The arrangement of electrons across energy levels and orbitals is called electron configuration. The configuration can be depicted in the form of so-called electron formulas, in which the number in front indicates the number of the energy level, then the letter indicates the sublevel, and at the top right of the letter the number of electrons at this sublevel. The sum of the last numbers corresponds to the positive charge of the atomic nucleus. For example, the electronic formulas of sulfur and calcium will have the following form: S (+ 16) - ls22s22p63s23p\ Ca (+ 20) - ls22s22p63s23p64s2. The filling of electronic levels is carried out in accordance with the principle of lowest energy: the most stable state of an electron in an atom corresponds to the state with the minimum energy value. Therefore, the layers with the lowest energy values are filled first. The Soviet scientist V. Klechkovsky established that the energy of an electron increases as the sum of the main and orbital quantum numbers increases (n + /)> therefore, the filling of the electronic layers occurs in the order of increasing the sum of the main and orbital quantum numbers. If for two sublevels the sums (n -f1) are equal, then first the sublevels with the smallest n and the largest l9 are filled, and then the sublevels with larger n and smaller L. Let, for example, the sum (n + /) « 5. This sum corresponds to the following combinations I: n = 3; / 2; n *» 4; 1-1; l = / - 0. Based on this, first the d-sublevel of the third energy level should be filled, then the 4p-sublevel should be filled, and only after that the s-sublevel of the fifth energy level. All of the above determines the following order of filling electrons in atoms: Example 1 Draw the electronic formula of the sodium atom. Solution Based on the position in the periodic table, it is established that sodium is an element of the third period. This indicates that the electrons in the sodium atom are located at three energy levels. By the serial number of the element, the total number of electrons in these three levels is determined - eleven. At the first energy level (ls1, / = 0; s-sublevel) the maximum number of electrons is // « 2n2, N = 2. The distribution of electrons at the s-sublevel of the first energy level is represented by the notation - Is2, at the second energy level n = 2, I « 0 (s-sublevel) and I = 1 (p-sublevel), the maximum number of electrons is eight. Since the maximum 2е is located at the S-sublevel, there will be 6е at the p-sublevel. The distribution of electrons at the second energy level is represented by the notation - 2s22p6. At the third energy level, S-, p- and d-sublevels are possible. The sodium atom has only one electron at the third energy level, which, according to the principle of least energy, will occupy the Sv sublevel. Combining the records of the distribution of electrons on each layer into one, we obtain the electronic formula of the sodium atom: ls22s22p63s1. The positive charge of the sodium atom (+11) is compensated by the total number of electrons (11). In addition, the structure of electronic shells is depicted using energy or quantum cells (orbitals) - these are the so-called graphical electronic formulas. Each such cell is designated by a rectangle Q, the electron t> the direction of the arrow characterizes the spin of the electron. According to the Pauli principle, one (unpaired) or two (paired) electrons are placed in a cell (orbit). The electronic structure of a sodium atom can be represented by the diagram: When filling quantum cells, you need to know Hund's rule: the stable state of the atom corresponds to such a distribution of electrons within the energy sublevel (p, d, f), at which the absolute value of the total spin of the atom is maximum. So, if two electrons occupy one orbital\]j\ \ \, then their total spin will be zero. Filling two orbitals of 1 t 111 I with electrons will give a total spin equal to unity. Based on Hund's principle, the distribution of electrons over quantum cells, for example, for atoms 6C and 7N, will be as follows. Questions and tasks for independent solution 1. List all the basic theoretical principles necessary for filling electrons in atoms. 2. Show the validity of the principle of least energy using the example of filling electrons in atoms of calcium and scandium, strontium, yttrium and indium. 3. Which of the graphical electronic formulas of the phosphorus atom (unexcited state) is correct? Motivate your answer using Hund's rule. 4. Write all the quantum numbers for the electrons of the atoms: a) sodium, silicon; b) phosphorus, chlorine; c) sulfur, argon. 5. Make up electronic formulas for the atoms of the s-element of the first and third periods. 6. Create an electronic formula for the atom of the p-element of the fifth period, the outer energy level of which is 5s25p5. What are his Chemical properties? 7. Draw the distribution of electrons in orbitals in the atoms of silicon, fluorine, krypton. 8. Make up an electronic formula for an element in an atom of which the energy state of two electrons of the outer level is described by the following quantum numbers: n - 5; 0; t1 = 0; ta = + 1/2; ta « -1/2. 9. The outer and penultimate energy levels of atoms have the following form: a) 3d24s2; b) 4d105s1; c) 5s25p6. Write electronic formulas for atoms of elements. Specify the p- and d-elements. 10. Make up electronic formulas for d-element atoms that have 5 electrons at the d-sublevel. 11. Draw the distribution of electrons across quantum cells in the atoms of potassium, chlorine, and neon. 12. The outer electronic layer of an element is expressed by the formula 3s23p4. Determine the serial number and name of the element. 13. Write the electronic configurations of the following ions: 14. Do the O, Mg, Ti atoms contain M-level electrons? 15. Which particles of atoms are isoelectronic, i.e. contain the same number of electrons: 16. How many electronic levels do atoms have in the S2", S4+, S6+ state? 17. How many free d-orbitals are there in Sc, Ti, V atoms? Write the electronic formulas of the atoms of these elements. 18. Indicate the serial number of the element for which: a) the filling of the 4c1 sublevel with electrons ends; b) the filling of the 4p sublevel with electrons begins. 19. Indicate the features of the electronic configurations of copper and chromium atoms. What number Do the atoms of these elements contain 4b electrons in a stable state?20. How many vacant 3p orbitals does a silicon atom have in a stationary and excited state?
