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The atomic number of an element shows:
a) the number of elementary particles in an atom; b) the number of nucleons in an atom;
c) the number of neutrons in an atom; d) the number of protons in an atom.
The most accurate statement is that the chemical elements in PSE are arranged in ascending order:
a) the absolute mass of their atoms; b) relative atomic mass;
c) the number of nucleons in atomic nuclei; d) charge of the atomic nucleus.
Periodicity in changes in the properties of chemical elements is the result of:
a) increasing the number of electrons in atoms;
b) an increase in the charges of atomic nuclei;
c) increase in atomic mass;
d) periodicity in changes in the electronic structures of atoms.
Of the following, the characteristics of atoms of elements change periodically as the element's atomic number increases:
a) the number of energy levels in an atom;
b) relative atomic mass;
c) the number of electrons at the external energy level;
d) charge of the atomic nucleus.
Select pairs in which each characteristic of the atom changes periodically with increasing proton number of the element:
a) ionization energy and electron affinity energy;
b) radius and mass;
c) electronegativity and total number of electrons;
d) metallic properties and the number of valence electrons.
Select the correct statement for the elementsVAnd the groups:
a) all atoms have the same number of electrons;
b) all atoms have the same radius;
c) all atoms have the same number of electrons in the outer layer;
d) all atoms have a maximum valency equal to the group number.
A certain element has the following electron configuration:ns 2 (n-1) d 10 n.p. 4 . What group of the periodic table is this element in?
a) IVB group; b) VIB group; c) group IVA; d) VIA group.
During PSE periods with increasing charges of atomic nucleiNot changes: a) mass of atoms; b) number of electronic layers; c) the number of electrons in the outer electronic layer; d) radius of atoms. |
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In which series are the chemical elements arranged in order of increasing atomic radius? a) Li, Be, B, C; b) Be, Mg, Ca, Sr; c) N, O, F, Ne; d) Na, Mg, Al, Si. |
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The lowest ionization energy among stable atoms has: a) lithium; b) barium; c) cesium; d) sodium. |
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The electronegativity of elements increases in the series: a) P, Si, S, O; b) Cl, F, S, O; c) Te, Se, S, O; d) O, S, Se, Te. |
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In a row of elementsNa Mg Al Si P S Clfrom left to right: a) electronegativity increases; b) ionization energy decreases; c) the number of valence electrons increases; d) metallic properties decrease. |
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Indicate the most active metal of the fourth period: a) calcium; b) potassium; c) chromium; d) zinc. |
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Specify the most active metal of group IIA: a) beryllium; b) barium; c) magnesium; d) calcium. |
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Specify the most active Group VIIA nonmetal: a) iodine; b) bromine; c) fluorine; d) chlorine. |
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Choose the correct statements: a) in groups IA–VIIIA of PSE there are only s- and b) in groups IV–VIIIB only d-elements are located; c) all d-elements are metals; d) the total number of s-elements in the PSE is 13. |
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With an increase in the atomic number of an element in the VA group, the following increases: a) metallic properties; b) number of energy levels; c) total number of electrons; d) number of valence electrons. |
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P-elements include: a) potassium; b) sodium; c) magnesium; d) arsenic. |
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What family of elements does aluminum belong to? a) s-elements; b) p-elements; c) d-elements; d) f-elements. |
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Indicate the row that contains onlyd-elements: a) Al, Se, La; b) Ti, Ge, Sn; c) Ti, V, Cr; d) La, Ce, Hf. |
In which row are the symbols of the elements of the s, p and d-families shown? a) H, He, Li; b) H, Ba, Al; c) Be, C, F; d) Mg, P, Cu. |
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Which atom of period IV element contains the largest number of electrons? a) zinc; b) chromium; c) bromine; d) krypton. |
