AP Chemistry - Problem Drill 10: Atomic Structures and Electron Configuration

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1 AP Chemistry - Problem Drill 10: Atomic Structures and Electron Configuration No. 1 of 10 Instructions: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 1. A cation is a positive ion where its atom or group of atoms has lost one or more electrons. How many electrons does Sr 2+ have? (A) 38 (B) 36 (C) 40 (D) 88 This would be a neutral Sr atom. B. Correct. Good job! This would be for a +2 ion, with two less electrons than neutral Sr atom. Sr 2+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6. This would be a -2 ion. This is using mass number rather than atomic number to determine electrons. Sr s atomic number is 38, which means as a neutral atom it has 38 electrons. Sr +2 has had 2 electrons removed. It now has 36 electrons with the same electronic configuration as [Kr] or 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6. The correct answer is (B).

2 No. 2 of Box Orbital Notation is one way to show electron configurations where orbitals and spin states are clearly labeled. Which is the correct electronic configuration for 33As? (A) 1s 2s 2p 3s 3p (B) 1s 2s 2p 3s 3p _ _ _ (C) 1s 2s 2p 3s 3p 4s 3d 4p (D) 1s 2s 2p 3s 3p 4s 3d 4p _ _ This one doesn t have 33 electrons. This one doesn t have 33 electrons. This one didn t put one electron in each 4p orbital before doubling up. D. Correct! Good job! This one has the correct electrons configuration for 33 As the three outer shell p electrons are filled in each orbital with the same spin before doubling up. As has 33 electrons. This is the orbital box notation. According to the Hund s Rule, the last three p electrons must be filled in each p orbital with the same spin, not to double up. Its long spectroscopic notation is: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3. The correct answer is (D).

3 No. 3 of The Aufbau principle is the build-up method to determine the electron configuration of an atom, molecule or ion. Which of the following is correct for 14 Si? (A) 1s 2 2s 2 2p 6 3s 2 3p 2 (B) 1s 2 1p 6 2s 2 2p 4 (C) 1s 2 2s 2 2p 4 3s 2 3p 4 (D) 1s 2 2s 2 2p 2 3s 2 3p 2 A. Correct! Good job! There are total of 14 electrons, filled in sequentially. The outer shell 2p with 2 electrons on each orbital with the same spin. There is no 1p subshell. This one doesn t have 14 electrons. This one doesn t have 14 electrons. Si has 14 electrons. Fill them in order of 1s > 2s > 2p > 3s > 3p. The outer shell 2p with 2 electrons on each orbital with the same spin. The correct answer is (A).

4 No. 4 of The noble gases have the most stable electron configurations since all their subshells are filled. They are relatively inert. With their subshells all filled, noble gases can be used as a short hand way to represent the core electrons. The valence electrons are still explicitly written out. Write the noble gas notation for 34 Se. (A) [Ar] 1s 2 2s 2 2p 6 3s 2 3p 4 (B) [Ar] 4s 2 3d 10 4p 4 (C) [Kr] 4s 2 3d 10 4p 4 (D) [Kr] 1s 2 2s 2 2p 6 3s 2 3p 4 When using noble gas notation, you don t start at 1s. B. Correct! Good job! The 34 Se s core electrons are represented by [Ar] (1s 2 2s 2 2p 6 3s 2 3p 6 ) and the outer shells are 4s 2 3d 10 4p 4. Remember chose the noble gas that is closest without going over to the number of electrons needed. When using noble gas notation, you don t start at 1s. Se has 34 electrons. The noble gas closest to 34 without going over is Ar (18). The core electrons are represented by [Ar] (1s 2 2s 2 2p 6 3s 2 3p 6 ) and the outer shells are 4s 2 3d 10 4p 4. The correct answer is (B).

5 No. 5 of The typical quantum mechanics model uses four quantum numbers (n, l, m l and m s ) to describe any unique electronic state of an electron. There is one invalid quantum number set below, which one? (A) 1, 1, -1, -½ (B) 2, 0, 0, +½ (C) 3, 2, 2, +½ (D) 2, 1, 1, +½ A. Correct. Good job! This is for a 1p subshell (1,1), which doesn t exist. This is for a 2s on line 0 with an up arrow, which is all possible. This for a 3d on line 2 with an up arrow, which is all possible. This is for a 2p on line 1 with an up arrow, which is possible. Set A is for a 1p subshell, which does not exist. The rest of the sets are permissible. The correct answer is (A).

No. 6 of 10 6. The electronic configuration is the distribution of electrons of an atom or molecule in atomic or molecule orbitals. Sodium oxide is used in ceramics and glasses.

