Example 1... Determine the charge of the complexing agent in the NO 2 compound. Name this compound.
Solution
The outer sphere of the CS consists of one NO anion, therefore, the charge of the entire inner sphere is +1, that is, +. The inner sphere contains two groups of ligands NH 3 and Cl -. The oxidation state of the complexing agent is denoted by X and solve the equation
1 = 1X+ 0 · 4 + 2 · (–1). From here X = +1.
Thus, KS is a complex cation. Compound name: cobalt dichlorotetraammine nitrite (+1).
Example 2... Why is the + ion linear?
Solution
Determine the charge of the complexing agent in a given complex ion
1 = 1X+ 0 2. From here X = +1.
The electronic structure of the valence sublevels of the Cu + ion corresponds to configuration 3 d 10 4s 0 4R 0. Since 3 d - the sublevel does not contain vacancies, then one 4 s and one 4 p orbitals that hybridize as sp... This type of hybridization (see Table 1) corresponds to the linear structure of the complex.
Example 3... Determine the type of hybridization of the central ion AO and the geometric structure of complex 2–.
Solution
Electronic configuration of the central ion Hg 2+: 5 d 10 6s 0 6R 0, and the electronic diagram can be represented as follows
The chemical bond is formed according to the donor-acceptor mechanism, where each of the four donor ligands (Cl - ions) provides one lone pair of electrons (dotted arrows), and the complexing agent (Hg 2+ ion) provides free AOs: one 6 s and three 6 p JSC
Thus, in this complex ion, sp3 hybridization ao takes place, as a result of which the bonds are directed to the vertices of the tetrahedron and ion 2– has a tetrahedral structure.
Example 4... Make an energy diagram of the formation of bonds in complex 3– and indicate the type of hybridization of the orbitals of the central atom. What are the magnetic properties of the complex?
Solution
Electronic configuration of the central Fe 3+ ion:… 3 d 5 4s 0 4p 0 4d 0. Six monodentate ligands CN - create a strong octahedral field and form six σ-bonds, providing lone pairs of electrons of the carbon atom to the free AO of the complexing agent Fe 3+, while the degeneracy of AO 3 is removed d sublevel complexing agent. The energy diagram of the complex has the form
E 
dγ series
Fe 3+:… 3 d 5
dε series
Five 3 d-electrons are completely distributed in orbitals 3 dε series, since the splitting energy, which arises upon interaction with ligands of a strong field, turns out to be sufficient for the maximum pairing of electrons. Available 3 d, 4s and 4 R- orbitals exposed d 2 sp 3 - hybridization and determine the octahedral structure of the complex. The complex is paramagnetic, because there is one unpaired electron
d 2 sp 3
Example 5... Make an energy diagram of the formation of bonds in a complex - and indicate the type of hybridization.
Solution
Electronic formula Cr 3+:… 3 d 3 4s 0 4p 0 4d 0. Monodentate ligands F - form four σ-bonds, are weak field ligands and create a tetrahedral field
E 
dε series
dγ series
Free two 3 d, one 4 s and one 4 R AO complexing agents are hybridized according to the type d 2 sp, as a result, a paramagnetic complex of tetrahedral configuration is formed.
Example 6... Explain why ion 3 is paramagnetic and ion 3 is diamagnetic.
Solution
Electronic formula of the complexing agent Co 3+: ... 3 d 6. In the octahedral field of the F - ligands (weak field ligand), there is a slight splitting d– sublevel; therefore, electrons fill the AO in accordance with Gund's rule (see Fig. 3). In this case, there are four unpaired electrons; therefore, the 3-ion is paramagnetic. When the 3– ion is formed with the participation of a strong field ligand (CN - ion), the splitting energy d– sublevel will be so significant that it will exceed the energy of electron-electron repulsion of paired electrons. The electrons will fill the AO of the Co 3+ ion in violation of Gund's rule (see Fig. 4). In this case, all electrons are paired, the ion itself is diamagnetic.
Example 7.For the 3+ ion, the splitting energy is 167.2 kJ · mol –1. What is the color of chromium (III) compounds in aqueous solutions?
Solution
To determine the color of a substance, we determine the wavelength at which light is absorbed
or nm.
Thus, the 3+ ion absorbs light in the red part of the spectrum, which corresponds to the green color of the chromium (III) compound.
