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Chapter 10: Problem 97

Without referring to your textbook or a periodic table, write the full electron configuration, the orbital box diagram, and the noble gas shorthand configuration for the elements with the following atomic numbers. a. \(Z=21\) b. \(Z=15\) c. \(Z=36\) d. \(Z=38\) e. \(Z=30\)

Short Answer

Expert verified
a. \(Z=21\) 1. Electron Configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿 2. Orbital Box Diagram: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫� 3. Noble Gas Shorthand Configuration: [Ar] 4s虏 3d鹿 b. \(Z=15\) 1. Electron Configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鲁 2. Orbital Box Diagram: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫� 鈫� 鈫� 3. Noble Gas Shorthand Configuration: [Ne] 3s虏 3p鲁 c. \(Z=36\) 1. Electron Configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4p鈦� 5s虏 4d虏 2. Orbital Box Diagram: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 | 4p: 鈫戔啌 鈫戔啌 鈫戔啌 | 5s: 鈫戔啌 | 4d: 鈫� 鈫� 3. Noble Gas Shorthand Configuration: [Kr] d. \(Z=38\) 1. Electron Configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4p鈦� 5s虏 4d鲁 2. Orbital Box Diagram: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 | 4p: 鈫戔啌 鈫戔啌 鈫戔啌 | 5s: 鈫戔啌 | 4d: 鈫� 鈫� 鈫� 3. Noble Gas Shorthand Configuration: [Kr] 5s虏 4d鲁 e. \(Z=30\) 1. Electron Configuration: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 2. Orbital Box Diagram: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 3. Noble Gas Shorthand Configuration: [Ar] 4s虏 3d鹿鈦�

Step by step solution

01

1. Electron Configuration

For an element with \(Z = 21\), the electron configuration would be: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿
02

2. Orbital Box Diagram

The orbital box diagram for \(Z=21\) can be represented as: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫�
03

3. Noble Gas Shorthand Configuration

The noble gas preceding \(Z=21\) is Argon (\(Ar\), \(Z=18\)). So the shorthand configuration would be: [Ar] 4s虏 3d鹿 b. \(Z=15\)
04

1. Electron Configuration

For an element with \(Z = 15\), the electron configuration would be: 1s虏 2s虏 2p鈦� 3s虏 3p鲁
05

2. Orbital Box Diagram

The orbital box diagram for \(Z=15\) can be represented as: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫� 鈫� 鈫�
06

3. Noble Gas Shorthand Configuration

The noble gas preceding \(Z=15\) is Neon (\(Ne\), \(Z=10\)). So the shorthand configuration would be: [Ne] 3s虏 3p鲁 c. \(Z=36\)
07

1. Electron Configuration

For an element with \(Z = 36\), the electron configuration would be: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4p鈦� 5s虏 4d虏
08

2. Orbital Box Diagram

The orbital box diagram for \(Z=36\) can be represented as: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 | 4p: 鈫戔啌 鈫戔啌 鈫戔啌 | 5s: 鈫戔啌 | 4d: 鈫� 鈫�
09

3. Noble Gas Shorthand Configuration

The noble gas preceding \(Z=36\) is Krypton (\(Kr\), \(Z=36\)). So the shorthand configuration would be: [Kr] d. \(Z=38\)
10

1. Electron Configuration

For an element with \(Z = 38\), the electron configuration would be: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦� 4p鈦� 5s虏 4d鲁
11

2. Orbital Box Diagram

The orbital box diagram for \(Z=38\) can be represented as: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 | 4p: 鈫戔啌 鈫戔啌 鈫戔啌 | 5s: 鈫戔啌 | 4d: 鈫� 鈫� 鈫�
12

3. Noble Gas Shorthand Configuration

The noble gas preceding \(Z=38\) is Krypton (\(Kr\), \(Z=36\)). So the shorthand configuration would be: [Kr] 5s虏 4d鲁 e. \(Z=30\)
13

1. Electron Configuration

For an element with \(Z = 30\), the electron configuration would be: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� 4s虏 3d鹿鈦�
14

2. Orbital Box Diagram

The orbital box diagram for \(Z=30\) can be represented as: 1s: 鈫戔啌 | 2s: 鈫戔啌 | 2p: 鈫戔啌 鈫戔啌 鈫戔啌 | 3s: 鈫戔啌 | 3p: 鈫戔啌 鈫戔啌 鈫戔啌 | 4s: 鈫戔啌 | 3d: 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌 鈫戔啌
15

3. Noble Gas Shorthand Configuration

The noble gas preceding \(Z=30\) is Argon (\(Ar\), \(Z=18\)). So the shorthand configuration would be: [Ar] 4s虏 3d鹿鈦�

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Most popular questions from this chapter

When students first see a drawing of the \(p\) orbitals, they often question how the electron is able to jump through the nucleus to get from one lobe of the \(p\) orbital to the other. How would you explain this?

Using the symbol of the previous noble gas to indicate the core electrons, write the valence shell electron configuration for each of the following elements. a. phosphorus, \(Z=15\) b. chlorine, \(Z=17\) c. magnesium, \(Z=12\) d. \(\operatorname{zinc}, Z=30\)

Though all the elements in a given period (horizontal row) of the periodic table have their valence electrons in the same types of orbitals, the sizes of the atoms decrease from left to right within a period. Explain why.

Write the full electron configuration \(\left(1 s^{2} 2 s^{2}, \text { etc. }\right)\) for each of the following elements. a. sodium, \(Z=11\) b. cesium, \(Z=55\) c. nitrogen, \(Z=7\) d. beryllium, \(Z=4\)

Arrange the following sets of elements in order of increasing atomic size. a. \(\mathrm{Sn}, \mathrm{Xe}, \mathrm{Rb}, \mathrm{Sr}\) b. \(\mathrm{Rn}, \mathrm{He}, \mathrm{Xe}, \mathrm{Kr}\) c. \(\mathrm{Pb}, \mathrm{Ba}, \mathrm{Cs}, \mathrm{At}\)
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