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Chapter 14: Problem 14

Fluorine lies between oxygen and neon in Period \(2 .\) Whereas atomic sizes and ionization energies of these three elements change smoothly, their electronegativities display a dramatic change. What is this change, and how do their electron configurations explain it?

Short Answer

Expert verified
Fluorine has the highest electronegativity, oxygen is less electronegative, and neon has no electronegativity. This is due to their electron configurations and valence electron completion.

Step by step solution

01

- Identify the elements and their positions

Fluorine is located between oxygen and neon in Period 2 of the periodic table.
02

- Understand atomic size and ionization energy trends

Atomic size decreases and ionization energy increases as you move from left to right across Period 2.
03

- Define electronegativity

Electronegativity measures an atom's ability to attract and hold onto electrons.
04

- Examine the electronegativity values

Oxygen has an electronegativity of 3.44, fluorine has the highest at 3.98, and neon is generally considered to have no electronegativity.
05

- Compare electron configurations

Oxygen: 1s² 2s² 2p⁴, Fluorine: 1s² 2s² 2p⁵, Neon: 1s² 2s² 2p⁶. Notice that as you move from oxygen to fluorine, they tend to gain an increasing number of protons and electrons, filling their 2p orbitals.
06

- Explain the dramatic change in electronegativity

Fluorine's almost complete outer shell (2s² 2p⁵) makes it extremely electronegative because it needs just one more electron to fill its valence shell. In contrast, neon has a full valence shell (2s² 2p⁶), making it stable and non-electronegative.

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Key Concepts

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

Atomic Size
Atomic size, or atomic radius, refers to the distance from the nucleus of an atom to the outermost electrons. As you move from left to right across Period 2 of the periodic table, atomic size decreases.
This is because electrons are added to the same energy level (shell) while the number of protons in the nucleus increases.
The increased positive charge from the protons pulls the electrons closer to the nucleus, resulting in a smaller atomic radius. For example:
  • Oxygen: 1s² 2s² 2p⁴
  • Fluorine: 1s² 2s² 2p⁵
  • Neon: 1s² 2s² 2p⁶
Despite more electrons being added, the increased nuclear charge creates a stronger attraction, drawing the outer electrons closer and making the overall size smaller.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom. As atomic size decreases across Period 2, ionization energy increases.
This happens because the electrons are held more tightly by the nucleus due to increased nuclear charge. Therefore, more energy is needed to remove one of these tightly held electrons. In Period 2:
  • Oxygen has a lower ionization energy compared to fluorine.
  • Fluorine has a higher ionization energy compared to both oxygen and neon.
  • Neon has the highest ionization energy in this group since it has a full outer shell and is very stable.
The trend signifies that removing an electron becomes more difficult as you move from left to right.
Electron Configurations
Electron configurations describe how electrons are distributed in an atom's orbitals. This is crucial in understanding elements' chemical properties. For our elements:
  • Oxygen: 1s² 2s² 2p⁴
  • Fluorine: 1s² 2s² 2p⁵
  • Neon: 1s² 2s² 2p⁶
These configurations reveal the filling of the 2p orbital. Oxygen has four electrons in its 2p orbital, fluorine has five, and neon completes its 2p orbital with six electrons. The closer an element's electron configuration is to a full shell, the more stable it becomes and the more pronounced its chemical characteristics, like electronegativity, appear.
Periodic Table Trends
Periodic table trends help predict various properties of elements, such as atomic size, ionization energy, and electronegativity. Key trends across Period 2 include:
  • Decreasing atomic size: Moving left to right, atoms become smaller due to an increase in nuclear charge.
  • Increasing ionization energy: Elements require more energy to remove an electron as you go across the period.
  • Increasing electronegativity: The tendency of atoms to attract electrons increases, with fluorine being the most electronegative.
Understanding these trends helps explain why elements behave the way they do and interact in predictable patterns based on their placement on the periodic table.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a key role in chemical reactions and bond formation. In Period 2, the valence electrons increase as you move from left to right. Consider the elements:
  • Oxygen: 6 valence electrons
  • Fluorine: 7 valence electrons
  • Neon: 8 valence electrons (a full valence shell)
Fluorine is highly electronegative because it is one electron short of a full valence shell, making it eager to gain an electron. In contrast, neon has a complete valence shell, meaning it is stable and not inclined to gain or lose electrons. This property explains the relatively stable configuration of noble gases like neon.

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Most popular questions from this chapter

What is a disproportionation reaction, and which of the following fit the description? (a) \(\mathrm{I}_{2}(s)+\mathrm{KI}(a q) \longrightarrow \mathrm{KI}_{3}(a q)\) (b) \(2 \mathrm{ClO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{HClO}_{3}(a q)+\mathrm{HClO}_{2}(a q)\) (c) \(\mathrm{Cl}_{2}(g)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{NaClO}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{2}(s) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (e) \(3 \mathrm{MnO}_{4}^{2-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$$$ 2 \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s)+4 \mathrm{OH}^{-}(a q)$$ (f) \)3 \mathrm{AuCl}(s) \longrightarrow \mathrm{AuCl}_{3}(s)+2 \mathrm{Au}(s)$

lime \((\mathrm{CaO})\) is one of the most abundantly produced chemicals in the world. Write balanced equations for these reactions: (a) The preparation of lime from natural sources (b) The use of slaked lime to remove \(\mathrm{SO}_{2}\) from flue gases (c) The reaction of lime with arsenic acid \(\left(\mathrm{H}_{3} \mathrm{~A} \mathrm{~s} \mathrm{O}_{4}\right)\) to manufacture the insecticide calcium arsenate (d) The regeneration of \(\mathrm{NaOH}\) in the paper industry by reaction of lime with aqueous sodium carbonate

Complete and balance the following equations. If no reaction occurs, write NR: (a) \(\mathrm{Rb}(s)+\mathrm{Br}_{2}(l) \longrightarrow\) (b) \(\mathrm{I}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (c) \(\mathrm{Br}_{2}(l)+\mathrm{I}^{-}(a q) \longrightarrow\) (d) \(\mathrm{CaF}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow\)

In addition to those in Table \(14.3,\) other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following: (a) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with an \(\mathrm{N}-\mathrm{N}\) bond (b) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with no \(\mathrm{N}-\mathrm{N}\) bond (c) \(\mathrm{N}_{2} \mathrm{O}_{3}\) with no \(\mathrm{N}-\mathrm{N}\) bond (d) \(\mathrm{NO}^{+}\) and \(\mathrm{NO}_{3}^{-}\), products of the ionization of liquid \(\mathrm{N}_{2} \mathrm{O}_{4}\)

(a) Give the physical state and color of each halogen at STP. (b) Explain the change in physical state down Group \(7 \mathrm{~A}(17)\) in terms of molecular properties.
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