Example 8... Establish whether a precipitate of silver (I) sulfide will precipitate at a temperature of 25 ° C, if you mix equal volumes of 0.001 M solution - containing the same ligand CN - with a concentration of 0.12 mol / dm 3, and a solution of the precipitating ion S 2– with a concentration 3.5 · 10 –3 M.
Solution
The dissociation process for a given ion can be represented by the scheme
- ↔ Ag + + 2CN -,
and the deposition process can be written as
2Ag + + S 2– ↔ Ag 2 S.
To determine whether a precipitate will form, it is necessary to calculate the solubility product of silver sulfide PR (Ag 2 S) according to the formula
To determine the concentration of silver ions, we write the expression for the instability constant of the complex ion
... From here 
According to the reference book, we select the value of the instability constant of the complex - ( TO nest = 1 · 10 -21). Then
mol / dm 3.
Calculate the product of the solubility of the resulting precipitate
According to the reference book, we select the tabular value of the solubility product of silver sulfide (PR (Ag 2 S) table = 5.7 · 10 -51) and compare it with the calculated one. Since the PR table< ПР расчет, то из данного раствора осадок выпадает, так как соблюдается условие выпадения осадка.
Example 9... Calculate the concentration of zinc ions in a solution of sodium tetracyanozincate with a concentration of 0.3 mol / dm 3 with an excess of cyanide ions in the solution equal to 0.01 mol / dm 3.
Solution
Primary dissociation proceeds almost entirely according to the scheme
Na 2 → 2Na 2+ + 2–
Secondary dissociation proceeds according to the equation
2– ↔ Zn 2+ + 4CN -
Let us write the expression for the instability constant for this process
... From here 
Using the reference book, we find the value of the instability constant of a given ion ( TO nest = 1.3 · 10 -17). The concentration of cyanide ions formed as a result of the dissociation of the complex is much lower than the concentration of the introduced excess, and it can be assumed that 0.01 mol / dm 3, that is, the concentration of CN - ions formed as a result of dissociation can be neglected. Then
mol / dm 3.
The most important achievement of TKP is a good explanation of the reasons for a particular color of complex compounds. Before we try to explain the reason for the appearance of color in complex compounds, we recall that visible light is electromagnetic radiation, the wavelength of which is in the range from 400 to 700 nm. The energy of this radiation is inversely proportional to its wavelength:
E = h × n = h × c / l
Energy 162 193 206 214 244 278 300
E, kJ / mol
Wavelength 760 620 580 560 490 430 400
It turns out that the energy of splitting of the d-level by the crystal field, denoted by the symbol D, is of the same order of magnitude as the energy of a photon of visible light. Therefore, transition metal complexes can absorb light in the visible region of the spectrum. The absorbed photon excites an electron from the lower energy level of the d-orbitals to a higher level. Let us explain the above with the example of 3+. Titanium (III) has only 1 d-electron, the complex has only one absorption peak in the visible region of the spectrum. The maximum intensity is 510 nm. Light with this wavelength causes the d-electron to move from the lower energy level of the d-orbitals to the upper one. As a result of absorption of radiation, the molecule of the absorbed substance passes from the ground state with a minimum energy E 1 to a higher energy state E 2. The excitation energy is distributed over the individual vibrational energy levels of the molecule, turning into thermal energy. Electronic transitions caused by the absorption of strictly defined quanta of light energy are characterized by the presence of strictly defined absorption bands. Moreover, the absorption of light occurs only when the energy of the absorbed quantum coincides with the energy difference DE between the quantum energy levels in the final and initial states of the absorbing molecule:
DE = E 2 - E 1 = h × n = h × c / l,
where h is Planck's constant; n is the frequency of absorbed radiation; c is the speed of light; l is the wavelength of the absorbed light.
When a sample of a substance is illuminated with light, rays of all colors not absorbed by the sample are reflected into our eye. If the sample absorbs light of all wavelengths, the rays are not reflected from it, and such an object appears black to us. If the sample does not absorb light at all, we perceive it as white or colorless. If the sample absorbs all rays except orange, then it appears orange. Another option is also possible - the sample may appear orange even when rays of all colors except blue fall into our eye. Conversely, if the sample absorbs only orange rays, it appears blue. Cyan and orange are called complementary colors.
Sequence of spectral colors: To every O hotnik f wants s nat, G de With walking f adhan - To red, O rank, f yellow, s green , G blue, With inium , f iolety.
