The molar mass of technetium is 95 43. Technetium. Molar mass of elements and compounds
Here we must make a small, purely physical digression, otherwise it will not be clear why Segre needed this piece of molybdenum so much. The “tooth” of the deflection plate of the world’s first cyclotron, low-power by today’s standards, was made from molybdenum. A cyclotron is a machine that accelerates the movement of charged particles, for example deuterons - nuclei of heavy hydrogen, deuterium. The particles are accelerated by a high-frequency electric field in a spiral and become more powerful with each turn. Anyone who has ever worked at a cyclotron knows well how difficult it can be to conduct an experiment if the target is installed directly in the vacuum chamber of the cyclotron. It is much more convenient to work on an extracted beam, in a special chamber where all the necessary equipment can be placed. But getting the beam out of the cyclotron is far from easy. This is done using a special deflection plate to which high voltage is applied. The plate is installed in the path of the already accelerated particle beam and deflects it in the desired direction. Calculating the best plate configuration is a science. But despite the fact that cyclotron plates are manufactured and installed with maximum precision, its frontal part, or “tooth,” absorbs about half of the accelerated particles. Naturally, the “tooth” heats up from impacts, which is why it is now made from refractory molybdenum.
But it is also natural that particles absorbed by the tooth material should cause nuclear reactions in it, more or less interesting to physicists. Segre believed that an extremely interesting nuclear reaction was possible in molybdenum, as a result of which element No. 43 (technetium), which had been discovered many times and invariably “closed” before, could finally be truly discovered.
From Ilmenia to Masuria
Element No. 43 has been sought for a long time. And for a long time. They looked for it in ores and minerals, mainly manganese. Mendeleev, leaving an empty cell for this element in the table, called it ekamanganese. However, the first contenders for this cell appeared even before the discovery of the periodic law. In 1846, an analogue of manganese, ilmenium, was allegedly isolated from the mineral ilmenite. After Ilmenium was “closed”, new candidates appeared: Davy, Lucium, Nipponium. But they also turned out to be “false elements.” The forty-third cell of the periodic table continued to be empty.
In the 20s of our century, the problem of ekamanganese and dwimanganese (eka means “one”, dvi - “two”), i.e. elements No. 43 and 75, was taken up by the excellent experimenters spouses Ida and Walter Noddak. Having traced the patterns of changes in the properties of elements across groups and periods, they came to the seemingly seditious, but essentially correct idea that the similarity of manganese and its eka- and di-analogs is much less than previously thought, and that it is more reasonable to look for these elements not in manganese ores, and in crude platinum and molybdenum ores.
The Noddack couple's experiments continued for many months. In 1925, they announced the discovery of new elements - masurium (element no. 43) and rhenium (element no. 75). The symbols of new elements occupied the empty cells of the periodic table, but it later turned out that only one of the two discoveries was actually made. Ida and Walter Noddak mistook impurities for masurium that had nothing in common with element No. 43 technetium.
The symbol Ma stood in the table of elements for more than 10 years, although back in 1934 two theoretical works appeared that claimed that element No. 43 could not be found in manganese, platinum, or any other ores. We are talking about the prohibition rule, formulated almost simultaneously by the German physicist G. Matthauch and the Soviet chemist S. A. Shchukarev.
Technetium - "Forbidden" element and nuclear reactions
Soon after the discovery of isotopes, the existence of isobars was established. Note that isobar and isobar are concepts as distant as decanter and countess. Isobars are atoms with the same mass numbers belonging to different elements. Example of several isobars: 93 Zr, 93 Nb, 93 Mo.
The meaning of the Mattauch-Shchukarev rule is that stable isotopes with odd numbers cannot have stable isobars. So, if the isotope of element No. 41, niobium-93, is stable, then the isotopes of neighboring elements - zirconium-93 and molybdenum-93 - must necessarily be radioactive. The rule applies to all elements, including element No. 43.
This element is located between molybdenum (atomic weight 95.92) and ruthenium (atomic weight 101.07). Consequently, the mass numbers of isotopes of this element should not go beyond the range of 96-102. But all stable “vacancies” in this range are filled. Molybdenum has stable isotopes with mass numbers 96, 97, 98 and 100, and ruthenium has stable isotopes with mass numbers 99, 101, 102 and some others. This means that element number 43 cannot have a single non-radioactive isotope. However, it does not at all follow from this that it cannot be found in the earth’s crust: radium, uranium, and thorium exist.
Uranium and thorium have been preserved on the globe due to the enormous lifetime of some of their isotopes. Other radioactive elements are products of their radioactive decay. Element No. 43 could only be detected in two cases: either if it has isotopes whose half-lives are measured in millions of years, or if its long-lived isotopes are formed (and quite often) from the decay of elements No. 90 and 92.
Segre did not count on the first: if long-lived isotopes of element No. 43 existed, they would have been found earlier. The second is also unlikely: most thorium and uranium atoms decay by emitting alpha particles, and the chain of such decays ends with stable isotopes of lead, an element with atomic number 82. Lighter elements cannot be formed by alpha decay of uranium and thorium.
