Electronic formulas of atoms and ions. Electronic structure of homonuclear diatomic molecules and ions Distribution of electrons using the periodic system D

Electronic configuration an atom is a numerical representation of its electron orbitals. Electron orbitals are regions of various shapes located around the atomic nucleus, in which it is mathematically probable that an electron will be found. The electronic configuration helps to quickly and easily tell the reader how many electron orbitals an atom has, as well as to determine the number of electrons in each orbital. After reading this article, you will master the method of compiling electronic configurations.

Steps

Distribution of electrons using the periodic system of D. I. Mendeleev

Find the atomic number of your atom. Each atom has a certain number of electrons associated with it. Find the symbol for your atom in the periodic table. The atomic number is a positive integer starting from 1 (for hydrogen) and increasing by one for each subsequent atom. The atomic number is the number of protons in an atom, and therefore it is also the number of electrons in an atom with zero charge.

Determine the charge of an atom. Neutral atoms will have the same number of electrons as shown in the periodic table. However, charged atoms will have more or fewer electrons, depending on the magnitude of their charge. If you are working with a charged atom, add or subtract electrons as follows: add one electron for every negative charge and subtract one for every positive charge.

• For example, a sodium atom with a charge of -1 will have an extra electron in addition to its base atomic number of 11. In other words, an atom will have 12 electrons in total.
• If we are talking about a sodium atom with a charge of +1, one electron must be subtracted from the base atomic number 11. So the atom will have 10 electrons.

1. Memorize the basic list of orbitals. As the number of electrons increases in an atom, they fill the various sublevels of the electron shell of the atom according to a certain sequence. Each sublevel of the electron shell, when filled, contains an even number of electrons. There are the following sublevels:

   Understand the electronic configuration record. Electronic configurations are written down in order to clearly reflect the number of electrons in each orbital. Orbitals are written sequentially, with the number of atoms in each orbital written as a superscript to the right of the orbital name. The completed electronic configuration has the form of a sequence of sublevel designations and superscripts.

   • Here, for example, is the simplest electronic configuration: 1s 2 2s 2 2p 6 . This configuration shows that there are two electrons in the 1s sublevel, two electrons in the 2s sublevel, and six electrons in the 2p sublevel. 2 + 2 + 6 = 10 electrons in total. This is the electronic configuration of the neutral neon atom (neon atomic number is 10).

2. Remember the order of the orbitals. Keep in mind that electron orbitals are numbered in ascending order of electron shell number, but arranged in ascending energy order. For example, a filled 4s 2 orbital has less energy (or less mobility) than a partially filled or filled 3d 10, so the 4s orbital is written first. Once you know the order of the orbitals, you can easily fill them in according to the number of electrons in the atom. The order in which the orbitals are filled is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.

   • The electronic configuration of an atom in which all orbitals are filled will have the following form: 10 7p 6
   • Note that the above notation, when all orbits are filled, is the electronic configuration of the element Uuo (ununoctium) 118, the highest numbered atom in the Periodic Table. Therefore, this electronic configuration contains all currently known electronic sublevels of a neutrally charged atom.

3. Fill in the orbitals according to the number of electrons in your atom. For example, if we want to write down the electronic configuration of a neutral calcium atom, we must start by looking up its atomic number in the periodic table. Its atomic number is 20, so we will write the configuration of an atom with 20 electrons according to the above order.

   • Fill in the orbitals in the above order until you reach the twentieth electron. The first 1s orbital will have two electrons, the 2s orbital will also have two, the 2p orbital will have six, the 3s orbital will have two, the 3p orbital will have 6, and the 4s orbital will have 2 (2 + 2 + 6 +2 +6 + 2 = 20 .) In other words, the electronic configuration of calcium has the form: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 .
   • Note that the orbitals are in ascending order of energy. For example, when you are ready to move to the 4th energy level, then first write down the 4s orbital, and then 3d. After the fourth energy level, you move on to the fifth, where the same order is repeated. This happens only after the third energy level.