For the aqua complex 3+ the numerical value of D dist. = 163 kJ / mol corresponds to the limit of visible red radiation, therefore, aqueous solutions of Fe 3+ salts are practically colorless. Hexacyanoferrate (III) has D spl. = 418 kJ / mol, which corresponds to absorption in the blue-violet part of the spectrum and reflection in the yellow-orange part. Solutions containing hexacyanoferrate (III) ions are yellow with an orange tint. The value of D dist. 3+ is small compared to 3-, which reflects not very high binding energy Fe 3+ -OH 2. The large splitting energy of 3- indicates that the Fe 3+ -CN bond energy is higher, and, therefore, more energy is needed to split off CN. It is known from experimental data that H2O molecules in the 3+ coordination sphere have an average lifetime of about 10 -2 s, and complex 3- cleaves off CN - ligands extremely slowly.
Let's consider a few examples that allow solving problems using the TCH.
Example: complex ion trans- + absorbs light mainly in the red region of the spectrum - 640 nm. What is the color of this complex?
Solution: since the complex under consideration absorbs red light, its color must be complementary to the red color - green.
Example: ions A1 3+, Zn 2+ and Co 2+ are in the octahedral environment of the ligands. Which of these ions can absorb visible light and, as a result, appears to us to be colored?
Solution: the A1 3+ ion has an electronic configuration. Since it has no outer d-electrons, it is not colored. The Zn 2+ ion has an electronic configuration - 3d 10. In this case, all d-orbitals are filled with electrons. The d x 2– y2 and d x 2 orbitals cannot accept an electron excited from the lower energy level of the d xy, d yz, d xz orbitals. Therefore, the Zn 2+ complex is also colorless. The Co 2+ ion has an electronic configuration - d 7. In this case, it is possible to move one d-electron from the lower energy level of the d xy, d yz, d xz orbitals to the upper energy level of the d x 2– y2 and d x 2 orbitals. Therefore, the complex of the Co 2+ ion is colored.
Example: how to explain why the color of the diamagnetic complexes 3+, 3+, 3– is orange, while the color of the paramagnetic complexes 3–, 0 is blue?
Solution: orange coloration of the complexes indicates absorption in the blue-violet part of the spectrum, i.e. in the region of short wavelengths. Thus, the splitting for these complexes is large, which ensures that they belong to low-spin complexes (D> P). Pairing of electrons (d 6 -configuration, all six electrons are at the t 2g sublevel) is due to the fact that the ligands NH 3, en, NO 2 - belong to the right side of the spectrochemical series. Therefore, upon complexation, they create a strong field. Coloration of the second group of complexes in blue means that they absorb energy in yellow-red, i.e. long-wavelength part of the spectrum. Since the wavelength at which the complex absorbs light determines the amount of splitting, we can say that the value of D in this case is relatively small (D<Р). Это и понятно: лиганды F – и H 2 O находятся в левой части спектрохимического ряда и образуют слабое поле. Поэтому энергии расщепления D в данном случае недостаточно для спаривания электронов кобальта (III) и электронная конфигурация в этом случае - t 4 2g ,е 2 g , а не t 6 2g e 0 g .
Example: Using crystal field theory, explain why the complex ion is colorless in aqueous solution, and 2 is colored green?
Solution : complex - is formed by a copper cation Cu + with an electronic configuration 3d 10 4s 0, all d-orbitals are filled, the transition of electrons is impossible, therefore the solution is not colored. Complex 2- is formed by the cation Cu 2+, the electronic configuration of which is 3d 9 4s 0, therefore there is a vacancy on the d– sublevel. The transition of electrons upon absorption of light at the d-sublevel determines the color of the complex. Aqua complexes of copper (II) have a blue color in an aqueous solution, the introduction of chloride ions into the inner sphere of the complex leads to the formation of a mixed-ligand complex, which causes a change in the color of the solution to green.
Example: Using the method of valence bonds, taking into account the theory of the crystal field, determine the type of hybridization of the central atom and predict the geometric shape of the complexes:
- + -
Solution: Let us choose among the indicated complexes the compounds formed by E +, these are:
+ - 3-
- + .
The chemical bond in these complexes is formed by the donor-acceptor mechanism, the electron donors are ligands: ammonia molecules and cyanide ions (monodentate ligands) and thiosulfate ions (bidentate ligand). The electron acceptor is the E + cation. Electronic configuration (n-1) d 10 ns 0 np 0. External ns and np orbitals participate in the formation of two bonds with monodentate ligands, the type of hybridization of the central atom is sp, the geometric shape of the complexes is linear, there are no unpaired electrons, the ion is diamagnetic. One s-orbital and three p-orbitals of the central atom participate in the formation of four donor-acceptor bonds with the bidentate ligand along the MBC, the type of hybridization is sp 3, the geometric shape of the complex is tetrahedral, there are no unpaired electrons.
