How many unpaired electrons does aluminum have. unpaired electron
Chemical element- a certain type of atoms, denoted by a name and a symbol and characterized by a serial number and relative atomic mass.
In table. 1 lists common chemical elements, the symbols with which they are denoted are given (in brackets - pronunciation), ordinal numbers, relative atomic masses, characteristic oxidation states.
Zero the oxidation state of an element in its simple matter(substances) is not listed in the table.
All atoms of one element have the same number of protons in the nucleus and the same number of electrons in the shell. Thus, in the element atom hydrogen H is 1 p+ in the core and on the periphery 1 e- ; in the element atom oxygen Oh located 8 p+ in the core and 8 e- in the shell; element atom aluminum Al contains 13 R+ in the core and 13 e- in the shell.
Atoms of the same element can differ in the number of neutrons in the nucleus, such atoms are called isotopes. So, the element hydrogen H three isotopes: hydrogen-1 (special name and symbol protium 1 H) with 1 p+ in the core and 1 e- in the shell; hydrogen-2 (deuterium 2 H, or D) with 1 p+ and 1 P 0 in the core and 1 e- in the shell; hydrogen-3 (tritium 3 N, or T) with 1 p+ and 2 P 0 in the core and 1 e- in the shell. In the symbols 1 H, 2 H, and 3 H, the superscript indicates mass number is the sum of the numbers of protons and neutrons in the nucleus. Other examples:
Electronic formula an atom of any chemical element in accordance with its location in the Periodic system of elements of D. I. Mendeleev can be determined from Table. 2.
The electron shell of any atom is divided into energy levels(1, 2, 3rd, etc.), levels are divided into sublevels(marked with letters s, p, d, f). The sublevels are made up of atomic orbitals– areas of space where electrons are likely to stay. Orbitals are designated as 1s (orbital of the 1st level of the s-sublevel), 2 s, 2R, 3s, 3p, 3d, 4s… Number of orbitals in sublevels:
The filling of atomic orbitals with electrons occurs in accordance with three conditions:
1) minimum energy principle
Electrons fill the orbitals starting from the lower energy sublevel.
The sequence of increasing energy of sublevels:
1s < 2c < 2p < 3s < 3p < 4s ? 3d < 4p < 5s ? 4d < 5p < 6s…
2)prohibition rule (Pauli principle)
Each orbital can hold at most two electrons.
One electron in the orbital is called unpaired, two electrons - electron pair:
3) principle of maximum multiplicity (Hund's rule)
Within a sublevel, electrons first fill all the orbitals by half, and then completely.
Each electron has its own characteristic - spin (conventionally represented by an up or down arrow). The spins of the electrons add up as vectors, the sum of the spins given number electrons at the sublevel should be maximum(multiplicity):
Filling of levels, sublevels and orbitals of atoms of elements from H with electrons (Z= 1) to Kr (Z= 36) shown in energy diagram(the numbers correspond to the filling sequence and coincide with the serial numbers of the elements):
From the completed energy diagrams are derived electronic formulas element atoms. The number of electrons in the orbitals of a given sublevel is indicated in the superscript to the right of the letter (for example, 3 d 5 is 5 electrons per 3 d-sublevel); electrons of the 1st level go first, then the 2nd, 3rd, etc. The formulas can be complete and short, the latter contain the symbol of the corresponding noble gas in brackets, which conveys its formula, and, moreover, starting with Zn , the filled inner d-sublevel. Examples:
3 Li = 1s 2 2s 1 = [ 2 He]2s 1
8O=1s2 2s 2 2p 4= [ 2He] 2s 2 2p 4
13 Al = 1s 2 2s 2 2p 6 3s 2 3p 1= [ 10 Ne] 3s 2 3p 1
17 Cl = 1s 2 2s 2 2p 6 3s 2 3p 5= [ 10 Ne] 3s 2 3p 5
2O Ca = 1s 2 2s 2 2p 6 3s 2 3p 4s 2= [ 18 Ar] 4s 2
21 Sc = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 1 4s 2= [ 18 Ar] 3d 1 4s 2
25 Mn = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 2= [ 18 Ar] 3d 5 4s 2
26 Fe = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2= [ 18 Ar] 3d 6 4s 2
3O Zn = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2= [ 18 Ar, 3d 10 ] 4s 2
33 As = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 3= [ 18 Ar, 3d 10 ] 4s 2 4p 3
36 Kr = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6= [ 18 Ar, 3d 10 ] 4s 2 4p 6
Electrons outside brackets are called valence. They are involved in the formation of chemical bonds.