True, there is another type of decay - spontaneous fission, in which heavy nuclei spontaneously divide into two fragments of approximately the same mass. During the spontaneous fission of uranium, nuclei of element No. 43 could be formed, but there would be very few such nuclei: on average, one uranium nucleus out of two million fissions spontaneously, and out of a hundred spontaneous fission events of uranium nuclei, element No. 43 is formed in only two. However, Emilio Segre did not know this then. Spontaneous fission was discovered only two years after the discovery of element No. 43.
Segre was carrying a piece of irradiated molybdenum across the ocean. But there was no certainty that a new element would be discovered in it, and there could not be. There were “for” and “against”.
Falling on a molybdenum plate, a fast deuteron penetrates quite deeply into its thickness. In some cases, one of the deuterons can merge with the nucleus of a molybdenum atom. For this, first of all, it is necessary that the energy of the deuteron be sufficient to overcome the forces of electrical repulsion. This means that the cyclotron must accelerate the deuteron to a speed of about 15 thousand km/sec. The compound nucleus formed by the fusion of a deuteron and a molybdenum nucleus is unstable. It must get rid of excess energy. Therefore, as soon as the merger occurs, a neutron flies out of such a nucleus, and the former nucleus of the molybdenum atom turns into the nucleus of an atom of element No. 43.
Natural molybdenum consists of six isotopes, which means that, in principle, an irradiated piece of molybdenum could contain atoms of six isotopes of the new element. This is important because some isotopes can be short-lived and therefore chemically elusive, especially since more than a month has passed since the irradiation. But other isotopes of the new element could “survive.” These are what Segre hoped to find. That's where all the pros ended, actually. There were much more “against” ones.
Ignorance of the half-lives of the isotopes of element No. 43 worked against the researchers. It could also happen that not a single isotope of element No. 43 exists for more than a month. “Accompanying” nuclear reactions, in which radioactive isotopes of molybdenum, niobium and some other elements were formed, also worked against the researchers.
It is very difficult to isolate the minimum amount of an unknown element from a radioactive multicomponent mixture. But this is exactly what Segre and his few assistants had to do.
The work began on January 30, 1937. First of all, they found out what particles were emitted by molybdenum that had been in the cyclotron and crossed the ocean. It emitted beta particles - fast nuclear electrons. When about 200 mg of irradiated molybdenum was dissolved in aqua regia, the beta activity of the solution was approximately the same as that of several tens of grams of uranium.
Previously unknown activity was discovered; it remained to determine who the “culprit” was. First, radioactive phosphorus-32, formed from impurities that were in molybdenum, was chemically isolated from the solution. The same solution was then “cross-examined” by row and column of the periodic table. Carriers of unknown activity could be isotopes of niobium, zirconium, rhenium, ruthenium, and finally molybdenum itself. Only by proving that none of these elements were involved in the emitted electrons could we talk about the discovery of element number 43.
Two methods were used as the basis for the work: one is the logical method of exclusion, the other is the “carrier” method, widely used by chemists for separating mixtures, when a compound of this element or another, similar to it in chemical properties. And if a carrier substance is removed from the mixture, it carries away “related” atoms from there.
First of all, niobium was excluded. The solution was evaporated, and the resulting precipitate was dissolved again, this time in potassium hydroxide. Some elements remained in the undissolved part, but unknown activity went into solution. And then potassium niobate was added to it so that the stable niobium would “take away” the radioactive one. If, of course, it was present in the solution. Niobium is gone, but the activity remains. Zirconium was subjected to the same test. But the zirconium fraction also turned out to be inactive. Molybdenum sulfide was then precipitated, but the activity still remained in solution.
After this, the most difficult part began: it was necessary to separate the unknown activity and rhenium. After all, the impurities contained in the “tooth” material could turn not only into phosphorus-32, but also into radioactive isotopes of rhenium. This seemed all the more likely since it was the rhenium compound that brought the unknown activity out of the solution. And as the Noddacks found out, element No. 43 should be more similar to rhenium than to manganese or any other element. Separating the unknown activity from rhenium meant finding a new element, because all other "candidates" had already been rejected.
Emilio Segre and his closest assistant Carlo Perier were able to do this. They found that in hydrochloric acid solutions (0.4-5 normal), a carrier of unknown activity precipitates when hydrogen sulfide is passed through the solution. But rhenium also falls out at the same time. If precipitation is carried out from a more concentrated solution (10-normal), then rhenium precipitates completely, and the element carrying unknown activity only partially.
Finally, for control purposes, Perrier conducted experiments to separate a carrier of unknown activity from ruthenium and manganese. And then it became clear that beta particles could only be emitted by the nuclei of a new element, which was called technetium (from the Greek “artificial”).
These experiments were completed in June 1937. Thus, the first of the chemical “dinosaurs” was recreated - elements that once existed in nature, but were completely “extinct” as a result of radioactive decay.
Later, extremely small amounts of technetium, formed as a result of the spontaneous fission of uranium, were discovered in the ground. The same thing, by the way, happened with neptunium and plutonium: first the element was obtained artificially, and only then, after studying it, they were able to find it in nature.