>> Chemistry: Electronic configurations atoms of chemical elements
The Swiss physicist W. Pauli in 1925 established that in an atom in one orbital there can be no more than two electrons having opposite (antiparallel) spins (translated from English as “spindle”), that is, having such properties that can be conventionally imagined itself as the rotation of an electron around its imaginary axis: clockwise or counterclockwise. This principle is called the Pauli principle.
If there is one electron in the orbital, then it is called unpaired; if there are two, then these are paired electrons, that is, electrons with opposite spins.
Figure 5 shows a diagram of the division of energy levels into sublevels.
The s-orbital, as you already know, has a spherical shape. The electron of the hydrogen atom (s = 1) is located in this orbital and is unpaired. Therefore, its electronic formula or electronic configuration will be written as follows: 1s 1. In electronic formulas, the number of the energy level is indicated by the number preceding the letter (1 ...), the Latin letter indicates the sublevel (type of orbital), and the number, which is written to the upper right of the letter (as an exponent), shows the number of electrons in the sublevel.
For a helium atom He, which has two paired electrons in one s-orbital, this formula is: 1s 2.
The electron shell of the helium atom is complete and very stable. Helium is a noble gas.
At the second energy level (n = 2) there are four orbitals: one s and three p. The electrons of the s-orbital of the second level (2s-orbitals) have higher energy, since they are at a greater distance from the nucleus than the electrons of the 1s-orbital (n = 2).
In general, for each value of n there is one s orbital, but with a corresponding supply of electron energy on it and, therefore, with a corresponding diameter, growing as the value of n increases.
The p-Orbital has the shape of a dumbbell or a three-dimensional figure eight. All three p-orbitals are located in the atom mutually perpendicular along the spatial coordinates drawn through the nucleus of the atom. It should be emphasized once again that each energy level (electronic layer), starting from n = 2, has three p-orbitals. As the value of n increases, electrons occupy p-orbitals located at large distances from the nucleus and directed along the x, y, z axes.
For elements of the second period (n = 2), first one b-orbital is filled, and then three p-orbitals. Electronic formula 1l: 1s 2 2s 1. The electron is more loosely bound to the nucleus of the atom, so the lithium atom can easily give it up (as you remember, this process is called oxidation), turning into a Li+ ion.
In the beryllium atom Be 0, the fourth electron is also located in the 2s orbital: 1s 2 2s 2. The two outer electrons of the beryllium atom are easily detached - Be 0 is oxidized into the Be 2+ cation.
In the boron atom, the fifth electron occupies the 2p orbital: 1s 2 2s 2 2p 1. Next, the C, N, O, E atoms are filled with 2p orbitals, which ends with the noble gas neon: 1s 2 2s 2 2p 6.
For elements of the third period, the Sv and Sr orbitals are filled, respectively. Five d-orbitals of the third level remain free:
11 Na 1s 2 2s 2 Sv1; 17С11в22822р63р5; 18Аг П^Ёр^Зр6.
Sometimes in diagrams depicting the distribution of electrons in atoms, only the number of electrons at each energy level is indicated, that is, abbreviated electronic formulas of atoms of chemical elements are written, in contrast to the full electronic formulas given above.
For elements of large periods (fourth and fifth), the first two electrons occupy the 4th and 5th orbitals, respectively: 19 K 2, 8, 8, 1; 38 Sr 2, 8, 18, 8, 2. Starting from the third element of each major period, the next ten electrons will enter the previous 3d and 4d orbitals, respectively (for elements of side subgroups): 23 V 2, 8, 11, 2; 26 Tr 2, 8, 14, 2; 40 Zr 2, 8, 18, 10, 2; 43 Tg 2, 8, 18, 13, 2. As a rule, when the previous d-sublevel is filled, the outer (4p- and 5p-respectively) p-sublevel will begin to fill.
For elements of large periods - the sixth and the incomplete seventh - electronic levels and sublevels are filled with electrons, as a rule, like this: the first two electrons will go to the outer b-sublevel: 56 Va 2, 8, 18, 18, 8, 2; 87Gg 2, 8, 18, 32, 18, 8, 1; the next one electron (for Na and Ac) to the previous one (p-sublevel: 57 La 2, 8, 18, 18, 9, 2 and 89 Ac 2, 8, 18, 32, 18, 9, 2.
Then the next 14 electrons will enter the third outer energy level in the 4f and 5f orbitals of the lanthanides and actinides, respectively.
Then the second external energy level (d-sublevel) will begin to build up again: for elements of side subgroups: 73 Ta 2, 8.18, 32.11, 2; 104 Rf 2, 8.18, 32, 32.10, 2, - and, finally, only after the current level is completely filled with ten electrons will the outer p-sublevel be filled again:
86 Rn 2, 8, 18, 32, 18, 8.
Very often, the structure of the electronic shells of atoms is depicted using energy or quantum cells - so-called graphical electronic formulas are written. For this notation, the following notation is used: each quantum cell is designated by a cell that corresponds to one orbital; Each electron is indicated by an arrow corresponding to the spin direction. When writing a graphical electronic formula, you should remember two rules: the Pauli principle, according to which there can be no more than two electrons in a cell (orbital), but with antiparallel spins, and F. Hund’s rule, according to which electrons occupy free cells (orbitals) and are located in At first, they are one at a time and have the same spin value, and only then they pair, but the spins will be oppositely directed according to the Pauli principle.
In conclusion, let us once again consider the display of the electronic configurations of atoms of elements according to the periods of the D.I. Mendeleev system. Diagrams of the electronic structure of atoms show the distribution of electrons across electronic layers (energy levels). 