The second group of complexes:
- - - 3+
formed by a gold (III) ion, the electronic configuration of which is 5d 8 6s 0. The ligands participating in the formation of complexes can be divided into weak ligands: chloride and bromide ions and strong ones: ammonia and cyanide ions in accordance with the spectrochemical range of ligands. In 5d orbitals, in accordance with Hund's rule, there are two unpaired electrons, and they are retained during the formation of donor-acceptor bonds with weak-field ligands. For bond formation, the gold cation provides one 6s and three 6p orbitals. The type of hybridization of the central atom is sp 3. The spatial structure of a complex ion is tetrahedral. There are two unpaired electrons, the complex is paramagnetic.
Under the influence of ligands of a strong field, the electrons of the gold (III) ion are spraying with the release of one 5d orbital. One 5d, one 6s, and two 6p orbitals of the central atom are involved in the formation of four donor-acceptor bonds. Hybridization type dsp 2. This results in a square-planar structure of the complex ion. There are no unpaired electrons, the complexes are diamagnetic.
The color of the solution of the complex depends on its composition, structure and is determined by the wavelength l max corresponding to the maximum of the absorption band, the intensity of the band, depending on whether the quantum-chemically corresponding electronic transition is forbidden, the smearing of the absorption band, which depends on a number of parameters, such as the electronic structure of the complex , the intensity of thermal movement in the system, the degree of distortion of the correct geometric shape of the coordination polyhedron, etc.
Dizinc tetrafluoride
Zn 2 F 4 (d). The thermodynamic properties of gaseous dizinc tetrafluoride in the standard state in the temperature range 100 - 6000 K are given in table. Zn 2 F 4.
The molecular constants used to calculate the thermodynamic functions of Zn 2 F 4 are given in table. Zn. 8. The structure of the Zn 2 F 4 molecule has not been studied experimentally. By analogy with Be 2 F 4 [82SOL / OZE], Mg 2 F 4 [81SOL / SAZ] (see also [94GUR / VEY]) and Al 2 F 4 [82ZAK / CHA] for Zn 2 F 4 in the main electronic condition 1 A g, a planar cyclic structure is adopted (the symmetry group D 2h). The static weight of the ground electronic state of Zn 2 F 4 is recommended to be equal to I, based on the fact that the Zn 2+ ion has ... d 10 electronic configuration. The product of the moments of inertia, given in table. Zn.8, calculated from the estimated structural parameters: r(Zn- F t) = 1.75 ± 0.05 Å (terminal Zn-F bond), r(Zn- F b) = 1.95 ± 0.05 Å (bridging Zn-F bond) and Р F b- Zn- F b= 80 ± 10 o. The bond length Zn-F t is taken the same as r(Zn-F) in the ZnF 2 molecule, the r (Zn-F b) value is recommended to be 0.2 Å larger than the terminal bond, as is observed in dimers of Al, Ga, In, Tl, Be, and Fe halides. Angle value F b- Zn- F b estimated from the corresponding values in the molecules Be 2 F 4, Mg 2 F 4 and Al 2 F 4. Calculated value error I A I B I C is 3 · 10 ‑113 g 3 · cm 6.
The frequencies of stretching vibrations of the terminal bonds of Zn-F n 1 and n 2 were taken from the work of Givan and Levenschuss [80GIV / LOE], who studied the IR and Raman spectra of Zn 2 F 4 molecules isolated in a krypton matrix. The vibration frequencies of all Zn-F (n 3) bridging bonds were taken to be the same, and their values were estimated under the assumption that (n b/ n t) cp = 0.7, as in the dimers of Fe, Al, Ga, and In halides. The values of the frequencies of the deformation vibrations of the terminal bonds (n 4 - n 5) Zn 2 F 4 are recommended, assuming that the ratio of their values in Zn 2 F 4 and Zn 2 Cl 4 is the same as for ZnF 2 and ZnCl 2. The frequency of the out-of-plane bending cycle vibration (n 7) is assumed to be slightly higher than the corresponding frequency for Zn 2 Cl 4. The value of the frequency of the bending vibration of the cycle in the plane (n 6) was estimated by comparison with the value adopted for Zn 2 Cl 4, and taking into account the ratio of the vibration frequencies of the bridging bonds of Zn-F and Zn-Cl in Zn 2 F 4 and Zn 2 Cl 4 ... The errors of the experimentally observed vibration frequencies are 20 cm –1, estimated at 20% of their value.
