Download Worksheet Key - UCSB C.L.A.S.

4
Chem 1B - CLAS – Ch 12 - Key
1. A green photon has a frequency (v) of 5.45 x 108 MHz
a. Calculate the photons wavelength (λ) in nm
𝑐
c = λv ⇒ λ = =
𝑣
3 𝑥 108 𝑚𝑠 −1 109 𝑛𝑚
5.45 𝑥 1014 𝑠 −1
= 550 nm
𝑚
b. Calculate the energy (in J) of one photon and for one mole of photons (kJ/mol)
E photon = hv ⇒ E photon = (6.626 x 10–34Js) (5.45 x 1014s–1) = 3.61 x 10–19 J
6.022 𝑥 1023 𝑘𝐽
E mol photons = (3.61 x 10–19 J)
2. Determine the following:
a. Higher frequency:
b. Higher Energy:
c. Longer wavelength:
𝑚𝑜𝑙
IR
or
X-Rays or
yellow or
103𝐽
= 217 kJ/mol
UV
Microwaves
purple
3. According to Heisenberg uncertainty principle;
a. the momentum of a particle cannot be measured precisely
b. neither the position nor the momentum can be measured precisely
c. the position and the momentum of a particle can be measured precisely, but not at the same time
d. the positon of a particle cannot be measured precisely
4. From the following list of observations, choose the one that most clearly supports the conclusion that electrons have
wave properties.
a. the emission spectrum of hydrogen
b. the photoelectric effect
c. the scattering of alpha particles by metal foil
d. diffraction
e. cathode "rays"
5. Calculate the wavelength of a beryllium atom traveling at 15% the speed of light.
mass of a beryllium atom = 1 atom
λ=
ℎ
𝑚𝑣
=
1 𝑚𝑜𝑙
6.022 𝑥 1023 𝑎𝑡𝑜𝑚𝑠
−34
6.626 𝑥 10
x
9.012 𝑔
𝑚𝑜𝑙
𝐽𝑠
= 1.5 x 10–23 g
= 9.82 x 10–19
(1.5 𝑥 10−23 𝑔)(0.15)(3 𝑥 108 𝑚𝑠 −1 )
m
6. It takes 208.4 kJ of energy to remove 1 mol of electrons from the atoms on the surface of rubidium metal. If rubidium
metal is irradiated with 254-nm light, what is the maximum kinetic energy (kJ/mol) the released electrons can
have?
KEe- = Ephoton - Ebinding
KEe- =
KEe- =
ℎ𝑐
𝜆
- Ebinding
(6.626 𝑥 10−34 𝐽𝑠)(3 𝑥 108 𝑚𝑠 −1 )
𝑥
𝑘𝐽
x
6.022 𝑥 1023
𝑚𝑜𝑙
(2.54 𝑥 10−7 𝑚)
103 𝐽
7. Consider the following transitions. Which will emit light with a longer wavelength?
a. n = 4  n = 2
or
n=3  n=2
b. n = 3  n = 1
or
n=1  n=3
c. n = 5  n = 3
or
n=3  n=1
– 208.4 KJ/mol = 262.9kJ/mol
8. Calculate the wavelengths (nm) emitted for the following electronic transitions in a Li2+ ion.
a. n = 5  n = 3
𝑍2
𝑍2
𝑓
𝑖
𝑍2
𝑍2
𝑓
𝑖
ΔE = -2.178 x 10–18J(𝑛2 − 𝑛2 ) and E photon = | ΔE | =
ℎ𝑐
𝜆
= |2.178 x 10–18J(𝑛2 − 𝑛2) |
ℎ𝑐
𝜆
(6.626
𝑥 10
−34
8
𝐽𝑠)(3 𝑥 10 𝑚𝑠−1)
𝜆
b. n = 3  n = 1
(6.626
𝑥 10
−34
8
𝐽𝑠)(3 𝑥 10 𝑚𝑠−1)
𝜆
12
12
3
52
12
12
1
32
= |2.178 x 10–18J(
= |2.178 x 10–18J(
−
2
−
2
) ⇒ λ =1283 nm
) ⇒ λ =103 nm
9. An excited hydrogen atom with an electron in n = 5 state emits light having a frequency of 6.9 x 1014 s-1. Determine
the principal quantum level (n) for the final state in this electronic transition.
𝑍2
𝑛𝑓2
hv = |2.178 x 10–18J(
−
𝑍2
)|
𝑛𝑖2
12
12
𝑥
52
(6.626 𝑥 10−34 𝐽𝑠)( 6.9 x 1014 s-1) = |2.178 x 10–18J(
−
2
) ⇒ nf = 2
10. The figure below represents part of the emission spectrum for a one-electron ion in the gas phase. All of the lines
result from electronic transitions from excited states to the n = 3 state.
a. What electronic transitions correspond to lines A and B?
Wave C is the longest so it has the lowest energy ⇒ n=4 → n=3
Wave B is the second longest so second lowest energy ⇒ n=5 → n=3
Wave A is the third longest so third lowest energy ⇒ n=6 → n=3
b. If the wavelength of line B is 142.5 nm, calculate the wavelength of line A (nm).
1
𝑛𝑓2
Since Z, h and c are constants ⇒ (
1
1
1
−
1
) 𝜆𝐴
𝑛𝑖2
1
𝑛𝑓2
=(
−
1
) 𝜆𝐵
𝑛𝑖2
1
(32 − 62 ) 𝜆𝐴 = (32 − 52 ) (142.5 𝑛𝑚) ⇒ 𝜆𝐴 = 121.6 𝑛𝑚
𝑓
𝑓
𝑖
𝑖
11. An electron is excited from the ground state to the n = 3 state in a hydrogen atom. Which of the following
statement(s) is/are true?
a. It takes more energy to ionize the electron from n= 3 than from the ground state.
b. The electron is farther from the nucleus on average in the n = 3 state than in the ground state
c. The wavelength of light emitted if the electron drops from n = 3 to n = 2 is shorter than the wavelength of light
emitted if the electron falls from n = 3 to n = 1.
d. The wavelength of light emitted when the electron returns to the ground state from n = 3 is the same as the
wavelength absorbed to go from n = 1 to n = 3.
e. The ground state ionization energy of He+ is four times the ground state ionization of H.
12. Calculate the ground state ionization energy (in kJ/mol) and the wavelength (in nm) required for B4+.
𝑍2
𝑍2
ΔE = -2.178 x 10–18J(𝑛2 − 𝑛2 ) and E photon = | ΔE | =
𝑓
ℎ𝑐
𝜆
𝑖
= |2.178 x 10
For ionization nf = ∞ ⇒
–18
ℎ𝑐
𝑍2
J(𝑛2
𝑓
−
𝜆
𝑍2
)|
𝑛𝑖2
2
= 2.178 x 10–18J(𝑍𝑛2) ⇒
𝜆
(6.626 𝑥 10−34 𝐽𝑠)(3 𝑥 108 𝑚𝑠 −1 )
𝜆
ℎ𝑐
𝑖
2
= 2.178 x 10–18J(512 ) = 3.65 nm
13. How many electrons in any one atom can have the following quantum numbers?
a. n = 5 ⇒ 50
b. n = 6, l = 0 ⇒ 2
c. n = 4, l = 2 ⇒ 10
d. n = 4, l = 3, ml = -2 ⇒ 4
e. n = 2, l = 0, ml = 0, ms = -1/2 ⇒ 1
14. Fill in the following table:
Cl ⇒ 1s22s22p63s23p5 ⇒ 7 valence ⇒ 1 unpaired
Ni ⇒ [Ar]4s23d8 ⇒ 2 valence ⇒ 2 unpaired
Cr ⇒ [Ar]4s13d5 ⇒ 1 valence ⇒ 6 unpaired
Ag ⇒ [Kr]5s14d10 ⇒ 1 valence ⇒ 1 unpaired
Te2– ⇒ [Kr] 5s24d105p6⇒ 8 valence ⇒ 0 unpaired
Fe2+⇒ [Ar]3d6⇒ 14 valence ⇒ 4 unpaired
15. Determine if each of the following corresponds with an excited state or ground state electron configuration.
a. [Ar]4s24p5 ⇒ excited
b. [Kr]6s1⇒ excited
c. [Ne]3s23p4⇒ ground
16. Which of the following has the largest radius?
a. Al or Si
b. F or Cl
c. S or S2d. K or K+
17. Which of the following has the greatest ionization energy?
a. K or Ca
b. P or As
c. Sr or Sr2+
18. Which of the following has the most negative electron affinity?
a. Br or Kr
b. C or Si
19. The successive ionization energies for an unknown element are:
I1 = 896 kJ/mol
I2 = 1752 kJ/mol
I3 = 14,807 kJ/mol
I4 = 17,948 kJ/mol
Which family does the unknown element most likely belong? 2
Equations:
c = λv
λ=
E = hv
KEe- = Ephoton - Ebinding
E = mc2
𝑍2
𝑍2
𝑓
𝑖
ΔE = -2.178 x 10–18J(𝑛2 − 𝑛2 )
ℎ
𝑚𝑣
Quantum Numbers
1. n ⇒ principal quantum number ⇒ proportional to the size and energy of the orbital ⇒ range is 1 to ∞
2. l ⇒ angular momentum quantum number ⇒ shape of the subshell ⇒ range is 0 to ∞
3. m ⇒ magnetic quantum number ⇒ orbital orientation ⇒ range is –l to +l
l
4. m ⇒ electron spin quantum number ⇒ +1/2 or -1/2
s
Electron configuration