4. Use the periodic table as a visual cue. You have probably already noticed that the shape of the periodic table corresponds to the order of electronic sublevels in electronic configurations. For example, atoms in the second column from the left always end in "s 2 ", while atoms on the right edge of the thin middle section always end in "d 10 ", and so on. Use the periodic table as a visual guide to writing configurations - as the order in which you add to the orbitals corresponds to your position in the table. See below:

   • In particular, the two leftmost columns contain atoms whose electronic configurations end in s orbitals, the right block of the table contains atoms whose configurations end in p orbitals, and at the bottom of the atoms end in f orbitals.
   • For example, when you write down the electronic configuration of chlorine, think like this: "This atom is located in the third row (or "period") of the periodic table. It is also located in the fifth group of the orbital block p of the periodic table. Therefore, its electronic configuration will end in. ..3p 5
   • Note that the elements in the d and f orbital regions of the table have energy levels that do not correspond to the period in which they are located. For example, the first row of a block of elements with d-orbitals corresponds to 3d orbitals, although it is located in the 4th period, and the first row of elements with f-orbitals corresponds to the 4f orbital, despite the fact that it is located in the 6th period.

5. Learn the abbreviations for writing long electronic configurations. The atoms on the right side of the periodic table are called noble gases. These elements are chemically very stable. To shorten the process of writing long electron configurations, simply write in square brackets the chemical symbol for the nearest noble gas with fewer electrons than your atom, and then continue to write the electronic configuration of subsequent orbital levels. See below:

   • To understand this concept, it will be helpful to write an example configuration. Let's write the configuration of zinc (atomic number 30) using the noble gas abbreviation. The complete zinc configuration looks like this: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 . However, we see that 1s 2 2s 2 2p 6 3s 2 3p 6 is the electronic configuration of argon, a noble gas. Simply replace the electronic configuration part of zinc with the chemical symbol for argon in square brackets (.)
   • So, the electronic configuration of zinc, written in abbreviated form, is: 4s 2 3d 10 .
   • Note that if you are writing the electronic configuration of a noble gas, say argon, you cannot write! One must use the abbreviation of the noble gas in front of this element; for argon it will be neon ().

   Using ADOMAH Periodic Table

   1. Master the ADOMAH periodic table. This method of recording the electronic configuration does not require memorization, however, it requires a modified periodic table, since in the traditional periodic table, starting from the fourth period, the period number does not correspond to the electron shell. Find the ADOMAH periodic table, a special type of periodic table designed by scientist Valery Zimmerman. It is easy to find with a short internet search.

      • In the ADOMAH periodic table, the horizontal rows represent groups of elements such as halogens, noble gases, alkali metals, alkaline earth metals, etc. Vertical columns correspond to electronic levels, and so-called "cascades" (diagonal lines connecting blocks s, p, d and f) correspond to periods.
      • Helium is moved to hydrogen, since both of these elements are characterized by a 1s orbital. The period blocks (s,p,d and f) are shown on the right side and the level numbers are given at the bottom. Elements are represented in boxes numbered from 1 to 120. These numbers are the usual atomic numbers, which represent the total number of electrons in a neutral atom.

   2. Find your atom in the ADOMAH table. To write down the electronic configuration of an element, find its symbol in the ADOMAH periodic table and cross out all elements with a higher atomic number. For example, if you need to write down the electronic configuration of erbium (68), cross out all the elements from 69 to 120.

      • Pay attention to the numbers from 1 to 8 at the base of the table. These are the electronic level numbers, or column numbers. Ignore columns that contain only crossed out items. For erbium, columns with numbers 1,2,3,4,5 and 6 remain.

   3. Count the orbital sublevels up to your element. Looking at the block symbols shown to the right of the table (s, p, d, and f) and the column numbers shown at the bottom, ignore the diagonal lines between the blocks and break the columns into block-columns, listing them in order from bottom to top. And again, ignore the blocks in which all the elements are crossed out. Write the column blocks starting from the column number followed by the block symbol, thus: 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 6s (for erbium).

