同位素、电子结构及电离能证据
引入| English | 释义 | 拼音 |
|---|---|---|
| isotopes/ˈaɪsətəʊps/ | 同位素 | tóng wèi sù |
| atomic number/əˈtɒmɪk ˈnʌmbə/ | 原子序数 | yuán zi xù shù |
| relative abundance/ˈrelətɪv əˈbʌndəns/ | 相对丰度 | xiāng duì fēng dù |
| ionisation energy/ˌaɪənaɪˈzeɪʃn ˈenədʒi/ | 电离能 | diàn lí néng |
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.
某同位素的原子序数Z = 29，质量数A = 65。其原子核中含有多少个中子？
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
中子数 = A − Z = 65 − 29 = 36。若存在离子电荷，仅会影响电子数。
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.
质量分别为10和11的同位素丰度分别为20%和80%。计算相对原子质量。
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
加权平均值 = (10 × 20 + 11 × 80) / 100 = 10.8。该值更接近11，因其对应的同位素丰度更高。
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.
下列哪个方程式代表第一电离能？
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第一电离能起始于气态中性原子并移去一个电子。从1+离子开始则描述的是第二电离能。
在第二次连续电离后出现显著跃升，可唯一确定该元素所在的周期。
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该跃升表明有两个价电子。多个周期均含具有两个价电子的元素；需额外证据进一步确认。
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.
为何硫的第一电离能低于磷，尽管其核电荷更大？
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磷有三个单占据的 3p 轨道；硫在一个 3p 轨道中配对两个电子。增大的电子排斥力解释了电离能下降的原因。