Now technetium is obtained from fission fragments of uranium-35 in nuclear reactors. True, it is not easy to separate it from the mass of fragments. Per kilogram of fragments there are about 10 g of element No. 43. This is mainly the isotope technetium-99, the half-life of which is 212 thousand years. Thanks to the accumulation of technetium in reactors, it was possible to determine the properties of this element, obtain it in its pure form, and study quite a few of its compounds. In them, technetium exhibits valency 2+, 3+ and 7+. Just like rhenium, technetium is a heavy metal (density 11.5 g/cm3), refractory (melting point 2140°C), and chemically resistant.
Although technetium- one of the rarest and most expensive metals (much more expensive than gold), it has already brought practical benefits.
The damage caused to humanity by corrosion is enormous. On average, every tenth blast furnace operates to “cover the costs” of corrosion. There are inhibitor substances that slow down the corrosion of metals. The best inhibitors turned out to be pertechnates - salts of technicic acid HTcO 4. Addition of one ten-thousandth mole of TcO 4 -
prevents corrosion of iron and low-carbon steel - the most important structural material.
The widespread use of pertechnates is hampered by two circumstances: the radioactivity of technetium and its high cost. This is especially unfortunate because similar compounds of rhenium and manganese do not prevent corrosion.
Element No. 43 has another unique property. The temperature at which this metal becomes a superconductor (11.2 K) is higher than that of any other pure metal. True, this figure was obtained on samples of not very high purity - only 99.9%. Nevertheless, there is reason to believe that alloys of technetium with other metals will prove to be ideal superconductors. (As a rule, the temperature of transitions to the state of superconductivity in alloys is higher than in commercially pure metals.)
Although not so utilitarian, technetium has served useful purposes for astronomers. Technetium was discovered by spectral methods on some stars, for example on the star and constellation Andromeda. Judging by the spectra, element No. 43 is no less widespread there than zirconium, niobium, molybdenum, and ruthenium. This means that the synthesis of elements in the Universe continues today.
Technetium(lat. technetium), Te, radioactive chemical element of group VII of the periodic system of Mendeleev, atomic number 43, atomic mass 98, 9062; metal, malleable and ductile.
The existence of an element with atomic number 43 was predicted by D. I. Mendeleev. T. was obtained artificially in 1937 by Italian scientists E. Segre and K. Perrier during the bombardment of molybdenum nuclei with deuterons; received its name from the Greek. technet o s - artificial.
T. has no stable isotopes. Of the radioactive isotopes (about 20), two are of practical importance: 99 Tc and 99m tc with half-lives, respectively T 1/2 = 2,12 ? 10 5 years and t 1/2 = 6,04 h. In nature, the element is found in small quantities - 10 -10 G in 1 T uranium tar.
Physical and chemical properties . Metal T. in powder form is gray in color (reminiscent of re, mo, pt); compact metal (fused metal ingots, foil, wire) silver-gray. T. in the crystalline state has a hexagonal lattice of close packing ( A= 2.735 å, c = 4.391 å); in thin layers (less than 150 å) - a cubic face-centered lattice ( a = 3.68 ± 0.0005 å); T. density (with hexagonal lattice) 11.487 g/cm 3,t pl 2200 ± 50 °C; t kip 4700 °C; electrical resistivity 69 10 -6 oh? cm(100 °C); temperature of transition to the state of superconductivity Tc 8.24 K. T. paramagnetic; its magnetic susceptibility at 25°C is 2.7 10 -4 . Configuration of the outer electron shell of the Tc 4 atom d 5 5 s 2 ; atomic radius 1.358 å; ionic radius Tc 7+ 0.56 å.
In terms of chemical properties, tc is close to mn and especially to re; in compounds it exhibits oxidation states from -1 to +7. Tc compounds in the oxidation state +7 are the most stable and well studied. When T. or its compounds interact with oxygen, the oxides tc 2 o 7 and tco 2 are formed, with chlorine and fluorine - halides TcX 6, TcX 5, TcX 4, the formation of oxyhalides is possible, for example TcO 3 X (where X is a halogen), with sulfur - sulfides tc 2 s 7 and tcs 2. T. also forms technetic acid htco 4 and its pertechnate salts mtco 4 (where M is a metal), carbonyl, complex, and organometallic compounds. In the series of voltages, T. is to the right of hydrogen; it does not react with hydrochloric acid of any concentration, but easily dissolves in nitric and sulfuric acids, aqua regia, hydrogen peroxide, and bromine water.
Receipt. The main source of T. is waste from the nuclear industry. The yield of 99 tc when dividing 235 u is about 6%. T. is extracted from a mixture of fission products in the form of pertechnates, oxides, and sulfides by extraction with organic solvents, ion exchange methods, and precipitation of poorly soluble derivatives. The metal is obtained by reduction with hydrogen nh 4 tco 4, tco 2, tc 2 s 7 at 600-1000 °C or by electrolysis.
Application. T. is a promising metal in technology; it can find applications as a catalyst, high temperature and superconducting material. T. compounds are effective corrosion inhibitors. 99m tc is used in medicine as a source of g-radiation . T. is radiation hazardous; working with it requires special sealed equipment .
