Coloring of complex compounds. Examples of problem solving Electronic configuration of the zn2 ion
The most important achievement of TCT is a good explanation of the reasons for the particular color of complex compounds. Before we try to explain the reason for the appearance of color in complex compounds, 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 d-level splitting by a 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 the electron from the lower energy level of the d-orbitals to a higher level. Let us explain this using the example 3+. Titanium (III) has only 1 d-electron; the complex has only one absorption peak in the visible region of the spectrum. Maximum intensity 510 nm. Light at this wavelength causes a d electron to move from the lower energy level of the d orbitals to the upper one. As a result of the absorption of radiation, the molecule of the absorbed substance passes from the ground state with minimal 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 in the case 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 absorbed light.
When a sample of a substance is illuminated by light, the reflected rays from all the colors not absorbed by the sample enter our eye. If a 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 a sample absorbs all rays except orange, it appears orange. Another option is possible - the sample may appear orange even when rays of all colors except blue enter our eye. Conversely, if a sample absorbs only orange rays, it appears blue. Blue and orange are called complementary colors.
Sequence of spectral colors: To every O hunter and wants h nah, G de With goes f adhan - To red, O range, and yellow, h green , G blue, With blue , f purple
For an aqua complex 3+ the numerical value of D calc. = 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 dist. = 418 kJ/mol, which corresponds to absorption in the blue-violet part of the spectrum and reflection in the yellow-orange. Solutions containing hexacyanoferrate (III) ions are yellow with an orange tint. D value 3+ is small compared to 3-, which reflects the not very high binding energy of Fe 3+ -OH 2. The high energy of cleavage of 3- indicates that the binding energy of Fe 3+ -CN is greater, and, therefore, more energy is needed for the elimination of CN. It is known from experimental data that H 2 O molecules in the 3+ coordination sphere have an average lifetime of about 10 -2 s, and complex 3- extremely slowly removes CN - ligands.
Let's look at several examples that allow us to solve problems using TCP.
Example: the trans‑+ complex ion absorbs light mainly in the red region of the spectrum - 640 nm. What is the color of this complex?
Solution: since the complex in question absorbs red light, its color should be green, complementary to red.
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, appear colored to us?
Solution: the A1 3+ ion has an electronic configuration of . Since it has no external 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 Co 2+ ion complex is colored.
Example: how to explain why the color of diamagnetic complexes 3+, 3+, 3– is orange, while the color of paramagnetic complexes 3–, 0 is blue?
Solution: the orange color of the complexes indicates absorption in the blue-violet part of the spectrum, i.e. in the short wavelength region. Thus, the splitting for these complexes is a large value, which ensures their belonging to low-spin complexes (D>P). Electron pairing (d 6 configuration, all six electrons on 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, when complexing, they create a strong field. Coloring the second group of complexes blue means that they absorb yellow-red energy, i.e. long-wave 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 an aqueous solution, and 2 is colored green?
Solution : complex - formed by the copper cation Cu + with the electronic configuration 3d 10 4s 0, all d-orbitals are filled, electron transfer is impossible, therefore the solution is not colored. Complex 2- is formed by the Cu 2+ cation, the electronic configuration of which is 3d 9 4s 0, therefore there is a vacancy at the d– sublevel. The transition of electrons upon absorption of light at the d-sublevel determines the color of the complex. Copper (C) aqua complexes 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 the solution to change color to green.
Example: Using the valence bond method, taking into account crystal field theory, 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 a donor-acceptor mechanism; 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 . In the formation of two bonds with monodentate ligands, external ns- and np-orbitals take part, 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. When four donor-acceptor bonds are formed with a bidentate ligand, one s-orbital and three p-orbitals of the central atom take part in 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 involved in the formation of complexes can be divided, in accordance with the spectrochemical series of ligands, into weak: chloride and bromide ions and strong: ammonia and cyanide ions. In accordance with Hund's rule, there are two unpaired electrons in the 5d orbitals and they are retained during the formation of donor-acceptor bonds with weak-field ligands. To form bonds, the gold cation provides one 6s and three 6p orbitals. Type of hybridization of the central sp 3 atom. The spatial structure of the complex ion is tetrahedral. There are two unpaired electrons, the complex is paramagnetic.
Under the influence of strong field ligands, the electrons of the gold (III) ion are paired with the release of one 5d orbital. One 5d-, one 6s-, and two 6p-orbitals of the central atom take part in the formation of four donor-acceptor bonds. Hybridization type dsp 2. This results in a planar square structure of the complex ion. There are no unpaired electrons, the complexes are diamagnetic.
The color of a solution of a 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, which depends on whether the quantum-chemically corresponding electronic transition is prohibited, and the blurring of the absorption band, which depends on a number of parameters, such as the electronic structure of the complex , intensity of thermal movement in the system, degree of distortion of the regular geometric shape of the coordination polyhedron, etc.
