Problem 67
Question
Why is the radius of \(L i^{+}\) so much smaller than the radius of Li? Why is the radius of \(F^{-}\) so much larger than the radius of \(\mathrm{F} ?\)
Step-by-Step Solution
Verified Answer
Li^{+}
is smaller due to losing an electron and increased nuclear pull, while
F^{-}
is larger due to gaining an electron and increased electron repulsion.
1Step 1: Understanding Atomic and Ionic Structure
First, let's understand the difference between atoms and their ions. When an atom loses or gains electrons, it becomes an ion. In the case of a lithium ion (
Li^{+}
), the atom loses one electron, while a fluoride ion (
F^{-}
) gains one electron.
2Step 2: Analyze Lithium Ion Formation
A lithium atom has three electrons and a neutral charge. When it loses one electron to become
Li^{+}
, the remaining electrons are held more tightly by the nucleus, which contains three protons. This increased effective nuclear charge results in the
Li^{+}
ion being smaller than the neutral Li atom.
3Step 3: Analyze Fluoride Ion Formation
A fluorine atom has seven valence electrons and gains one electron to become
F^{-}
, reaching a stable electronic configuration similar to neon, with a full octet. The additional electron causes increased electron-electron repulsion in the outer shell, which results in a larger atomic radius for
F^{-}
compared to neutral F.
Key Concepts
Ion FormationEffective Nuclear ChargeElectron-Electron Repulsion
Ion Formation
Atoms transform into ions through the process of ion formation, which involves the gain or loss of electrons. An atom like lithium (
Li
) can become a positively charged ion (
Li^+
) by losing an electron. On the other hand, fluorine (
F
) can turn into a negatively charged ion (
F^-
) by gaining an electron. This change affects the electron configuration and the charge of the atom considerably.
When lithium loses an electron, it is left with more protons than electrons, creating a Li^{+} ion. This loss results in a smaller ionic radius as the remaining electrons are drawn closer to the nucleus, due to the positive charge inside the atom exceeding the negative one outside.
Fluorine, however, gains an extra electron to become F^{-} . This addition brings about more negative charges compared to positive ones, and the electrons experience more space to move around. Hence, this leads to an increase in the ionic radius.
When lithium loses an electron, it is left with more protons than electrons, creating a Li^{+} ion. This loss results in a smaller ionic radius as the remaining electrons are drawn closer to the nucleus, due to the positive charge inside the atom exceeding the negative one outside.
Fluorine, however, gains an extra electron to become F^{-} . This addition brings about more negative charges compared to positive ones, and the electrons experience more space to move around. Hence, this leads to an increase in the ionic radius.
Effective Nuclear Charge
The effective nuclear charge (ENC) is an essential concept to understand ion formation and size differences. It is the net positive charge experienced by an electron within an atom. The more protons there are in the nucleus, the stronger the ENC, pulling the electrons closer to the nucleus.
In the case of a lithium ion ( Li^{+} ), after losing an electron, the ENC experienced by the remaining electrons increases. This is because there are now more protons than electrons in the ion, amplifying the attractive force exerted by the nucleus. The increased ENC in Li^{+} results in a smaller ion compared to the neutral lithium atom.
In the case of a lithium ion ( Li^{+} ), after losing an electron, the ENC experienced by the remaining electrons increases. This is because there are now more protons than electrons in the ion, amplifying the attractive force exerted by the nucleus. The increased ENC in Li^{+} results in a smaller ion compared to the neutral lithium atom.
- A higher ENC means electrons are pulled in closer.
- A lower ENC can mean electrons spread out, increasing size.
Electron-Electron Repulsion
Electron-electron repulsion is an important factor that influences the size of ions, especially when an atom gains additional electrons to form an ion. Electrons naturally repel each other because they carry the same negative charge, which can affect the atom's size.
When fluorine gains an extra electron to become F^{-} , there is an increase in electron-electron repulsion. This repulsion leads to an expansion of the electron cloud, making the ion's radius larger than that of the neutral atom. The additional electron increases the number of electron-electron interactions within the same shell, pushing existing electrons further apart.
When fluorine gains an extra electron to become F^{-} , there is an increase in electron-electron repulsion. This repulsion leads to an expansion of the electron cloud, making the ion's radius larger than that of the neutral atom. The additional electron increases the number of electron-electron interactions within the same shell, pushing existing electrons further apart.
- Increased electron repulsion leads to a larger ion.
- A balance between ENC and repulsion determines final size.
Other exercises in this chapter
Problem 60
A Two elements in the second transition series (Y through Cd) have four unpaired electrons in their \(3+\) ions. What elements fit this description?
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Answer the questions below about the elements \(A\) and \(\mathrm{B},\) which have the ground state electron configurations shown. $$ \mathrm{A}=[\mathrm{Kr}] 5
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Which ions in the following list are not likely to be found in chemical compounds: \(\mathbf{K}^{2+}, \mathbf{C s}^{+}, \mathbf{A} \mathbf{l}^{4+}, \mathbf{F}^{
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Answer the following questions about first ionization energies. (a) Generally ionization energies increase on proceeding across a period, but this is not true f
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