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: the stable state of the atom corresponds to such a distribution of electrons within the 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

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 the same as 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

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

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, you 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;

The side (orbital, azimuthal) quantum number l is encoded by the letter designation of the 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 is a set of energy sublevels that participate in the formation of 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 the 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

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 of the Periodic Table of 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 chlorine in the ground state 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- by 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 1 n 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 1 n 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 1 n 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. Since 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 an atom to an excited state is not typical.

The nitrogen atom is not able to go into an excited state because the 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- by 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 1 n p 3 .

Write down the numbers of the selected elements in the answer field.

Answer: 24

Explanation:

s 1 n 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 2 n p 2, and in excited n s 1 n 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 2 n p 4 .

Write down the numbers of the selected elements in the answer field.

Answer: 25

Explanation:

Formula for external energy level n s 2 n 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 2 n 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 2 n 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.

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 electron energy 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 the 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 its 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 begins; 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?

Electronic configurations of atoms of elements of the Periodic Table.

The distribution of electrons over various AOs is called electronic configuration of an atom. The lowest energy electronic configuration corresponds to basic state atom, the remaining configurations refer to excited states.

The electronic configuration of an atom is depicted in two ways - in the form of electronic formulas and electron diffraction diagrams. When writing electronic formulas, the principal and orbital quantum numbers are used. The sublevel is designated using the principal quantum number (number) and the orbital quantum number (corresponding letter). The number of electrons in a sublevel is characterized by the superscript. For example, for the ground state of the hydrogen atom the electronic formula is: 1 s 1 .

The structure of electronic levels can be more fully described using electron diffraction diagrams, where the distribution among sublevels is represented in the form of quantum cells. In this case, the orbital is conventionally depicted as a square with a sublevel designation next to it. The sublevels at each level should be slightly offset in height, since their energy is slightly different. Electrons are represented by arrows or ↓ depending on the sign of the spin quantum number. Electron diffraction diagram of a hydrogen atom:

The principle of constructing electronic configurations of multi-electron atoms is to add protons and electrons to the hydrogen atom. The distribution of electrons across energy levels and sublevels is subject to the rules discussed earlier: the principle of least energy, the Pauli principle and Hund's rule.

Taking into account the structure of the electronic configurations of atoms, all known elements, in accordance with the value of the orbital quantum number of the last filled sublevel, can be divided into four groups: s-elements, p-elements, d-elements, f-elements.

In a helium atom He (Z=2) the second electron occupies 1 s-orbital, its electronic formula: 1 s 2. Electron diffraction diagram:

Helium ends the first shortest period of the Periodic Table of Elements. The electronic configuration of helium is denoted by .

The second period is opened by lithium Li (Z=3), its electronic formula: Electron diffraction diagram:

The following are simplified electron diffraction diagrams of atoms of elements whose orbitals of the same energy level are located at the same height. Internal, fully filled sublevels are not shown.

After lithium comes beryllium Be (Z=4), in which an additional electron populates 2 s-orbital. Electronic formula of Be: 2 s 2

In the ground state, the next boron electron B (z=5) occupies 2 r-orbital, V:1 s 2 2s 2 2p 1 ; its electron diffraction diagram:

The following five elements have electronic configurations:

C (Z=6): 2 s 2 2p 2 N (Z=7): 2 s 2 2p 3

O (Z=8): 2 s 2 2p 4 F (Z=9): 2 s 2 2p 5

Ne (Z=10): 2 s 2 2p 6

The given electronic configurations are determined by Hund's rule.

The first and second energy levels of neon are completely filled. Let us denote its electronic configuration and will use it in the future for brevity in writing the electronic formulas of atoms of elements.

Sodium Na (Z=11) and Mg (Z=12) open the third period. Outer electrons occupy 3 s-orbital:

Na (Z=11): 3 s 1

Mg (Z=12): 3 s 2

Then, starting with aluminum (Z=13), fill 3 r-sublevel. The third period ends with argon Ar (Z=18):

Al (Z=13): 3 s 2 3p 1

Ar (Z=18): 3 s 2 3p 6

The elements of the third period differ from the elements of the second in that they have free 3 d-orbitals that can participate in the formation of a chemical bond. This explains the valence states exhibited by elements.