Lit.: Kotegov K.V., Pavlov O.N., Shvedov V.P., Technetius, M., 1965; Obtaining Tc 99 in the form of metal and its compounds from nuclear industry waste, in the book: Production of Isotopes, M., 1973.
DEFINITION
Technetium located in the fifth period of the VII group of the secondary (B) subgroup of the Periodic table.
Refers to elements d-families. Metal. Designation - Tc. Serial number - 43. Relative atomic mass - 99 amu.
Electronic structure of the technetium atom
A technetium atom consists of a positively charged nucleus (+43), inside of which there are 43 protons and 56 neutrons, and 43 electrons move around in five orbits.
Fig.1. Schematic structure of a technetium atom.
The distribution of electrons among orbitals is as follows:
43Tc) 2) 8) 18) 13) 2 ;
1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 5 5s 2 .
The outer energy level of the technetium atom contains 7 electrons, which are valence electrons. The energy diagram of the ground state takes the following form:
The valence electrons of a technetium atom can be characterized by a set of four quantum numbers: n(main quantum), l(orbital), m l(magnetic) and s(spin):
|
Sublevel |
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Examples of problem solving
EXAMPLE 1
| Exercise | Which element of the fourth period - chromium or selenium - has more pronounced metallic properties? Write down their electronic formulas. |
| Answer | Let us write down the electronic configurations of the ground state of chromium and selenium: 24 Cr 1 s 2 2s 2 2p 6 3s 2 3p 6 3 d 5 4 s 1 ; 34 Se 1 s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4 s 2 4 p 4 . Metallic properties are more pronounced in selenium than in chromium. The veracity of this statement can be proven using the Periodic Law, according to which, when moving in a group from top to bottom, the metallic properties of an element increase, and non-metallic ones decrease, which is due to the fact that when moving down the group in an atom, the number of electronic layers in an atom increases, as a result of which the valence electrons are weaker held by the core. |
Technetium (lat. Technetium), Tc, radioactive chemical element of group VII of the periodic system of Mendeleev, atomic number 43, atomic mass 98, 9062; metal, malleable and ductile.
Technetium has no stable isotopes. Of the radioactive isotopes (about 20), two are of practical importance: 99 Tc and 99m Tc with half-lives, respectively T 1/2= 2.12 ×10 5 years and T 1/2 = 6,04 h. In nature, the element is found in small quantities - 10 -10 G in 1 T uranium tar.
Physical and chemical properties.
Technetium metal in powder form is gray in color (reminiscent of Re, Mo, Pt); compact metal (fused metal ingots, foil, wire) silver-gray. Technetium in the crystalline state has a close-packed hexagonal lattice ( A = 2,735
, с = 4.391); in thin layers (less than 150) - a cubic face-centered lattice ( a = 3.68? 0.0005); T. density (with hexagonal lattice) 11.487 g/cm 3, t pl 2200? 50?C; t kip 4700?C; electrical resistivity 69 * 10 -6 ohm×cm(100? C); temperature of transition to the state of superconductivity Tc 8.24 K. Technetium is paramagnetic; its magnetic susceptibility at 25 0 C is 2.7 * 10 -4 . Configuration of the outer electron shell of the Tc 4 atom d 5 5s 2 ; atomic radius 1.358; ionic radius Tc 7+ 0.56.According to chemical properties Tc is close to Mn and especially to Re; in compounds it exhibits oxidation states from -1 to +7. Tc compounds in the oxidation state +7 are the most stable and well studied. When Technetium or its compounds interact with oxygen, the oxides Tc 2 O 7 and TcO 2 are formed, with chlorine and fluorine - halides TcX 6, TcX 5, TcX 4, the formation of oxyhalides is possible, for example TcO 3 X (where X is a halogen), with sulfur - sulfides Tc 2 S 7 and TcS 2. Technetium also forms technetium acid HTcO 4 and its pertechnate salts MeTcO 4 (where Me is a metal), carbonyl, complex and organometallic compounds. In the voltage series, Technetium is to the right of hydrogen; it does not react with hydrochloric acid of any concentration, but easily dissolves in nitric and sulfuric acids, aqua regia, hydrogen peroxide, and bromine water.
Receipt.
The main source of Technetium is waste from the nuclear industry. The yield of 99 Tc from fission of 235 U is about 6%. Technetium in the form of pertechnates, oxides, and sulfides is extracted from a mixture of fission products by extraction with organic solvents, ion exchange methods, and precipitation of poorly soluble derivatives. The metal is obtained by reduction of NH 4 TcO 4, TcO 2, Tc 2 S 7 with hydrogen at 600-1000 0 C or by electrolysis.
Application.
Technetium is a promising metal in technology; it can find applications as a catalyst, high temperature and superconducting material. Technetium compounds. - effective corrosion inhibitors. 99m Tc is used in medicine as a source of g-radiation . Technetium is radiation hazardous; working with it requires special sealed equipment.
History of discovery.