In a helium atom, the first electron layer is complete - it has 2 electrons.
Hydrogen and helium are s-elements; the s-orbital of these atoms is filled with electrons.
Elements of the second period
For all elements of the second period, the first electron layer is filled and electrons fill the e- and p-orbitals of the second electron layer in accordance with the principle of least energy (first s-, and then p) and the Pauli and Hund rules (Table 2).
In the neon atom, the second electron layer is complete - it has 8 electrons.
Table 2 Structure of electronic shells of atoms of elements of the second period
End of table. 2 
Li, Be - b-elements.
B, C, N, O, F, Ne are p-elements; these atoms have p-orbitals filled with electrons.
Elements of the third period
For atoms of elements of the third period, the first and second electronic layers are completed, so the third electronic layer is filled, in which electrons can occupy the 3s, 3p and 3d sublevels (Table 3).
Table 3 Structure of electronic shells of atoms of elements of the third period
The magnesium atom completes its 3s electron orbital. Na and Mg-s-elements. 
An argon atom has 8 electrons in its outer layer (third electron layer). As an outer layer, it is complete, but in total in the third electron layer, as you already know, there can be 18 electrons, which means that the elements of the third period have unfilled 3d orbitals.
All elements from Al to Ar are p-elements. The s- and p-elements form the main subgroups in the Periodic Table.
A fourth electron layer appears in the potassium and calcium atoms, and the 4s sublevel is filled (Table 4), since it has lower energy than the 3d sublevel. To simplify the graphical electronic formulas of atoms of elements of the fourth period: 1) let us denote the conventional graphical electronic formula of argon as follows:
Ar;
2) we will not depict sublevels that are not filled in these atoms.
Table 4 Structure of electronic shells of atoms of elements of the fourth period
K, Ca - s-elements included in the main subgroups. In atoms from Sc to Zn, the 3rd sublevel is filled with electrons. These are Zy elements. They are included in side subgroups, their outermost electronic layer is filled, and they are classified as transition elements.
Pay attention to the structure of the electronic shells of chromium and copper atoms. In them there is a “failure” of one electron from the 4th to the 3rd sublevel, which is explained by the greater energy stability of the resulting electronic configurations Zd 5 and Zd 10: 
In the zinc atom, the third electron layer is complete - all sublevels 3s, 3p and 3d are filled in it, with a total of 18 electrons.
In the elements following zinc, the fourth electron layer, the 4p-sublevel, continues to be filled: Elements from Ga to Kr are p-elements. 
The krypton atom has an outer layer (fourth) that is complete and has 8 electrons. But in total in the fourth electron layer, as you know, there can be 32 electrons; the krypton atom still has unfilled 4d and 4f sublevels.
For elements of the fifth period, sublevels are filled in in the following order: 5s-> 4d -> 5p. And there are also exceptions associated with the “failure” of electrons in 41 Nb, 42 MO, etc.
In the sixth and seventh periods, elements appear, that is, elements in which the 4f- and 5f-sublevels of the third outside electronic layer are being filled, respectively.
4f elements are called lanthanides.
5f-Elements are called actinides.
The order of filling the electronic sublevels in the atoms of elements of the sixth period: 55 Сs and 56 Ва - 6s elements;
57 La... 6s 2 5d 1 - 5d element; 58 Ce - 71 Lu - 4f elements; 72 Hf - 80 Hg - 5d elements; 81 Tl- 86 Rn - 6p-elements. But here, too, there are elements in which the order of filling the electron orbitals is “violated,” which, for example, is associated with the greater energy stability of half and completely filled f sublevels, that is, nf 7 and nf 14.
Depending on which sublevel of the atom is filled with electrons last, all elements, as you already understood, are divided into four electronic families or blocks (Fig. 7).
1) s-Elements; the b-sublevel of the outer level of the atom is filled with electrons; s-elements include hydrogen, helium and elements of the main subgroups of groups I and II;
2) p-elements; the p-sublevel of the outer level of the atom is filled with electrons; p elements include elements of the main subgroups of groups III-VIII;
3) d-elements; the d-sublevel of the pre-external level of the atom is filled with electrons; d-elements include elements of secondary subgroups of groups I-VIII, that is, elements of plug-in decades of large periods located between s- and p-elements. They are also called transition elements;
4) f-elements, the f-sublevel of the third outer level of the atom is filled with electrons; these include lanthanides and actinides.
1. What would happen if the Pauli principle were not observed?
2. What would happen if Hund's rule were not followed?
3. Make diagrams of the electronic structure, electronic formulas and graphic electronic formulas of atoms of the following chemical elements: Ca, Fe, Zr, Sn, Nb, Hf, Pa.
4. Write the electronic formula for element #110 using the appropriate noble gas symbol.
Lesson content lesson notes supporting frame lesson presentation acceleration methods interactive technologies Practice tasks and exercises self-test workshops, trainings, cases, quests homework discussion questions rhetorical questions from students Illustrations audio, video clips and multimedia photographs, pictures, graphics, tables, diagrams, humor, anecdotes, jokes, comics, parables, sayings, crosswords, quotes Add-ons abstracts articles tricks for the curious cribs textbooks basic and additional dictionary of terms other Improving textbooks and lessonscorrecting errors in the textbook updating a fragment in a textbook, elements of innovation in the lesson, replacing outdated knowledge with new ones Only for teachers perfect lessons calendar plan for the year guidelines discussion programs Integrated LessonsElectronic configuration of an atom is a formula showing the arrangement of electrons in an atom by levels and sublevels. After studying the article, you will learn where and how electrons are located, get acquainted with quantum numbers and be able to construct the electronic configuration of an atom by its number; at the end of the article there is a table of elements.
