Guidelines for solving problems

Practical work

DETERMINATION OF THE QUANTUM CHARACTERISTICS OF ELECTRONS AND ATOMS AND THE CALCULATION OF THE MULTIPLENETICITY

Purpose: formation of ideas about the structure of the atom and the corpuscular-wave nature of the electron and the ability to predict the chemical properties of elements.

Tasks: to gain knowledge about quantum numbers and the principles of filling atomic orbitals with electrons; be able to write formulas for electronic configurations in the stationary and excited states of the outer shells of atoms and ions of s-, p- and d-elements; comprehend the periodic law and understand the principles of the periodicity of changes in the properties of elements.

Theoretical part

Quantum numbers of a multielectron atom

The quantum numbers of individual electrons lose their meaning when it comes to a multi-electron atom. In this case, the total orbital L and total spin S quantum numbers of the set of electrons are used, as well as the quantum number of the total angular momentum of the atom J.

1. The value of L is determined through the values of l i individual electrons. Closed electron shells s 2 , p 6 , d 10 have L = 0. To calculate the total orbital quantum number, it is necessary to consider electrons only in unfilled shells. For two electrons with orbital quantum numbers l 1 and l 2 ( l 1 ≥ l 2 ), the quantum number L takes the values:

L u003d (l 1 + l 2 ), (l 1 + l 2 – 1), …, (l 1 – l 2 + 1), (l 1 – l 2 ),

total 2 l 2 + 1 value. For example, for a carbon atom 1s 2 2s 2 p 2 l 1 u003d l 2 u003d 1 (two p-electrons) L u003d 0, 1, 2. Similarly to the notation of orbitals in a hydrogen atom, states with different L are denoted as follows:

L value
State of the atom S P D F G H

2. The total spin quantum number S is found by the same rules. Depending on whether the number of electrons in an open shell is even or odd, the values of S will be integer or half-integer, respectively. So, for a carbon atom, S = 0 or 1.

3. The quantum number of the total angular momentum of the atom J takes on positive integer or half-integer values.

For L>S, the number of possible values of J is 2S + 1, for L

The value 2S+1 is called the multiplicity of the term , it determines the multiplicity of the state of the atom:

S 2S+1 State
singlet
1/2 doublet
triplet
3/2 quartet, etc.

So, the state of a multielectron atom is described by quantum numbers:

L = |l 1 – l 2 |, … ,|l 1 + l 2 |;

S = |s 1 – s 2 |, … ,|s 1 + s 2 |;

J = |L – S|, … ,|L + S|.

The above considerations are of great importance in atomic spectroscopy, in particular, for the classification of atomic states and their energy sequence. A certain energy state of an atom is called an atomic term and is denoted 2 S +1 L J .

The order of terms in terms of energy is determined by Hund’s empirical rules:

1. The ground state term (of the lowest energy) always has the highest spin multiplicity value.

2. If several terms have the same multiplicity, then the one with the maximum L will be the most stable.

3. Among the terms with the same L and S, the terms with the smallest J for shells filled up to half inclusive, and those with the largest J for shells more than half filled, have the least energy.

For example, a carbon atom (p 2 -shell) is characterized by S=1/2 + 1/2 = 1;2S+1=3. Possible values M L = 1 + 0; eleven; 0 – 1, i.e. L = 1 (P-state). In accordance with Hund’s third rule, J min = S – L = 0. Thus, the most stable state of the carbon atom is 3 P 0 .

To classify all possible terms of an atom, it is necessary to enumerate all possible combinations of microstates, i.e. all possible combinations of M L and M S in combination with possible values of m and ms of all electrons. Then the designation of this state is determined and, according to Hund’s rules, its relative stability.

As an example, consider the sequence of actions for determining the term of the most stable state of a multielectron atom.

For this you should:

A) Write the electronic configuration of the atom;

B) Distribute electrons in a partially filled shell so that:

– get the maximum multiplicity of the atom;

– fill in the cells with the maximum value of the quantum number m;

C) Determine L by the sum of singly filled cells;

D) Calculate the total spin and 2S + 1;

E) Calculate L – S for shells less than half filled, and L + S for other cases.

The results of determining the most stable term of some atoms are summarized in the table:

Atom Configuration L S J Term
mg …3s 2 1S0 _
P …3p 3 3/2 3/2 4 S 3/2
S …3p 4 3P2 _
Cl …3p 5 ½ 3/2 2 P 3/2
Ti …4s 2 3d 2 3 F 2
V …4s 2 3d 3 3/2 3/2 4 F 3/2
Cr …4s 1 ½
(4s)
Cr …3d 5 5/2
(3d)
Cr …4s 1 3d 5 7S3 _
In the latter case, unpaired electrons are in two shells. Therefore, first find L and S in each, and then the total value in accordance with Hund’s rules.
Fe …4s 2 3d 6 5D4 _
co …4s 2 3d 7 3/2 9/2 4 F 9/2
Ni …4s 2 3d 8 3 F 4
etc.

