Catch the answer.
a) Consider the scheme for the formation of an ionic bond between sodium and
oxygen.
1. Sodium is an element of the main subgroup of group I, a metal. It is easier for its atom to give away the first outer electron than to accept the missing 7:
2. Oxygen is an element of the main subgroup of group VI, a non-metal.
It is easier for its atom to accept 2 electrons, which are not enough to complete the outer level, than to give up 6 electrons from the outer level. 
3. First, let’s find the least common multiple between the charges of the formed ions; it is equal to 2(2∙1). In order for Na atoms to give up 2 electrons, they need to take 2 (2:1), so that oxygen atoms can take 2 electrons, they need to take 1.
4. Schematically, the formation of an ionic bond between sodium and oxygen atoms can be written as follows: 
b) Consider the scheme for the formation of an ionic bond between lithium and phosphorus atoms.
I. Lithium is an element of group I of the main subgroup, a metal. It is easier for its atom to give away 1 outer electron than to accept the missing 7: 
2. Chlorine is an element of the main subgroup of group VII, a non-metal. His
It is easier for an atom to accept 1 electron than to give up 7 electrons: 
2. The least common multiple of 1, i.e. In order for 1 lithium atom to give up and a chlorine atom to receive 1 electron, we must take them one at a time.
3. Schematically, the formation of an ionic bond between lithium and chlorine atoms can be written as follows: 
c) Consider the scheme for the formation of an ionic bond between atoms
magnesium and fluorine.
1. Magnesium is an element of group II of the main subgroup, metal. His
It is easier for an atom to give away 2 outer electrons than to accept the missing 6: 
2. Fluorine is an element of the main subgroup of group VII, a non-metal. His
It is easier for an atom to accept 1 electron, which is not enough to complete the outer level, than to give away 7 electrons: 
2. Let's find the smallest common multiple between the charges of the formed ions; it is equal to 2(2∙1). In order for magnesium atoms to give up 2 electrons, only one atom is needed; for fluorine atoms to accept 2 electrons, they need to take 2 (2: 1).
3. Schematically, the formation of an ionic bond between lithium and phosphorus atoms can be written as follows: 
Part I
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.
4. Complete the table “Ionic Bonding”.
Part II
1. Complete the schemes for the formation of positively charged ions. From the letters corresponding to the correct answers, you will form the name of one of the oldest natural dyes: indigo.
2. Play tic-tac-toe. Show the winning path of formulas for substances with ionic chemical bonds.
3. Are the following statements true?
3) only B is correct
4. Underline the pairs of chemical elements between which an ionic chemical bond is formed.
1) potassium and oxygen
3) aluminum and fluorine
Make diagrams of the formation of chemical bonds between selected elements.
5. Create a comic-style drawing that depicts the process of forming an ionic chemical bond.
6. Make a diagram of the formation of two chemical compounds with an ionic bond using the conventional notation:
Select chemical elements "A" and "B" from the following list:
calcium, chlorine, potassium, oxygen, nitrogen, aluminum, magnesium, carbon, bromine.
Suitable for this scheme are calcium and chlorine, magnesium and chlorine, calcium and bromine, magnesium and bromine.
7. Write a short literary work (essay, short story or poem) about one of the substances with ionic bonds that a person uses in everyday life or at work. To complete the task, use the Internet.
Sodium chloride is a substance with an ionic bond, without it there is no life, although when there is a lot of it, this is also not good. There is even a folk tale that says that the princess loved her father the king as much as salt, for which she was expelled from the kingdom. But when the king one day tried food without salt and realized that it was impossible to eat, he then realized that his daughter loved him very much. This means that salt is life, but its consumption should be in
measure. Because excessive salt consumption is very harmful to health. Excess salt in the body leads to kidney disease, changes skin color, retains excess fluid in the body, which leads to swelling and strain on the heart. Therefore, you need to control your salt intake. 0.9% sodium chloride solution is a saline solution used to infuse medications into the body. Therefore, it is very difficult to answer the question: is salt good or bad? We need it in moderation.