Why study the electronic configuration of elements?
Atoms are like a construction set: there is a certain number of parts, they differ from each other, but two parts of the same type are absolutely the same. But this construction set is much more interesting than the plastic one and here’s why. The configuration changes depending on who is nearby. For example, oxygen next to hydrogen Maybe turn into water, when near sodium it turns into gas, and when near iron it completely turns it into rust. To answer the question of why this happens and predict the behavior of an atom next to another, it is necessary to study the electronic configuration, which will be discussed below.
How many electrons are in an atom?
An atom consists of a nucleus and electrons rotating around it; the nucleus consists of protons and neutrons. In the neutral state, each atom has the number of electrons equal to the number of protons in its nucleus. The number of protons is designated serial number element, for example, sulfur, has 16 protons - the 16th element of the periodic table. Gold has 79 protons - the 79th element of the periodic table. Accordingly, sulfur has 16 electrons in the neutral state, and gold has 79 electrons.
Where to look for an electron?
By observing the behavior of the electron, certain patterns were derived; they are described by quantum numbers, there are four in total:
- Principal quantum number
- Orbital quantum number
- Magnetic quantum number
- Spin quantum number
Orbital
Further, instead of the word orbit, we will use the term “orbital”; an orbital is the wave function of an electron; roughly, it is the region in which the electron spends 90% of its time.
N - level
L - shell
M l - orbital number
M s - first or second electron in the orbital
Orbital quantum number l
As a result of studying the electron cloud, they found that depending on the energy level, the cloud takes four main forms: a ball, dumbbells and two other, more complex ones. In order of increasing energy, these forms are called the s-, p-, d- and f-shell. Each of these shells can have 1 (on s), 3 (on p), 5 (on d) and 7 (on f) orbitals. The orbital quantum number is the shell in which the orbitals are located. The orbital quantum number for the s,p,d and f orbitals takes the values 0,1,2 or 3, respectively.
There is one orbital on the s-shell (L=0) - two electrons
There are three orbitals on the p-shell (L=1) - six electrons
There are five orbitals on the d-shell (L=2) - ten electrons
There are seven orbitals on the f-shell (L=3) - fourteen electrons
Magnetic quantum number m l
There are three orbitals on the p-shell, they are designated by numbers from -L to +L, that is, for the p-shell (L=1) there are orbitals “-1”, “0” and “1”. The magnetic quantum number is denoted by the letter m l.
Inside the shell, it is easier for electrons to be located in different orbitals, so the first electrons fill one in each orbital, and then a pair of electrons is added to each one.
Consider the d-shell:
The d-shell corresponds to the value L=2, that is, five orbitals (-2,-1,0,1 and 2), the first five electrons fill the shell taking the values M l =-2, M l =-1, M l =0 , M l =1,M l =2.
Spin quantum number m s
Spin is the direction of rotation of an electron around its axis, there are two directions, so the spin quantum number has two values: +1/2 and -1/2. One energy sublevel can only contain two electrons with opposite spins. The spin quantum number is denoted m s
Principal quantum number n
The main quantum number is the energy level at this moment seven energy levels are known, each designated Arabic numeral: 1,2,3,...7. The number of shells at each level is equal to the level number: there is one shell on the first level, two on the second, etc.
Electron number
So, any electron can be described by four quantum numbers, the combination of these numbers is unique for each position of the electron, take the first electron, the lowest energy level is N = 1, at the first level there is one shell, the first shell at any level has the shape of a ball (s -shell), i.e. L=0, the magnetic quantum number can take only one value, M l =0 and the spin will be equal to +1/2. If we take the fifth electron (in whatever atom it is), then the main quantum numbers for it will be: N=2, L=1, M=-1, spin 1/2.
Determine which atoms of the elements indicated in the series have four electrons at the outer energy level.
Answer: 35
Explanation:
The number of electrons in the outer energy level (electronic layer) of the elements of the main subgroups is equal to the group number.
Thus, from the presented answer options, silicon and carbon are suitable, because they are in the main subgroup of the fourth group of the D.I. table. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.
Determine which atoms of the elements indicated in the series in the ground state have the number of unpaired electrons in the outer level equal to 1.
Write down the numbers of the selected elements in the answer field.
Answer: 24
Explanation:
Barium is an element of the main subgroup of the second group and the sixth period Periodic table D.I. Mendeleev, therefore, the electronic configuration of its outer layer will be 6 s 2. On the outside 6 s s-orbital, barium atom contains 2 paired electrons with opposite spins (complete filling of the sublevel).
Aluminum is an element of the main subgroup of the third group and the third period of the Periodic Table, and the electronic configuration of the outer layer of the aluminum atom is 3 s 2 3p 1: by 3 s-sublevel (consists of one s-orbitals) there are 2 paired electrons with opposite spins (full occupancy), and 3 p-sublevel - one unpaired electron. Thus, in aluminum in the ground state the number of unpaired electrons in the outer energy level is 1.
Nitrogen is an element of the main subgroup of the fifth group and the second period of the Periodic Table, the electronic configuration of the outer layer of the nitrogen atom is 2 s 2 2p 3 : by 2 s-sublevel there are 2 paired electrons with opposite spins, and on 2 p p-orbitals ( p x, p y, p z) - three unpaired electrons, each of which is in each orbital. Thus, in aluminum in the ground state the number of unpaired electrons in the outer energy level is 1.
