Isótopos, estrutura eletrônica e evidências de ionização
Introduzido| English | Significado |
|---|---|
| isotopes/ˈaɪsətəʊps/ | Isótopos |
| atomic number/əˈtɒmɪk ˈnʌmbə/ | número atômico |
| relative abundance/ˈrelətɪv əˈbʌndəns/ | abundância relativa |
| ionisation energy/ˌaɪənaɪˈzeɪʃn ˈenədʒi/ | energia de ionização |
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.
Um isótopo tem Z = 29 e A = 65. Quantos nêutrons seu núcleo contém?
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Nêutrons = A − Z = 65 − 29 = 36. Carga iônica, se presente, afetaria apenas o número de elétrons.
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.
Isótopos de massas 10 e 11 têm abundâncias 20% e 80%. Calcule a massa atômica relativa.
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Média ponderada = (10 × 20 + 11 × 80)/100 = 10.8. É mais próxima de 11, o isótopo mais abundante.
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.
Qual equação representa a primeira energia de ionização?
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A primeira ionização parte de átomos neutros gasosos e remove um elétron. Começar de um íon 1+ descreve a segunda ionização.
Um grande salto após a segunda ionização sucessiva identifica unicamente o período.
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O salto suporta dois elétrons externos. Vários períodos podem conter elementos com dois elétrons externos; evidências adicionais são necessárias.
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.
Por que a primeira energia de ionização do enxofre é menor que a do fósforo apesar de sua maior carga nuclear?
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O fósforo tem três orbitais 3p semipreenchidos; o enxofre emparelha dois elétrons em um orbital 3p. A maior repulsão eletrônica explica a diminuição.