双语化学辞典Chem Lexicon
概念 · 原子结构 · 周期性 Atomic structure · Periodicity

屏蔽效应Shielding effect

高中High school
拼音Pinyin píng bì xiào yìng 别名Also 屏蔽作用;screening effect; electron shielding

定义Definition

屏蔽效应是多电子原子中,内层电子(以及同层其他电子)削弱原子核对某一电子吸引的作用。被屏蔽的电子感受到的是有效核电荷 Zeff = Z − S,小于实际核电荷。

The shielding effect is the reduction, by inner electrons (and to a lesser extent other electrons in the same shell), of the nucleus's pull on a given electron in a many-electron atom. The shielded electron feels an effective nuclear charge Zeff = Z − S, smaller than the true charge.

规律How it works

内层电子位于外层电子与原子核之间,屏蔽作用强;同层电子相互屏蔽较弱。因此同周期从左到右增加的电子在同一层,屏蔽增加不多,有效核电荷增大;同族从上到下增加了新的电子层,外层电子受到更多屏蔽且离核更远。

Inner electrons lie between the outer electrons and the nucleus and shield strongly; electrons in the same shell shield each other weakly. Across a period the added electrons go into the same shell, so shielding grows little and the effective charge rises; down a group a new shell is added and the outer electrons are more shielded and further away.

钻穿效应Penetration

同一电子层中,s 电子比 p 电子更能钻入内层电子区域靠近原子核,受屏蔽较少、能量较低,所以 2s 低于 2p。这也解释了为什么 B 的第一电离能比 Be 低:硼失去的是能量较高的 2p 电子。

Within a shell, s electrons penetrate the inner electron region more than p electrons, are shielded less and lie lower in energy, so 2s is below 2p. This also explains why boron's first ionisation energy is lower than beryllium's: boron loses a higher-energy 2p electron.

年表Timeline

  • 1930

    斯莱特提出定量估算屏蔽常数的规则

    Slater proposes rules for estimating shielding constants

相关公式Formulas

斯莱特规则与有效核电荷 Slater's rules and effective nuclear charge→
\[Z_\text{eff} = Z - S,\qquad E_n \approx -13.6\ \mathrm{eV}\left(\frac{Z_\text{eff}}{n^*}\right)^2,\quad r \approx \frac{n^{*2}a_0}{Z_\text{eff}}\]
例题Worked example

N 的 2p 电子:S = 4×0.35 + 2×0.85 = 3.10,Z_eff = 3.90。Zn 的 4s:S = 0.35 + 18×0.85 + 10×1.00 = 25.65,Z_eff = 4.35。

N 2p: Z_eff = 3.90. Zn 4s: Z_eff = 4.35.

化学位移 Chemical shift→
\[\delta = \frac{\nu - \nu_\text{ref}}{\nu_0}\times10^6\ \text{ppm} = (\sigma_\text{ref} - \sigma)\times10^6\]
例题Worked example

400 MHz 谱中某峰比 TMS 高 2908 Hz:δ = 2908/400 = 7.27 ppm(芳香 H)。

A peak 2908 Hz from TMS at 400 MHz: δ = 7.27 ppm (aromatic H).