Chapter 2: Problem 19
Write the electron configuration for the following elements, using both the complete and the shorthand notations. Indicate which electrons are the valence electrons. (a) F (b) \(\mathrm{Al}\) (c) \(\mathrm{As}\)
Short Answer
Expert verified
(a) F: complete: \(1s^2 \ 2s^2 \ 2p^5\), shorthand: \([\mathrm{He}] \ 2s^2 \ 2p^5\); (b) Al: complete: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1\), shorthand: \([\mathrm{Ne}] \ 3s^2 \ 3p^1\); (c) As: complete: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^3\), shorthand: \([\mathrm{Ar}] \ 4s^2 \ 3d^{10} \ 4p^3\).
Step by step solution
01
Identify the Atomic Number
To write the electron configuration, we first determine the atomic number of each given element. The atomic number represents the number of protons and also the number of electrons in a neutral atom. \(\mathrm{F}\), or Fluorine, has an atomic number of 9. \(\mathrm{Al}\), or Aluminum, has an atomic number of 13. \(\mathrm{As}\), or Arsenic, has an atomic number of 33.
02
Write the Complete Electron Configuration for F
Starting with \(\mathrm{F}\), which has an atomic number of 9, we fill the electron orbitals in order of increasing energy levels: \(1s^2 \ 2s^2 \ 2p^5\). The electrons fill the orbitals in the order starting from \(1s\) to \(2p\). Valence electrons: the electrons in the highest energy level, \(2s^2 \ 2p^5\).
03
Write the Shorthand Electron Configuration for F
The shorthand configuration uses the previous noble gas as a reference. For Fluorine, the previous noble gas is Helium (\([\mathrm{He}]\)). Therefore, Fluorine's shorthand electron configuration is \([\mathrm{He}] \ 2s^2 \ 2p^5\).
04
Write the Complete Electron Configuration for Al
Aluminum has an atomic number of 13. The complete electron configuration is \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1\). The electrons in the highest principal quantum number (3) are the valence electrons: \(3s^2 \ 3p^1\).
05
Write the Shorthand Electron Configuration for Al
For Aluminum, the shorthand configuration uses Neon (\([\mathrm{Ne}]\)) as the previous noble gas. The shorthand electron configuration for Aluminum is \([\mathrm{Ne}] \ 3s^2 \ 3p^1\).
06
Write the Complete Electron Configuration for As
Arsenic, with an atomic number of 33, has the complete electron configuration: \(1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^{10} \ 4p^3\). The highest energy level, \(4\), contains the valence electrons: \(4s^2 \ 4p^3\).
07
Write the Shorthand Electron Configuration for As
The shorthand configuration for Arsenic uses Argon (\([\mathrm{Ar}]\)) as the noble gas. The shorthand electron configuration is \([\mathrm{Ar}] \ 4s^2 \ 3d^{10} \ 4p^3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Valence Electrons
Valence electrons are like the keys to the kingdom when it comes to chemical reactions. They are the electrons located in the outermost energy level of an atom. These special electrons determine how an atom will interact or bond with other atoms.
- For Fluorine (F), with the electron configuration of \(1s^2 \, 2s^2 \, 2p^5\), the valence electrons are in the 2s and 2p orbitals, totaling 7.
- For Aluminum (Al), which has the configuration \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^1\), the valence electrons are in the 3s and 3p orbitals, adding up to 3.
- Arsenic (As), with its configuration \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^3\), has 5 valence electrons in the 4s and 4p orbitals.
Atomic Number
An element's atomic number is fundamental in defining its identity. It corresponds to the number of protons in the nucleus of an atom and by extension, in a neutral atom, equals the number of electrons.
- Fluorine's atomic number is 9, meaning it has 9 protons and 9 electrons.
- Aluminum's atomic number is 13, so it contains 13 protons and 13 electrons.
- Arsenic has an atomic number of 33, indicating 33 protons and 33 electrons.
Shorthand Notation
Shorthand notation is a concise way to express the electron configuration of an element by using the noble gas from the preceding row of the periodic table as a reference point. This method simplifies the notation, focusing only on the outer electron layers by representing them after the noble gas.
- For Fluorine (F), the shorthand notation is \([\text{He}] \, 2s^2 \, 2p^5\), with Helium ([\text{He}]) as the noble gas configuration before it.
- Aluminum (Al) uses Neon ([\text{Ne}]) for its shorthand: \([\text{Ne}] \, 3s^2 \, 3p^1\).
- Arsenic (As) is represented with Argon ([\text{Ar}]): \([\text{Ar}] \, 4s^2 \, 3d^{10} \, 4p^3\).
Complete Electron Configuration
The complete electron configuration provides a detailed map of where each electron resides around the nucleus of an atom. It lists all the electrons in an atom, showing how they are distributed across different energy levels and sublevels.
- Fluorine (F) is written as \(1s^2 \, 2s^2 \, 2p^5\), indicating each level filled in sequence up to 9 electrons.
- Aluminum (Al) has the configuration \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^1\), totaling up to 13 electrons.
- Arsenic (As) is represented as \(1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^{10} \, 4p^3\), which adds up to 33 electrons.