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In an atom of which element, the electrons of the outer energy level are most tightly bound to the nucleus? a) potassium; b) carbon; c) fluorine; d) French. |
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The force of attraction of valence electrons to the nucleus of an atom decreases in the series of elements: a) Na, Mg, Al, Si; b) Rb, K, Na, Li; c) Sr, Ca, Mg, Be; d) Li, Na, K, Rb. |
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The element with serial number 31 is located: a) in group III; b) short period; c) long period; d) in group A. |
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From the electronic formulas below, select those that correspond to p-elementsVperiod: a) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 4p 6 4d 1 5s 2 5p 1 ; b) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 4p 6 5s 2 ; c) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 4p 2 ; d) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 4p 6 4d 1 5s 2 5p 6 . |
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From the given electronic formulas, select those that correspond to the chemical elements that form the higher oxide of composition E 2 ABOUT 3 : a) 1s 2 2s 2 2p 6 3s 2 3p 1 ; b) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 4p 3 ; c) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2 ; d) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2. |
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Determine the element whose atom contains 4 electrons in the 4p sublevel. What period and group is it in? a) arsenic, period IV, group VA; b) tellurium, period V, group VI; c) selenium, period IV, group VI; d) tungsten, period VI, group VIB. |
The calcium and scandium atoms differ from each other: a) the number of energy levels; b) radius; c) the number of valence electrons; d) formula of the higher oxide. |
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For sulfur and chromium atoms the same: a) number of valence electrons; b) number of energy levels; c) higher valence; d) formula of the higher oxide. |
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Nitrogen and phosphorus atoms have: a) the same number of electronic layers; b) the same number of protons in the nucleus; c) the same number of valence electrons; d) identical radii. |
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The formula of the highest oxide of an element of period III, the atom of which in the ground state contains three unpaired electrons: a) E 2 O 3; b) EO 2; c) E 2 O 5; d) E 2 O 7. |
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The formula of the highest oxide of the element is EO 3. Give the formula of its hydrogen compound: a) EN 2; b) EN; c) EN 3; d) EN 4. |
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The nature of the oxides from basic to acidic changes in the series: a) Na 2 O, MgO, SiO 2; b) Cl 2 O, SO 2, P 2 O 5, NO 2; c) BeO, MgO, B 2 O 3, Al 2 O 3,; d) CO 2, B 2 O 3, Al 2 O 3, Li 2 O; e) CaO, Fe 2 O 3, Al 2 O 3, SO 2. |
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Select the rows in which the formulas are arranged in increasing order of the acidic properties of the compounds: a) N 2 O 5, P 2 O 5, As 2 O 5; c) H 2 SeO 3, H 2 SO 3, H 2 SO 4; b) HF, HBr, HI; d) Al 2 O 3, P 2 O 5, Cl 2 O 7. |
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Indicate the series in which the hydroxides are arranged in increasing order of their basic properties: a) LiOH, KOH, NaOH; c) LiOH, Ca(OH) 2, Al(OH) 3; b) LiOH, NaOH, Mg(OH) 2; d) LiOH, NaOH, KOH. |
Tasks
The phosphorus sample contains two nuclides: phosphorus-31 and phosphorus-33. The mole fraction of phosphorus-33 is 10%. Calculate the relative atomic mass of phosphorus in this sample.
Natural copper consists of nuclides Cu 63 and Cu 65. The ratio of the number of Cu 63 atoms to the number of Cu 65 atoms in the mixture is 2.45:1.05. Calculate the relative atomic mass of copper.
The average relative atomic mass of natural chlorine is 35.45. Calculate the mole fractions of its two isotopes if it is known that their mass numbers are 35 and 37.
The oxygen sample contains two nuclides: 16 O and 18 O, whose masses are 4.0 g and 9.0 g, respectively. Determine the relative atomic mass of oxygen in this sample.
A chemical element consists of two nuclides. The nucleus of the first nuclide contains 10 protons and 10 neutrons. There are 2 more neutrons in the nucleus of the second nuclide. For every 9 atoms of a lighter nuclide there is one atom of a heavier nuclide. Calculate the average atomic mass of the element.