The exception is:
24 Cr = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1= [ 18 Ar] 3d 5 4s 1(not 3d 4 4s 2 !),
29 Cu = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1= [ 18 Ar] 3d 10 4s 1(not 3d 9 4s 2 !).
Examples of tasks of part A1. Name, not related to hydrogen isotopes
1) deuterium
2) oxonium
2. The formula for the valence sublevels of a metal atom is
3. The number of unpaired electrons in the ground state of an iron atom is
4. In the excited state of the aluminum atom, the number of unpaired electrons is
5. The electronic formula 3d 9 4s 0 corresponds to the cation
6. The electronic formula of the anion E 2- 3s 2 3p 6 corresponds to the element
7. The total number of electrons in the Mg 2+ cation and F anion is
Paired electrons
If there is one electron in an orbital, then it is called unpaired, and if there are two, then it is paired electrons.
Four quantum numbers n, l, m, m s fully characterize the energy state of an electron in an atom.
Considering the structure of the electron shell of multielectron atoms of various elements, it is necessary to take into account three main provisions:
· Pauli Principle,
The principle of least energy
Gund's rule.
According to Pauli principle An atom cannot have two electrons with the same values of all four quantum numbers.
The Pauli principle determines the maximum number of electrons in one orbital, level and sublevel. Since AO is characterized by three quantum numbers n, l, m, then the electrons of a given orbital can differ only in the spin quantum number m s. But the spin quantum number m s can only have two values + 1 / 2 and - 1 / 2 . Therefore, no more than two electrons can be in one orbital with different meanings spin quantum numbers.
Rice. 4.6. The maximum capacity of one orbital is 2 electrons.
The maximum number of electrons in an energy level is defined as 2 n 2 , and at the sublevel as 2(2 l+ 1). The maximum number of electrons located at different levels and sublevels is given in Table. 4.1.
Table 4.1.
Maximum number of electrons at quantum levels and sublevels
Energy level | Energy sublevel | Possible values of the magnetic quantum number m | Number of orbitals per | Maximum number of electrons per | ||
sublevel | level | sublevel | level | |||
K (n=1) | s (l=0) | |||||
L (n=2) | s (l=0) p (l=1) | –1, 0, 1 | ||||
M (n=3) | s (l=0) p (l=1) d (l=2) | –1, 0, 1 –2, –1, 0, 1, 2 | ||||
N (n=4) | s (l=0) p (l=1) d (l=2) f (l=3) | –1, 0, 1 –2, –1, 0, 1, 2 –3, –2, –1, 0, 1, 2, 3 |
The sequence of filling orbitals with electrons is carried out in accordance with principle of least energy .
According to the principle of least energy, electrons fill orbitals in order of increasing energy.
The order in which the orbitals are filled is determined Klechkovsky's rule: the increase in energy and, accordingly, the filling of the orbitals occurs in the order of increasing sum of the main and orbital quantum numbers (n + l), and if the sum is equal (n + l), in the order of increasing of the main quantum number n.
For example, the energy of an electron at sublevel 4s is less than that at sublevel 3 d, since in the first case the sum n+ l = 4 + 0 = 4 (recall that for s-sublevel value of the orbital quantum number l= = 0), and in the second n+ l = 3 + 2= 5 ( d- sublevel, l= 2). Therefore, sublevel 4 is filled first s and then 3 d(see figure 4.8).
At sublevels 3 d (n = 3, l = 2) , 4R (n = 4, l= 1) and 5 s (n = 5, l= 0) sum of values P and l are the same and equal to 5. In the case of equality of the values of the sums n and l the sublevel with the minimum value is filled first n, i.e. sublevel 3 d.
In accordance with the Klechkovsky rule, the energy of atomic orbitals increases in the series:
1s < 2s < 2R < 3s < 3R < 4s < 3d < 4R < 5s < 4d < 5p < 6s < 5d »
"four f < 6p < 7s….
Depending on which sublevel in the atom is filled last, all chemical elements are divided into 4 electronic family : s-, p-, d-, f-elements.
4f
4 4d
3 4s
3p
3s
1 2s
Levels Sublevels
Rice. 4.8. Energy of atomic orbitals.
Elements whose atoms have the s-sublevel of the outer level filled last are called s-elements . At s-elements are valence s-electrons of the outer energy level.
At p-elements the p-sublevel of the outer level is filled last. They have valence electrons in p- and s-sub-levels of the outer level. At d-elements are last filled d-sublevel of the preexternal level and valence are s-electrons of the outer and d-electrons of the preexternal energy levels.
