Name:_____________
Chemistry 112
Final Exam (Non comprehensive part)
Constants
h = 6.626x10-34 J . s (Planck’s constant) R= 8.3145 J/(K . mol) (Gas Constant)
c=2.998x10+8 m/s (speed of light)
R=0.08206 L . atm/(K .mol) (Gas Constant)
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N=6.022x1023 mol-1 (Avogadro’s number)
mn =1.675x10 kg (mass of neutron)
Remember- Show all work for partial credit.
Note: Special rules for final
1.) I will give you 5 points for putting your name at the top of this page.
This makes the test worth 215 points total. To make the test only worth 200 points
2.) You may skip one 15 point question.
1. (15 points) Violet light has a wavelength of about 400 nm.
A. What is the frequency of this light in units of Hertz (sec-1)
c = 8<
2.998x108 m/s = 400x10-9 m X
X = 2.998x108/400x10-9
=7.495x1014 s-1 (or Hz)
B. What is the energy of a single photon of this light?
E = h<
= 6.626x10-34 J@s ×7.495x1014s-1
= 4.97x10-19 J
C. What is the energy of one mole of these photons?
E = E/photon × Avogadro’s #
= 4.97x10-19 J × 6.022x1023
= 2.99x105 J/mol
= 299 kJ/mol
2. (15 points) Using the Bohr model, the energy available to an electron in the hydrogen atom can be
found using the equation:
E = − 2.178x10
−18
 Z2 
J 2 
n 
A. Find the energy associated with a transition from n=2 to n=1.
E = Efinal - E initial
E = -2.178x10-18(12/12) - [-2.178x10-18(12/22)
= -2.178x10-18(1-1/4)
= -1.634x10-18 J
B. In the lab I mentioned that this transition does exist but isn’t visible. Using your answer to
1B above explain this.
This transition has a much higher E than the transition in 1A, therefor it is in the ultraviolet region
2
of the spectrum and outside our visual range.
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3. (15 points) For each of the four quantum numbers: (1) give the name of the quantum number, (2)
give the abbreviation of the quantum number, (3) give a short explanation of the physical attributes of
the quantum number (energy, shape , etc.), and (4) tell the range in values for this quantum number.
(Assume a principle quantum number of n.)
Principal Quantum Number (n) ; n = integer >0, gives the overall energy level
Angular Momentum Quantum # (l); l = integers between o and n, inclusive;
gives the shape of the orbital
Magnetic Quantum Number (ml) ; ml = an integer between -l and +l inclusive;
gives the orientation of the orbital
Electron Spin Quantum Number ms; ms = + ½ or - ½ ; spin of the electron
4.(15 points) What is the maximum number of electrons that can have the following designations:
(NOTE: If the designation is not allowed write NA for your answer)
_18__ A. n=3
_NA_ B. 3f
_NA_ C. n=1, l=1, ml = -1
__1__ D. n=2, l=1, ml = -1, ms = +1/2
__6__ E. 2p
5.(15 points) Write the expected electron configuration for the following elements or ions
Ca
[Ar] 4s2
Fe3+
[Ar] 4s23d3
O
[He] 2s22p4
O2-
[Ne]
6. (10 points) The electronic configurations of N is 1s2 2s2 2p3.
How many valence shell p orbitals are there? ___3___
Would these orbitals be filled with paired or unpaired electrons? unpaired
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7. (15 points) What is the difference between an electron affinity, an ionization energy, and a reduction
potential?
Electron affinity = X(g) + e- - X-(g)
Ionization E = X(g) -X+(g) + eReduction Potential = X(aq) + e- -X-(aq)
8.(15 points) Rank the following
Largest atom to smallest atom
Si, Rb, Sn
Rb>Sn>Si
As, Br, F
Most negative electron affinity to least negative electron affinity
(Most Neg) F Br As (Least negative)
K, Ga, Cs
Smallest first ionization energy to largest ionization energy
Cs<K<Ga
O2-, F-, Li+
Largest ion to smallest ion
O2- >F- >Li+
Most electronegative to least electronegative
S, Sn, Te
S>Te>Sn
9. (15 points) Give the name and formula of each of the binary compounds formed with the elements:
Li and N
Li3N Lithium nitride
Al and F
AlF 3 Aluminum fluoride
Na and O
Na2O Sodium oxide
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10. (20 points) Calculate the lattice energy for the formation of LiF(s) from Li+(g) and F-(g). Some
energies that might be useful are:
Li(s) + 1/2F2 (g) 6 LiF(s)
Li(s) 6Li(g)
Li(g)6Li+ + eF2(g) 62F(g)
F(g) + e- 6 F-
-617 kJ/mol (heat of formation)
161 kJ/mol (sublimation)
520 kJ/mol (ionization )
154 kJ/mol (bond energy)
-328 kJ/mol (electron affinity)
Lattice Energy = Li+(g) + F-(g) - LiF(s)
Li(s) + 1/2F2 (g) 6 LiF(s)
Li(g) - Li(s)
Li+(g)+ e--Li(g)
1/2[2F(g) -F2(g)
F- (g)- F(g) + eNet:
-617 kJ/mol
-161 kJ/mol
-520 kJ/mol
½(-154) kJ/mol
+328 kJ/mol
Li+(g) + F-(g) - LiF(s) -617 -161 -520 -77 + 328 = -1047 kJ/mol