What relative atomic mass would oxygen have if in a natural mixture for every 4 atoms of oxygen-16 there were 3 atoms of oxygen-17 and 1 atom of oxygen-18?
Answers:1. 31,2. 2. 63,6. 3. 35 Cl: 77.5% and 37 Cl: 22.5%. 4. 17,3. 5. 20,2. 6. 16,6.
Chemical bond
The main volume of educational material:
Nature and types of chemical bonds. Basic parameters of a chemical bond: energy, length.
Covalent bond. Exchange and donor-acceptor mechanisms of covalent bond formation. Directionality and saturation of covalent bonds. Polarity and polarizability of covalent bonds. Valency and oxidation state. Valence possibilities and valence states of atoms of A-group elements. Single and multiple bonds. Atomic crystal lattices. The concept of hybridization of atomic orbitals. Basic types of hybridization. Angles of connections. Spatial structure of molecules. Empirical, molecular and structural (graphical) formulas of molecules.
Ionic bond. Ionic crystal lattices. Chemical formulas of substances with molecular, atomic and ionic structures.
Metal connection. Crystal lattices of metals.
Intermolecular interaction. Molecular crystal lattice. Energy of intermolecular interaction and state of aggregation of substances.
Hydrogen bond. The importance of hydrogen bonding in natural objects.
As a result of studying the topic, students should know:
what is a chemical bond?
main types of chemical bonds;
mechanisms of covalent bond formation (exchange and donor-acceptor);
main characteristics of a covalent bond (saturation, directionality, polarity, multiplicity, s- and p-bonds);
basic properties of ionic, metallic and hydrogen bonds;
main types of crystal lattices;
how the energy reserve and the nature of the movement of molecules change during the transition from one state of aggregation to another;
How do substances with a crystalline structure differ from substances with an amorphous structure?
As a result of studying the topic, students should acquire the skills:
determining the type of chemical bond between atoms in various compounds;
comparing the strength of chemical bonds by their energy;
determination of oxidation states using the formulas of various substances;
establishing the geometric shape of some molecules based on the theory of hybridization of atomic orbitals;
predicting and comparing the properties of substances depending on the nature of the bonds and the type of crystal lattice.
Having finished studying the topic, students should have an idea:
– about the spatial structure of molecules (direction of covalent bonds, bond angle);
– about the theory of hybridization of atomic orbitals (sp 3 -, sp 2 -, sp-hybridization)
After studying the topic, students should remember:
elements with a constant oxidation state;
compounds of hydrogen and oxygen, in which these elements have oxidation states that are not characteristic of them;
the size of the angle between bonds in a water molecule.
Section 1. Nature and types of chemical bonds
The formulas of the substances are given: Na 2 O, SO 3, KCl, PCl 3, HCl, H 2, Cl 2, NaCl, CO 2, (NH 4) 2 SO 4, H 2 O 2, CO, H 2 S, NH 4 Cl, SO 2, HI, Rb 2 SO 4, Sr(OH) 2, H 2 SeO 4, He, ScCl 3, N 2, AlBr 3, HBr, H 2 Se, H 2 O, OF 2, CH 4, NH 3, KI, CaBr 2, BaO, NO, FCl, SiC. Select connections:
molecular and non-molecular structure;
only with covalent polar bonds;
only with covalent non-polar bonds;
only with ionic bonds;
combining ionic and covalent bonds in the structure;
combining covalent polar and covalent nonpolar bonds in the structure;
capable of forming hydrogen bonds;
having bonds in the structure formed according to the donor-acceptor mechanism;
How does the polarity of bonds in rows change?
a) H 2 O; H2S; H2Se; H 2 Te b) PH 3; H2S; HCl.
In what state - ground or excited - are the atoms of isolated elements in the following compounds:
B Cl 3; P Cl 3; Si O2; Be F 2 ; H 2 S; C H4; H Cl O4?