In the fourth period, in accordance with the rule ( n+l), potassium K (Z=19) and calcium Ca (Z=20) have 4 electrons s-sublevel, not 3 d.Starting with scandium Sc (Z=21) and ending with zinc Zn (Z=30), filling occurs3 d-sublevel:

Electronic formulas d-elements can be represented in ionic form: the sublevels are listed in increasing order of the main quantum number, and at a constant n– in order of increasing orbital quantum number. For example, for Zn such an entry would look like this: Both of these entries are equivalent, but the previously given formula for zinc correctly reflects the order in which the sublevels are filled.

In row 3 d-elements in chromium Cr (Z=24) there is a deviation from the rule ( n+l). In accordance with this rule, the configuration of Cr should look like this: It has been established that its real configuration is - Sometimes this effect is called the “failure” of the electron. Such effects are explained by half the increased resistance ( p 3 , d 5 , f 7) and completely ( p 6 , d 10 , f 14) filled sublevels.

Deviations from the rule ( n+l) are also observed in other elements (Table 6). This is due to the fact that as the principal quantum number increases, the differences between the energies of sublevels decrease.

Next comes filling 4 p-sublevel (Ga - Kr). The fourth period contains only 18 elements. Filling 5 occurs in the same way s-, 4d- and 5 p- sublevels of 18 elements of the fifth period. Note that the energy is 5 s- and 4 d-sublevels are very close, and the electron with 5 s-sublevels can easily move to 4 d-sublevel. At 5 s-sublevel Nb, Mo, Tc, Ru, Rh, Ag has only one electron. In ground state 5 s-Pd sublevel is not filled. A “failure” of two electrons is observed.

In the sixth period after filling 6 s-sublevel of cesium Cs (Z=55) and barium Ba (Z=56) the next electron, according to the rule ( n+l), should take 4 f-sublevel. However, in lanthanum La (Z=57) the electron goes to 5 d-sublevel. Half filled (4 f 7) 4f-sublevel has increased stability, so gadolinium has Gd (Z=64), next to europium Eu (Z=63), by 4 f- the sublevel retains the same number of electrons (7), and a new electron arrives at 5 d-sublevel, breaking the rule ( n+l). In terbium Tb (Z=65) the next electron occupies 4 f-sublevel and the transition of the electron from 5 occurs d-sublevel (configuration 4 f 9 6s 2). Filling 4 f-sublevel ends at ytterbium Yb (Z=70). The next electron of the lutetium atom Lu occupies 5 d-sublevel. Its electronic configuration differs from that of the lanthanum atom only in that it is completely filled 4 f-sublevel.

Table 6

Exceptions from ( n+l) – rules for the first 86 elements

Currently, in the Periodic Table of Elements D.I. Mendeleev under scandium Sc and yttrium Y are sometimes located lutetium (and not lanthanum) as the first d-element, and all 14 elements in front of it, including lanthanum, are placed in a special group lanthanides beyond the Periodic Table of Elements.

The chemical properties of elements are determined mainly by the structure of the outer electronic levels. Change in the number of electrons on the third outside 4 f-sublevel has little effect on the chemical properties of elements. Therefore all 4 f-elements are similar in their properties. Then in the sixth period the filling of 5 occurs d-sublevel (Hf – Hg) and 6 p-sublevel (Tl – Rn).

In the seventh period 7 s-sublevel is filled with francium Fr (Z=87) and radium Ra (Z=88). Sea anemone exhibits a deviation from the rule ( n+l), and the next electron populates 6 d-sublevel, not 5 f. Next comes a group of elements (Th – No) with a filling 5 f-sublevels that form a family actinides. Note that 6 d- and 5 f- sublevels have such close energies that the electronic configuration of actinide atoms often does not obey the rule ( n+l). But in this case the exact configuration value is 5 f t 5d m is not so important, since it has a rather weak effect on the chemical properties of the element.

In lawrencium Lr (Z=103), a new electron arrives at 6 d-sublevel. This element is sometimes placed under lutetium in the Periodic Table. The seventh period is not completed. Elements 104 – 109 are unstable and their properties are little known. Thus, as the charge of the nucleus increases, similar electronic structures of the outer levels are periodically repeated. In this regard, periodic changes in various properties of elements should also be expected.

Note that the described electronic configurations refer to isolated atoms in the gas phase. The configuration of an element's atom can be completely different if the atom is in a solid or solution.

