Atomic and ionic radius
Введено| English | Значение |
|---|---|
| atomic radius/əˈtɒmɪk ˈreɪdɪəs/ | атомный радиус |
| nuclear charge/ˈnjuːklɪə tʃɑːdʒ/ | заряд ядра |
| shielding/ˈʃiːldɪŋ/ | экранирование |
| cation/ˈkætaɪən/ | катион |
| anion/ˈænaɪən/ | анион |
| isoelectronic/ˌaɪsəʊlɪkˈtrɒnɪk/ | изоэлектронный |
A larger charge does not always mean a larger atom
- Atomic radius 原子半径 describes atomic size. To explain a comparison, identify the occupied shells, the attraction from the nucleus and the repulsion between electrons.
- Compare like measurements and similar species. A trend for neutral atoms cannot simply be applied to a mixture of positive and negative ions.
Explain the neutral-atom trends
- Across a period, nuclear charge 核电荷 increases while the added electrons enter the same main shell. Inner-electron shielding 屏蔽 changes relatively little, so greater attraction draws the outer electrons closer and atomic radius generally decreases.
- Down a group, a new occupied shell increases the distance of the outer electrons and adds inner shielding. These effects outweigh the increased nuclear charge, so radius increases.
Use attraction, shielding and occupied shells to explain the directions.
Atomic and ionic radius trends
Atomic radius falls across a period (rising nuclear charge pulls the same shell in) and rises down a group (an extra shell each time). Step across Period 3 to see it.
Across a period, the atomic radius decreases because:
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Electrons go into the same outer shell, so rising nuclear charge with similar shielding pulls them closer.
Down a group, the atomic radius increases because:
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Adding shells increases distance and shielding, so the radius grows down a group.
Match each change to its effect on radius.
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Pulling power vs number of shells sets the size; ions differ from their atoms by the electrons gained or lost.
Compare an atom with its ion
- A cation 阳离子 is smaller than its corresponding atom: fewer electrons give less electron-electron repulsion and a stronger attraction for those remaining. Loss of the whole outer occupied shell can make the contraction especially large, but not every electron loss removes a whole shell.
- An anion 阴离子 is larger than its atom: added electrons increase repulsion while the number of protons stays the same. Do not explain either change by a change in nuclear charge.
Na is $2,8,1$ and Na⁺ is $2,8$: its outer occupied shell is lost. Cl is $2,8,7$ and Cl⁻ is $2,8,8$: the extra electron enters the same outer shell. These two changes need different explanations.
A positive ion (cation) is larger than its parent atom.
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False. A cation is smaller: losing the outer-shell electrons removes a shell or reduces repulsion, so the remaining electrons are pulled closer.
Compared with its atom, a cation (positive ion) is:
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Proton number stays unchanged. Electron loss can remove an outer occupied shell, but does not always do so; reduced repulsion and greater attraction for remaining electrons explain contraction.
A student says that forming any positive ion changes the proton number and always removes an entire shell. Correct both claims and explain why a typical cation is smaller than its atom.
Ion formation removes electrons, not protons, so the element and nuclear charge stay the same. A whole outer occupied shell may be emptied, but losing one electron does not always empty a shell. Fewer electrons reduce repulsion and allow the remaining electrons to be attracted more strongly; losing an outer shell can make the contraction especially large.
Order isoelectronic 等电子 species
- Isoelectronic species have the same number of electrons. When they have the same electron arrangement, the species with more protons attracts those electrons more strongly and has a smaller radius.
- Worked: S²⁻, Cl⁻, K⁺ and Ca²⁺ all have 18 electrons. Nuclear charges increase $16,17,19,20$, so increasing radius is Ca²⁺ < K⁺ < Cl⁻ < S²⁻. Count the electrons before using this rule.
Among ions with the same number of electrons, the smallest is the one with the most ______.
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More protons pull the same number of electrons in more tightly, giving a smaller ion.
Arrange O²⁻, F⁻, Na⁺ and Mg²⁺ in order of increasing radius. State the electron count and compare the nuclear charges before explaining the order.
Each has 10 electrons. Proton numbers are 8, 9, 11 and 12, respectively. With the same electron arrangement, greater nuclear charge gives stronger attraction and a smaller radius: Mg²⁺ < Na⁺ < F⁻ < O²⁻.
Ionic radii across and down the table
- Na⁺, Mg²⁺ and Al³⁺ each have 10 electrons: their radii decrease as nuclear charge rises. P³⁻, S²⁻ and Cl⁻ each have 18 electrons and also decrease in radius within that separate series.
- The two series do not form one smooth decrease: the anions have an extra occupied shell compared with those cations. Down a group of same-charge ions, occupied shells increase, so ionic radius increases.
“More protons means smaller” works as an isolated explanation only when the electron arrangements being compared are suitably matched. Down a group there are more protons and more occupied shells: name which effect dominates.
K⁺ has more protons than Na⁺. A student therefore predicts that K⁺ is smaller. Evaluate this argument using their occupied shells and shielding.
Na⁺ has the shell arrangement 2,8; K⁺ has 2,8,8. The extra occupied shell in K⁺ increases distance from the nucleus and inner shielding. These effects outweigh the increased nuclear charge, so K⁺ is larger. The more-protons rule cannot be used alone when the electron arrangements differ.
You've got it
- Across a period: stronger nuclear attraction within the same main shell generally contracts atoms.
- Down a group: added shells, distance and shielding increase size.
- Electron loss and gain change repulsion and sometimes the outer occupied shell; they do not change proton number.
- Establish equal electron arrangements before ranking an isoelectronic series by nuclear charge.