6 No. 6 of The electronic configuration is the distribution of electrons of an atom or molecule in atomic or molecule orbitals. Sodium oxide is used in ceramics and glasses. Which of the following is the electron configuration for Na in Na 2 O? (A) 1s 2 2s 2 2p 6 3s 1 (B) 1s 2 2s 2 2p 6 3s 2 (C) 1s 2 2s 2 2p 6 (D) 1s 1 2s 1 2p 3 3s 1 3p 3 4s 1 This doesn t have 10 electrons. This doesn t have 10 electrons. C. Correct! Good job! Sodium has 11 electrons in total. With +1, one electron is removed. Therefore Na + has 11-1 = 10 electrons. This one has 10 electrons correctly placed. This one only puts half the number of electrons in each orbital that are necessary. Na in Na 2 O is Na +. Na has 11 electrons, so Na + will have 10 electrons. Only answers C & D have 10 electrons. Answer D uses only half as many electrons as possible in each subshell. The correct answer is (C).

7 No. 7 of Aluminum ( 13 Al) is the third most abundant element (beside oxygen and silicon) and the most abundant metal in the Earth s crust. Which is the correct electronic configuration for 13 Al? (A) [Ne] 3s 2 3p 1 (B) [He] 3s 2 3p 1 (C) [Ne] 1s 2 2s 1 (D) [He] 1s 2 2s 1 A. Correct. Good job! This one has 13 electrons with the correct noble gas configuration. This one doesn t have 13 electrons. The core electrons are mis-represented by [He]. When using noble gas configuration, you don t begin with 1s. This one doesn t have 13 electrons. The core electrons are mis-represented by [He]. Al has 13 electrons. Ne has 10 electrons & H has 2 electrons. Only answers A & C have 13 electrons. Answer D starts over with 1s, not where Ne left off. Choice E has three electrons in an s subshell when s subshells can only hold 2 electrons. The correct answer is (A).

8 No. 8 of Electrons are built up onto the orbitals one level at a time according to the Aufbau Principle. There are also exceptions to this rule. Which are the correct configurations for Cu and Cu +? (A) [Ar]3d 9 4s 2 and [Ar]3d 9 4s 1 (B) [Ar]3d 9 4s 2 and [Ar]3d 8 4s 2 (C) [Ar]3d 10 4s 1 and [Ar]3d 10 (D) [Ar]3d 10 4s 1 and [Ar]3d 9 4s 1 The 4s should not be doubled up since half-filled is more favorable. The 4s should not be doubled up since half-filled is more favorable. C. Correct! Good job! Use the half-filled rule to build up the outer shells. The removal of one electron is from the highest occupied sublevel. The neutral species is correct but the removal of electron should be on 4s first, not 3d. This is an exception to the rule. Half-filled sublevels have lower energy since unpaired electrons are more stable. 29 Cu has 29 electrons with the core electrons represented by [Ar] and the outer shell electrons are piled up as half-filled 3d 10 4s 1 for more stability. To make a cation Cu +, we need to take away one electron from the highest energy sublevel, 4s. Therefore, Cu + = [Ar]3d 10. The correct answer is (C).

9 No. 9 of Isoelectric elements and/or ions have the same number of electrons and configuration. Isoelectric chemical species tend to have similar chemical properties. All pairs below are isoelectric except. (A) Be 2+ and He (B) S 2- and Cl (C) Mg 2+ and Na + (D) Ca + and K These two species are isoelectric with the same configuration of 1s 2. B. Correct. Good job! S 2- has the configuration of 1s 2 2s 2 2p 6 3s 2 3p 6 and Cl has a slightly different configuration 1s 2 2s 2 2p 6 3s 2 3p 5. They are not isoelectric pair. These two species are isoelectric with the same configuration of 1s 2 2s 2 2p 6. These two species are isoelectric with the same configuration of [Ne]4s 1. S 2- has the configuration of 1s 2 2s 2 2p 6 3s 3 3p 6 and Cl has a slightly different configuration 1s 2 2s 2 2p 6 3s 3 3p 5. Each has different number of electrons. The correct answer is (B).

10 No. 10 of The four quantum numbers described an electron are the principal (n), angular or azimuthal (l), magnetic (m l ) and spin (m s ). The first three quantum numbers describe the size, shape, and orientation in space of the orbitals on an atom. The last one (m s ) describes the spin state of an electron. The possible values of each quantum number correlate to one another. Which of the following sets of quantum numbers (n,l,m l,m s ) is not possible? (A) 1, 0, 0, -½ (B) 2, 0, -1, +½ (C) 2, 1, -1, +½ (D) 3, 2, 1, +½ 1s with a down arrow on line 0 is allowed. B. Correct. Good job! There is no -1 line in the 2s subshell. This configuration is not possible. 2p with an up arrow on the -1 line is allowed. An up arrow on the 1 line in a 3d subshell is allowed. The second set is an 2s subshell. The 2s subshell only has 1 orbital, which means that the 3 rd number has to be 0. The correct answer is (B).

CLEP Chemistry - Problem Drill 10: Atomic Structure and Electron Configuration

CLEP Chemistry - Problem Drill 10: Atomic Structure and Electron Configuration No. 1 of 10 Instructions: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 1.