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Chapter 8: Problem 25

Write the full ground-state electron configuration for each: (a) Cl (b) Si (c) Sr

Short Answer

Expert verified
\( \text{Cl: } [1s^2 2s^2 2p^6 3s^2 3p^5] \) \( \text{Si: } [1s^2 2s^2 2p^6 3s^2 3p^2] \) \( \text{Sr: } [1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2] \)

Step by step solution

01

- Determine the Atomic Number

Identify the atomic number for each element from the periodic table. (a) Cl has an atomic number of 17. (b) Si has an atomic number of 14. (c) Sr has an atomic number of 38.
02

- Use the Aufbau Principle

According to the Aufbau principle, electrons fill the lowest energy orbitals first. Use the order of orbital filling: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.
03

- Write Electron Configuration for Cl

Add electrons to the orbitals for Cl until you reach the atomic number:\[1s^2 2s^2 2p^6 3s^2 3p^5\]
04

- Write Electron Configuration for Si

Add electrons to the orbitals for Si until you reach the atomic number:\[1s^2 2s^2 2p^6 3s^2 3p^2\]
05

- Write Electron Configuration for Sr

Add electrons to the orbitals for Sr until you reach the atomic number:\[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Number
The atomic number of an element is a fundamental property that defines the number of protons in the nucleus of an atom. It also indicates the number of electrons in a neutral atom. For example, chlorine (Cl) has an atomic number of 17, silicon (Si) has an atomic number of 14, and strontium (Sr) has an atomic number of 38. With these values, we can determine how many electrons each element has, which is the first step in writing their electron configurations.
  • Cl: 17 electrons
  • Si: 14 electrons
  • Sr: 38 electrons
Aufbau Principle
The Aufbau principle is a key guideline in chemistry for determining the electron configuration of an atom. According to this principle, electrons occupy the lowest energy orbitals available. This principle translates into a specific order in which orbitals are filled:
  • 1s
  • 2s
  • 2p
  • 3s
  • 3p
  • 4s
  • 3d
  • 4p
  • 5s
    • and so on.
For example, to write the electron configuration for chlorine, knowing it has 17 electrons, you fill the orbitals starting from the lowest energy level:
  • 1s (2 electrons)
  • 2s (2 electrons)
  • 2p (6 electrons)
  • 3s (2 electrons)
  • 3p (5 electrons, totaling 17 electrons)
Electron Orbitals
Electron orbitals are regions around the nucleus where electrons are likely to be found. They are sub-divided into different types based on their shape: s, p, d, and f orbitals.
  • s orbitals: Spherical in shape, each s orbital can hold a maximum of 2 electrons.
  • p orbitals: Dumbbell-shaped, each set of p orbitals can hold a total of 6 electrons (2 in each of the 3 p orbitals).
  • d orbitals: More complex shapes, can hold up to 10 electrons (in 5 d orbitals).
  • f orbitals: Even more complex shapes, can hold up to 14 electrons (in 7 f orbitals).
Understanding this breakdown helps in filling out the electron configurations. For Strontium (Sr) with 38 electrons, the electron configuration extends to the 5s orbital: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2.

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Most popular questions from this chapter

Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element: (a) \([\mathrm{Ar}] 4 s^{2} 3 d^{5}\) (b) \([\mathrm{Kr}] 5 s^{2} 4 d^{2}\)

Draw a partial (valence-level) orbital diagram, and write the condensed ground-state electron configuration for each: (a) Ti (b) Cl (c) V

The energy difference between the \(5 d\) and \(6 s\) sublevels in gold accounts for its color. Assuming this energy difference is about \(2.7 \mathrm{eV}\) (electron volts; \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\) ), explain why gold has a warm yellow color.

A fundamental relationship of electrostatics states that the energy required to separate opposite charges of magnitudes \(Q_{1}\) and \(Q_{2}\) that are a distance \(d\) apart is proportional to \(\frac{Q_{1} \times Q_{2}}{d} .\) Use this relationship and any other relevant factors to explain the following: (a) The IE \(_{2}\) of He \((Z=2)\) is more than twice the \(\mathrm{IE}_{1}\) of \(\mathrm{H}(Z=1)\). (b) The IE \(_{1}\) of He is less than twice the \(\mathrm{IE}_{1}\) of \(\mathrm{H}\).

Identify each element below, and give the symbols of the other elements in its group: (a) \([\mathrm{Ar}] 4 s^{2} 3 d^{10} 4 p^{2}\) (b) \([\mathrm{Ar}] 4 s^{2} 3 d^{7}\) (c) \([\mathrm{Kr}] 5 s^{2} 4 d^{5}\)
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