Examples of problem solving. Structure of complex compounds Magnetic quantum number m l
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 find out where and how electrons are located, get acquainted with quantum numbers and be able to build 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 constructor: there are a certain number of parts, they differ from each other, but two parts of the same type are exactly the same. But this constructor 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, next to sodium into gas, and being next to iron completely turns it into rust. To answer the question why this happens and to 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 revolving around it, the nucleus consists of protons and neutrons. In the neutral state, each atom has the same number of electrons as the number of protons in its nucleus. The number of protons was indicated by the element's serial number, for example, sulfur has 16 protons - the 16th element of the periodic system. Gold has 79 protons - the 79th element of the periodic table. Accordingly, there are 16 electrons in sulfur in the neutral state, and 79 electrons in gold.
Where to look for an electron?
Observing the behavior of an electron, certain patterns were derived, they are described by quantum numbers, there are four of them 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", the orbital is the wave function of the electron, roughly - this is the area in which the electron spends 90% of the time.
N - level
L - shell
M l - orbital number
M s - the first or second electron in the orbital
Orbital quantum number l
As a result of the study of the electron cloud, it was found that depending on the level of energy, the cloud takes four main forms: a ball, dumbbells and the other two, more complex. In ascending order of energy, these forms are called s-, p-, d- and f-shells. 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 on which the orbitals are located. The orbital quantum number for s, p, d and f orbitals, respectively, takes the values 0,1,2 or 3.
On the s-shell one orbital (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 (L=3) on the f-shell - fourteen electrons
Magnetic quantum number m l
There are three orbitals on the p-shell, they are denoted 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 for each orbital, and then its pair is added to each.
Consider a 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. Only two electrons with opposite spins can be on the same energy sublevel. The spin quantum number is denoted m s
Principal quantum number n
The main quantum number is the energy level, at the moment seven energy levels are known, each is denoted 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, and so on.
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, let's take the first electron, the lowest energy level is N=1, one shell is located on the first level, 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.
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/OSE ], 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 (symmetry group D 2h). The static weight of the ground electronic state of Zn 2 F 4 is recommended to be 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 (bridged Zn-F bond) and Ð F b-Zn-F b= 80±10o. The bond length Zn-F t is taken to be the same as r(Zn-F) in the ZnF 2 molecule, the r(Zn-F b) value is recommended to be larger by 0.2 Å of the terminal bond, as is observed in Al, Ga, In, Tl, Be, and Fe halide dimers. Angle value F b-Zn-F b estimated from the corresponding values in 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 stretching frequencies of the Zn-F n 1 and n 2 terminal bonds were taken from the work of Givan and Levenshuss [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) bridge bonds are assumed to be the same, and their values are estimated under the assumption that (n b/n t) cp = 0.7, as in dimers of Fe, Al, Ga, and In halides. The bending vibration frequencies of the end 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 out-of-plane deformation vibration of the cycle (n 7) is assumed to be slightly higher than the corresponding frequency for Zn 2 Cl 4 . The value of the frequency of deformation vibrations 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 bridge bonds Zn-F and Zn-Cl in Zn 2 F 4 and Zn 2 Cl 4 . The errors of the experimentally observed oscillation frequencies are 20 cm -1 , estimated at 20% of their value.
Excited electronic states of Zn 2 F 4 were not taken into account in the calculation of thermodynamic functions.
The thermodynamic functions of Zn 2 F 4 (r) are 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 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 (d) 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) \u003d 1760 ± 30 kJ × mol -1.
The value is estimated 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 performed. These estimates are based on the following considerations:
1. for Fe, Co and Ni compounds, a small move in the F-Cl-Br-I series and the absence of such a move in the Fe-Co-Ni series 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 a value is taken, the average of the other values;
3. for Cu compounds, a small move was adopted in the series F-Cl-Br-I, by analogy with compounds of the iron group, based on the proximity of the values; the move itself is taken somewhat smaller.