Example 1. Determine the charge of the complexing agent in the NO 2 compound. Give this connection a name.
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 degree of oxidation of the complexing agent is denoted by X and solve the equation
1 = 1X+ 0·4 + 2·(–1). From here X = +1.
Thus, CS is a complex cation. Compound name: cobalt dichlorotetraammine nitrite (+1).
Example 2. Why does the + ion have a linear structure?
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 by type 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 graphic circuit 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 (dashed arrows), and the complexing agent (Hg 2+ ion) provides free AO: one 6 s and three 6 p JSC
Thus, in this complex ion, sp3 hybridization of the ao takes place, as a result of which the bonds are directed towards the vertices of the tetrahedron and the 2– ion has a tetrahedral structure.
Example 4. Draw up an energy diagram for the formation of bonds in complex 3– and indicate the type of hybridization of the orbitals of the central atom. What magnetic properties does the complex have?
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 complexing agent sublevel. The energy diagram of the complex looks like
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 that arises during interaction with high-field ligands turns out to be sufficient for maximum electron pairing. Available 3 d, 4s and 4 R- orbitals are 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. Draw up an energy diagram for the formation of bonds in the 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 hybridize 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), slight splitting occurs d– sublevel, therefore electrons fill the AO in accordance with Hund’s rule (see Fig. 3). In this case, there are four unpaired electrons, so the ion is 3– paramagnetic. When the 3– ion is formed with the participation of a high-field ligand (CN– ion), the splitting energy d– sublevel will be so significant that it will exceed the energy of interelectron repulsion of paired electrons. Electrons will fill the AO of the Co 3+ ion in violation of Hund’s rule (see Fig. 4). In this case, all electrons are paired, and 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. Determine whether a precipitate of silver (I) sulfide will form at a temperature of 25°C if you mix equal volumes of a 0.001 M solution - containing the ligand of the same name 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 diagram
– ↔ Ag + + 2CN – ,
and the deposition process can be written as follows
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) using the formula
To determine the concentration of silver ions, we write the expression for the instability constant of the complex ion
. From here 
Using the reference book, we select the value of the instability constant of the complex - ( TO nest = 1·10 -21). Then
mol/dm 3 .
Let us calculate the solubility product of the resulting precipitate
Using the reference book, we select the tabulated value of the product of silver sulfide solubility (PR(Ag 2 S) tab = 5.7·10 –51) and compare it with the calculated value. Since 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 completely according to the scheme
Na 2 → 2Na 2+ + 2–
Secondary dissociation follows the equation
2– ↔ Zn 2+ + 4CN –
Let us write down 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 dissociation of the complex is much less 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 .
Electronic configuration of an atom is a formula showing the arrangement of electrons in an atom by levels and sublevels. After studying the article, you will learn where and how electrons are located, get acquainted with quantum numbers and be able to construct the electronic configuration of an atom by its number; at the end of the article there is a table of elements.
Why study the electronic configuration of elements?
Atoms are like a construction set: there is a certain number of parts, they differ from each other, but two parts of the same type are absolutely the same. But this construction set is much more interesting than the plastic one and here’s why. The configuration changes depending on who is nearby. For example, oxygen next to hydrogen Maybe turn into water, when near sodium it turns into gas, and when near iron it completely turns it into rust. To answer the question of why this happens and predict the behavior of an atom next to another, it is necessary to study the electronic configuration, which will be discussed below.
How many electrons are in an atom?
An atom consists of a nucleus and electrons rotating around it; the nucleus consists of protons and neutrons. In the neutral state, each atom has the number of electrons equal to the number of protons in its nucleus. The number of protons is designated by the atomic number of the element, for example, sulfur has 16 protons - the 16th element of the periodic table. Gold has 79 protons - the 79th element of the periodic table. Accordingly, sulfur has 16 electrons in the neutral state, and gold has 79 electrons.
Where to look for an electron?
By observing the behavior of the electron, certain patterns were derived; they are described by quantum numbers, there are four in total:
- Principal quantum number
- Orbital quantum number
- Magnetic quantum number
- Spin quantum number
Orbital
Further, instead of the word orbit, we will use the term “orbital”; an orbital is the wave function of an electron; roughly, it is the region in which the electron spends 90% of its time.
N - level
L - shell
M l - orbital number
M s - first or second electron in the orbital
Orbital quantum number l
As a result of studying the electron cloud, they found that depending on the energy level, the cloud takes four main forms: a ball, dumbbells and two other, more complex ones. In order of increasing energy, these forms are called the s-, p-, d- and f-shell. Each of these shells can have 1 (on s), 3 (on p), 5 (on d) and 7 (on f) orbitals. The orbital quantum number is the shell in which the orbitals are located. The orbital quantum number for the s,p,d and f orbitals takes the values 0,1,2 or 3, respectively.