      • Please note: The above electronic configuration Er is written in ascending order of the electronic sublevel number. It can also be written in the order in which the orbitals are filled. To do this, follow the cascades from bottom to top, not columns, when you write column blocks: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 12 .

   4. Count the electrons for each electronic sublevel. Count the elements in each column block that have not been crossed out by attaching one electron from each element, and write their number next to the block symbol for each column block as follows: 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 4f 12 5s 2 5p 6 6s 2 . In our example, this is the electronic configuration of erbium.

   5. Be aware of incorrect electronic configurations. There are eighteen typical exceptions related to the electronic configurations of atoms in the lowest energy state, also called the ground energy state. They do not obey the general rule only in the last two or three positions occupied by electrons. In this case, the actual electronic configuration assumes that the electrons are in a state of lower energy compared to the standard configuration of the atom. Exception atoms include:

      • Cr(..., 3d5, 4s1); Cu(..., 3d10, 4s1); Nb(..., 4d4, 5s1); Mo(..., 4d5, 5s1); Ru(..., 4d7, 5s1); Rh(..., 4d8, 5s1); Pd(..., 4d10, 5s0); Ag(..., 4d10, 5s1); La(..., 5d1, 6s2); Ce(..., 4f1, 5d1, 6s2); Gd(..., 4f7, 5d1, 6s2); Au(..., 5d10, 6s1); AC(..., 6d1, 7s2); Th(..., 6d2, 7s2); Pa(..., 5f2, 6d1, 7s2); U(..., 5f3, 6d1, 7s2); Np(..., 5f4, 6d1, 7s2) and cm(..., 5f7, 6d1, 7s2).
   • To find the atomic number of an atom when it is written in electronic form, simply add up all the numbers that follow the letters (s, p, d, and f). This only works for neutral atoms, if you are dealing with an ion, then nothing will work - you will have to add or subtract the number of extra or lost electrons.
   • The number following the letter is a superscript, do not make a mistake in the control.
   • The "stability of a half-filled" sublevel does not exist. This is a simplification. Any stability that pertains to "half-full" sublevels is due to the fact that each orbital is occupied by one electron, so repulsion between electrons is minimized.
   • Each atom tends to a stable state, and the most stable configurations have filled sublevels s and p (s2 and p6). Noble gases have this configuration, so they rarely react and are located on the right in the periodic table. Therefore, if a configuration ends in 3p 4 , then it needs two electrons to reach a stable state (it takes more energy to lose six, including s-level electrons, so four is easier to lose). And if the configuration ends in 4d 3 , then it needs to lose three electrons to reach a stable state. In addition, half-filled sublevels (s1, p3, d5..) are more stable than, for example, p4 or p2; however, s2 and p6 will be even more stable.
   • When you're dealing with an ion, that means the number of protons is not the same as the number of electrons. The charge of the atom in this case will be shown at the top right (usually) of the chemical symbol. Therefore, an antimony atom with a charge of +2 has the electronic configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 1 . Note that 5p 3 has changed to 5p 1 . Be careful when the configuration of a neutral atom ends at sublevels other than s and p. When you take electrons, you can only take them from valence orbitals (s and p orbitals). Therefore, if the configuration ends with 4s 2 3d 7 and the atom gets +2 charge, then the configuration will end with 4s 0 3d 7 . Please note that 3d 7 not changes, instead electrons of the s-orbital are lost.
   • There are conditions when an electron is forced to "move to a higher energy level." When a sublevel lacks one electron to be half or full, take one electron from the nearest s or p sublevel and move it to the sublevel that needs an electron.
   • There are two options for writing an electronic configuration. They can be written in ascending order of the numbers of energy levels or in the order in which the electron orbitals are filled, as was shown above for erbium.
   • You can also write the electronic configuration of an element by writing only the valence configuration, which is the last s and p sublevel. Thus, the valence configuration of antimony will be 5s 2 5p 3 .
   • Ions are not the same. It's much more difficult with them. Skip two levels and follow the same pattern depending on where you started and how high the number of electrons is.

The process of H2+ particle formation can be represented as follows:

H + H+ H2+.