Back in 1846, the chemist and mineralogist R. Herman, who worked in Russia, found a previously unknown mineral in the Ilmen Mountains in the Urals, which he called yttroilmenite. The scientist did not rest on his laurels and tried to isolate from it a new chemical element, which he believed was contained in the mineral. But before he had time to open his ilmenium, the famous German chemist G. Rose “closed” it, proving the fallacy of Herman’s work.
A quarter of a century later, ilmenium again appeared on the forefront of chemistry - it was remembered as a contender for the role of “eka-manganese”, which was supposed to take the empty place in the periodic table at number 43. But the reputation of ilmenium was greatly “tarnished” by the works of G. Rose, and, despite the fact that many of its properties, including atomic weight, were quite suitable for element No. 43, D.I. Mendeleev did not register it in his table. Further research finally convinced the scientific world that , that ilmenium can go down in the history of chemistry only with the sad glory of one of the many false elements.
Since a holy place is never empty, claims for the right to occupy it appeared one after another. Davy, Lucium, Nipponium - they all burst like soap bubbles, barely having time to be born.
But in 1925, the German scientific couple Ida and Walter Noddack published a message that they had discovered two new elements - masurium (No. 43) and rhenium (No. 75). Fate turned out to be favorable to Renius: he was immediately legitimized and immediately occupied the residence prepared for him. But fortune turned its back on masurium: neither its discoverers nor other scientists could scientifically confirm the discovery of this element. True, Ida Noddak said that “soon masurium, like rhenium, will be able to be bought in stores,” but chemists, as you know, do not believe the words, and the Noddak spouses could not provide other, more convincing evidence - a list of “false forty-thirds” added another loser.
During this period, some scientists began to be inclined to believe that not all of the elements predicted by Mendeleev, in particular element No. 43, exist in nature. Maybe they simply don’t exist and there’s no need to waste time and break spears? Even the prominent German chemist Wilhelm Prandtl, who vetoed the discovery of masurium, came to this conclusion.
The younger sister of chemistry, nuclear physics, which by that time had already gained strong authority, made it possible to clarify this issue. One of the laws of this science (noted in the 20s by the Soviet chemist S.A. Shchukarev and finally formulated in 1934 by the German physicist G. Mattauch) is called the Mattauch-Shchukarev rule, or the prohibition rule.
Its meaning is that in nature two stable isobars cannot exist, the nuclear charges of which differ by one. In other words, if any chemical element has a stable isotope, then its nearest neighbors in the table are “categorically prohibited” from having a stable isotope with the same mass number. In this sense, element No. 43 was clearly unlucky: its neighbors to the left and right - molybdenum and ruthenium - made sure that all stable vacancies in nearby “territories” belonged to their isotopes. And this meant that element No. 43 had a hard fate: no matter how many isotopes it had, they were all doomed to instability, and thus they had to continuously - day and night - decay, whether they wanted to or not.
It is reasonable to assume that element No. 43 once existed on Earth in noticeable quantities, but gradually disappeared, like the morning fog. So why, in this case, have uranium and thorium survived to this day? After all, they are also radioactive and, therefore, from the very first days of their life they decay, as they say, slowly but surely? But this is precisely where the answer to our question lies: uranium and thorium have been preserved only because they decay slowly, much more slowly than other elements with natural radioactivity (and yet, during the existence of the Earth, uranium reserves in its natural storehouses have decreased by about a hundred once). Calculations by American radiochemists have shown that an unstable isotope of one or another element has a chance of surviving in the earth’s crust from the “creation of the world” to the present day only if its half-life exceeds 150 million years. Looking ahead, we will say that when various isotopes of element No. 43 were obtained, it turned out that the half-life of the longest-living of them was only a little more than two and a half million years, and, therefore, its last atoms ceased to exist, apparently even long before their appearance on Earth. The Earth of the first dinosaur: after all, our planet has been “functioning” in the Universe for about 4.5 billion years.
Therefore, if scientists wanted to “touch” element No. 43 with their own hands, they had to create it with the same hands, since nature had long ago included it in the list of missing ones. But is science up to such a task?
Yes, on the shoulder. This was first experimentally proven back in 1919 by the English physicist Ernest Rutherford. He subjected the nucleus of nitrogen atoms to a fierce bombardment, in which the constantly decaying radium atoms served as the weapons, and the resulting alpha particles served as the projectiles. As a result of prolonged shelling, the nuclei of nitrogen atoms were replenished with protons and it turned into oxygen.
Rutherford's experiments armed scientists with extraordinary artillery: with its help it was possible not to destroy, but to create - to transform some substances into others, to obtain new elements.
So why not try to get element No. 43 this way? The young Italian physicist Emilio Segre took up the solution to this problem. In the early 30s he worked at the University of Rome under the leadership of the then famous Enrico Fermi. Together with other “boys” (as Fermi jokingly called his talented students), Segre took part in experiments on neutron irradiation of uranium and solved many other problems of nuclear physics. But the young scientist received a tempting offer - to head the department of physics at the University of Palermo. When he arrived in the ancient capital of Sicily, he was disappointed: the laboratory that he was to lead was more than modest and its appearance was not at all conducive to scientific exploits.