The above 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 is not included in this publication. s H° (CuI 2, k, 0) \u003d 180 ± 10 kJ × mol -1. (see the text on the enthalpy of sublimation of CuBr 2).
The accuracy of the performed 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 f H° (Zn 2 F 4. g, 0) \u003d -1191.180 ± 30.0 kJ × mol -1.
Osina E.L. [email protected]
Gusarov A.V. [email protected]
Consider tasks No. 1 of the USE options 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 the unexcited and excited states, that is, the electronic configuration, for example, of a potassium ion does not correspond to its position in the periodic system. 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 - only argon has a suitable electron configuration. Consider answer 2 - Cl‾ and K+. K+: 1s2 2s2 2p6 3s2 4s0; Cl‾: 1s2 2s2 2p6 3s2 3p6. Consequently, 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. For potassium: 1s2 2s2 2p6 3s2 3p6 4s0. The correct answer is 2.
Task number 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 system, such an electronic configuration has a sulfur atom. The correct answer is 4.
Task number 4.
A similar configuration of the external energy level have magnesium atoms and
1. Calcium 2. Chromium 3. Silicon 4. Aluminum
Explanation: magnesium has an external energy level configuration: 3s2. Calcium: 4s2, chromium: 4s2 3d4, silicon: 3s2 2p2, 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 and
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.
Aluminum: 3s2 3p1;
For argon: 3s2 3p6;
For chlorine: 3s2 3p5;
For nitrogen: 2s2 2p3.
The correct answer is 4.
Task number 7.
The electronic 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. 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 a 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 on the last two sublevels 4s and 3d (here there is a jump of one electron from sublevel s to d). The correct answer is 3.
Task number 9.
Three unpaired electrons in the outer 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. Consequently, 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 in the outer level. Consequently, 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 there are 6 electrons in the outer level), but we need the highest oxide to have the formula RO2, and such an element will have the highest 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 serial number of the element, the electronic structure of the atom of which is 1s2 2s2 2p3, is equal to
1. 5 2. 6 3. 7 4. 4
4. The number of electrons in the Cu2+ copper ion is
1. 64 2. 66 3. 29 4. 27
5. Nitrogen atoms and
1. Sulfur 2. Chlorine 3. Arsenic 4. Manganese
6. Which compound contains a cation and an anion with an electronic configuration of 1s2 2s2 2p6 3s3 3p6?
1. NaCl 2. NaBr 3. KCl 4. KBr
7. The number of electrons in the Fe2+ iron ion 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. The fluorine and
1. Oxygen 2. Lithium 3. Bromine 4. Neon
10. An element whose electronic formula is 1s2 2s2 2p6 3s2 3p4 corresponds to a hydrogen compound
1. HCl 2. PH3 3. H2S 4. SiH4
This note used assignments from the USE collection of 2016, edited by A.A. Kaverina.
Theory of the valence bond method
1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 6 4p 0 4d 0
In accordance with Hund's rule electrons in the outer energy level are arranged as follows:
complexing agent has a coordination number c.h. = 6, therefore, it can attach 6 ligands, each of which has an unshared electron pair and is thus an electron donor. An acceptor (complexing agent) to accommodate six electron pairs must provide six vacant orbitals. When a 3+ complex ion is formed, four unpaired electrons in the d-state of Co 3+ first form electron pairs, as a result of which two 3d-orbitals are released:
Then the complex ion itself 3+ is formed, having the following structure:
The formation of this complex ion involves the inner 3d orbitals and the outer 4s and 4p orbitals. type of hybridization d 2 sp 3 .
The presence of only paired electrons indicates the diamagnetic properties of the ion.
Crystal field theory
Crystal field theory is based on the assumption that the relationship between the complexing agent and the ligands is partially . However, the influence of the electrostatic field of the ligands on the energy state of the electrons of the central ion is taken into account.
Consider two complex salts: K 2 and K 3 .