Thus, one electron is located on the bonding molecular s-orbital.

The multiplicity of the bond is equal to the half-difference of the number of electrons in the bonding and loosening orbitals. Hence, the multiplicity of the bond in the H2+ particle is equal to (1 – 0):2 = 0.5. The VS method, in contrast to the MO method, does not explain the possibility of bond formation by one electron.

The hydrogen molecule has the following electronic configuration:

The H2 molecule has two bonding electrons, which means that the bond in the molecule is single.

The molecular ion H2- has an electronic configuration:

H2- [(s 1s)2(s *1s)1].

The multiplicity of the bond in H2- is (2 - 1): 2 = 0.5.

Let us now consider homonuclear molecules and ions of the second period.

The electronic configuration of the Li2 molecule is as follows:

2Li(K2s)Li2 .

The Li2 molecule contains two bonding electrons, which corresponds to a single bond.

The process of formation of the Be2 molecule can be represented as follows:

2 Be(K2s2) Be2 .

The number of bonding and loosening electrons in the Be2 molecule is the same, and since one loosening electron destroys the action of one bonding electron, the Be2 molecule in the ground state was not found.

In a nitrogen molecule, 10 valence electrons are located in orbitals. Electronic structure of the N2 molecule:

Since there are eight bonding and two loosening electrons in the N2 molecule, this molecule has a triple bond. The nitrogen molecule is diamagnetic because it does not contain unpaired electrons.

On the orbitals of the O2 molecule, 12 valence electrons are distributed, therefore, this molecule has the configuration:

Rice. 9.2. Scheme of the formation of molecular orbitals in the O2 molecule (only 2p electrons of oxygen atoms are shown)

In the O2 molecule, in accordance with Hund's rule, two electrons with parallel spins are placed one at a time in two orbitals with the same energy (Fig. 9.2). According to the VS method, the oxygen molecule does not have unpaired electrons and should have diamagnetic properties, which is inconsistent with the experimental data. The molecular orbital method confirms the paramagnetic properties of oxygen, which are due to the presence of two unpaired electrons in the oxygen molecule. The multiplicity of bonds in an oxygen molecule is (8–4):2 = 2.

Let us consider the electronic structure of the O2+ and O2- ions. In the O2+ ion, 11 electrons are placed in its orbitals, therefore, the configuration of the ion is as follows:

The multiplicity of the bond in the O2+ ion is (8–3):2 = 2.5. In the O2- ion, 13 electrons are distributed in its orbitals. This ion has the following structure:

O2-.

The multiplicity of bonds in the O2- ion is (8 - 5): 2 = 1.5. O2- and O2+ ions are paramagnetic, as they contain unpaired electrons.

The electronic configuration of the F2 molecule has the form:

The bond multiplicity in the F2 molecule is 1, since there is an excess of two bonding electrons. Since there are no unpaired electrons in the molecule, it is diamagnetic.

In the N2, O2, F2 series, the energies and bond lengths in molecules are:

An increase in the excess of binding electrons leads to an increase in the binding energy (bond strength). When passing from N2 to F2, the bond length increases, which is due to the weakening of the bond.

In the O2-, O2, O2+ series, the bond multiplicity increases, the bond energy also increases, and the bond length decreases.

The number of electrons in an atom is determined by the atomic number of the element in the periodic system. Using the rules for placing electrons in an atom, for a sodium atom (11 electrons), we can obtain the following electronic formula:

11 Na: 1s 2 2s 2 2p 6 3s 1

The electronic formula of the titanium atom:

22 Ti: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2

If before full or half filling d-sublevel ( d 10 or d 5-configuration) one electron is missing, then " electron slip " - go to d- sublevel of one electron from the neighboring s-sublevel. As a result, the electronic formula of the chromium atom has the form 24 Cr: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5, and not 24 Cr: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4, and copper atoms - 29 Cu: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 and not 29 Cu: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 9 .

The number of electrons in a negatively charged ion - anion - exceeds the number of electrons of a neutral atom by the charge of the ion: 16 S 2– 1s 2 2s 2 2p 6 3s 2 3p 6 (18 electrons).