Which pair of the indicated elements during chemical interaction has the maximum tendency to form an ionic bond:
Ca, C, K, O, I, Cl, F?
In which of the chemical substances proposed below will the cleavage of bonds be more likely to occur with the formation of ions, and in which with the formation of free radicals: NaCl, CS 2, CH 4, K 2 O, H 2 SO 4, KOH, Cl 2?
The hydrogen halides are given: HF, HCl, HBr, HI. Select hydrogen halide:
an aqueous solution of which is the strongest acid (the weakest acid);
with the most polar bond (least polar bond);
with the longest connection length (with the shortest connection length);
with the highest boiling point (lowest boiling point).
When one fluorine–fluorine chemical bond is formed, 2.64 ´
10–19 J of energy. Calculate the chemical quantity of fluorine molecules that must be formed in order for 1.00 kJ of energy to be released.
TEST 6.
When a molecule is formed from two isolated atoms, the energy in the system is: a) increases; b) decreases; c) does not change; d) both a decrease and an increase in energy is possible. |
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Indicate in which pair of substances the common electron pairs are shifted towards the oxygen atom: a) OF 2 and CO; b) Cl 2 O and NO; c) H 2 O and N 2 O 3; d) H 2 O 2 and O 2 F 2. |
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Specify compounds with a covalent nonpolar bond: a) O 2; b) N 2; c) Cl 2; d) PCl 5 . |
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Specify compounds with polar covalent bonds: a) H 2 O; b) Br 2; c) Cl 2 O; d) SO 2. |
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Select a pair of molecules in which all bonds are covalent: a) NaCl, HCl; b) CO 2, Na 2 O; c) CH 3 Cl, CH 3 Na; d) SO 2, NO 2. |
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Compounds with covalent polar and covalent nonpolar bonds are, respectively: a) water and hydrogen sulfide; b) potassium bromide and nitrogen; c) ammonia and hydrogen; d) oxygen and methane. |
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None of the covalent bonds are formed by the donor-acceptor mechanism in the particle: a) CO 2; b) CO; c) BF 4 – ; d) NH 4 + . |
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As the difference in electronegativity between bonded atoms increases, the following occurs: a) decreasing the polarity of the bond; b) strengthening the polarity of the connection; c) increasing the degree of ionicity of the bond; d) decreasing the degree of ionicity of the bond. |
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In which row are the molecules arranged in order of increasing bond polarity? a) HF, HCl, HBr; b) NH 3, PH 3, AsH 3; c) H 2 Se, H 2 S, H 2 O; d) CO 2, CS 2, CSe 2. |
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Highest binding energy in a molecule: a) H 2 Te; b) H 2 Se; c) H 2 S; d) H 2 O. |
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The chemical bond is the weakest in a molecule: a) hydrogen bromide; b) hydrogen chloride; c) hydrogen iodide; d) hydrogen fluoride. |
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The bond length increases in a number of substances having the formulas: a) CCl 4, CBr 4, CF 4; b) SO 2, SeO 2, TeO 2; c) H 2 S, H 2 O, H 2 Se; d) HBr, HCl, HF. |
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Maximum numbers-bonds that can exist between two atoms in a molecule: a) 1; b) 2; c) 3; d) 4. |
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A triple bond between two atoms involves: a) 2 s-bonds and 1 π-bond; b) 3 s-bonds; c) 3 π bonds; d) 1s bond and 2π bond. |
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CO molecule 2 contains chemical bonds: a) 1s and 1π; b) 2s and 2π; c) 3s and 1π; d) 4s. |
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Sums- Andπ- connections (s + π) in a moleculeSO 2 Cl 2 is equal to: a) 3 + 3; b) 3 + 2; c) 4 + 2; d) 4 + 3. |
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Specify compounds with ionic bonds: a) sodium chloride; b) carbon monoxide (II); c) iodine; d) potassium nitrate. |
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Only ionic bonds support the structure of a substance: a) sodium peroxide; b) slaked lime; c) copper sulfate; d) sylvinite. |
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Indicate which element atom can participate in the formation of a metallic and ionic bond: a) As; b) Br; c) K; d) Se. |