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table, the electronic configuration of the outer layer of the chlorine atom is 3 s 2 3p5: by 3 s-sublevel contains 2 paired electrons with opposite spins, and 3 p-sublevel, consisting of three p-orbitals ( p x, p y, p z) - 5 electrons: 2 pairs of paired electrons in orbitals p x, p y and one unpaired one - in orbital p z. Thus, in the ground state of chlorine the number of unpaired electrons in the outer energy level is 1.
Calcium is an element of the main subgroup of the second group and the fourth period of the Periodic Table of D. I. Mendeleev. The electronic configuration of its outer layer is similar to the electronic configuration of a barium atom. On the outside 4 s-sublevel, consisting of one s-orbitals, the calcium atom contains 2 paired electrons with opposite spins (complete filling of the sublevel).
Determine which atoms of the elements indicated in the series have all valence electrons located on 4 s-energy sublevel.
Write down the numbers of the selected elements in the answer field.
Answer: 25
Explanation:
s 2 3p 5, i.e. valence electrons of chlorine are located on 3 s- and 3 p-sublevels (3rd period).
Potassium is an element of the main subgroup of the first group and the fourth period of the Periodic Table, and the electronic configuration of the outer layer of the potassium atom is 4 s 1, i.e. The only valence electron of the potassium atom is located at 4 s-sublevel (4th period).
Bromine is an element of the main subgroup of the seventh group and the fourth period of the Periodic Table, the electronic configuration of the outer layer of the bromine atom is 4 s 2 4p 5, i.e. the valence electrons of the bromine atom are located on 4 s- and 4 p-sublevels (4th period).
Fluorine is an element of the main subgroup of the seventh group and the second period of the Periodic Table, the electronic configuration of the outer layer of the fluorine atom is 2 s 2 2p5, i.e. The valence electrons of the fluorine atom are located on 2s- And 2p- sublevels. However, due to the high electronegativity of fluorine, only a single electron located on 2p- sublevel, participates in the formation of chemical bonds.
Calcium is an element of the main subgroup of the second group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of its outer layer is 4 s 2, i.e. valence electrons are located on 4 s-sublevel (4th period).
Determine which atoms of the elements indicated in the series have valence electrons located at the third energy level.
Write down the numbers of the selected elements in the answer field.
Answer: 15
Explanation:
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of chlorine is 3 s 2 3p 5, i.e. The valence electrons of chlorine are located at the third energy level (3rd period).
s 2 2p 3, i.e. The valence electrons of nitrogen are located at the second energy level (2nd period).
Carbon is an element of the main subgroup of the fourth group and the second period of the Periodic Table, the electronic configuration of the outer layer of the carbon atom is 2 s 2 2p 2, i.e. The valence electrons of the carbon atom are located at the second energy level (2nd period).
Beryllium is an element of the main subgroup of the second group and the second period of the Periodic Table, the electronic configuration of the outer layer of the beryllium atom is 2 s 2, i.e. The valence electrons of the beryllium atom are located at the second energy level (2nd period).
Phosphorus is an element of the main subgroup of the fifth group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of its outer layer is 3 s 2 3p 3, i.e. The valence electrons of the phosphorus atom are located at the third energy level (3rd period).
Determine which atoms of the elements indicated in the series have d-there are no electrons in sublevels.
Write down the numbers of the selected elements in the answer field.
Answer: 12
Explanation:
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the chlorine atom is 1 s 2 2s 2 2p 6 3s 2 3p 5, i.e. d-sublevel does not exist for the chlorine atom.
Fluorine is an element of the main subgroup of the seventh group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the fluorine atom is 1 s 2 2s 2 2p 5, i.e. d There is also no -sublevel for the fluorine atom.
Bromine is an element of the main subgroup of the seventh group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the bromine atom is 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5, i.e. the bromine atom has a completely filled 3 d-sublevel.
Copper is an element of the secondary subgroup of the first group and the fourth period of the Periodic Table, the electronic configuration of the copper atom is 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10, i.e. the copper atom has a completely filled 3d-sublevel.
Iron is an element of the side subgroup of the eighth group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the iron atom is 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6, i.e. the iron atom has an unfilled 3d-sublevel.
Determine which atoms of the elements indicated in the series belong to s-elements.
Write down the numbers of the selected elements in the answer field.
Answer: 15
Explanation:
Helium is an element of the main subgroup of the second group and the first period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the helium atom is 1 s 2, i.e. The valence electrons of the helium atom are located only on 1s-sublevel, therefore, helium can be classified as s-elements.
Phosphorus is an element of the main subgroup of the fifth group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the phosphorus atom is 3 s 2 3p 3, therefore, phosphorus belongs to p-elements.
s 2 3p 1, therefore, aluminum belongs to p-elements.
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table of D.I. Mendeleev, the electronic configuration of the outer layer of the chlorine atom is 3s 2 3p 5, therefore, chlorine belongs to p-elements.
Lithium is an element of the main subgroup of the first group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the lithium atom is 2 s 1, therefore, lithium belongs to s-elements.
Determine which atoms of the elements indicated in the series in the excited state have the electronic configuration of the outer energy level ns 1 np 2.
Write down the numbers of the selected elements in the answer field.
Answer: 12
Explanation:
Boron is an element of the main subgroup of the third group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the boron atom in the ground state is 2 s 2 2p 1 . When a boron atom transitions to an excited state, the electronic configuration becomes 2 s 1 2p 2 due to electron hopping from 2 s- on 2 p- orbital.
Aluminum is an element of the main subgroup of the third group and the third period of the Periodic Table, the electronic configuration of the outer layer of the aluminum atom is 3 s 2 3p 1 . When an aluminum atom transitions to an excited state, the electronic configuration becomes 3 s 1 3 p 2 due to electron hopping from 3 s- by 3 p- orbital.