During the formation of a positively charged ion - a cation - electrons first of all leave sublevels with a large value of the main quantum number: 24 Cr 3+: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 3 (21 electrons).

Electrons in an atom can be divided into two types: internal and external (valence). Internal electrons occupy fully completed sublevels, have low energy values ​​and do not participate in chemical transformations of elements.

Valence electrons are all electrons of the last energy level and electrons of incomplete sublevels.

Valence electrons take part in the formation of chemical bonds. Unpaired electrons have a special activity. The number of unpaired electrons determines the valence of a chemical element.

If there are empty orbitals at the last energy level of the atom, then it is possible to pair valence electrons on them (formation excited state atom).

For example, the valence electrons of sulfur are the electrons of the last level (3 s 2 3p 4). Graphically, the scheme of filling these orbitals with electrons looks like:

In the ground (unexcited) state, the sulfur atom has 2 unpaired electrons and can exhibit valency II.

At the last (third) energy level, the sulfur atom has free orbitals (3d sublevel). With the expenditure of some energy, one of the paired electrons of sulfur can be transferred to an empty orbital, which corresponds to the first excited state of the atom

In this case, the sulfur atom has four unpaired electrons, and its valency is IV.

The paired 3s electrons of the sulfur atom can also be paired into a free 3d orbital:

In this state, the sulfur atom has 6 unpaired electrons and exhibits a valency equal to VI.

Task 1. Write the electronic configurations of the following elements: N, Si, F e, Kr , Te, W .

Decision. The energy of atomic orbitals increases in the following order:

1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d .

On each s-shell (one orbital) there can be no more than two electrons, on the p-shell (three orbitals) - no more than six, on the d-shell (five orbitals) - no more than 10 and on the f-shell (seven orbitals) - no more than 14.

In the ground state of an atom, electrons occupy orbitals with the lowest energy. The number of electrons is equal to the charge of the nucleus (the atom as a whole is neutral) and the atomic number of the element. For example, a nitrogen atom has 7 electrons, two of which are in 1s orbitals, two are in 2s orbitals, and the remaining three electrons are in 2p orbitals. The electronic configuration of the nitrogen atom:

7 N : 1s 2 2s 2 2p 3 . Electronic configurations of other elements:

14 Si: 1s 2 2s 2 2p 6 3s 2 3p 2 ,

26 F e : 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 ,

36 K r: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 3p 6 ,

52 Those : 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 3p 6 5s 2 4d 10 5p 4 ,

74 Those : 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 3p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 4 .

Task 2. What inert gas and ions of what elements have the same electronic configuration as the particle resulting from the removal of all valence electrons from the calcium atom?

Decision. The electron shell of the calcium atom has the structure 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 . When two valence electrons are removed, a Ca 2+ ion is formed with the configuration 1s 2 2s 2 2p 6 3s 2 3p 6 . An atom has the same electronic configuration Ar and ions S 2-, Cl -, K +, Sc 3+, etc.

Task 3. Can the electrons of the Al 3+ ion be in the following orbitals: a) 2p; b) 1r; c) 3d?

Decision. Electronic configuration of aluminum atom: 1s 2 2s 2 2p 6 3s 2 3p 1 . The Al 3+ ion is formed upon the removal of three valence electrons from an aluminum atom and has the electronic configuration 1s 2 2s 2 2p 6 .

a) electrons are already in the 2p orbital;

b) in accordance with the restrictions imposed on the quantum number l (l = 0, 1, ... n -1), at n = 1, only the value l = 0 is possible, therefore, the 1p orbital does not exist;

c) electrons can be in the 3d orbital if the ion is in an excited state.

Task 4. Write the electronic configuration of the neon atom in the first excited state.

Decision. The electronic configuration of the neon atom in the ground state is 1s 2 2s 2 2p 6 . The first excited state is obtained by the transition of one electron from the highest occupied orbital (2p) to the lowest free orbital (3s). The electronic configuration of the neon atom in the first excited state is 1s 2 2s 2 2p 5 3s 1 .