Electronic configuration of an atom – shows the distribution of ē by energy. levels and sublevels.

1s 1 ←number ē with a given cloud shape

↖ electron cloud shape

energy level

Graphic electronic formulas (images of the electronic structure of an atom) –

shows the distribution of ē by energy. levels, sublevels and orbitals.

I period:+1 N

Where - ē, ↓ - ē with antiparallel spins, orbital.

When writing a graphical electronic formula, you should remember the Pauli rule and Hundd's rule “If within one sublevel there are several free orbitals, then ē are placed each on a separate orbital and only in the absence of free orbitals are combined into pairs.”

(Working with electronic and graphic electronic formulas).

For example, H +1 1s 1 ; He +2 1s 2 ; Li +3 1s 2 2s 1 ; Na +11 1s 2 2s 2 2p 6 3s 1 ; Ar +18 1s 2 2s 2 2p 6 3s 2 3p 6 ;

I period: hydrogen and helium – s-elements, their s-orbital is filled with electrons.

II period: Li and Be are s-elements

B, C, N, O, F, Ne – p-elements

Depending on which sublevel of the atom is filled with electrons last, all elements are divided into 4 electron families or blocks:

1) s-elements – their s-sublevel of the outer layer of the atom is filled with ē; these include hydrogen, helium and el-you gl.p/gr. Groups I and II.

2) p-elements – their electron world, the sublevel of the outer level of the atom, is filled; these include elements of the main group. III - VIII groups.

3) d-elements – in them the d-sublevel of the outermost level of the atom is filled with electrons; These include el-you side.p/gr. . I-VIII groups, i.e. elements of plug-in decades of large periods, located between the s- and p-elements, they are also called transition elements.

4) f-elements- their f-sublevel of the third outer level of the atom is filled with electrons; these include lanthanides (4f elements) and actinides (5f elements).

Copper and chromium atoms have “failure” ē from the 4s to the 3d sublevel, which is explained by the greater energy stability of the resulting electronic configurations 3d 5 and 3d 10:

29 Cu 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 24 Cr 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5

It has been experimentally proven that states of atoms in which the p-, d-, f-orbitals are half filled (p 3, d 5, f 7), completely (p 6, d 10, f 14) or free, have increased stability. This explains the transitions - “dips” - of electrons between closely spaced orbitals. The same deviations are observed in the analogue of chromium - molybdenum, as well as in elements of the copper subgroup - silver and gold. Palladium is unique in this regard, the atom of which has no 5s electrons at all and which has a trace. Configuration: 46 Pd 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 0 4d 10 .

Questions for self-control

1. What is an electron cloud?

2. What is the difference between a 1s orbital and a 2s orbital?

3. What is the principal quantum number? How does it relate to the period number?

4. What is a sublevel and how does this concept relate to the period number?

5. Create electronic configurations of atoms of elements of the 4th-6th period of the PSCE.

6. Create the electronic configuration of magnesium and neon atoms.

7. Determine which atom has the electronic configuration 1S 2 2S 2 2p 6 3S 1, 1S 2 2S 2 2p 6 3S 2, 1S 2 2S 2 2p 4, 1S 2 2S 1

LESSON PLAN No. 7

Discipline: Chemistry.

Subject:

Purpose of the lesson: Study the mechanisms of formation of ionic and covalent bonds, consider ionic, atomic and molecular crystal lattices.

Planned results

Subject: mastery of fundamental chemical concepts: chemical bonds, ions, crystal lattices, confident use of chemical terminology and symbols; developed ability to give quantitative estimates and make calculations using chemical formulas and equations;

Metasubject: the use of various types of cognitive activity and basic intellectual operations: compiling electronic configurations of atoms of chemical elements.

Personal: the ability to use the achievements of modern chemical science and chemical technologies to improve one’s own intellectual development in the chosen professional activity;

Standard time: 2 hours

Type of lesson: Lecture.

Lesson plan:

1. Cations, their formation from atoms as a result of the oxidation process. Anions, their formation from atoms as a result of the reduction process. Ionic bonding is a bond between cations and anions due to electrostatic attraction.

