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[CHEM 1120]Midterm 1
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[CHEM 1120] All questions and answers provided
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Test Bank for Chemistry A Molecular Approach 2nd Edition by Tro Chapter 10 Chemical Bonding II Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory
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CHEM 1120 – MIDTERM 1 -- 36 MARKS1. Fill in the blank or multiple choice (7 marks):
a) What are the possible values of ml when L = 1? When l=1, it’s the p orbital. Ml= -1, 0,1
b) Which of the following orbitals can have 5 different, but equal-in-energy orientations? (choose one)
I. 3p
II. 3s
III. 4d
IV. 4f
c) Gamma ray radiation has LONGER / SHORTER wavelength than infrared radiation.
d) What is the name of the quantum number that characterizes the shape of an atomic orbital?
L = angular momentum quantum number
e) Hund’s rule states:
I. No two electrons in an atom can have the same 4 quantum numbers
II. The mass of an electron depends on how fast it is travelling
III. Electrons will occupy orbitals singly before pairing
IV. It is impossible to know the exact location and momentum of a particle at any given time
f) As the principle quantum number for a p orbital increases, the number of radial nodes _________?
Increases
g) The two experiments that demonstrated that light could act as both a wave and a particle were
______________ & ________________ Young’s Slit light experiment & de Broglie
2. The work function for an unknown metal is 312 kJ mol-1 . What wavelength of light is required to eject an
electron with a velocity of 3.13 x 105 m/s? The mass of an electron is 9.11 x 10-31 kg. (5 marks)
Note: 1 Joule = 1 kg m2 / s2 1 ℎ𝑐
Formula 1: KE = 2 m v2 Formula 2: KE = ( 𝜆 ) - Φ
1 ℎ𝑐
KE = mv KE = ( ) – Φ
2 𝜆
1 6.626 𝑥 10−34 Js)(3.00 x 108 𝑚/𝑠) kJ
KE = 2(9.11 x 10-31 kg)(3.13 x 105 m/s)2 4.46 x 10-20 J = ( 𝜆
– 312 mol
1 1.99 𝑥 10−25 𝐽𝑚 𝑘𝐽 1000 𝐽 1 𝑚𝑜𝑙
KE = 2(8.93 x 10-20 J) 4.46 x 10-20 J = ( 𝜆
) – 312 𝑚𝑜𝑙 x 𝑘𝐽 6.022 𝑥 1023
1.99 𝑥 10−25 𝐽𝑚
KE = 4.46 x 10-20 J 4.46 x 10-20 J = ( ) – 5.18 x 10-19 J
𝜆
1 𝑛𝑚
λ = 3.54 x 10-7 metres *can convert to nm by x
1 𝑥 109 𝑚
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