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The ionization cnergies for the removal of the first electron from atoms of \(\mathrm{Si}, \mathrm{P}, \mathrm{S},\) and \(\mathrm{Cl}\) are listed below. Briefly rationalize this trend.

Short Answer

Expert verified
Ionization energy increases from Si to Cl due to increasing effective nuclear charge across the period.

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the amount of energy required to remove an electron from an atom in its gaseous state. It generally increases across a period on the periodic table from left to right due to increasing effective nuclear charge.
02

Observation of Ionization Energies

In the elements given (Si, P, S, Cl), note that ionization energy should increase from Si to Cl, because these elements are in the same period and move from left to right on the periodic table.
03

Si to P Comparison

Phosphorus (P) has a higher ionization energy than silicon (Si). This is because phosphorus has more protons, leading to a stronger attraction between the nucleus and electrons, requiring more energy to remove the first electron.
04

P to S Comparison

Sulfur (S) has a slightly higher ionization energy than phosphorus (P), but the increase is not as pronounced. This is due to electron-electron repulsion in the 3p orbital, which slightly offsets the higher nuclear charge.
05

S to Cl Comparison

Chlorine (Cl) has a significantly higher ionization energy compared to sulfur (S). This increase is due to the even higher effective nuclear charge relative to the distance of the outer electrons from the nucleus, intensifying the pull on its electrons.
06

Conclusion of Trend

The trend of increasing ionization energy from Si to Cl is due to an increasing number of protons in the nucleus, which enhances the effective nuclear charge and hence the attraction between the nucleus and valence electrons across the period.

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Key Concepts

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

Periodic Table
The periodic table is a powerful tool that helps us predict the properties of elements, including their ionization energy. It is arranged primarily by increasing atomic number, and elements are grouped in rows and columns based on their chemical properties.
As you move across a row from left to right, the atomic number increases, meaning each element has more protons and electrons than the one before it. This increase in protons leads to a greater attraction between the nucleus and the valence electrons.
  • Elements in the same row are called a period.
  • Ionization energy usually increases across each period.
The elements silicon (Si), phosphorus (P), sulfur (S), and chlorine (Cl) are all in the same period, showcasing this trend as you move from left to right.
Effective Nuclear Charge
Effective nuclear charge is a key factor influencing ionization energy. Simply put, it is the net positive charge experienced by an electron in a multi-electron atom.
As electrons fill orbitals, they experience attraction to the positive charge of the nucleus but are also repelled by other electrons present in the atom. This repulsion is called shielding or screening effect.
Effective nuclear charge increases across a period because additional electrons are added to the same energy level without substantial shielding, while the number of protons increases:
  • Higher proton count increases the nuclear charge felt by the valence electrons.
  • Less effective shielding in the same period means a stronger attraction to the nucleus.
This is why as we move from Si to Cl, the effective nuclear charge gets stronger, making it harder to remove an electron.
Electron Configuration
Electron configuration describes how electrons are distributed in an atom's orbitals. Each electron configuration contributes to an element’s chemical properties, including its ionization energy.
Electrons fill orbitals according to the Aufbau principle, filling lower energy levels first.
  • For silicon: [Ne] 3s² 3p²
  • For phosphorus: [Ne] 3s² 3p³
  • For sulfur: [Ne] 3s² 3p⁴
  • For chlorine: [Ne] 3s² 3p⁵
The distribution of electrons, especially in the 3p orbitals for these elements, affects their ionization energies. The arrangement leads to stronger nuclear attraction, influencing how readily each element gives up electrons as you move across the period.
Trends in Ionization Energy
Ionization energy trends are important in understanding the chemical behavior of elements. Across a period on the periodic table from left to right, ionization energy generally increases. This is primarily due to the increase in effective nuclear charge that occurs with each additional proton in the nucleus.
As the atomic radius decreases, the outer electrons are held more tightly to the nucleus. For the given elements:
  • Silicon to phosphorus: The increase is consistent with added protons and stronger nuclear pull.
  • Phosphorus to sulfur: Marginal increase due to electron pairing and repulsion in the same orbital.
  • Sulfur to chlorine: A more pronounced increase due to fewer shielding effects and continued strong nuclear charge.
These trends help predict and rationalize the ionization energy sequence in elements like Si, P, S, and Cl.

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Most popular questions from this chapter

When sulfur dioxide reacts with chlorine, the products are thionyl chloride, \(\mathrm{SOCl}_{2}\), and dichlorine monoxide, \(\mathrm{OCl}_{2}\) \(\mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SOCl}_{2}(\mathrm{~g})+\mathrm{OCl}_{2}(\mathrm{~g})\) (a) In what period of the periodic table is \(\mathrm{S}\) located? (b) Give the complete electron configuration of \(S\). Do not use the noble-gas notation. (c) \(\Lambda n\) atom of which element involved in this reaction (O. S, or C.1) should have the smallest first ionization energy? The smallest radius? (d) You want to make 675 g \(\mathrm{SOCl}_{2}\). Calculate what mass, in grams, of \(\mathrm{Cl}_{2}\) is required. (e) If you use \(10.0 \mathrm{~g} \mathrm{SO}_{2}\) and \(20.0 \mathrm{~g} \mathrm{Cl}_{2}\), determine the theoretical yield of \(\mathrm{SOCl}_{2}\)

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