电子构型

M. Hautefeuille, J. Hernández-Cordero
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引用次数: 0

摘要

这是确定原子中电子构型的一种方法。如果你知道原子中电子的数目,你就可以写出电子排布来确定元素的身份。例如,考虑元素硅Si。通过查看元素周期表,你可以看到硅的原子序数是14,所以硅中有14个质子和14个电子。我可以把14个电子放在原子核周围的轨道上。每个轨道有两个电子。2自我在1 s轨道的剩余部分12电子(14 - 2 = 12)2自我2 s轨道的剩余10电子(12 - 2 = 10)2自我到每个3 2 p轨道余数的四个电子(10 - 6 = 4)2自我在3 s轨道的剩余部分两个电子(4 - 2 = 2)最后2自我的3 p轨道(2 - 2 = 0)完整的电子排布是1 s 2 2 2 2 p 6 3年代2 3 p 2惰性气体核心符号(Ne) 3 2 3 p 2因为1年代2 2 2 2 p 6的配置霓虹灯,Ne。然而,如果你只知道电子排布,你可以数上标,得到14个电子的总数。然后你可以看看元素周期表,看到有14个电子的元素是硅,Si, 1s 2s 2s 3s 3s 1s 2s 2s 3s 3s 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Electron Configuration
Here is one way to determine the configuration of electrons in an atom. If you know the number of electrons in the atom, you can write the electron configuration and determine the identity of the element. For example, consider the element silicon, Si. By looking at the periodic table, you can see that the atomic number of silicon is 14, so there are 14 protons and 14 electrons in silicon. I can place the 14 electrons in orbitals around the nucleus of the atom. Two electrons fill each orbital. 2 ego in the 1s orbital with a remainder of 12 electrons (14 – 2 = 12) 2 ego in the 2s orbital with a remainder of 10 electrons (12 – 2 = 10) 2 ego into each of the three 2p orbitals with a remainder of four electrons (10 – 6 = 4) 2 ego in the 3s orbital with a remainder of two electrons (4 – 2 = 2) The last 2 ego in the 3p orbitals (2 – 2 = 0) The full electron configuration is 1s 2 2s 2 2p 6 3s 2 3p 2 The noble gas core notation is [Ne]3s 2 3p 2 because 1s 2 2s 2 2p 6 is the configuration of neon, Ne. However, if you only had the electron configuration, you could count the superscripts to get 14 as the total number of electrons. Then you could look at the periodic table and see that the element with 14 electrons was silicon, Si. 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p 1s 2s 2p 3s 3p
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