The excited electronic states of Zn 2 F 4 were not taken into account in the calculation of the thermodynamic functions.
The thermodynamic functions Zn 2 F 4 (r) are calculated in the "rigid rotator - harmonic oscillator" approximation according to equations (1.3) - (1.6), (1.9), (1.10), (1.122) - (1.124), (1.128), ( 1.130). The errors of the calculated thermodynamic functions are due to the inaccuracy of the accepted values of the molecular constants, as well as the approximate nature of the calculation and are 6, 16 and 20 J × K ‑1 × mol ‑1 in the values of Φº ( T) at 298.15, 3000 and 6000 K.
The table of thermodynamic functions of Zn 2 F 4 (g) is published for the first time.
The equilibrium constant Zn 2 F 4 (g) = 2Zn (g) + 4F (g) is calculated using the accepted value
D atHº (Zn 2 F 4. g, 0) = 1760 ± 30 kJ × mol ‑1.
The value is estimated by comparing the enthalpies of sublimation and dimerization of the dihalides included in this edition. Table Zn.12 shows the values of the ratios D sHº (MeHal 2. k, 0) / D rHº (MeHal 2 - MeHal 2, 0), corresponding to the values adopted in this edition.
In 9 cases out of a total of 20, experimental data are absent. The estimates for these compounds are given in the table in square brackets. These estimates were made based on the following considerations:
1. for the Fe, Co and Ni compounds, a small move is adopted in the F-Cl-Br-I series and the absence of such a move in the Fe-Co-Ni series;
2. for Zn compounds, the course of values in the series F-Cl-Br-I cannot be noticed, and for fluoride the average value of the remaining values is taken;
3. for Cu compounds, a small course is adopted in the F-Cl-Br-I series by analogy with compounds of the iron group, based on the proximity of values; the move itself is assumed to be somewhat smaller.
The outlined approach leads to the values of the enthalpies of atomization Me 2 Hal 4 given in Table. Zn. 13.
When calculating the atomization energy of Cu 2 I 4, the value D not included in this edition was used s H° (CuI 2, k, 0) = 180 ± 10 kJ × mol ‑1. (see the text on the enthalpy of sublimation of CuBr 2).
The accuracy of the estimates performed can be estimated equal to 50 kJ × mol ‑1 for Cu 2 I 4 and 30 kJ × mol ‑1 in other cases.
The accepted value of the enthalpy of atomization of Zn 2 F 4 corresponds to the value of the enthalpy of formation:
D f H° (Zn 2 F 4. g, 0) = -1191.180 ± 30.0 kJ × mol ‑1.
Osina E.L. [email protected]
A.V. Gusarov [email protected]
Consider the task number 1 of the options for the exam for 2016.
Task number 1.
The electronic formula of the outer electron layer 3s²3p6 corresponds to the structure of each of the two particles:
1. Arº and Kº 2. Cl‾ and K + 3. S²‾ and Naº 4. Clº and Ca2 +
Explanation: Among the answer options are atoms in unexcited and excited states, that is, the electronic configuration, for example, the potassium ion does not correspond to its position in the periodic table. Consider option 1 Arº and Kº. Let's write their electronic configurations: Arº: 1s2 2s2 2p6 3s2 3p6; Kº: 1s2 2s2 2p6 3s2 3p6 4s1 is a suitable electron configuration only for argon. Consider answer # 2 - Cl‾ and K +. K +: 1s2 2s2 2p6 3s2 4s0; Cl‾: 1s2 2s2 2p6 3s2 3p6. Hence, the correct answer is 2.
Task number 2.
1. Caº 2. K + 3. Cl + 4. Zn2 +
Explanation: for we write the electronic configuration of argon: 1s2 2s2 2p6 3s2 3p6. Calcium is not suitable because it has 2 more electrons. In potassium: 1s2 2s2 2p6 3s2 3p6 4s0. The correct answer is 2.
Task number 3.
The element, the electronic configuration of the atom of which is 1s2 2s2 2p6 3s2 3p4, forms a hydrogen compound
1. CH4 2. SiH4 3. H2O 4. H2S
Explanation: let's look at the periodic table, the sulfur atom has such an electronic configuration. The correct answer is 4.
