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Tutorial – 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

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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

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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

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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

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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

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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 βˆ’ 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

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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»

𝐸 = 𝐸 =

β„Žβˆ—π‘ Ξ»

3.00x108 π‘š βˆ— ) 𝑠 300 π‘š

(6.626x10βˆ’34 𝐽𝑠

𝐸 = 6.626x10βˆ’28 𝐽 𝐸 = 6.63x10βˆ’28 𝐽

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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 =𝑅 2βˆ’ 2 Ξ» 𝑛1 𝑛2

1 1 1 =𝑅 2βˆ’ 2 Ξ» 𝑛1 𝑛2 1 1 1 = 𝑅( 2 βˆ’ 2 ) Ξ» 2 3 1 1.096776x107 =( )(0.1388888) Ξ» π‘š 1 1523300 = Ξ» π‘š 1π‘š =Ξ» 1523300 6.56469507x10βˆ’7 π‘š = Ξ» 656 π‘›π‘š = Ξ»

In this case, the sign indicates a gain of energy, which means the photon Is being absorbed

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Practice Radial Probability Distribution (sum of ψ2)

7. Draw the radial probability distribution for px orbital.

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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 Chester’s 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„Ž Ξ»= π‘šπ‘£

Ξ»= 2 π‘˜π‘š 1β„Ž 1 π‘šπ‘–π‘› 1000 π‘š 𝑣= βˆ— βˆ— βˆ— β„Ž 60 π‘šπ‘–π‘› 60 𝑠 1 π‘˜π‘š 𝑣 = 0.56 π‘š/𝑠

β„Ž π‘šπ‘£

6.626x10βˆ’34 𝐽 Ξ»= π‘š 40 π‘˜π‘” βˆ— (0.56 𝑠 ) Ξ» = 2.9817x10βˆ’35 π‘š Ξ» = 2.98x10βˆ’26 π‘›π‘š

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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

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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

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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

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Practice 11. Show the condensed electronic configuration for these elements that don’t follow the typical rules. a) Mo [Kr]5s24d4 [Kr]

[Kr]5s14d5

b) Cu

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Practice 11. Show the condensed electronic configuration for these elements that don’t follow the typical rules. a) Mo

b) Cu

[Ar]4s23d9 [Ar]

[Ar]4s13d10

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