrons can be found in Gd?
Gd is in row 6 Row 4 provides 10 d electrons Row 5 provides 10 d electrons We do not take any from row 6, as the f block (l=3) is filled first)
Therefore, there are 20 d electrons in Gd
14
Practice 11. Show an energy level diagram (show splitting of p, d, and f orbitals) for Se24p
Se has 34 electrons. 2- means two extra electrons 36 total electrons
15
Practice 11. Show the condensed electronic configuration for these elements that donβt follow the typical rules. a) Mo [Kr]5s24d4 [Kr]
[Kr]5s14d5
b) Cu
16
Practice 11. Show the condensed electronic configuration for these elements that donβt follow the typical rules. a) Mo
b) Cu
[Ar]4s23d9 [Ar]
[Ar]4s13d10
17
Have a question, concern or comment? Head on over to the Wizedemy Forums and Iβll be happy to clarify any questions you may have. Happy Studying!
18