There is one orbital on the s-shell (L=0) - two electrons
There are three orbitals on the p-shell (L=1) - six electrons
There are five orbitals on the d-shell (L=2) - ten electrons
There are seven orbitals on the f-shell (L=3) - fourteen electrons
Magnetic quantum number m l
There are three orbitals on the p-shell, they are designated by numbers from -L to +L, that is, for the p-shell (L=1) there are orbitals “-1”, “0” and “1”. The magnetic quantum number is denoted by the letter m l.
Inside the shell, it is easier for electrons to be located in different orbitals, so the first electrons fill one in each orbital, and then a pair of electrons is added to each one.
Consider the d-shell:
The d-shell corresponds to the value L=2, that is, five orbitals (-2,-1,0,1 and 2), the first five electrons fill the shell taking the values M l =-2, M l =-1, M l =0 , M l =1,M l =2.
Spin quantum number m s
Spin is the direction of rotation of an electron around its axis, there are two directions, so the spin quantum number has two values: +1/2 and -1/2. One energy sublevel can only contain two electrons with opposite spins. The spin quantum number is denoted m s
Principal quantum number n
The main quantum number is the energy level; currently seven energy levels are known, each indicated by an Arabic numeral: 1,2,3,...7. The number of shells at each level is equal to the level number: there is one shell on the first level, two on the second, etc.
Electron number
So, any electron can be described by four quantum numbers, the combination of these numbers is unique for each position of the electron, take the first electron, the lowest energy level is N = 1, at the first level there is one shell, the first shell at any level has the shape of a ball (s -shell), i.e. L=0, the magnetic quantum number can take only one value, M l =0 and the spin will be equal to +1/2. If we take the fifth electron (in whatever atom it is), then the main quantum numbers for it will be: N=2, L=1, M=-1, spin 1/2.
Let's look at task No. 1 from the Unified State Exam options for 2016.
Task No. 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 of, say, a potassium ion does not correspond to its position in the periodic table. Let's 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 - suitable electronic configuration only for argon. Let's consider answer option No. 2 - Cl‾ and K+. K+: 1s2 2s2 2p6 3s2 4s0; Cl‾: 1s2 2s2 2p6 3s2 3p6. Hence, the correct answer is 2.
Task No. 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. For potassium: 1s2 2s2 2p6 3s2 3p6 4s0. The correct answer is 2.
Task No. 3.
An element whose atomic electronic configuration 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 this electronic configuration. The correct answer is 4.
Task No. 4.
The atoms of magnesium and
1. Calcium 2. Chromium 3. Silicon 4. Aluminum
Explanation: Magnesium has an external energy level configuration: 3s2. For calcium: 4s2, for chromium: 4s2 3d4, for silicon: 3s2 2p2, for aluminum: 3s2 3p1. The correct answer is 1.
Task No. 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 the electronic configuration: 1s2 2s2 2p6 3s2 3p(4+2). The correct answer is 1.
Task No. 6.
Phosphorus and phosphorus atoms have a similar configuration of the outer energy level.
1. Ar 2. Al 3. Cl 4. N
Explanation: Let's write the electronic configuration of the outer level of the phosphorus atom: 3s2 3p3.
For aluminum: 3s2 3p1;
For argon: 3s2 3p6;
For chlorine: 3s2 3p5;
For nitrogen: 2s2 2p3.
The correct answer is 4.
Task No. 7.
The electron configuration 1s2 2s2 2p6 3s2 3p6 corresponds to the particle
1. S4+ 2. P3- 3. Al3+ 4. O2-
Explanation: this electronic configuration corresponds to the argon atom in the ground state. Let's consider the answer options:
S4+: 1s2 2s2 2p6 3s2 2p0
P3-: 1s2 2s2 2p6 3s2 3p(3+3)
The correct answer is 2.
Task No. 8.
Which electronic configuration corresponds to the distribution of valence electrons in the chromium atom:
1. 3d2 4s2 2. 3s2 3p4 3. 3d5 4s1 4. 4s2 4p6
Explanation: Let's write the electronic configuration of chromium in the ground state: 1s2 2s2 2p6 3s2 3p6 4s1 3d5. Valence electrons are located in the last two sublevels 4s and 3d (here one electron jumps from the s to d sublevel). The correct answer is 3.
Task No. 9.
The atom contains three unpaired electrons in the outer electronic level in the ground state.
1. Titanium 2. Silicon 3. Magnesium 4. Phosphorus
Explanation: In order to have 3 unpaired electrons, the element must be in group 5. Hence, the correct answer is 4.
Task No. 10.