Task number 4.
Magnesium atoms and
1. Calcium 2. Chromium 3. Silicon 4. Aluminum
Explanation: magnesium has the configuration of the external energy level: 3s2. For calcium: 4s2, for chromium: 4s2 3d4, for silicon: 3s2 2p2, for aluminum: 3s2 3p1. The correct answer is 1.
Task number 5.
The argon atom in the ground state corresponds to the electron configuration of the particle:
1. S²‾ 2. Zn2 + 3. Si4 + 4. Seº
Explanation: the electronic configuration of argon in the ground state is 1s2 2s2 2p6 3s2 3p6. S²‾ has an electronic configuration: 1s2 2s2 2p6 3s2 3p (4 + 2). The correct answer is 1.
Task number 6.
The phosphorus atoms and
1. Ar 2. Al 3. Cl 4. N
Explanation: we write the electronic configuration of the outer level of the phosphorus atom: 3s2 3p3.
For aluminum: 3s2 3p1;
For argon: 3s2 3p6;
Chlorine: 3s2 3p5;
For nitrogen: 2s2 2p3.
The correct answer is 4.
Task number 7.
Electronic configuration 1s2 2s2 2p6 3s2 3p6 corresponds to a particle
1. S4 + 2. P3- 3. Al3 + 4. O2-
Explanation: this electronic configuration corresponds to an argon atom in the ground state. Consider the answer options:
S4 +: 1s2 2s2 2p6 3s2 2p0
P3-: 1s2 2s2 2p6 3s2 3p (3 + 3)
The correct answer is 2.
Task number 8.
What electronic configuration corresponds to the distribution of valence electrons in the chromium atom:
1.3d2 4s2 2.3s2 3p4 3.d5 4s1 4.4s2 4p6
Explanation: we write the electronic configuration of chromium in the ground state: 1s2 2s2 2p6 3s2 3p6 4s1 3d5. Valence electrons are located on the last two sublevels 4s and 3d (here one electron hops from the s to d sublevel). The correct answer is 3.
Task number 9.
Three unpaired electrons at the external electronic level in the ground state contains an atom
1. Titanium 2. Silicon 3. Magnesium 4. Phosphorus
Explanation: in order to have 3 unpaired electrons, the element must be in the fifth group. Hence, the correct answer is 4.
Task number 10.
An atom of a chemical element, the highest oxide of which is RO2, has an external level configuration:
1.ns2 np4 2.ns2 np2 3.ns2 4.ns2 np1
Explanation: this element has an oxidation state (in this compound) +4, that is, it must have 4 valence electrons at the external level. Hence, the correct answer is 2.
(you might think that the correct answer is 1, but such an atom will have a maximum oxidation state of +6 (since 6 electrons are at the outer level), but we need the higher oxide to have the formula RO2, and such an element will have the higher oxide RO3)
Assignments for independent work.
1. The electronic configuration 1s2 2s2 2p6 3s2 3p5 corresponds to the atom
1. Aluminum 2. Nitrogen 3. Chlorine 4. Fluorine
2. The eight-electron outer shell has a particle
1. P3 + 2. Mg2 + 3. Cl5 + 4. Fe2 +
3. The ordinal number of the element, the electronic structure of the atom of which is 1s2 2s2 2p3, is
1. 5 2. 6 3. 7 4. 4
4. The number of electrons in the copper ion Cu2 + is
1. 64 2. 66 3. 29 4. 27
5. A similar configuration of the external energy level has nitrogen atoms and
1. Sulfur 2. Chlorine 3. Arsenic 4. Manganese
6. Which compound contains a cation and an anion with the electronic configuration 1s2 2s2 2p6 3s3 3p6?
1. NaCl 2. NaBr 3. KCl 4. KBr
7. The number of electrons in the iron ion Fe2 + is
1. 54 2. 28 3. 58 4. 24
8. The electronic configuration of an inert gas has an ion
1. Cr2 + 2. S2- 3. Zn2 + 4. N2-
9. A similar configuration of the external energy level has fluorine atoms and
1. Oxygen 2. Lithium 3. Bromine 4. Neon
10. The element, the electronic formula of the atom of which is 1s2 2s2 2p6 3s2 3p4, corresponds to a hydrogen compound
1. HCl 2. PH3 3. H2S 4. SiH4
In this note, we used tasks from the collection of the USE 2016, edited by A.A. Kaverina.