Fluorine is an element of the main subgroup of the seventh group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the fluorine atom is 3 s 2 3p 5 . In this case, in the excited state it is impossible to obtain the electronic configuration of the external electronic level n s 1n p 2 .
Iron is an element of the side subgroup of the eighth group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the iron atom is 4 s 2 3d 6. In this case, in the excited state it is also impossible to obtain the electronic configuration of the external electronic level n s 1n p 2 .
Nitrogen is an element of the main subgroup of the fifth group and the second period of the Periodic Table, and the electronic configuration of the outer layer of the nitrogen atom is 2 s 2 2p 3. In this case, in the excited state it is also impossible to obtain the electronic configuration of the external electronic level n s 1n p 2 .
Determine for which atoms of the elements indicated in the series a transition to an excited state is possible.
Write down the numbers of the selected elements in the answer field.
Answer: 23
Explanation:
Rubidium and cesium - elements of the main subgroup of the first group of the Periodic Table of D.I. Mendeleev, are alkali metals, the atoms of which have one electron at the outer energy level. Because the s-the orbital for atoms of these elements is external, it is impossible for an electron to jump from s- on p-orbital, and therefore, the transition of the atom to an excited state is not typical.
The nitrogen atom is not able to go into an excited state because its 2nd energy level is filled and there are no free orbitals at this energy level.
Aluminum is an element of the main subgroup of the third group of the Periodic Table of Chemical Elements, the electronic configuration of the outer layer of the aluminum atom is 3 s 2 3p 1 . When an aluminum atom transitions to an excited state, an electron jumps from 3 s- by 3 p- orbital, and the electron configuration of the aluminum atom becomes 3 s 1 3 p 2 .
Carbon is an element of the main subgroup of the fourth group of the Periodic Table, the electronic configuration of the outer layer of the carbon atom is 2 s 2 2p2. When a carbon atom transitions to an excited state, an electron jumps from 2 s- on 2 p- orbital, and the electron configuration of the carbon atom becomes 2s 1 2p 3 .
Determine which atoms of the elements indicated in the series correspond to the electronic configuration of the outer electron layer ns 2 n.p. 3 .
Write down the numbers of the selected elements in the answer field.
Answer: 23
Explanation:
Electronic configuration of the outer electron layer ns 2 n.p. 3 indicates that the element to be filled in is p sublevel, i.e. This p-elements. All p-elements are located in the last 6 cells of each period in a group whose number is equal to the sum of electrons per s And p sublevels of the outer layer, i.e. 2+3 = 5. Thus, the elements we are looking for are nitrogen and phosphorus.
Determine which atoms of the elements indicated in the series have a similar configuration of the external energy level.
Write down the numbers of the selected elements in the answer field.
Answer: 34
Among the listed elements, bromine and fluorine have a similar electronic configuration. The electronic configuration of the outer layer has the form ns 2 np 5
Determine which atoms of the elements indicated in the series have a completely completed second electronic level.
Write down the numbers of the selected elements in the answer field.
Answer: 13
Explanation:
The filled 2nd electronic level has the noble gas neon, as well as any chemical element, located after it in the periodic table.
Determine which atoms of the elements indicated in the series lack 2 electrons to complete the outer energy level.
Write down the numbers of the selected elements in the answer field.
Answer: 34
An electron is missing before the outer electron level 2 is completed p-elements of the sixth group. Let us remind you that everything p-elements are located in the last 6 cells of each period.
Determine which atoms of the elements indicated in the series in the excited state have the electronic formula of the outer energy level n s 1n p 3 .
Write down the numbers of the selected elements in the answer field.
Answer: 24
Explanation:
s 1n p 3 tells us that there are 4 electrons (1+3) at the outer energy level (electronic layer). Among these elements, only silicon and carbon atoms have 4 electrons in the outer level.
The electronic configuration of the external energy level of these elements in the ground state has the form n s 2n p 2, and in excited n s 1n p 3 (when carbon and silicon atoms are excited, the electrons of the s-orbital are paired and one electron falls on the free p-orbital).
Determine which atoms of the elements indicated in the series in the ground state have the electronic formula of the outer energy level n s 2n p 4 .
Write down the numbers of the selected elements in the answer field.
Answer: 25
Explanation:
Formula for external energy level n s 2n p 4 tells us that there are 6 electrons (2+4) at the outer energy level (electronic layer). The number of electrons in the outer electronic level for elements of the main subgroups is always equal to the group number. Thus, the electronic configuration n s 2n p 4 among the indicated elements there are atoms of selenium and sulfur, since these elements are located in the VIA group.
Determine which atoms of the elements indicated in the series have only one unpaired electron in the ground state.
Write down the numbers of the selected elements in the answer field.
Answer: 25
Determine which atoms of the elements have the configuration of the outer electronic level n s 2n p 3 .
Answer: 45
Determine which atoms of the elements indicated in the series do not contain unpaired electrons in the ground state.
Write down the numbers of the selected elements in the answer field.
Lecture 2. Electronic configuration of an element
At the end of the last lecture, based on Klechkovsky’s rules, we constructed the order of filling energy sublevels with electrons
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 5d1 4f14 5d9 6p6 7s2 6d1 5f14 6d9 7p6 …
The distribution of an atom's electrons across energy sublevels is called electronic configuration. First of all, when looking at the filling row, a certain periodicity-pattern catches the eye.
The filling of energy orbitals with electrons in the ground state of an atom follows the principle of least energy: first, the more favorable low-lying orbitals are filled, and then successively higher-lying orbitals according to the order of filling.
