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Practice Problems on Electron Configurations and Pauli Exclusion Principle - Prof. Glake A, Assignments of Chemistry

Practice problems related to electron configurations and the pauli exclusion principle. The problems cover topics such as filling subshells according to the pauli principle, identifying incorrect electron configurations, and determining the number of unpaired electrons in various ions. Students studying atomic structure and quantum mechanics will find these problems useful for reinforcing their understanding of these concepts.

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Pre 2010

Uploaded on 08/08/2009

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Chapter 8 Practice Problems [Watts, Fall 2008.]
Problem 22
(a) Having two electrons with the same spin in one orbital (the 2s orbital) is a violation of
the Pauli exclusion principle: if two electrons in the same orbital have the same spin
quantum number, they have all four quantum numbers the same, which is not allowed.
(b) Again, the Pauli exclusion principle is violated. This example has 3 electrons in one
of the 2p orbitals, and two of them have all four quantum numbers the same.
(c) There is a problem with the 3p subshell. In the most stable arrangement, i.e. the
ground state, all three 3p electrons will have the same spin: Hund’s rule says the most
stable arrangement maximizes the number of parallel spins.
Problem 24
(a) The 2p subshell should be filled (i.e. have 6 electrons) before any electrons are put in
the 3s subshell. This example only has 2 electrons in the 2p subshell.
(b) The 3d subshell should be filled (i.e. have 10 electrons) before any electrons are put in
the 4p subshell. Also, it is customary to write the 4s occupation before the 3d.
(c) The 2p subshell should be filled before electrons are put in the 3s subshell.
Problem 26
(a) There is no such thing as a 2d orbital. Since the maximum value of l is n – 1, the first
d subshell is 3d.
(b) There is no such thing as a 3f orbital. The f subshell is 4f.
(c) The 5s subshell is filled after Kr. Then comes 4d, and then 5p. Electrons enter the 4f
subshell after the 6s subshell is filled.
Problem 28
Ar: 1s22s22p63s23p6
Al: 1s22s22p63s23p1
Ne: 1s22s22p6
pf3
pf4

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Chapter 8 Practice Problems [Watts, Fall 2008.] Problem 22 (a) Having two electrons with the same spin in one orbital (the 2s orbital) is a violation of the Pauli exclusion principle: if two electrons in the same orbital have the same spin quantum number, they have all four quantum numbers the same, which is not allowed. (b) Again, the Pauli exclusion principle is violated. This example has 3 electrons in one of the 2p orbitals, and two of them have all four quantum numbers the same. (c) There is a problem with the 3p subshell. In the most stable arrangement, i.e. the ground state, all three 3p electrons will have the same spin: Hund’s rule says the most stable arrangement maximizes the number of parallel spins. Problem 24 (a) The 2p subshell should be filled (i.e. have 6 electrons) before any electrons are put in the 3s subshell. This example only has 2 electrons in the 2p subshell. (b) The 3d subshell should be filled (i.e. have 10 electrons) before any electrons are put in the 4p subshell. Also, it is customary to write the 4s occupation before the 3d. (c) The 2p subshell should be filled before electrons are put in the 3s subshell. Problem 26 (a) There is no such thing as a 2d orbital. Since the maximum value of l is n – 1, the first d subshell is 3d. (b) There is no such thing as a 3f orbital. The f subshell is 4f. (c) The 5s subshell is filled after Kr. Then comes 4d, and then 5p. Electrons enter the 4f subshell after the 6s subshell is filled. Problem 28 Ar: 1s^2 2s^2 2p^6 3s^2 3p^6 Al: 1s^2 2s^2 2p^6 3s^2 3p^1 Ne: 1s^2 2s^2 2p^6

B: 1s^2 2s^2 2p^1 Ca: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 N: 1s^2 2s^2 2p^3 Cl: 1s^2 2s^2 2p^6 3s^2 3p^5 S: 1s^2 2s^2 2p^6 3s^2 3p^4 Zn: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 Problem 32 I don’t have the software to draw orbital diagrams right now. Here are the electron configurations: C: 1s^2 2s^2 2p^2 [C has 2 unpaired electrons in the 2p subshell, each with parallel spins] O: 1s^2 2s^2 2p^4 [O has 4 electrons in the 2p subshell. 2 are in the same orbital, the other 2 are unpaired and have the same spin] K: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 [K has one unpaired electron in the 4s subshell] Al: 1s^2 2s^2 2p^6 3s^2 3p^1 [Al has one unpaired electron in the 3p subshell] S: 1s^2 2s^2 2p^6 3s^2 3p^4 [S is isovalent with O] Mg: 1s^2 2s^2 2p^6 3s^2 [Mg has no unpaired electrons] Problem 36 Ga3+^ The configuration is [Ar]3d^10 V3+^ The configuration is [Ar]3d^1 I-^ The configuration is [Kr]5s^2 4d^10 5p^6 [same as Xe] Sb3+^ The configuration is [Kr]5s^2 4d^10 In V3+^ there is one unpaired electron in one of the 3d orbitals. In the other ions all electrons are paired.

The order one might predict is Ba < Sc < Zn < Al < Br < Kr [clearly, we expect Ba to have a relatively small I.E., while Br and Kr have large I.E.s]. The middle is a little trickier. In fact the order is Ba < Al < Sc < Zn < Br < Kr. Problem 68 Exactly how one quantifies non-metallic character is debatable. The official solutions manual correlates the electron affinity with non-metallic character, giving the order Pb < Sb < As < N < Si < O < Br < F Some aspects of this order make good sense, some could be strongly criticized. Pb is clearly a metal. Generally, going down a group, the metallic character increases. Therefore, Sb is more metallic than As, which is more metallic than N. Similarly, one could say that F is more non-metallic than Br. [I’m not sure if this is very meaningful: both F and Br are clearly non-metals]. Saying Si is more non-metallic than N is, in my opinion, not a very sensible or useful statement. N is more electronegative than Si. Si is a semi-conductor. Si forms polymeric solid oxides (e.g. quartz). N forms acidic oxides.