Examples of problem solving

Note to problems on quantum numbers. In such problems, there is a certain convention – we believe that the very first electron that begins to populate the electron cloud in any atom has a spin s = +1/2. In fact, for electrons with spin +1/2 and -1/2, this is equally probable. So, when hydrogen atoms are formed in the interiors of stars, they all have an electron shell 1s 1 , but 50% of hydrogen atoms have electrons with s = +1/2, and 50% – electrons with s = -1/2. The problems presented here are proposed to be solved within the framework of the convention introduced by us.

Task 1. Find the corresponding spectral designation of the atom term:

a) sodium, the valence electron of which has a principal quantum number n = 4;

b) with electronic configuration 1s 2 2p3d.

Solution: a) for a valence electron at n = 4, the orbital quantum number l can take on the values l = 0, 1, 2, 3 , then l max = 3, which means that for the atom L max = 3.

The valence electron has a spin s = 1/2, the spin quantum number of a given atom is S = s, then the maximum value of the quantum number of the total mechanical moment J max will be equal to J max = | Lmax + S | = 3 + ½ = 7/2.

The multiplicity of such an atom is equal to k = 2s + 1 = 2.

The term will have the designation: 2 F 7/2 .

b) the configuration 1s 2 2p3d means that for two s-electrons n = 1, l = 0;

for a valence p-electron n = 2, l = 0, 1, then l 1 max = 1;

for a valence d-electron n=3, l = 0, 1, 2, then l 2 max = 2.

The orbital quantum number L for an atom can take the values L= | l 1l 2 |,…, | l 1 + l 2 |; L=1, 2, 3, so L max = 3.

For s-electrons, the total spin is zero, for valence electrons, the spin is s 1 =s 2 =1/2.

Spin quantum number of an atom S = 1;

quantum number J max = | Lmax + S | = 4.

The multiplicity of the atom is k = 2s + 1 = 3.

Therm – 3 F 4 .

Answer: a) 2 F 7/2 ; b) 3 F 4 .

Problem 2. The system consists of a d-electron and an atom in the 2 P 3/2 state. Find possible spectral terms of this system.

Decision. The system is an electron and an atom. For a d-electron, the orbital and spin quantum numbers are equal, respectively: l = 2, s = 1/2.

For atom 2 P 3/2 : L = 1, S = 1/2 (because k = 2s + 1 = 2).

Then for the system:

L=| l 1l 2 |, …, | l 1 + l 2 | = 1, 2, 3;

S = |s 1 – s 2 |, … , | s1 + s2 | = 0, 1.

1) for S = 0, possible values of the quantum number of the total mechanical moment:

J= |L–S|,…, |L+S| =1, 2, 3.

System multiplicity: k = 2S + 1 = 1.

The corresponding terms will have designations: 1 P 1 ; 1D2 ; 1 F 3 ;

2) for S=1 the multiplicity of the system is k = 2S + 1 = 3.

a) for L=1: J= |L–S|,…, |L+S| =0, 1, 2.

Term designations: 3 P 0 ; 3 P 1 ; 3P2 ;

b) for L=2: J= |L–S|,…, |L+S| =1, 2, 3.

Terms notation: 3 D 1 ; 3D2 ; 3 D 3 ;

c) for L=3: J= |L–S|,…, |L+S| =2, 3, 4.

Terms notation: 3 F 2 ; 3 F 3 ; 3 F 4 .

Answer: 1 P 1 ; 1D2 ; 1 F 3 ; 3P0 ; 3 P 1 ; 3P2 ; 3D1 ; 3D2 ; 3 D 3 ; 3 F 2 ; 3 F 3 ; 3 F 4 .

Problem 3. Which atom has a K-, L- and M-shell, a 4s-subshell and a half-filled 4p-subshell?

Decision. The filled K-shell contains two s-electrons; L-shell – two s – and six p-electrons; M-shell – two s -, six p – and ten d-electrons. A filled 4s subshell contains two electrons, and a half filled 4p subshell contains three electrons.

The electronic configuration of such an atom is as follows:

1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 3 .

Thus, the total number of electrons in an atom is 33:

N e u003d 2 + 2 + 6 + 2 + 6 + 10 + 2 + 3 u003d 33.

The number of electrons in an atom is equal to the serial number of the element in the periodic table.

Using the table, we determine that atom number 33 is an arsenic atom.

Answer: arsenic atom.

Guidelines for solving problems

When solving problems, it is advisable to be guided by the following rules:

1. Starting to solve the problem, understand its condition well.

2. If the nature of the task allows, make a schematic drawing explaining its essence.

3. As a rule, the problem is first solved in a general form (i.e., in alphabetical notation), and the desired value must be expressed in terms of given values.

4. Having received the solution in a general form, check whether it has the correct dimension. An incorrect dimension is a clear sign of an erroneous decision.

5. After making sure that the general solution is correct, substitute the numerical values of the quantities indicated by them instead of letters, taking all these values in the same system of units. It should be remembered that the numerical values of physical quantities are always approximate. Therefore, when calculating, it is necessary to be guided by the rules of actions with approximate numbers. In particular, in the obtained value of the calculated value, it is necessary to save the last sign, the unit of which exceeds the error of this value. All following significant digits must be discarded.

6. Having received a numerical answer, you need to evaluate its plausibility.

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