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The most pronounced character of the ionic bond in the compound is: a) calcium chloride; b) potassium fluoride; c) aluminum fluoride; d) sodium chloride. |
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Indicate substances whose state of aggregation under normal conditions is determined by hydrogen bonds between molecules: a) hydrogen; b) hydrogen chloride; c) liquid hydrogen fluoride; d) water. |
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Indicate the strongest hydrogen bond: a) –N....H–; b) –O....H–; c) –Cl....H–; d) –S....H–. |
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Which chemical bond is the least strong? a) metal; b) ionic; c) hydrogen; d) covalent. |
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Indicate the type of bond in the NF molecule 3 : a) ionic; b) non-polar covalent; c) polar covalent; d) hydrogen. |
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Chemical bond between atoms of elements with atomic numbers 8 and 16: a) ionic; b) covalent polar; c) covalent nonpolar; d) hydrogen. |
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Lesson 2
The quantum numbers discussed above may seem like abstract concepts and far from chemistry. Indeed, they can be used to calculate the structure of real atoms and molecules only with special mathematical training and a powerful computer. However, if we add one more principle to the schematically outlined concepts of quantum mechanics, quantum numbers “come to life” for chemists.
In 1924, Wolfgang Pauli formulated one of the most important postulates of theoretical physics, which did not follow from known laws: more than two electrons cannot simultaneously be in one orbital (in one energy state), and even then only if their spins are in opposite directions. . Other formulations: two identical particles cannot be in the same quantum state; One atom cannot have two electrons with the same values of all four quantum numbers.
Let's try to “create” the electron shells of atoms using the latest formulation of the Pauli principle.
The minimum value of the principal quantum number n is 1. It corresponds to only one value of the orbital number l, equal to 0 (s-orbital). The spherical symmetry of s-orbitals is expressed in the fact that at l = 0 in a magnetic field there is only one orbital with m l = 0. This orbital can contain one electron with any spin value (hydrogen) or two electrons with opposite spin values (helium) . Thus, with n = 1, no more than two electrons can exist.
Now let's start filling the orbitals with n = 2 (there are already two electrons in the first level). The value n = 2 corresponds to two values of the orbital number: 0 (s-orbital) and 1 (p-orbital). At l = 0 there is one orbital, at l = 1 there are three orbitals (with m l values: -1, 0, +1). Each orbital can contain no more than two electrons, so the value n = 2 corresponds to a maximum of 8 electrons. The total number of electrons in a level with a given n can thus be calculated using the formula 2n 2:
Let us denote each orbital by a square cell, the electrons by oppositely directed arrows. For further “construction” of the electronic shells of atoms, it is necessary to use one more rule, formulated in 1927 by Friedrich Hund (Hund): the most stable states for a given l are those with the largest total spin, i.e. the number of filled orbitals at a given sublevel should be maximum (one electron per orbital).
The beginning of the periodic table will look like this:
Scheme of filling the external level of elements of the 1st and 2nd periods with electrons.
Continuing the “construction”, you can reach the beginning of the third period, but then you will have to introduce the order of filling the d and f orbitals as a postulate.
From the diagram constructed on the basis of minimal assumptions, it is clear that quantum objects (atoms of chemical elements) will relate differently to the processes of giving and receiving electrons. He and Ne objects will be indifferent to these processes due to a fully occupied electron shell. The F object will most likely actively accept the missing electron, and the Li object will be more likely to give up the electron.
Object C must have unique properties - it has the same number of orbitals and the same number of electrons. Perhaps he will strive to form connections with himself due to such high symmetry of the external level.
It is interesting to note that the concepts of the four principles of constructing the material world and the fifth that connects them have been known for at least 25 centuries. In Ancient Greece and Ancient China, philosophers spoke of four first principles (not to be confused with physical objects): “fire”, “air”, “water”, “earth”. The connecting principle in China was “wood”, in Greece it was “quintessence” (fifth essence). The relationship of the “fifth element” with the other four is demonstrated in the science fiction film of the same name.
Game "Parallel World"
In order to better understand the role of “abstract” postulates in the world around us, it is useful to move to the “Parallel World”. The principle is simple: the structure of quantum numbers is slightly distorted, then, based on their new values, we build a periodic system of a parallel world. The game will be successful if only one parameter changes, which does not require additional assumptions about the relationship between quantum numbers and energy levels.
For the first time, a similar problem-game was offered to schoolchildren at the All-Union Olympiad in 1969 (9th grade):
“What would a periodic system of elements look like if the maximum number of electrons in a layer was determined by the formula 2n 2 -1, and the outer level could not have more than seven electrons? Draw a table of such a system for the first four periods (designating the elements by their atomic numbers). What oxidation states could element N 13 exhibit? What properties of the corresponding simple substance and compounds of this element could you assume?
This task is too difficult. In the answer, it is necessary to analyze several combinations of postulates establishing the values of quantum numbers with postulates about the relationship between these values. After a detailed analysis of this problem, we came to the conclusion that the distortions in the “parallel world” are too large, and we cannot correctly predict the properties of the chemical elements of this world.
We at the Scientific Research Center of Moscow State University usually use a simpler and more visual problem, in which the quantum numbers of the “parallel world” are almost no different from ours. In this parallel world live analogues of people - homozoids(the description of the homozoids themselves should not be taken seriously).
Periodic law and atomic structure
Task 1.
Homozoids live in a parallel world with the following set of quantum numbers:
n = 1, 2, 3, 4, ...
l= 0, 1, 2, ... (n – 1)
m l = 0, +1, +2,...(+ l)
m s = ± 1/2
Construct the first three periods of their periodic table, keeping our names for the elements with corresponding numbers.
1. How do homozoids wash themselves?
2. What do homozoids get drunk on?
3. Write the equation for the reaction between Their sulfuric acid and aluminum hydroxide.
Solution Analysis
Strictly speaking, you cannot change one of the quantum numbers without affecting the others. Therefore, everything described below is not the truth, but an educational task.
The distortion is almost imperceptible - the magnetic quantum number becomes asymmetric. However, this means the existence of unipolar magnets in a parallel world and other serious consequences. But let's get back to chemistry. In the case of s-electrons, no changes occur ( l= 0 and m 1 = 0). Therefore, hydrogen and helium are the same there. It is useful to remember that according to all data, hydrogen and helium are the most common elements in the Universe. This allows us to assume the existence of such parallel worlds. However, for p-electrons the picture changes. At l= 1 we get two values instead of three: 0 and +1. Therefore, there are only two p orbitals that can accommodate 4 electrons. The length of the period has decreased. We build “arrow cells”:
Construction of the Periodic Table of a Parallel World:
The periods, naturally, have become shorter (in the first there are 2 elements, in the second and third - 6 instead of 8. The changed roles of the elements are perceived very cheerfully (we deliberately keep the names behind the numbers): inert gases O and Si, alkali metal F. In order not to get confused, we will denote their elements are symbols only, and our- in words.
Analysis of the questions in the problem makes it possible to analyze the significance of the distribution of electrons at the external level for the chemical properties of the element. The first question is simple - hydrogen = H, and C becomes oxygen. Everyone immediately agrees that the parallel world cannot exist without halogens (N, Al, etc.). The answer to the second question is related to solving the problem - why carbon is an “element of life” for us and what will be its parallel analogue. During the discussion, we find out that such an element should give the “most covalent” bonds with analogues of oxygen, nitrogen, phosphorus, and sulfur. We have to go ahead a little and analyze the concepts of hybridization, ground and excited states. Then the element of life becomes an analogue of our carbon in symmetry (B) - it has three electrons in three orbitals. The result of this discussion is an analogue of ethyl alcohol BH 2 BHCH.
At the same time, it becomes obvious that in the parallel world we have lost direct analogues of our 3rd and 5th (or 2nd and 6th) groups. For example, period 3 elements correspond to:
Maximum oxidation states: Na (+3), Mg (+4), Al (+5); however, the priority is the chemical properties and their periodic change, and the length of the period has decreased.
Then the answer to the third question (if there is no analogue of aluminum):
Sulfuric acid + aluminum hydroxide = aluminum sulfate + water
H 2 MgC 3 + Ne(CH) 2 = NeMgC 3 + 2 H 2 C
Or as an option (there is no direct analogue of silicon):
H 2 MgC 3 + 2 Na(CH) 3 = Na 2 (MgC 3) 3 + 6 H 2 C
The main result of the described “journey to a parallel world” is the understanding that the infinite diversity of our world stems from a not very large set of relatively simple laws. An example of such laws are the analyzed postulates of quantum mechanics. Even a small change in one of them dramatically changes the properties of the material world.
Test yourself
Select the correct answer (or answers)
Atomic structure, periodic law
1. Eliminate the unnecessary concept:
1) proton; 2) neutron; 3) electron; 4) ion
2. The number of electrons in an atom is equal to:
1) the number of neutrons; 2) the number of protons; 3) period number; 4) group number;
3. Of the following, the characteristics of atoms of elements change periodically as the atomic number of the element increases:
1) the number of energy levels in an atom; 2) relative atomic mass;
3) the number of electrons at the external energy level;
4) charge of the atomic nucleus
4. At the outer level of an atom of a chemical element, there are 5 electrons in the ground state. What element could this be:
1) boron; 2) nitrogen; 3) sulfur; 4) arsenic
5. The chemical element is located in the 4th period, group IA. The distribution of electrons in an atom of this element corresponds to a series of numbers:
1) 2, 8, 8, 2 ; 2) 2, 8, 18, 1 ; 3) 2, 8, 8, 1 ; 4) 2, 8, 18, 2
6. The p-elements include:
1) potassium; 2) sodium; 3) magnesium; 4) aluminum
7. Can the electrons of the K+ ion be in the following orbitals?
1) 3p; 2) 2f ; 3) 4s; 4) 4p
8. Select the formulas of particles (atoms, ions) with the electron configuration 1s 2 2s 2 2p 6:
1) Na+; 2) K + ; 3) Ne; 4) F –
9. How many elements would there be in the third period if the spin quantum number had a single value of +1 (the remaining quantum numbers have the usual values)?
1) 4 ; 2) 6 ; 3) 8 ; 4) 18
10. In which series are the chemical elements arranged in order of increasing atomic radius?
1) Li, Be, B, C;
2) Be, Mg, Ca, Sr;
3) N, O, F, Ne;
4) Na, Mg, Al, Si
© V.V.Zagorsky, 1998-2004
ANSWERS
- 4) ion
- 2) number of protons
- 3) the number of electrons in the outer energy level
- 2) nitrogen; 4) arsenic
- 3) 2, 8, 8, 1
- 4) aluminum
- 1) 3p; 3) 4s; 4) 4p
- 1) Na+; 3) Ne; 4) F –
- 2) Be, Mg, Ca, Sr
- Zagorsky V.V. A version of the presentation in the physics and mathematics school of the topic “Structure of the Atom and the Periodic Law”, Russian Chemical Journal (ZhRKhO named after D.I. Mendeleev), 1994, v. 38, N 4, p. 37-42
- Zagorsky V.V. The structure of the atom and the Periodic Law / "Chemistry" N 1, 1993 (supplement to the newspaper "First of September")