K 2 - has a tetrahedral spatial structure ( sp 3 – hybridization)
K 3 - has an octahedral spatial structure ( sp 3 d 2 -hybridization)
Complexing agents have the following electronic configuration:
d - electrons of the same energy level are the same in the case of a free atom or ion. But the action of the electrostatic field of ligands contributes to the splitting of the energy levels of d-orbitals in the central ion. And the splitting is the greater (with the same complexing agent), the stronger the field created by the ligands. According to their ability to cause splitting of energy levels, ligands are arranged in a row:
CN - > NO 2 - > NH 3 > SCN - > H 2 O > OH - > F - > Cl - > Br - > I -
The structure of the complex ion affects the nature of the splitting of the energy levels of the complexing agent.
At octahedral structure complex ion, d γ -orbitals (d z 2 -, d x 2 - y 2 -orbitals) are subject to strong interaction field of ligands, and the electrons of these orbitals can have a higher energy than the electrons of the d ε orbitals (d xy , d xz , d yz are orbitals).
The splitting of energy levels for electrons in the d-state in the octahedral field of ligands can be represented as diagram form:
Here Δ oct is the splitting energy in the octahedral field of ligands.
With a tetrahedral structure complex ion d γ -orbitals have a lower energy than d ε -orbitals:
Here Δtetr is the splitting energy in the tetrahedral field of ligands.
Splitting energy Δ determined experimentally from the absorption spectra of light quanta by the substance, the energy of which is equal to the energy of the corresponding electronic transitions. The absorption spectrum, as well as the color of complex compounds of d-elements, are due to the transition of electrons from the lower-energy d-orbital to the higher-energy d-orbital.
Thus, in the case of the K 3 salt, upon absorption of a light quantum, an electron transition from the d ε orbital to the d γ orbital is probable. This explains why this salt has an orange-red color. And salt K 2 cannot absorb light and, as a result, it is colorless. This is explained by the fact that the transition of electrons from the d γ -orbital to the d ε -orbital is not feasible.
Theory of molecular orbitals
MO method was previously discussed in the section.
Using this method, we will depict the electronic configuration of a high-spin 2+ complex ion.
Electronic configuration of the Ni 2+ ion:
1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 8 4p 0 4d 0 or …4s 0 3d 8 4p 0 4d 0
In a complex ion 2+ take part in the formation of a chemical bond 8 electrons central Ni 2+ ion and 12 electrons of six NH 3 ligands.
complex ion It has octahedral structure. The formation of MO is possible only when the energies of the initial interacting particles are close in their values, and also oriented in space in an appropriate way.
In our case, the 4s orbital of the Ni 2+ ion overlaps equally with the orbitals of each of the six ligands. As a result, molecular orbitals are formed: bonding σ s St and loosening σ s res.
Overlapping three 4p orbitals of the complexing agent with ligand orbitals leads to the formation of six σp-orbitals: bonding σ x sv, σ y sv, σ z sv and loosening σ x res, σ y res, σ z res.
Overlapping d z 2 and d x 2 - y 2 complexing agents with ligand orbitals promotes the formation of four molecular orbitals: two bonding σ sv x 2 - y 2 , σ sv z 2 and two loosening σ raz x 2 - y 2 , σ raz z 2 .
The orbitals d xy , d xz , d yz of the Ni 2+ ion do not bond with the orbitals of the ligands, because not directed towards them. As a result, they do not take part in the formation of the σ-bond, and are non-bonding orbitals: π xz , π xy , π yz .
Total the 2+ complex ion contains 15 molecular orbitals. The arrangement of electrons can be depicted as follows:
(σ s sv) 2 (σ x sv) 2 (σ y sv) 2 (σ z sv) 2 (σ sv x 2 - y 2) 2 (σ sv z 2) 2 (π xz) 2 (π xy) 2 (π yz) 2 (σ raz x 2 - y 2) (σ raz z 2)
Schematically, the formation of molecular orbitals is shown in the diagram below:
