Quiz 2 - GENCHEM

2017 Spring Semester Quiz 2
For General Chemistry I (CH101)
Date: Apr 3 (Mon),
Professor Name
Time: 19:00 ~ 19:45
Class
Student I.D. Number
Name
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Use the following constants to solve problems.
(Planck constant 𝒉 = 6.626 × 10-34 J s)
(Mass of electron π’Žπ’† = 9.109 × 10-31
kg)
(Permittivity of the vacuum 𝜺𝟎 = 8.854 × 10-12 C2 J-1 m-1)
(Charge of the electron 𝒆 = 1.602 × 1019
C)
(Avogadro’s number 𝑡𝑨 = 6.022 × 1023 mol-1) (Ratio of a circle's circumference to its diameter 𝝅 =
3.14)
(Speed of light c = 𝟐. πŸ—πŸ—πŸ– × 108 m s-1
(Bohr radius π’‚πŸŽ =
(1 Rydberg =
𝜺𝟎 π’‰πŸ
π…π’†πŸ π’Žπ’†
π’†πŸ’ π’Žπ’†
𝟐
πŸ–πœΊπŸ
πŸŽπ’‰
= 2.18 × 10-18 J)
= 𝟎. πŸ“πŸπŸ— × 10-10 m)
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1. (Total 8 pts, each 2 pts) Read the following statements or equations, and verify whether these
are β€œTRUE (T)” or β€œFALSE (F)”. (2 pts for a right answer, -2 pts for a wrong answer, 0 pt for no answer)
(a) The probability for locating an electron in the volume element 𝒅𝑽 = π’“πŸ 𝐬𝐒𝐧 𝜽 π’…π’“π’…πœ½π’…π“ when
the electron is in the specific quantum state (𝒏, 𝒍, π’Ž) is given by
(ππ’π’π’Ž )𝟐 𝒅𝑽 = [𝑹𝒏𝒍 (𝒓)]𝟐 [π’€π’π’Ž (𝜽, 𝝓)]𝟐 𝒅𝑽
Answer:
T
(b) Hartree atomic orbitals are the exact solutions of Schrodinger equation for many-electron
atoms.
Answer:
F
(c) Among four quantum numbers (𝒏, 𝒍, π’Ž, π’Žπ’” ), the energy levels of electron in one-electron
atoms depend only on the principal quantum number n.
Answer:
T
(d) The Pauli exclusion principle states that no two electrons in an atom can have the same set
of four quantum numbers (𝒏, 𝒍, π’Ž, π’Žπ’” )
Answer:
T
2. (Total 10 pts) Here is a wave function of one-electron atoms.
πŸ‘
ππ’π’π’Ž (𝒓, 𝜽, 𝝓) =
𝟏
𝒁 𝟐
πŸ“πŸπŸβˆšπŸ“π… π’‚πŸŽ
𝝈
𝒁𝒓
πŸ’
π’‚πŸŽ
( ) 𝝈(πŸ–πŸŽ βˆ’ 𝟐𝟎𝝈 + 𝝈𝟐 )𝒆𝒙𝒑 (βˆ’ ) 𝐜𝐨𝐬 𝜽, 𝝈 =
(a) (3 pts) Find the values of 3 quantum numbers 𝒏, 𝒍, and π’Ž.
각 numberλ‹Ή 1점, λΆ€λΆ„μ μˆ˜ μ—†μŒ
n: 4
------ +1 pt
l: 1
------ +1 pt
m: 0
------ +1 pt
(b) (2 pts) How many the number of radial nodes and angular nodes does ππ’π’π’Ž (𝒓, 𝜽, 𝝓) have?
각각의 λ…Έλ“œλ‹Ή 1점, λΆ€λΆ„μ μˆ˜ μ—†μŒ
# of Radial node: 2
------ +1 pt
# of Angular node: 1
------ +1 pt
(c) (3 pts) For existed nodes in this given wave function, find the conditions of variants for each
nodes. Suppose that this atom is Hydrogen (𝒁 = 1).
[Boundary condition: 𝟎 ≀ 𝐫 < ∞, 𝟎 ≀ 𝜽 ≀ 𝝅, 𝟎 ≀ 𝝓 ≀ πŸπ…]
2 radial nodes-> 𝝈(πŸ–πŸŽ βˆ’ 𝟐𝟎𝝈 + 𝝈𝟐 ) = 𝟎
r = (𝟏𝟎 ± πŸβˆšπŸ“ )π’‚πŸŽ or r = 5.53π’‚πŸŽ , 14.47π’‚πŸŽ
------ +1 pt
---------------------------------------- +1 pt(ν•˜λ‚˜λ‹Ή 0.5 pt)
r = 0 μ“°λ©΄ 감점 [𝝈 = 𝟎(not a root for radial node)]
--------------- -0.5 pt
1 angular node-> 𝐜𝐨𝐬 𝜽 = 𝟎
--------------- +0.5 pt
𝜽 = 𝝅/2
--------------- +0.5 pt
(d) (2 pts) Which orbital does ππ’π’π’Ž (𝒓, 𝜽, 𝝓) represents?