At f-elements last to be filled f-sublevel of the third outside energy level.
The order in which electrons are placed within one sublevel is determined by Gund's rule:
within a sublevel, electrons are arranged in such a way that the sum of their spin quantum numbers would have a maximum value in absolute value.
In other words, the orbitals of a given sublevel are filled first by one electron with the same value of the spin quantum number, and then by the second electron with the opposite value.
For example, if it is necessary to distribute 3 electrons in three quantum cells, then each of them will be located in a separate cell, i.e. occupy a separate orbital:
∑m s= ½ – ½ + ½ = ½.
The order of distribution of electrons over energy levels and sublevels in the shell of an atom is called its electronic configuration, or electronic formula. Composing electronic configuration room energy level (principal quantum number) are denoted by numbers 1, 2, 3, 4…, sublevel (orbital quantum number) - by letters s, p, d, f. The number of electrons in a sublevel is indicated by a number, which is written at the top of the sublevel symbol.
The electronic configuration of an atom can be represented as the so-called electronic graphic formula. This is the arrangement of electrons in quantum cells, which are a graphic representation of an atomic orbital. Each quantum cell can contain no more than two electrons with different values of spin quantum numbers.
To make an electronic or electronic graphic formula of any element, you should know:
1. Ordinal number of the element, i.e. the charge of its nucleus and the corresponding number of electrons in the atom.
2. The number of the period, which determines the number of energy levels of the atom.
3. Quantum numbers and connection between them.
For example, a hydrogen atom with atomic number 1 has 1 electron. Hydrogen is an element of the first period, so a single electron occupies the first energy level. s orbital with the lowest energy. The electronic formula of the hydrogen atom will look like:
1 H 1 s 1 .
The electron-graphic formula of hydrogen will look like:
Electronic and electron-graphic formulas of the helium atom:
2 Not 1 s 2
2 Not 1 s
reflect the completeness of the electron shell, which determines its stability. Helium is a noble gas characterized by high chemical stability (inertness).
The lithium atom 3 Li has 3 electrons, this is an element of the II period, which means that the electrons are located on 2 energy levels. Two electrons fill s- sublevel of the first energy level and the 3rd electron is located on s- sublevel of the second energy level:
3 Li 1 s 2 2s 1
Valence I
The lithium atom has an electron located on 2 s-sublevel, is less firmly bound to the nucleus than the electrons of the first energy level, therefore, in chemical reactions a lithium atom can easily donate this electron, turning into an Li + ion ( and he -electrically charged particle ). In this case, the lithium ion acquires a stable complete shell of the noble gas helium:
3 Li + 1 s 2 .
It should be noted that, the number of unpaired (single) electrons determines element valence , i.e. its ability to form chemical bonds with other elements.
Thus, the lithium atom has one unpaired electron, which determines its valency equal to one.
The electronic formula of the beryllium atom:
4 Be 1s 2 2s 2 .
The electronic graphic formula of the beryllium atom:
2 Valency is basically
State is 0
Sublevel 2 electrons are detached from beryllium more easily than others s 2, forming the Be +2 ion:
It can be seen that the helium atom and lithium 3 Li + and beryllium 4 Be +2 ions have the same electronic structure, i.e. characterized isoelectronic structure.
Determine which atoms of the elements indicated in the series have four electrons on the external energy level.
Answer: 35
Explanation:
The number of electrons in the outer energy level (electronic layer) of the elements of the main subgroups is equal to the group number.
Thus, from the presented answers, silicon and carbon are suitable, because. they are in the main subgroup of the fourth group of the table D.I. Mendeleev (IVA group), i.e. Answers 3 and 5 are correct.
Determine the atoms of which of the elements indicated in the row in the ground state have the number of unpaired electrons at the outer level equal to 1.
Write down the numbers of the selected elements in the answer field.
Answer: 24
Explanation:
Barium is an element of the main subgroup of the second group and the sixth period Periodic system D. I. Mendeleev, therefore, the electronic configuration of its outer layer will be 6 s 2. On the outside 6 s s-orbitals, the barium atom has 2 paired electrons with opposite spins (complete filling of the sublevel).
Aluminum is an element of the main subgroup of the third group and the third period of the Periodic system, and the electronic configuration of the outer layer of the aluminum atom is 3 s 2 3p 1:3 s sublevel (consists of one s-orbitals) there are 2 paired electrons with opposite spins (complete filling), and 3 p sublevel - one unpaired electron. Thus, in aluminum in the ground state, the number of unpaired electrons in the outer energy level is 1.