But Segre’s desire to penetrate deeper into the secrets of the atom was great. In the summer of 1936, he crosses the ocean to visit the American city of Berkeley. Here, in the radiation laboratory of the University of California, the cyclotron, an atomic particle accelerator invented by Ernest Lawrence, had been operating for several years. Today this small device would seem to physicists something like a children's toy, but at that time the world's first cyclotron aroused the admiration and envy of scientists from other laboratories (in 1939, E. Lawrence was awarded the Nobel Prize for its creation).
Task 1.Write the electronic formula of the technetium atom. How many electrons are in the d sublevel of the penultimate electron layer? Which electron family does the element belong to?
Solution: The Tc atom in the periodic table has serial number 43. Consequently, its shell contains 43 electrons. In the electronic formula, we distribute them into sublevels according to the order of filling (in accordance with Klechkovsky’s rules) and taking into account the capacity of the sublevels: Tc 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 5 5s 2. In this case, the order of filling the sublevels is as follows: 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d. The last electron is located in the 4d sublevel, which means technetium belongs to the d-element family. There are 5 electrons at the d-sublevel of the penultimate (4th) layer.
Answer: 5, d.
Task 2.An atom of which element has the electronic configuration 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 1?
Solution:
The number of electrons in the shell of a neutral atom is 49. Therefore, its nuclear charge and, therefore, the serial number are also 49. In the periodic table of D.I. Mendeleev, we find that this element is indium.
Task 3.Which of the following compounds has the least acidic properties? a) HNO 3, b) H 3 PO 4, c) H 3 AsO 4, d) H 3 SbO 4.
Solution:
The given oxygen-containing compounds are hydroxides of elements of the main subgroup of group V of the periodic table. It is known that the acidic properties of hydroxides weaken from top to bottom in the subgroup. Therefore, in this series, H 3 SbO 4 has the least pronounced acidic properties.
Answer: H 3 SbO 4.
Task 4.Indicate the type of hybridization of boron orbitals in the BBr 3 molecule.
Solution:
The formation of three covalent bonds between boron and bromine atoms involves one s- and two p-orbitals of the boron atom, the properties of which differ. Since all chemical bonds in the BBr 3 molecule are equivalent, the boron atom undergoes hybridization. The above three orbitals of the outer electron layer take part in it. Therefore, the type of hybridization is sp 2.
Answer: sp 2.
Task 5.Using the periodic table data, create an empirical formula for higher lead oxide. What is its molar mass?
Solution:
Lead is in group 4 of the periodic table, so its highest oxidation state is +4. The oxygen atom in oxides has an oxidation state of –2, therefore in an oxide molecule there are two oxygen atoms for every lead atom. The formula of the highest oxide is PbO 2. Let's calculate its molar mass: 207+2·16=239.
Answer: 239 g/mol.
Task 6.What types of chemical bonds are there in the NH 4 I molecule?
Solution:
The NH 4 I molecule consists of NH 4 + and I – ions, between which there is an ionic bond. In the NH 4 + ion, four bonds are polar covalent, and one of them is formed according to the donor-acceptor type (see section 3.2.3).
Answer: ionic, covalent polar, donor-acceptor.
Task 7.Binding Energy Calculation.
Calculate the H-S bond energy in the H 2 S molecule using the following data: 2H 2 (g) + S 2 (g) = 2 H 2 S (g) – 40.30 kJ; the energies of the D(H-H) and D(S-S) bonds are, respectively, –435.9 kJ/mol and –417.6 kJ/mol.
Solution: The formation of two H 2 S molecules can be represented as a sequential process of bond breaking H-H in a molecule H 2 and connections S-S in a molecule S 2:
2 H-H 4 H – 2D(H-H)
S-S 2 S – D(S-S)
4 H + 2 S 2 H 2 S+ 4D(S-H),
Where D(H-H), D(S-S) And D(S-H) – energy of bond formation H-H, S-S And S-H respectively. Summing the left and right sides of the above equations, we arrive at the thermochemical equation
2H 2 (g) + S 2 (g) = 2 H 2 S (g) –2D(H-H) – D(S-S) + 4D(S-H).
The thermal effect of this reaction is
Q =–2D(H-H) – D(S-S) + 4D(S-H), where D(S-H)= .
Task 8.Calculation of link length.
Calculate the bond length in the HBr molecule if the internuclear distance in the H 2 and Br 2 molecules is 0.74∙10 -10 and 2 ,28 ∙10 -10 m respectively.
Solution: The length of a covalent bond between two dissimilar atoms is equal to the sum of their covalent radii
l(H-Br) = r(H) + r(Br).
In turn, the covalent radius of an atom is defined as half the internuclear distance in molecules H 2 And BR 2:
Thus,
Answer: 1.51·10 -10 m.
Task 9.Determination of the type of hybridization of orbitals and spatial structure of the molecule.
What type of hybridization of electron clouds takes place in a silicon atom during the formation of a SiF 4 molecule? What is the spatial structure of this molecule?
