Isotopes, electron structure and ionisation evidence
導入済み| English | 意味 |
|---|---|
| isotopes/ˈaɪsətəʊps/ | 同位体 |
| atomic number/əˈtɒmɪk ˈnʌmbə/ | 原子番号 |
| relative abundance/ˈrelətɪv əˈbʌndəns/ | 相対的豊度 |
| ionisation energy/ˌaɪənaɪˈzeɪʃn ˈenədʒi/ | 電離エネルギー |
Why is chlorine not a whole-number mass?
- Chlorine atoms can have the same proton number but different masses. These isotopes 同位素 differ in neutron number, not proton number.
- Predict what a mass spectrum would show for a sample containing two isotopes before calculating its mean mass.
Count particles before using a mass
- Atomic number 原子序数 Z counts protons; mass number A counts protons plus neutrons. For chlorine-37, Z = 17, so neutrons = 37 − 17 = 20.
- A chloride ion Cl⁻ has 18 electrons. Its extra electron changes charge, not the proton or neutron count. Relative atomic mass is a mean relative to one twelfth of a carbon-12 atom's mass.
An isotope has Z = 29 and A = 65. How many neutrons does its nucleus contain?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Neutrons = A − Z = 65 − 29 = 36. Ion charge, if present, would affect electron count only.
Weight each isotope by its abundance
- Relative abundance 相对丰度 measures how much each isotope contributes. For 75% chlorine-35 and 25% chlorine-37, the weighted mean is (35 × 75 + 37 × 25)/100 = 35.5.
- A mean mass of 35.5 does not mean an individual chlorine atom contains half a neutron. If gallium-69 and gallium-71 have mean mass 69.8, let the fraction of gallium-71 be x: 69(1 − x) + 71x = 69.8; x = 0.40.
For an unfamiliar element with isotopes 63 and 65 at 70% and 30%, mean mass = 63 × 0.70 + 65 × 0.30 = 63.6. Check that the mean lies between the two isotope masses and nearer the more abundant isotope.
Isotopes of masses 10 and 11 have abundances 20% and 80%. Calculate relative atomic mass.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Weighted mean = (10 × 20 + 11 × 80)/100 = 10.8. It is nearer 11, the more abundant isotope.
Use ionisation evidence to explain structure
- Ionisation energy 电离能 is the energy to remove one electron from each atom in one mole of gaseous atoms, forming one mole of gaseous 1+ ions: X(g) → X⁺(g) + e⁻. Use gaseous species and the stated charge.
- A large jump between successive ionisation energies marks removal from an inner shell. A jump after the second removal supports two outer electrons; seven supplied energies alone do not identify the period.
Which equation represents first ionisation energy?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
First ionisation starts from gaseous neutral atoms and removes one electron. Starting from a 1+ ion describes second ionisation.
A large jump after the second successive ionisation uniquely identifies the period.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The jump supports two outer electrons. Several periods can contain elements with two outer electrons; additional evidence is needed.
Explain trends with the correct competition
- Across a period, increasing nuclear charge usually raises first ionisation energy. A new higher-energy subshell explains the Mg-to-Al decrease; repulsion between paired 3p electrons explains the P-to-S decrease.
- Down a group, greater distance and shielding usually outweigh greater nuclear charge. State which factor changes and how it affects attraction; do not call electron repulsion extra shielding by an inner shell.
A mean atomic mass of 35.5 does not describe an atom with half a neutron. It is the abundance-weighted mean for a sample of whole isotopes.
Count particles, weight isotopes, then explain electron removal with nuclear attraction, distance, shielding and subshell occupancy. Evidence about outer electrons does not automatically identify every feature of the atom.
Why is sulfur first ionisation energy lower than phosphorus despite its greater nuclear charge?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Phosphorus has three singly occupied 3p orbitals; sulfur pairs two electrons in one 3p orbital. The increased electron repulsion explains the decrease.