原子中电子的排列

元素的电子构型描述了电子在其原子轨道中的分布情况。它们遵循标准符号，其中所有含电子的原子子壳按顺序排列。这种方法是由 Bohr 和 Bury 提出的。写入不同能级或壳层的电子数遵循以下规则：

1. The maximum number of electrons in a shell is given by the formula 2n², where n is the orbit number i.e. 1,2,3 and so on. Therefore, the maximum number of electrons in different shells are as follows-
   1. First orbit or K-shell: 2 × 1² = 2
   2. Second orbit or L-shell: 2 × 2² = 8
   3. Third orbit or M-shell: 2 × 3² = 18
   4. Fourth orbit or N-shell: 2 × 4² = 32
2. The maximum number of electrons that can be accommodated in the outermost orbit is 8.
3. Electrons are not accommodated in a given shell unless the inner shells are filled. That is, the shells are filled in a step-wise manner.

电子配置可用于：

• 确定元素的化合价。
• 预测一组元素的属性。
• 解释原子光谱。

壳中的电子

电子占据不同的能级或壳层。每个壳可以容纳最大数量的电子。通过元素周期表中的元素，每个原子比上一个多一个电子，因为电子数与原子序数相同。电子按顺序占据壳层，从离核最近的壳层开始。只有当这个壳变满时，它们才开始占据下一个壳。

预测电子排列

从原子序数可以看出原子的电子排列。例如，钠的原子序数是 11，这意味着钠原子有 11 个质子和 11 个电子，其中：

• 2个电子占据第一层，
• 8 个电子占据第二层和
• 1 个电子占据第三层。

这种电子排列可以写成2, 8, 1。前 18 个元素的电子排列如下所示：

Orbit	Maximum Number of Electrons
K	2
L	8
M	18
N	32

价

The electrons present in the outermost shell of an atom are known as the valence electrons. 

从 Bohr-Bury 方案中，我们发现原子的最外层最多可以容纳 8 个电子，因此，观察到具有完全填充的最外层（意味着零化合价）的元素几乎没有化学活性。在这些惰性元素中，氦原子的最外层有两个电子，而其他元素的最外层有八个电子。具有八个电子的最外层具有一个八位组，因此原子会发生反应以在最外层中实现一个八位组。一个八位组可以通过共享、获得或失去电子来实现。为完成最外层电子的八位字节而获得、丢失或共享的电子数量为我们提供了它的化合价。

化合价的一些例子

1. 氢（H）、锂（Li）、钠（Na）原子的最外层各有一个电子，这意味着它们中的每一个都可以失去一个电子。因此，它们都具有一价。
2. 同样，镁 (Mg) 和铝 (Al) 的化合价分别为 2 和 3，因为它们的最外层有 2 和 3 个电子。
3. 如果原子最外层的电子数接近其全部容量，则以不同的方式确定化合价。例如，氟（F）在最外层有 7 个电子，它的化合价可能是 7，但氟更容易获得 1 个电子而不是失去 7 个电子。因此，它的化合价是通过从八位字节中减去七个电子来确定的，这使得它的化合价为一。
4. 对于氧 (O) 可以遵循类似的程序，这将使其化合价为 2。

示例问题

问题一：不同能级或壳层电子数的书写规则是什么？

回答：

The following rules are followed for writing the number of electrons in different energy levels or shells-

1. The maximum number of electrons in a shell is given by the formula 2n², where n is the orbit number i.e. 1,2,3 and so on.
2. The maximum number of electrons that can be accommodated in the outermost orbit is 8.
3. Electrons are not accommodated in a given shell, unless the inner shells are filled. That is, the shells are filled in a step-wise manner.

问题 2：为什么电子配置有用？

回答：

Electron Configurations are useful for:

1. Determining the valency of an element.
2. Predicting the properties of a group of elements.
3. Interpreting atomic spectra.

问题3：预测镁的电子排列。

回答：

The electron arrangement of an atom can be found out from its atomic number. The atomic number of Magnesium is 12 which means that it has 12 protons and 12 electrons where-

1. 2 electrons occupy the first shell
2. 8 electrons occupy the second shell
3. 2 electron occupies the third shell

This electron arrangement can be written as 2,8,2.

问题 4：预测硫的电子排列。

回答：

The electron arrangement of an atom can be found out from its atomic number. The atomic number of Sulphur is 16 which means that it has 16 protons and 16 electrons where-

1. 2 electrons occupy the first shell
2. 8 electrons occupy the second shell
3. 6 electron occupies the third shell

This electron arrangement can be written as 2,8,6

问题 5：写一个关于价电子和八位字节的简短笔记。

回答：

The electrons present in the outermost shell of an atom are known as the valence electrons. 

From the Bohr-Bury scheme, we find out the outermost shell of an atom can accommodate a maximum of 8 electrons and thus, it was observed that the elements, having a completely filled outermost shell (meaning zero valency) show little chemical activity. Of these inert elements, the helium atom has two electrons in its outermost shell while the other elements have eight electrons in the outermost shell. An outermost-shell, which had eight electrons possesses an octet and thus atoms would react to achieve an octet in the outermost shell. An octet can be achieved by sharing, gaining or losing electrons.