Hour Exam 3 2013 - HOUR EXAM 3 Chemistry 030.102 Section 1...

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HOUR EXAM 3 NAME (Print clearly)__________________ ____________
Chemistry 030.102 Last name First name
Section 1 – Prof. Dagdigian
April 26, 2013 jhed ID__________________
BE SURE TO PUT YOUR
NAME ON THE LAST PAGE
Calculators allowed, but do not use programmable features.
One sheet of 8 ½ × 11 in. paper with notes may be brought into the exam.
Miscellaneous information:
Speed of light c = 3.00 × 10
8
m s
–1
Avogadro's number N
0
= 6.02 × 10
23
mol
–1
Planck's constant h = 6.63 × 10
–34
J s Electron charge e = 1.60 × 10
–19
C
Particle in a box energies E
n
= n
2
h
2
/(8 mL
2
) Electron mass m
e
= 9.11 × 10
–31
kg
Bohr radius a
0
= 5.29 × 10
–11
m Proton mass m
p
= 1.67 × 10
–27
kg
Planck's formula E = h ν de Broglie's relation: λ = h / p
1 Joule = 1 kg m
2
s
–2
1 eV = 1.602 × 10
–19
J
Heisenberg uncertainty relation: ( Δ x )( Δ p ) ≥ h /4 π
Permittivity of vacuum ε
0
= 8.85 × 10
–12
C
2
J
–1
m
–1
Energies in one-electron atoms E
n
= –(2.18 × 10
–18
J) Z
2
/ n
2
Relationship of light frequency and wavelength: νλ = c
Beer-Lambert law (base 10): –log
10
[ I
S
/ I
R
] = A = ε Cl
First five rows of the periodic table
H He
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
Atomic numbers of the inert gases:
He 2
Ne 10
Ar 18
Kr 36
Xe 54
Rn 86
You may use the space below or on page 9 for calculations, etc .

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1. Orbital Correlation Diagram for Second-Row Diatomic Molecules
One of the oxides of nitrogen is the diatomic molecule nitric oxide (NO). This molecule is a by-
product of combustion in air and is an important signaling molecule in mammals.
(a) Draw the molecular orbital correlation diagram for NO. Please note that the ordering of the
orbitals is similar to that of the O
2
molecule.