An atom of a chemical element whose highest oxide is RO2 has the external level configuration:
1. ns2 np4 2. ns2 np2 3. ns2 4. ns2 np1
Explanation: this element has an oxidation state (in this compound) of +4, that is, it must have 4 valence electrons in the outer level. Hence, the correct answer is 2.
(you might think that the correct answer is 1, but such an atom would have a maximum oxidation state of +6 (since there are 6 electrons in the outer level), but we need the higher oxide to have the formula RO2, and such an element would have the higher oxide RO3)
Assignments for independent work.
1. Electronic configuration 1s2 2s2 2p6 3s2 3p5 corresponds to an atom
1. Aluminum 2. Nitrogen 3. Chlorine 4. Fluorine
2. The particle has an eight-electron outer shell
1. P3+ 2. Mg2+ 3. Cl5+ 4. Fe2+
3. The atomic number of an element whose atomic electronic structure is 1s2 2s2 2p3 is equal to
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. The nitrogen atoms and
1. Sulfur 2. Chlorine 3. Arsenic 4. Manganese
6. Which compound contains a cation and an anion with the electron 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 ion has the electronic configuration of an inert gas
1. Cr2+ 2. S2- 3. Zn2+ 4. N2-
9. The fluorine and fluorine atoms have a similar configuration of the outer energy level
1. Oxygen 2. Lithium 3. Bromine 4. Neon
10. An element whose atomic electronic formula is 1s2 2s2 2p6 3s2 3p4 corresponds to a hydrogen compound
1. HCl 2. PH3 3. H2S 4. SiH4
This note uses tasks from the 2016 Unified State Exam collection edited by A.A. Kaverina.
Dizinc tetrafluoride
Zn 2 F 4 (g). 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 mainly electronic state 1 A g a flat cyclic structure is adopted (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 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 (bridged Zn-F bond) and Ð F b- Zn-F b= 80 ± 10 o. The Zn-F t bond length is assumed to be the same as r(Zn-F) in the ZnF 2 molecule, the value r(Zn-F b), is recommended to be larger by 0.2 Å of 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 Be 2 F 4, Mg 2 F 4 and Al 2 F 4 molecules. 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 Zn-F n 1 and n 2 bonds were taken from the work of Givan and Levenschuss [80GIV/LOE], who studied the IR spectrum and Raman spectra of Zn 2 F 4 molecules isolated in a krypton matrix. The vibration frequencies of all Zn-F (n 3) bridge bonds are assumed to be the same, and their values are estimated under the assumption that (n b/n t) av = 0.7, as in dimers of Fe, Al, Ga and In halides. The frequencies of deformation vibrations of the terminal bonds (n 4 - n 5) of 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 non-plane deformation vibration of the cycle (n 7) is taken to be slightly higher than the corresponding frequency for Zn 2 Cl 4. The value of the frequency of deformation vibration of the cycle in the plane (n 6) is estimated by comparison with the value accepted for Zn 2 Cl 4, and taking into account the ratio of the vibration frequencies of the bridge bonds Zn-F and Zn-Cl in Zn 2 F 4 and Zn 2 Cl 4 . The errors in 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 thermodynamic functions.
Thermodynamic functions of Zn 2 F 4 (r) were calculated in the “rigid rotator - harmonic oscillator” approximation using equations (1.3) - (1.6) , (1.9) , (1.10) , (1.122) - (1.124) , (1.128) , ( 1.130) . The errors in 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 amount to 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) was calculated using the accepted value
D atHº(Zn 2 F 4. g, 0) = 1760 ± 30 kJ × mol ‑1.
The significance is assessed by comparing the enthalpies of sublimation and dimerization of the dihalides included in this publication. 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 accepted in this publication.
In 9 cases out of a total of 20, experimental data are missing. For these compounds, the estimates given in the table in square brackets were made. These estimates are made based on the following considerations:
1. for Fe, Co and Ni compounds, a small variation in the series F-Cl-Br-I and the absence of such a variation in the series Fe-Co-Ni are accepted;
2. for Zn compounds it is not possible to notice the variation of values in the series F-Cl-Br-I, and for fluoride the value taken is the average of the remaining values;
3. for Cu compounds, a small range in the series F-Cl-Br-I is accepted, by analogy with compounds of the iron group, based on the proximity of the values; the move itself was adopted somewhat smaller.
The described approach leads to the values of the enthalpies of atomization of 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 publication, was used s H° (CuI 2, k, 0) = 180 ± 10 kJ × mol ‑1. (See text on enthalpy of sublimation of CuBr 2).
The accuracy of the estimates can be estimated at 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 fH° (Zn 2 F 4. g, 0) = -1191.180 ± 30.0 kJ × mol ‑1.
Osina E.L. [email protected]
Gusarov A.V. [email protected]