This lesson is devoted to generalizing and systematizing knowledge about the types of chemical bonds. During the lesson, schemes for the formation of chemical bonds in various substances will be considered. The lesson will help reinforce the ability to determine the type of chemical bond in a substance based on its chemical formula.
Topic: Chemical bond. Electrolytic dissociation
Lesson: Schemes for the formation of substances with different types of bonds
Rice. 1. Scheme of bond formation in a fluorine molecule
The fluorine molecule consists of two atoms of the same nonmetal chemical element with the same electronegativity; therefore, a covalent nonpolar bond is realized in this substance. Let us depict a diagram of bond formation in a fluorine molecule. Rice. 1.
Around each fluorine atom, using dots, we will draw seven valence, that is, outer, electrons. Each atom needs one more electron to reach a stable state. Thus, one common electron pair is formed. Replacing it with a dash, we depict the graphical formula fluorine molecule F-F.
Conclusion:a covalent nonpolar bond is formed between molecules of one nonmetal chemical element. With this type of chemical bond, common electron pairs are formed that belong equally to both atoms, that is, there is no shift in electron density to any of the atoms of the chemical element
Rice. 2. Scheme of bond formation in a water molecule
A water molecule consists of hydrogen and oxygen atoms - two non-metallic elements with different relative electronegativity values, therefore, this substance has a polar covalent bond.
Since oxygen is a more electronegative element than hydrogen, the shared electron pairs are biased towards oxygen. A partial charge appears on the hydrogen atoms, and a partial negative charge appears on the oxygen atom. Replacing both common electron pairs with dashes, or rather arrows, showing the shift in electron density, we write down the graphic formula of water Fig. 2.
Conclusion:A covalent polar bond occurs between atoms of different nonmetal elements, that is, with different relative electronegativity values. With this type of bond, shared electron pairs are formed, which are shifted towards the more electronegative element.
1. Nos. 5,6,7 (p. 145) Rudzitis G.E. Inorganic and organic chemistry. 8th grade: textbook for general education institutions: basic level / G. E. Rudzitis, F.G. Feldman. M.: Enlightenment. 2011, 176 pp.: ill.
2. Indicate the particle with the largest and smallest radius: Ar atom, ions: K +, Ca 2+, Cl -. Justify your answer.
3. Name three cations and two anions that have the same electron shell as the F - ion.
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Lesson Objectives:
- Form a concept of chemical bonds using the example of an ionic bond. To achieve an understanding of the formation of ionic bonds as an extreme case of polar ones.
- During the lesson, ensure the mastery of the following basic concepts: ions (cation, anion), ionic bond.
- To develop the mental activity of students through the creation of a problem situation when learning new material.
Tasks:
- teach to recognize types of chemical bonds;
- repeat the structure of an atom;
- explore the mechanism of formation of ionic chemical bonds;
- teach how to draw up formation schemes and electronic formulas of ionic compounds, reaction equations with the designation of electron transitions.
Equipment: computer, projector, multimedia resource, periodic table of chemical elements D.I. Mendeleev, table “Ionic bonding”.
Lesson type: Formation of new knowledge.
Lesson type: Multimedia lesson.
X lesson od
I.Organizing time.
II . Checking homework.
Teacher: How can atoms take on stable electronic configurations? What are the ways to form a covalent bond?
Student: Polar and nonpolar covalent bonds are formed by an exchange mechanism. The exchange mechanism includes cases when one electron from each atom participates in the formation of an electron pair. For example, hydrogen: (slide 2)
Bonding occurs through the formation of a shared electron pair by combining unpaired electrons. Each atom has one s electron. The H atoms are equivalent and the pairs belong equally to both atoms. Therefore, the same principle occurs when common electron pairs are formed (overlapping p-electron clouds) during the formation of the F 2 molecule. (slide 3)
Record H · means that a hydrogen atom has 1 electron in its outer electron layer. The recording shows that there are 7 electrons on the outer electron layer of the fluorine atom.
When the N 2 molecule is formed. 3 common electron pairs are formed. The p-orbitals overlap. (slide 4)
The bond is called non-polar.
Teacher: We have now looked at cases when molecules of a simple substance are formed. But around us there are many substances with complex structures. Let's take a hydrogen fluoride molecule. How does the connection form in this case?
Student: When a hydrogen fluoride molecule is formed, the orbital of the s-electron of hydrogen and the orbital of the p-electron of fluorine H-F overlap. (slide 5)
The bonding electron pair is shifted to the fluorine atom, resulting in the formation dipole. Connection called polar.
III. Updating knowledge.
Teacher: A chemical bond arises as a result of changes that occur with the outer electron shells of the connecting atoms. This is possible because the outer electron layers are not complete in elements other than noble gases. The chemical bond is explained by the desire of atoms to acquire a stable electronic configuration similar to the configuration of the “closest” inert gas to them.
Teacher: Write down the diagram of the electronic structure of the sodium atom (at the board). (slide 6)
Student: To achieve stability of the electron shell, the sodium atom must either give up one electron or accept seven. Sodium will easily give up its electron, which is far from the nucleus and weakly bound to it.
Teacher: Make a diagram of electron release.
Na° - 1ē → Na+ = Ne
Teacher: Write down the diagram of the electronic structure of the fluorine atom (at the board).
Teacher: How to complete filling of the electronic layer?
Student: To achieve stability of the electron shell, the fluorine atom must either give up seven electrons or accept one. It is energetically more favorable for fluorine to accept an electron.
Teacher: Make a diagram for receiving an electron.
F° + 1ē → F- = Ne
IV. Learning new material.
The teacher asks a question to the class in which the task of the lesson is set:
Are there other possible ways in which atoms can take on stable electronic configurations? What are the ways to form such connections?
Today we will look at one type of bond - an ionic bond. Let us compare the structure of the electron shells of the already mentioned atoms and inert gases.
Conversation with the class.
Teacher: What charge did the sodium and fluorine atoms have before the reaction?
Student: The sodium and fluorine atoms are electrically neutral, because the charges of their nuclei are balanced by the electrons rotating around the nucleus.
Teacher: What happens between atoms when they give and take electrons?
Student: Atoms acquire charges.
The teacher gives explanations: In the formula of an ion, its charge is additionally written down. To do this, use a superscript. It indicates the amount of charge with a number (they do not write one), and then a sign (plus or minus). For example, a Sodium ion with a charge of +1 has the formula Na + (read “sodium-plus”), a Fluorine ion with a charge of -1 – F - (“fluorine-minus”), a hydroxide ion with a charge of -1 – OH - (“ o-ash-minus"), a carbonate ion with a charge -2 – CO 3 2- (“tse-o-three-two-minus”).
In the formulas of ionic compounds, the positively charged ions are first written, without indicating charges, and then the negatively charged ones. If the formula is correct, then the sum of the charges of all ions in it is zero.
Positively charged ion called a cation, and a negatively charged ion is an anion.
Teacher: We write down the definition in our workbooks:
And he is a charged particle into which an atom turns as a result of accepting or losing electrons.
Teacher: How to determine the charge value of the calcium ion Ca 2+?
Student: An ion is an electrically charged particle formed as a result of the loss or gain of one or more electrons by an atom. Calcium has two electrons in its last electron level; ionization of a calcium atom occurs when two electrons are lost. Ca 2+ is a doubly charged cation.
Teacher: What happens to the radii of these ions?
During the transition When an electrically neutral atom is transformed into an ionic state, the particle size changes greatly. The atom, giving up its valence electrons, turns into a more compact particle - a cation. For example, when a sodium atom transforms into a Na+ cation, which, as indicated above, has the structure of neon, the radius of the particle decreases greatly. The radius of an anion is always greater than the radius of the corresponding electrically neutral atom.
Teacher: What happens to differently charged particles?
Student: Oppositely charged sodium and fluorine ions, resulting from the transfer of an electron from a sodium atom to a fluorine atom, are mutually attracted and form sodium fluoride. (slide 7)
Na + + F - = NaF
The scheme for the formation of ions that we have considered shows how a chemical bond is formed between a sodium atom and a fluorine atom, which is called an ionic bond.
Ionic bond– a chemical bond formed by the electrostatic attraction of oppositely charged ions to each other.
The compounds that are formed in this case are called ionic compounds.
V. Consolidation of new material.
Assignments to consolidate knowledge and skills
1. Compare the structure of the electronic shells of a calcium atom and a calcium cation, a chlorine atom and a chloride anion:
Comment on the formation of ionic bonds in calcium chloride:
2. To complete this task, you need to divide into groups of 3-4 people. Each group member considers one example and presents the results to the whole group.
Student response:
1. Calcium is an element of the main subgroup of group II, a metal. It is easier for its atom to give away two outer electrons than to accept the missing six:
2. Chlorine is an element of the main subgroup of group VII, a non-metal. It is easier for its atom to accept one electron, which it lacks to complete the outer level, than to give away seven electrons from the outer level:
3. First, let's find the least common multiple between the charges of the resulting ions, it is equal to 2 (2x1). Then we determine how many calcium atoms need to be taken so that they give up two electrons, that is, we need to take one Ca atom and two CI atoms.
4. Schematically, the formation of an ionic bond between calcium and chlorine atoms can be written: (slide 8)
Ca 2+ + 2CI - → CaCI 2
Self-control tasks
1. Based on the scheme for the formation of a chemical compound, create an equation for the chemical reaction: (slide 9)
2. Based on the scheme for the formation of a chemical compound, create an equation for the chemical reaction: (slide 10)
3. A scheme for the formation of a chemical compound is given: (slide 11)
Select a pair of chemical elements whose atoms can interact in accordance with this scheme:
A) Na And O;
b) Li And F;
V) K And O;
G) Na And F
Answer to question 5.
The element with atomic number 35 is bromine (Br). The nuclear charge of its atom is 35. A bromine atom contains 35 protons, 35 electrons and 45 neutrons.
§ 7. Changes in the composition of the nuclei of atoms of chemical elements. Isotopes
Answer to question 1.
The isotopes 40 19 K and 40 18 Ar exhibit different properties because they have different nuclear charges and different numbers of electrons.
Answer to question 2.
The relative atomic mass of argon is close to 40, because in the nucleus of its atom there are 18 protons and 22 neutrons, and in the nucleus of the potassium atom there are 19 protons and 20 neutrons, so its relative atomic mass is close to 39. Since the number of protons in the nucleus of the potassium atom is greater, it appears in the table after argon.
Answer to question 3.
Isotopes are varieties of atoms of the same element that have the same number of protons and electrons and different numbers of neutrons.
Answer to question 4.
Isotopes of chlorine are similar in properties, because properties are determined by the charge of the nucleus, and not by its relative mass, even when the relative atomic mass of chlorine isotopes changes by 1 or 2 units, the mass changes slightly, unlike hydrogen isotopes, where when one or two neutrons are added, the mass of the nucleus changes by 2 or 3 times.
Answer to question 5.
Deuterium (heavy water) - a compound where 1 oxygen atom is bonded to two atoms of the hydrogen isotope 2 1 D, formula D2 O. Comparison of the properties of D2 O and H2 O
Answer to question 6.
The element with a large relative value is placed first
atomic mass in vapor:
Te-I (tellurium iodine) 128 Te and 127 I.
Th-Pa (thorium-protactinium) 232 90 Th and 231 91 Pa. U-Np (uranium-neptunium) 238 92 U and 237 93 Np.
§ 8 . Structure of electronic shells of atoms
Answer to question 1.
a) Al +13 |
b) P |
c) O |
|||||||
13 Al 2e– , 8e– , 3e– |
15 Р 2e–, 8e–, 5e– |
8 О 2e– , 6e– |
|||||||
a) - diagram of the structure of the aluminum atom; b) - diagram of the structure of the phosphorus atom; c) - diagram of the structure of the oxygen atom.