Solution: In the excited state, the structure of the outer energy level of the silicon atom is as follows:
| 3s | 3p | ||
| 3s | 3p x | 3p y | 3p z |
Electrons of the third energy level participate in the formation of chemical bonds in a silicon atom: one electron in the s-state and three electrons in the p-state. When a SiF 4 molecule is formed, four hybrid electron clouds (sp 3 hybridization) appear. The SiF 4 molecule has a spatial tetrahedral configuration.
Problem 10.Determination of the valences of elements in chemical compounds based on the analysis of graphical electronic formulas of the ground and excited states of the atoms of these elements.
What valence, due to unpaired electrons, can sulfur exhibit in the ground and excited states?
Solution: The electron distribution of the outer energy level of sulfur …3s 2 3p 4 taking into account Hund’s rule has the form:
| s | p | d | |||||||
| 16 S |
From the analysis of the ground and two excited states it follows that the valence (spinvalency) of sulfur in the normal state is two, in the first excited state - four, in the second - six.
Options for test tasks
Option 1
1. What information about an element can be learned based on its position in the PSE?
2. Write electronic formulas for atoms of elements with serial numbers 9 and 28. Show the distribution of electrons of these atoms across quantum cells. Which electron family does each of these elements belong to?
Option 2
1. Define: ionization energy, electron affinity and electronegativity of an atom? How do they change across periods and groups?
2. Write electronic formulas for atoms of elements with serial numbers 16 and 26. Distribute the electrons of these atoms among quantum cells. Which electron family does each of these elements belong to?
Option 3
1. Which covalent bond is called polar and which nonpolar? What is a quantitative measure of the polarity of a covalent bond?
2. What is the maximum number of electrons that can be occupied? s-, p-, d- And f-orbitals of a given energy level? Why? Write the electronic formula of an atom of an element with atomic number 31.
Option 4
1. How does the valence bond (BC) method explain the linear structure of the BeCI 2 molecule?
4s or 3d; 5s or 4p? Why? Write the electronic formula of an atom of an element with atomic number 21.
Option 5
1. Which bond is called a σ-bond and which is called a π-bond?
2. Which orbitals of the atom are filled with electrons first: 4d or 5s; 6s or 5p? Why? Write the electronic formula of an atom of an element with atomic number 43.
Option 6
1. What is called a dipole moment?
2. Write electronic formulas for atoms of elements with serial numbers 14 and 40. How many free 3d-orbitals of the atoms of the last element?
Option 7
1. What chemical bond is called ionic? What is the mechanism of its formation?
2. Write electronic formulas for atoms of elements with serial numbers 21 and 23. How many free 3d-orbitals in the atoms of these elements?
Option 8
1. Which version of the periodic table is most widely used and why?
2. How many free d- orbitals found in atoms Sc, Ti, V? Write the electronic formulas for the atoms of these elements.
Option 9
1. What properties of an ionic bond distinguish it from a covalent bond?
2. Using Hund’s rule, distribute electrons among quantum cells corresponding to the lowest energy state of atoms: chromium, phosphorus, sulfur, germanium, nickel.
2. Two different electronic states are possible for a boron atom 
And 
. What are these states called? How to move from the first state to the second?
Option 11
1. Which of the 4 different types of atomic orbitals has the most complex formula?
2. Which atom of the element corresponds to each of the given electronic formulas:
A) 
;b) 
;
Option 12
2. Using Hund’s rule, distribute electrons among quantum cells corresponding to the highest energy state of atoms: manganese, nitrogen, oxygen, silicon, cobalt.
Option 13
1. If there are 4 electrons in the p-orbitals of any layer, how many of them have unpaired spins and what is their total spin number 7?
2. What atoms of elements and what states of these elements correspond to the following electronic formulas 
And 
; 
And 
?
Option 14
1. What characteristics of an atom can be named, knowing: a) the serial number of the element in the periodic table; b) period number; c) number and type of group in which the element is located?
2. Write the electronic configuration of atoms using electronic formulas for elements with atomic numbers 12, 25, 31, 34, 45.
Option 15
1. How to determine, based on the position of an atom in the periodic table, the number of elementary particles in its composition? Determine the number of elementary particles in the composition of sulfur and zinc atoms.
2. Using Hund’s rule, distribute electrons into energy cells corresponding to the lowest energy state for atoms of elements with serial numbers 26, 39, 49, 74, 52.
Option 16
1. What are quantum numbers? What properties of orbitals and electrons do they reflect? What values do they take? Determine the maximum possible number of electrons in each energy level of aluminum and copper atoms.
2. Which of the electronic formulas reflecting the structure of an unexcited atom of some element are incorrect: a) 1s 2 2s 2 2p 5 3s 1 ; b) 1s 2 2s 2 2p 6; V) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 4 ; G) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2; d) 1s 2 2s 2 2p 6 3s 2 3d 2 ? Why? Which atoms of elements correspond to correctly composed electronic formulas?
Option 17
1. What principles form the basis of all modern theories of chemical bonding? What is an ionic bond? What properties does it have? Give examples of compounds with ionic bonds.