Let's analyze the filling sequence.
If an atom contains exactly 1 electron, it falls into the lowest-lying 1s-AO (AO – atomic orbital). Consequently, the resulting electronic configuration can be represented by the notation 1s1 or graphically (See below - arrow in a square).
It is not difficult to understand that if there is more than one electron in an atom, they sequentially occupy first 1s, then 2s, and finally move to the 2p sublevel. However, already for six electrons (a carbon atom in the ground state), two possibilities arise: filling the 2p sublevel with two electrons with the same spin or with the opposite one.
Let's give a simple analogy: let's assume that atomic orbitals are a kind of “room” for “tenants”, which are played by electrons. It is well known from practice that residents prefer, if possible, to occupy each separate room, rather than being crowded into one.
Similar behavior is typical for electrons, which is reflected in Hund’s rule:
Hund's rule: a stable state of an atom corresponds to a distribution of electrons within an energy sublevel at which the total spin is maximum.
The state of the atom with the minimum energy is called the ground state, and all the rest are called the excited states of the atom.
Lecture 2. Electronic configuration
Atoms of elements of periods I and II
1 electron | ||||||||||||||||||||||||||||
2 electrons | ||||||||||||||||||||||||||||
3 electrons | ||||||||||||||||||||||||||||
4 electrons | ||||||||||||||||||||||||||||
5 electrons | ||||||||||||||||||||||||||||
6 electrons | ||||||||||||||||||||||||||||
7 electrons | ||||||||||||||||||||||||||||
8 electrons | ||||||||||||||||||||||||||||
9 electrons | ||||||||||||||||||||||||||||
10Ne | 10 electrons | |||||||||||||||||||||||||||
Element of all e- | electronic configuration | electron distribution |
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Then, based on Hund’s rule, for nitrogen the ground state assumes the presence of three unpaired p-electrons (electronic configuration ...2p3). In oxygen, fluorine and neon atoms, electrons are sequentially paired and the 2p sublevel is filled.
Please note that the third period of the Periodic Table begins with the sodium atom,
the configuration of which (11 Na ... 3s1) is very similar to the configuration of lithium (3 Li ... 2s1)
except that the principal quantum number n is three, not two.
The filling of energy sublevels with electrons in atoms of elements of the III period is exactly similar to that observed for elements of the II period: the magnesium atom completes filling the 3s sublevel, then from aluminum to argon electrons are successively placed on the 3p sublevel according to Hund’s rule: first, individual electrons are placed on the AO ( Al, Si, P), then they are paired.
Atoms of elements of the III period
11Na | |||||||||||||||||||||||||||
12Mg | |||||||||||||||||||||||||||
13Al | |||||||||||||||||||||||||||
14Si | |||||||||||||||||||||||||||
17Cl | |||||||||||||||||||||||||||
18Ar | |||||||||||||||||||||||||||
abbreviated | distribution e- |
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Lecture 2. Electronic configuration
The fourth period of the Periodic Table begins with the filling of the 4s sublevel in the potassium and calcium atoms with electrons. As follows from the filling order, then comes the turn of the 3d orbitals.
Thus, we can conclude that the filling of d-AO with electrons is “late” by 1 period: in the IV period, 3(!) d-sublevels are filled).
So, from Sc to Zn the 3d sublevel is filled with electrons (10 electrons), then from Ga to Kr the 4p sublevel is filled.
Atoms of elements of the IV period
20Ca | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||
21Sc | 1s2 2s2 2p6 3s2 3p6 4s2 3d1 | 4s2 3d1 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
1s2 2s2 2p6 3s2 3p6 4s2 3d2 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||
22Ti | 4s2 3d2 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
30Zn | 1s2 2s2 2p6 3s2 3p6 4s2 3d10 | 4s2 3d10 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||
31Ga | 1s 2s 2p 3s 3p 4s 3d | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
36Kr | 1s 2s 2p 3s 3p 4s 3d | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
abbreviated | distribution e- | |||||||||||||||||||||||||||||||||||||||||||||||||||||||
The filling of energy sublevels in the atoms of period V elements with electrons is exactly similar to that observed for period IV elements
(take it apart yourself)
In the sixth period, the 6s sublevel is first filled with electrons (atoms 55 Cs and
56 Ba), and then one electron is located in the 5d orbital of lanthanum (57 La 6s2 5d1).
For the next 14 elements (from 58 to 71), the 4f sublevel is filled, i.e. the filling of f-orbitals is “late” by 2 periods, while the electron at the 5d sublevel is retained. For example, we should write down the electronic configuration of cerium
58 Ce 6s2 5d 1 4 f 1
Starting from the 72-element (72 Hf) and up to 80 (80 Hg), the 5d sublevel is “refilled”.
Consequently, the electronic configurations of hafnium and mercury have the form
72 Hf 6s2 5d 1 4 f 14 5d 1 or the entry 72 Hf 6s2 4 f 14 5d 2 80 Hg 6s2 5d 1 4 f 14 5d 9 or 80 Hg 6s2 4 f 14 5d 10 is acceptable
Lecture 2. Electronic configuration
In a similar way, electrons fill energy sublevels in the atoms of period VII elements.
Determining Quantum Numbers from Electronic Configuration
What are quantum numbers, how did they appear and why are they needed - see Lecture 1.