Answer to question 2.
a) compare the structure of nitrogen and phosphorus atoms.
7 N 2e– , 5e– |
15 Р 2e–, 8e–, 5e– |
|||||
The structure of the electron shell of these atoms is similar; both contain 5 electrons at the last energy level. However, nitrogen only has 2 energy levels, while phosphorus has 3.
b) Let's compare the structure of phosphorus and sulfur atoms.
15 Р 2e–, 8e–, 5e– |
16 S 2e– , 8e– , 6e– |
||||||
Phosphorus and sulfur atoms have 3 energy levels, each with an incomplete last level, but phosphorus has 5 electrons in its last energy level, and sulfur has 6.
Answer to question 3.
A silicon atom contains 14 protons and 14 neutrons in its nucleus. The number of electrons around the nucleus, as well as the number of protons, is equal to the atomic number of the element. The number of energy levels is determined by the period number and is equal to 3. The number of outer electrons is determined by the group number and is equal to 4.
Answer to question 4.
The number of elements contained in a period is equal to the maximum possible number of electrons at the external energy level and this number is determined by the formula 2n2, where n is the period number.
Therefore, the first period contains only 2 elements (2 12), and the second period contains 8 elements (2 22).
Answer to question 5.
IN astronomy - The period of rotation of the Earth around its axis is 24 hours.
IN Geography - Change of seasons with a period of 1 year.
IN physics - Periodic oscillations of a pendulum.
IN biology - Each yeast cell under optimal conditions once every 20 minutes. shares.
Answer to question 6.
Electrons and the structure of the atom were discovered at the beginning of the 20th century, a little later this poem was written, which largely reflects the nuclear, or planetary, theory of the structure of the atom, and the author also admits the possibility that electrons are also complex particles, the structure of which we simply do not yet understand studied.
Answer to question 7.
The 2 quatrains given in the textbook speak of V. Bryusov’s enormous poetic talent and flexible mind, since he could so easily understand and accept all the achievements of contemporary science, as well as, apparently, enlightenment and education in this area.
§ 9 . Change in the number of electrons at the external energy level of atoms of chemical elements
Answer to question 1.
a) Let’s compare the structure and properties of carbon and silicon atoms
6 C 2e– , 4e– |
14 Si 2e– , 8e– , 4e– |
||||
In terms of the structure of the electronic shell, these elements are similar: both have 4 electrons at the last energy level, but carbon has 2 energy levels, and silicon has 3. Because the number of electrons on the outer level is the same, then the properties of these elements will be similar, but the radius of the silicon atom is larger, therefore, compared to carbon, it will exhibit more metallic properties.
b) Let’s compare the structure and properties of silicon and phosphorus atoms:
14 Si 2e– , 8e– , 4e– |
15 Р 2e–, 8e–, 5e– |
|||||
Silicon and phosphorus atoms have 3 energy levels, and each has an incomplete last level, but silicon has 4 electrons at the last energy level, and phosphorus has 5, so the radius of the phosphorus atom is smaller and it exhibits non-metallic properties to a greater extent than silicon.
Answer to question 2.
a) Consider the scheme for the formation of an ionic bond between aluminum and oxygen.
1. Aluminum is an element of the main subgroup of group III, a metal. It is easier for its atom to give up 3 outer electrons than to accept the missing ones
Al0 – 3e– → Al+ 3
2. Oxygen is an element of the main subgroup of group VI, a non-metal. It is easier for its atom to accept 2 electrons, which are not enough to complete the outer level, than to give up 6 electrons from the outer level.
O0 + 2e– → O− 2
3. First, let's find the least common multiple between the charges of the resulting ions; it is equal to 6(3 2). For Al atoms to give up 6
electrons, they need to be taken 2(6:3), so that the oxygen atoms can accept 6 electrons, they need to be taken 3(6:2).