Task 5. What is the composition of nuclei of isotopes 12 C and 13 C , 14 N and 15 N ?

Decision. The number of protons in the nucleus is equal to the atomic number of the element and is the same for all isotopes of this element. The number of neutrons is equal to the mass number (indicated to the upper left of the element number) minus the number of protons. Different isotopes of the same element have different numbers of neutrons.

The composition of these nuclei:

12 C: 6p + 6n; 13 C: 6p + 7n; 14 N : 7p + 7n ; 15N: 7p + 8n.

The filling of orbitals in an unexcited atom is carried out in such a way that the energy of the atom is minimal (the principle of minimum energy). First, the orbitals of the first energy level are filled, then the second, and the orbital of the s-sublevel is filled first and only then the orbitals of the p-sublevel. In 1925, the Swiss physicist W. Pauli established the fundamental quantum-mechanical principle of natural science (the Pauli principle, also called the exclusion principle or the exclusion principle). According to the Pauli principle:

An atom cannot have two electrons that have the same set of all four quantum numbers.

The electronic configuration of an atom is conveyed by a formula in which the filled orbits are indicated by a combination of a number equal to the main quantum number and a letter corresponding to the orbital quantum number. The superscript indicates the number of electrons in these orbitals.

Hydrogen and helium

The electronic configuration of the hydrogen atom is 1s 1, and that of helium is 1s 2. A hydrogen atom has one unpaired electron, and a helium atom has two paired electrons. Paired electrons have the same values ​​of all quantum numbers, except for the spin. A hydrogen atom can give up its electron and turn into a positively charged ion - the H + cation (proton), which does not have electrons (electronic configuration 1s 0). A hydrogen atom can attach one electron and turn into a negatively charged H - ion (hydride ion) with an electronic configuration of 1s 2.

Lithium

Three electrons in a lithium atom are distributed as follows: 1s 2 1s 1 . In the formation of a chemical bond, only electrons of the outer energy level, called valence electrons, participate. In a lithium atom, the valence electron is the 2s sublevel, and the two electrons of the 1s sublevel are internal electrons. The lithium atom quite easily loses its valence electron, passing into the Li + ion, which has the configuration 1s 2 2s 0 . Note that the hydride ion, helium atom, and lithium cation have the same number of electrons. Such particles are called isoelectronic. They have a similar electronic configuration, but a different nuclear charge. The helium atom is very chemically inert, which is associated with the special stability of the 1s 2 electronic configuration. Orbitals that are not filled with electrons are called vacant orbitals. In the lithium atom, three orbitals of the 2p sublevel are vacant.

Beryllium

The electronic configuration of the beryllium atom is 1s 2 2s 2 . When an atom is excited, electrons from a lower energy sublevel move to vacant orbitals of a higher energy sublevel. The process of excitation of a beryllium atom can be represented by the following scheme:

1s 2 2s 2 (ground state) + hν→ 1s 2 2s 1 2p 1 (excited state).

A comparison of the ground and excited states of the beryllium atom shows that they differ in the number of unpaired electrons. In the ground state of the beryllium atom, there are no unpaired electrons; in the excited state, there are two of them. Despite the fact that during the excitation of an atom, in principle, any electrons from lower energy orbitals can move to higher orbitals, for the consideration of chemical processes, only transitions between energy sublevels with similar energies are essential.

This is explained as follows. When a chemical bond is formed, energy is always released, i.e., the aggregate of two atoms passes into an energetically more favorable state. The excitation process requires energy. When electrons are depaired within the same energy level, the costs of excitation are compensated by the formation of a chemical bond. When electrons are depaired within different levels, the excitation costs are so high that they cannot be compensated by the formation of a chemical bond. In the absence of a partner in a possible chemical reaction, an excited atom releases a quantum of energy and returns to the ground state - such a process is called relaxation.