2. Classification of ions: by composition, charge sign, presence of a hydration shell.

3. Ionic crystal lattices. Properties of substances with an ionic type of crystal lattice.

4. Mechanism of covalent bond formation (exchange and donor-acceptor).

5. Electronegativity. Covalent polar and nonpolar bonds. Multiplicity of covalent bond.

6. Molecular and atomic crystal lattices. Properties of substances with molecular and atomic crystal lattices.

Equipment: Models of crystal lattices, textbook, periodic system of chemical elements by D.I. Mendeleev.

Literature:

1. Chemistry 11th grade: textbook. for general education organizations G.E. Rudzitis, F.G. Feldman. – M.: Education, 2014. -208 p.: ill..

2. Chemistry for professions and technical specialties: a textbook for students. institutions prof. education / O.S. Gabrielyan, I.G. Ostroumov. – 5th ed., erased. – M.: Publishing Center “Academy”, 2017. – 272 pp., with colors. ill.

Teacher: Tubaltseva Yu.N.

Topic 7. Ionic and covalent chemical bonding.

1) Cations, their formation from atoms as a result of the oxidation process. Anions, their formation from atoms as a result of the reduction process. Ionic bonding is a bond between cations and anions due to electrostatic attraction.

2) Classification of ions: by composition, charge sign, presence of a hydration shell.

3) Ionic crystal lattices. Properties of substances with an ionic type of crystal lattice.

4) The mechanism of covalent bond formation (exchange and donor-acceptor).

5) Electronegativity. Covalent polar and nonpolar bonds. Multiplicity of covalent bond.

6) Molecular and atomic crystal lattices. Properties of substances with molecular and atomic crystal lattices.

Cations, their formation from atoms as a result of the oxidation process. Anions, their formation from atoms as a result of the reduction process. Ionic bonding is a bond between cations and anions due to electrostatic attraction.

A chemical bond is the interaction of atoms, which determines the stability of a chemical particle or crystal as a whole. A chemical bond is formed due to electrostatic interaction between charged particles: cations and anions, nuclei and electrons. When atoms come together, attractive forces begin to act between the nucleus of one atom and the electrons of another, as well as repulsive forces between nuclei and between electrons. At some distance, these forces balance each other, and a stable chemical particle is formed.

When a chemical bond is formed, a significant redistribution of the electron density of atoms in the compound can occur in comparison with free atoms. In the extreme case, this leads to the formation of charged particles - ions (from the Greek "ion" - going).

Ion interaction:

If an atom loses one or more electrons, then it turns into a positive ion - a cation (translated from Greek - “going down”). This is how the cations of hydrogen H +, lithium Li +, barium Ba 2+ are formed. By acquiring electrons, atoms turn into negative ions - anions (from the Greek "anion" - going up). Examples of anions are fluoride ion F −, sulfide ion S 2−.

Cations and anions are able to attract each other. In this case, a chemical bond occurs and chemical compounds are formed. This type of chemical bond is called an ionic bond:

Ionic bonds typically occur between atoms of typical metals and typical nonmetals. A characteristic property of metal atoms is that they easily give up their valence electrons, while non-metal atoms can easily attach them.

Consider the formation of an ionic bond, for example, between sodium atoms and chlorine atoms in sodium chloride NaCl.

The removal of an electron from a sodium atom leads to the formation of a positively charged ion - the sodium cation Na +.

The addition of an electron to a chlorine atom results in the formation of a negatively charged ion, the chlorine anion Cl - .

Between the resulting Na + and Cl - ions, which have opposite charges, electrostatic attraction occurs, as a result of which a compound is formed - sodium chloride with an ionic type of chemical bond.

Ionic bond is a chemical bond that occurs through the electrostatic interaction of oppositely charged ions.

Thus, the process of formation of an ionic bond is reduced to the transition of electrons from sodium atoms to chlorine atoms with the formation of oppositely charged ions that have completed electronic configurations of the outer layers.

1. Metal atoms, giving up external electrons, turn into positive ions:

where n is the number of electrons in the outer layer of the atom, corresponding to the group number of the chemical element.

2. Non-metal atoms, taking up electrons missing before completing the outer electron layer, turn into negative ions:

3. A bond occurs between oppositely charged ions, which is called ionic.

2. Classification of ions: by composition, charge sign, presence of a hydration shell.

Ion classification:

1. According to the sign of charge: cations (positive, K+, Ca2+, H+) and anions (negative, S2-, Cl-, I-).
2. By composition: complex ( , ) and simple (Na+, F-)

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