4. Schematically, the formation of an ionic bond between aluminum and oxygen atoms can be written as follows:
2Al0 + 3O0 → Al2 +3 O3 –2 → Al2 O3
6e–
b) Consider the scheme for the formation of an ionic bond between lithium and phosphorus atoms.
1. Lithium is an element of group I of the main subgroup, a metal. It is easier for its atom to give away 1 outer electron than to accept the missing 7:
Li0 – 1e– → Li+ 1
2. Phosphorus is an element of the main subgroup of group V, a non-metal. It is easier for its atom to accept 3 electrons, which are not enough to complete the outer level, than to give away 5 electrons:
Р0 + 3e– → Р− 3
3. Let's find the least common multiple between the charges of the formed ions; it is equal to 3(3 1). To give away lithium atoms
3 electrons, you need to take 3 (3:1), so that phosphorus atoms can take 5 electrons, you need to take only 1 atom (3:3).
4. Schematically, the formation of an ionic bond between lithium and phosphorus atoms can be written as follows:
3Li0 – + P0 → Li3 +1 P–3 → Li3 P
c) Consider the scheme for the formation of an ionic bond between magnesium and fluorine atoms.
1. Magnesium is an element of group II of the main subgroup, a metal. It is easier for its atom to give away 2 outer electrons than to accept the missing ones
Mg0 – 2e– → Mg+ 2
2. Fluorine is an element of the main subgroup of group VII, a non-metal. It is easier for its atom to accept 1 electron, which is missing until the completion of the outer level, than to give away 7 electrons:
F0 + 1e– → F− 1
3. Let us find the smallest common multiple between the charges of the formed ions; it is equal to 2(2 1). In order for magnesium atoms to give up 2 electrons, only one atom is needed; for fluorine atoms to accept 2 electrons, they need to take 2 (2: 1).
4. Schematically, the formation of an ionic bond between lithium and phosphorus atoms can be written as follows:
Mg0 +– 2F0 → Mg+2 F2 –1 → MgF2
Answer to question 3.
The most typical metals are arranged in the periodic table
V at the beginning of periods and at the end of groups, thus the most typical metal is francium (Fr). Typical nonmetals are located
V at the end of periods and at the beginning of groups. Thus, the most typical nonmetal is fluorine (F). (Helium does not show any chemical properties).
Answer to question 4.
Inert gases began to be called noble gases, just like metals, because in nature they are found exclusively in free form and form chemical compounds with great difficulty.
Answer to question 5.
The expression “The streets of the city at night were flooded with neon” is chemically incorrect, because... neon is an inert, rare gas; there is very little of it in the air. However, neon is filled with neon lamps and fluorescent lamps, which are often used to illuminate signs, posters, and advertisements at night.
§ 10 . Interaction of atoms of non-metal elements with each other
Answer to question 1.
The electronic scheme for the formation of a diatomic halogen molecule will look like this:
a + a→ aa
A structural formula
Answer to question 2.
a) Scheme of chemical bond formation for AlCl3:
Aluminum is a group III element. It is easier for its atom to give away 3 outer electrons than to accept the missing 5.
Al° - 3 e → Al+3
Chlorine is an element of group VII. It is easier for its atom to accept 1 electron, which is not enough to complete the outer level, than to give away 7 electrons.
Сl° + 1 e → Сl–1
Let's find the least common multiple between the charges of the formed ions; it is equal to 3(3:1). In order for aluminum atoms to give up 3 electrons, they need to take only 1 atom (3:3), so that chlorine atoms can take 3 electrons, they need to take 3 (3:1)
Al° + 3Сl° → Al+3 Cl–1 → AlСl3
3 e –
The bond between metal and non-metal atoms is ionic in nature. b) Scheme of chemical bond formation for Cl2:
Chlorine is an element of the main subgroup of group VII. Its atoms have 7 electrons in the outer level. The number of unpaired electrons is
→ClCl |
|||||||||
The bond between atoms of the same element is covalent.
Answer to question 3.
Sulfur is an element of the main subgroup of group VI. Its atoms have 6 electrons in the outer level. The number of unpaired electrons is (8–6)2. In S2 molecules, the atoms are connected by two shared electron pairs, so the bond is double.