2. Write electronic formulas for atoms of elements with serial numbers 24 and 33, taking into account that the first one has a “failure” of one 4s-electron to the 3d sublevel. What is the maximum spin? d-electrons in the atoms of the first and p-electrons in atoms of the second element?
Option 18
1. What is electronegativity? How does electronegativity change? R-elements in a period, in a group of the periodic system with increasing atomic number? Why?
2. Make up electronic formulas for atoms of elements with serial numbers 32 and 42, taking into account that the latter has a “failure” of one 5s-electron per 4d-sublevel. Which electron family does each of these elements belong to?
Option 19
1. What values can quantum numbers take? n, l, m l And m S, characterizing the state of electrons in an atom? What values do they take for the outer electrons of the magnesium atom?
2. How many free f-orbitals are found in atoms of elements with serial numbers 61, 62, 91, 92? Using Hund's rule, distribute the electrons among the energy cells for the atoms of these elements.
Option 20
1. What is ionization energy? In what units is it expressed? How does recovery activity change? s- And p-elements in groups of the periodic table with increasing atomic number? Why?
2. What is the Pauli principle? Could it be at some sublevel of the atom p 7 - or d 12 - electrons? Why? Compose an electronic formula for an atom of an element with atomic number 22 and indicate its valence electrons. .
Option 21
1. List the rules according to which orbitals are filled with electrons. What is the electronic formula of an atom? Write the electronic formulas for silicon and iron, emphasizing the valence electrons.
2. Quantum numbers for electrons of the outer energy level of atoms of some elements have the following values: n = 4; l = 0; m l= 0; m S= . Write electronic formulas for the atoms of these elements and determine how many are free 3d-orbitals contains each of them.
Option 22
1. What are isotopes? How can we explain that most elements of the periodic table have atomic masses expressed as fractions? Can atoms of different elements have the same mass? What are such atoms called?
2. Based on the position of the metal in the periodic table, give a motivated answer to the question: which of the two hydroxides is the stronger base: Ba(OH) 2 or Mg(OH) 2; Ca(OH) 2 or Fe(OH) 2; Cd(OH) 2 or Sr(OH) 2?
Option 23
1. What is electron affinity? In what units is it expressed? How does the oxidative activity of nonmetals change in a period and in a group of the periodic system with increasing atomic number? Motivate your answer with the atomic structure of the corresponding element.
2. Manganese forms compounds in which it exhibits an oxidation state of +2, +3, +4, +6, +7. Make up formulas for its oxides and hydroxides corresponding to these oxidation states. Write reaction equations proving the amphoteric nature of manganese (IV) hydroxide.
Option 24
1. How do the acid-base and redox properties of higher oxides and hydroxides of elements change with increasing charge of their nuclei: a) within a period; b) within a subgroup.
2. How many and what values can a magnetic quantum number take? m l at orbital number l= 0, 1, 2 and 3? What elements in the periodic table are called s-, p-, d- And f-elements? Give examples.
Option 25
1. Theory of hybridization. The mechanism of formation of donor-acceptor bonds. Connection examples
2. Which one R-elements of the fifth group of the periodic table - phosphorus or antimony - are non-metallic properties more pronounced? Which of the hydrogen compounds of these elements is the stronger reducing agent? Motivate your answer with the atomic structure of these elements.
Option 26
1. What is the lowest oxidation state of chlorine, sulfur, nitrogen and carbon? Why? Make up formulas for aluminum compounds with these elements in this oxidation state. What are the names of the corresponding compounds?
2. The energy state of the outer electron of an atom is described by the following values of quantum numbers: n=4, l=0, m l=0. What atoms of elements have such an electron? Write electronic formulas for the atoms of these elements. Write all the quantum numbers of the electrons of the atoms: a) lithium, beryllium, carbon; b) nitrogen, oxygen, fluorine.
Option 27
1. Metal connection. Mechanism of formation and properties. Examples of compounds and their properties.
2. Based on the position of germanium and technetium in the periodic table, create formulas for meta- and orthogermanic acids, and technetium oxide, corresponding to their highest oxidation state. Draw the formulas of these compounds graphically.
Option 28
1. Which element of the fourth period - chromium or selenium - has more pronounced metallic properties? Which of these elements forms a gaseous compound with hydrogen? Motivate your answer by the structure of chromium and selenium atoms.
2. The nickel-57 isotope is formed when alpha particles bombard the nuclei of iron-54 atoms. Create an equation for this nuclear reaction and write it in abbreviated form
Option 29
Write electronic formulas for atoms of elements and name them if the values of quantum numbers ( n, l, m l, m S) electrons of the outer (last) and penultimate electron layers are as follows:
a) 6, 0, 0, + ; 6, 0, 0, - ; 6, 1, -1, + ;
b) 3, 2, -2, + ; 3, 2, -1, + ; 4, 0, 0, + ; 4, 0, 0, - .
Option 30
1.Modern methods describing the formation of covalent bonds, their basic postulates. Properties of covalent bonds. Give examples of compounds with covalent bonds and their properties.
2. Make a comparative description of elements with serial numbers 17 and 25 based on their position in the PSE. Explain the reasons for the similarities and differences in the properties of these elements.
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