λ‹΅ +2 pts, λΆ€λΆ„μ μˆ˜ μ—†μŒ
4pz
--------------- +2 pts
3. (Total 8 pts) The wave function of an electron in the lowest (that is, ground) state of the
hydrogen-like He+ atom is
𝟏
πŸ‘πŸ 𝟐
πŸπ’“
𝟏
π’‚πŸŽ
π’‚πŸŽ
πŸ’π…
𝝍(𝒓) = 𝑹𝒏𝒍 (𝒓) βˆ™ π’€π’π’Ž (𝜽, 𝝓) = {( πŸ‘ ) 𝒆𝒙𝒑 (βˆ’ )} βˆ™ ( )
𝟏/𝟐
, π’‚πŸŽ = 𝟎. πŸ“πŸπŸ— × πŸπŸŽβˆ’πŸπŸŽ π’Ž,
(with the assumption that ΞΌ β‰ˆ π’Žπ’† )
(a) (4 pts) In this wave function, calculate the distance π’“π’Žπ’‚π’™ from the nucleus to the location that
the radial probability density is the largest. Express your answers in meter.
(Hint: radial probability density: π’“πŸ [𝑹𝒏𝒍 (𝒓)]𝟐 )
--- 미뢄값이 0μΌλ•Œ
값이 μ΅œλŒ€λΌλŠ” 식
적으면 + 1 pt
--- λ‹΅ +3 pts
(b) (2 pts) Estimate the energy, and 𝒓̅𝒏𝒍 (average distance of the electron from the nucleus) of
given wave function. Express your
answers in Rydberg or Joule for
energy, and meter for distance.
---
E식 +0.5 pt
---
Eλ‹΅ +0.5 pt
---
r식 +0.5 pt
---
rλ‹΅ +0.5 pt
(c) (2 pts) Estimate the energy, and 𝒓̅𝒏𝒍 of the 1s orbital of neutral He atom. The effective nuclear
charge for He is 𝒁𝒆𝒇𝒇 (𝟏) = 𝟏. πŸ”πŸ—. Express your answers in Rydberg or Joule for energy, and meter
for distance.
- E식+0.5 pt
-Eλ‹΅+0.5 pt
- r식 +0.5 pt
- rλ‹΅+0.5 pt
4. (Total 8 pts) Photoelectron spectroscopy studies of sodium atoms (Na, atomic number = 11)
excited by X-rays with wavelength 9.890 x 10-10 m show four peaks in which the electrons have
speeds 7.9924 x 106 m/s, 2.0421 x 107 m/s, 2.0712 x 107 m/s, and 2.0956 x 107 m/s.
(a) (4 pts) Calculate the ionization energy of the electrons in each peak. Express your answer in
kJ/mol.
- 식 +1 pt
- 식 +1 pt
-각
피크
IEλ‹Ή +0.5 pt
(b) (4 pts) Assign each peak to an orbital of the sodium atom.
각 피크당 + 1 pt. ν”Όν¬μ˜ μ „μžμ†λ„μ™€ 그에 λ§žλŠ” μ˜€λΉ„νƒˆμ΄ μΌμΉ˜ν•΄μ•Όλ§Œ μ •λ‹΅, λΆ€λΆ„μ μˆ˜ μ—†μŒ
Peak 1(ve = 7.9924 x 106 m/s) = 1s
------ +1 pt
Peak 2(ve = 2.0421 x 107 m/s) = 2s
------ +1 pt
Peak 3(ve = 2.0712 x 107 m/s) = 2p
------ +1 pt
Peak 4(ve = 2.0956 x 107 m/s) = 3s
------ +1 pt
5. (Total 6 pts) Consider following ions; Be+, C-, Ne2+, P2+, Cl-, and As+.
(Atomic number- Be: 4, C: 6, Ne: 10, P: 15, Cl: 17, As: 33)
(a) (3 pts) Write ground-state electron configurations for listed 6 ions.
각 μ΄μ˜¨λ‹Ή 0.5 pt, λΆ€λΆ„μ μˆ˜ μ—†μŒ
(b) (3 pts) Among them, choose the ion(s) expected to be paramagnetic due to the presence of
unpaired electrons.
Be+, C-. Ne2+, P2+, As+
----
각 μ΄μ˜¨λ‹Ή 0.6 pt, λΆ€λΆ„μ μˆ˜ μ—†μŒ