The formation scheme for the S2 molecule will look like this:
Answer to question 4.
In the S2 molecule there is a double bond, in the Cl molecule there is a single bond, in the N2 molecule there is a triple bond. Therefore, the strongest molecule will be N2, less strong S2, and even weaker Cl2.
The bond length is shortest in the N2 molecule, longer in the S2 molecule, and even longer in the Cl2 molecule.
§ eleven . Covalent polar chemical bond
Answer to question 1.
Since the EO values of hydrogen and phosphorus are the same, the chemical bond in the PH3 molecule will be covalent nonpolar.
Answer to question 2.
1. a) in the S2 molecule the bond is covalent nonpolar, because it is formed by atoms of the same element. The connection formation scheme will be as follows:
Sulfur is an element of the main subgroup of group VI. Its atoms have 6 electrons in their outer shell. There will be unpaired electrons: 8 – 6 = 2.
Let us denote the outer electrons S
b) in the K2 O molecule the bond is ionic, because it is formed by atoms of metal and non-metal elements.
Potassium is an element of group I of the main subgroup, a metal. It is easier for its atom to give away 1 electron than to accept the missing 7:
K0 – 1e– → K + 1
Oxygen is an element of the main subgroup of group VI, a non-metal. It is easier for his atom to accept 2 electrons, which are not enough to complete the level, than to give up 6 electrons:
O0 + 2e– → O− 2
Let's find the least common multiple between the charges of the formed ions; it is equal to 2(2 1). In order for potassium atoms to give up 2 electrons, they need to take 2, so that oxygen atoms can accept 2 electrons, only 1 atom is needed:
2K2e 0 – + O0 → K2 +1 O–2 → K2 O
c) in the H2 S molecule the bond is covalent polar, because it is formed by atoms of elements with different EO. The connection formation scheme will be as follows:
Sulfur is an element of the main subgroup of group VI. Its atoms have 6 electrons in their outer shell. There will be unpaired electrons: 8– 6=2.
Hydrogen is an element of the main subgroup of group I. Its atoms contain 1 electron in the outer shell. One electron is unpaired (for a hydrogen atom, the two-electron level is complete). Let us denote the outer electrons:
H + S + H → H |
|||
Common electron pairs are shifted to the sulfur atom, as it is more electronegative
H δ+ → S 2 δ− ← H δ+
1. a) in the N2 molecule the bond is covalent nonpolar, because it is formed by atoms of the same element. The connection formation scheme is as follows:
Nitrogen is an element of the main subgroup of group V. Its atoms have 5 electrons in the outer shell. Unpaired electrons: 8 – 5 = 3.
Let's denote the outer electrons: N
→ N N |
|||||||
N ≡ N
b) in the Li3 N molecule the bond is ionic, because it is formed by atoms of metal and non-metal elements.
Lithium is an element of the main subgroup of group I, a metal. It is easier for its atom to give away 1 electron than to accept the missing 7:
Li0 – 1e– → Li+ 1
Nitrogen is an element of the main subgroup of group V, a non-metal. It is easier for its atom to accept 3 electrons, which are not enough to complete the outer level, than to give up five electrons from the outer level:
N0 + 3e– → N− 3
Let's find the least common multiple between the charges of the formed ions; it is equal to 3(3 1). For lithium atoms to give up 3 electrons, 3 atoms are needed, for nitrogen atoms to accept 3 electrons, only one atom is needed:
3Li0 + N0 → Li3 +1 N–3 → Li3 N
3e–
c) in the NCl3 molecule the bond is covalent polar, because it is formed by atoms of non-metal elements with different EO values. The connection formation scheme is as follows:
Nitrogen is an element of the main subgroup of group V. Its atoms have 5 electrons in their outer shell. There will be unpaired electrons: 8– 5=3.
Chlorine is an element of the main subgroup of group VII. Its atoms contain 7 electrons in the outer shell. Remains unpaired