Bor

The electronic configurations of the atoms of the elements of the 3rd period of the Periodic Table of the Elements will be to a certain extent similar to those given above (the atomic number is indicated by the subscript):

11 Na 3s 1
12 Mg 3s 2
13 Al 3s 2 3p 1
14 Si 2s 2 2p2
15 P 2s 2 3p 3

However, the analogy is not complete, since the third energy level is split into three sublevels and all of the listed elements have vacant d-orbitals, to which electrons can pass during excitation, increasing the multiplicity. This is especially important for elements such as phosphorus, sulfur and chlorine.

The maximum number of unpaired electrons in a phosphorus atom can reach five:

This explains the possibility of the existence of compounds in which the phosphorus valency is 5. The nitrogen atom, which has the same configuration of valence electrons in the ground state as the phosphorus atom, cannot form five covalent bonds.

A similar situation arises when comparing the valence capabilities of oxygen and sulfur, fluorine and chlorine. The depairing of electrons in a sulfur atom leads to the appearance of six unpaired electrons:

3s 2 3p 4 (ground state) → 3s 1 3p 3 3d 2 (excited state).

This corresponds to the six-valence state, which is unattainable for oxygen. The maximum valency of nitrogen (4) and oxygen (3) requires a more detailed explanation, which will be given later.

The maximum valence of chlorine is 7, which corresponds to the configuration of the excited state of the atom 3s 1 3p 3 d 3 .

The presence of vacant 3d orbitals in all elements of the third period is explained by the fact that, starting from the 3rd energy level, there is a partial overlap of sublevels of different levels when filled with electrons. Thus, the 3d sublevel starts filling only after the 4s sublevel is filled. The energy reserve of electrons in atomic orbitals of different sublevels and, consequently, the order of their filling increases in the following order:

Orbitals are filled earlier for which the sum of the first two quantum numbers (n + l) is less; if these sums are equal, orbitals with a smaller principal quantum number are filled first.

This regularity was formulated by V. M. Klechkovsky in 1951.

Elements in whose atoms the s-sublevel is filled with electrons are called s-elements. These include the first two elements of each period: hydrogen. However, already in the next d-element - chromium - there is some “deviation” in the arrangement of electrons according to energy levels in the ground state: instead of the expected four unpaired electrons on the 3d sublevel in the chromium atom, there are five unpaired electrons in the 3d sublevel and one unpaired electron in the s sublevel: 24 Cr 4s 1 3d 5 .

The phenomenon of the transition of one s-electron to the d-sublevel is often called the "breakthrough" of the electron. This can be explained by the fact that the orbitals of the d-sublevel filled with electrons become closer to the nucleus due to an increase in the electrostatic attraction between the electrons and the nucleus. As a result, the state 4s 1 3d 5 becomes energetically more favorable than 4s 2 3d 4 . Thus, the half-filled d-sublevel (d 5) has an increased stability compared to other possible variants of the electron distribution. The electronic configuration corresponding to the existence of the maximum possible number of paired electrons, achievable in the previous d-elements only as a result of excitation, is characteristic of the ground state of the chromium atom. The electronic configuration d 5 is also characteristic of the manganese atom: 4s 2 3d 5 . For the following d-elements, each energy cell of the d-sublevel is filled with a second electron: 26 Fe 4s 2 3d 6 ; 27 Co 4s 2 3d 7 ; 28 Ni 4s 2 3d 8 .

At the copper atom, the state of a completely filled d-sublevel (d 10) becomes achievable due to the transition of one electron from the 4s-sublevel to the 3d-sublevel: 29 Cu 4s 1 3d 10 . The last element of the first row of d-elements has the electronic configuration 30 Zn 4s 23 d 10 .

The general trend, which manifests itself in the stability of the d 5 and d 10 configurations, is also observed for elements of lower periods. Molybdenum has an electronic configuration similar to chromium: 42 Mo 5s 1 4d 5, and silver - copper: 47 Ag5s 0 d 10. Moreover, the d 10 configuration is already achieved in palladium due to the transition of both electrons from the 5s orbital to the 4d orbital: 46Pd 5s 0 d 10 . There are other deviations from the monotonic filling of d- and also f-orbitals.