Given: record of electronic configuration “3p 4”
The main quantum number n is the first digit in the notation, i.e. "3". n = 3 "3 p4", principal quantum number;
Secondary (orbital, azimuthal) quantum number l is encoded letter designation sublevel. The letter p corresponds to the number l = 1.
cloud shape
l = 1 "3p 4",
"dumbbell"
Distribution of electrons within a sublevel according to the Pauli principle and Hund's rule
m Є [-1;+1] – the orbitals are identical (degenerate) in energyn = 3, l = 1, m Є [-1;+1] (m = -1); s = + ½
n = 3, l = 1, m Є [-1;+1] (m = 0); s = + ½n = 3, l = 1, m Є [-1;+1] (m = +1); s = + ½ n = 3, l = 1, m Є [-1;+1] (m = -1); s = - ½
Valence level and valence electrons
Valence level called a set of energy sublevels that participate in the formation chemical bonds with other atoms.
Electrons located at the valence level are called valence electrons.
PSHE elements are divided into 4 groups
s-elements. Valence electrons ns x. Two s-elements are found at the beginning of each period.
p-elements. Valence electrons ns 2 np x. Six p-elements are located at the end of each period (except the first and seventh).
Lecture 2. Electronic configuration
d-elements. Valence electrons ns 2 (n-1)d x. Ten d-elements form secondary subgroups, starting from period IV and are located between the s- and p-elements.
f -elements. Valence electrons ns 2 (n-1)d 1 (n-2)f x . Fourteen f elements form the lanthanide (4f) and actinide (5f) series, which are located below the table.
Electronic analogues- these are particles that are characterized by similar electronic configurations, i.e. distribution of electrons among sublevels.
For example
H 1s1 Li … 2s1 Na … 3s1 K … 4s1
Electronic analogues have similar electronic configurations, so their chemical properties are similar - and they are located in the same subgroup of the Periodic Table of Elements.
Electronic "failure" (or electronic "slip")
Quantum mechanics predicts that the state of a particle has the lowest energy when all levels are either completely or half filled with electrons.
That's why for chromium subgroup elements(Cr, Mo, W, Sg) and elements of the copper subgroup(Cu, Ag, Au) there is a movement of 1 electron cs - to the d-sublevel.
24 Cr 4s2 3d4 24 Cr 4s1 3d5 29 Cu 4s2 3d9 29 Cu 4s1 3d10
This phenomenon is called electronic "failure" and should be remembered.
A similar phenomenon is also typical for f-elements, but their chemistry is beyond the scope of our course.
Please note: for p-elements, electron dip is NOT observed!
To summarize, it should be concluded that the number of electrons in an atom is determined by the composition of its nucleus, and their distribution (electronic configuration) is determined by sets
Lecture 2. Electronic configuration
quantum numbers. In turn, the electronic configuration determines the chemical properties of the element.
Therefore, it is obvious that Properties of simple substances, as well as properties of compounds
elements are periodically dependent on the magnitude of the nuclear charge
atom (serial number).
Periodic law
Basic properties of atoms of elements
1. Atomic radius - the distance from the center of the nucleus to the outer energy level. IN
period, as the charge of the nucleus increases, the radius of the atom decreases; in Group,
on the contrary, as the number of energy levels increases, the radius of the atom increases.
Consequently, in the series O2-, F-, Ne, Na+, Mg2+ - the radius of the particle decreases, although their configuration is the same 1s2 2s2 2p6.
For non-metals we talk about the covalent radius, for metals - about the metallic radius, for ions - about the ionic radius.
2. Ionization potential is the energy that needs to be spent on detachment from atom 1
electron. According to the principle of lowest energy, the electron that is last in occupancy (for s and p elements) and the electron of the outer energy level (for d and f elements) is removed first.
In a period, as the charge of the nucleus increases, the ionization potential increases - at the beginning of the period there is an alkali metal with a low ionization potential, at the end of the period there is an inert gas. In the group, the ionization potentials weaken.
Ionization energy, eV
3. Electron affinity is the energy released when an electron is added to an atom, i.e. during the formation of an anion. 4. Electronegativity (EO) is the ability of atoms to attract electron density. Unlike the ionization potential, behind which there is a specific measurable physical quantity, EO is a certain quantity that can beonly calculated, it cannot be measured. In other words, people invented EO in order to use it to explain certain phenomena. For our educational purposes, we need to remember the qualitative order of change electronegativity: F > O > N > Cl > … > H > … > metals. EO is the ability of an atom to shift electron density towards itself, obviously increases in the period (since the charge of the nucleus increases - the force of attraction of the electron and the radius of the atom decreases) and, on the contrary, weakens in the group. It is not difficult to understand that since the period begins with an electropositive metal, and ends with a typical Group VII nonmetal ( inert gases are not taken into account), then the degree of change in EO in the period is greater than in the group.
Lecture 2. Electronic configuration 5. The oxidation state is the conditional charge of an atom in a chemical compound, calculated in the approximation that all bonds are formed by ions. The minimum oxidation state is determined by how many electrons an atom can accept per display the sequence of connections of atoms with each other. Let us consider each pair of atoms separately and denote with an arrow the displacement of electrons to the atom from the pair whose EO is greater (b). Consequently, the electrons shifted - and charges were formed - positive and negative: at the end of each arrow there is a charge (-1), corresponding to the addition of 1 electron; at the base of the arrow there is a charge (+1) corresponding to the removal of 1 electron. The resulting charges are the oxidation state of a particular atom.
That's all for today, thank you for your attention. Literature 1. S.G. Baram, M.A. Ilyin. Chemistry at Summer School. Textbook allowance / Novosibirsk. state University, Novosibirsk, 2012. 48 p. 2. A.V. Manuilov, V.I. Rodionov. Basics of chemistry for children and adults. – M.: ZAO Publishing House Tsentrpoligraf, 2014. – 416 p. – see p. 29-85. http://www.hemi.nsu.ru/ | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