Nitrogen is an element of the main subgroup of the fifth group and the second period of the Periodic system, the electronic configuration of the outer layer of the nitrogen atom is 2 s 2 2p 3 : by 2 s- sublevel has 2 paired electrons with opposite spins, and 2 p p-orbitals ( p x, py, pz) are three unpaired electrons, each of which is in each orbital. Thus, in aluminum in the ground state, the number of unpaired electrons in the outer energy level is 1.
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic system, the electronic configuration of the outer layer of the chlorine atom is 3 s 2 3p5: by 3 s- sublevel has 2 paired electrons with opposite spins, and 3 p sublevel, consisting of three p-orbitals ( p x, py, pz) - 5 electrons: 2 pairs of paired electrons in orbitals p x, py and one unpaired - on the orbitals pz. Thus, in chlorine in the ground state, the number of unpaired electrons in the outer energy level is 1.
Calcium is an element of the main subgroup of the second group and the fourth period of the Periodic Table of D. I. Mendeleev. The electronic configuration of its outer layer is similar to the electronic configuration of the barium atom. On the outside 4 s sublevel consisting of one s-orbitals, the calcium atom has 2 paired electrons with opposite spins (complete filling of the sublevel).
Determine which atoms of which of the elements indicated in the series have all valence electrons located on 4 s-energy sublevel.
Write down the numbers of the selected elements in the answer field.
Answer: 25
Explanation:
s 2 3p 5 , i.e. the valence electrons of chlorine are located on 3 s- and 3 p-sublevels (3rd period).
Potassium is an element of the main subgroup of the first group and the fourth period of the Periodic system, and the electronic configuration of the outer layer of the potassium atom is 4 s 1 , i.e. the only valence electron of the potassium atom is located at 4 s-sublevel (4th period).
Bromine is an element of the main subgroup of the seventh group and the fourth period of the Periodic system, the electronic configuration of the outer layer of the bromine atom is 4 s 2 4p 5 , i.e. the valence electrons of the bromine atom are located on 4 s- and 4 p-sublevels (4th period).
Fluorine is an element of the main subgroup of the seventh group and the second period of the Periodic system, the electronic configuration of the outer layer of the fluorine atom is 2 s 2 2p5, i.e. the valence electrons of a fluorine atom are located on 2s- and 2p- sublevels. However, due to the high electronegativity of fluorine, only a single electron located on 2p- sublevel, participates in the formation of a chemical bond.
Calcium is an element of the main subgroup of the second group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of its outer layer is 4 s 2 , i.e. valence electrons are located on 4 s-sublevel (4th period).
Determine which atoms of which of the elements indicated in the series have valence electrons located on the third energy level.
Write down the numbers of the selected elements in the answer field.
Answer: 15
Explanation:
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of chlorine is 3 s 2 3p 5 , i.e. the valence electrons of chlorine are located on the third energy level (3rd period).
s 2 2p 3 , i.e. valence electrons of nitrogen are located at the second energy level (2nd period).
Carbon is an element of the main subgroup of the fourth group and the second period of the Periodic system, the electronic configuration of the outer layer of the carbon atom is 2 s 2 2p 2 , i.e. the valence electrons of the carbon atom are located in the second energy level (2nd period).
Beryllium is an element of the main subgroup of the second group and the second period of the Periodic system, the electronic configuration of the outer layer of the beryllium atom is 2 s 2 , i.e. the valence electrons of the beryllium atom are located at the second energy level (2nd period).
Phosphorus is an element of the main subgroup of the fifth group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of its outer layer is 3 s 2 3p 3 , i.e. the valence electrons of the phosphorus atom are located on the third energy level (3rd period).
Determine which atoms of which of the elements indicated in the series have d There are no electron sublevels.
Write down the numbers of the selected elements in the answer field.
Answer: 12
Explanation:
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the chlorine atom is 1 s 2 2s 2 2p 6 3s 2 3p 5 , i.e. d There is no sublevel for the chlorine atom.
Fluorine is an element of the main subgroup of the seventh group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the fluorine atom is 1 s 2 2s 2 2p 5 , i.e. d-sublevel at the fluorine atom also does not exist.
Bromine is an element of the main subgroup of the seventh group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the bromine atom is 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 , i.e. the bromine atom has a completely filled 3 d-sublevel.
Copper is an element of a side subgroup of the first group and the fourth period of the Periodic Table, the electronic configuration of the copper atom is 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 , i.e. the copper atom has a completely filled 3d-sublevel.
Iron is an element of a side subgroup of the eighth group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the iron atom is 1 s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 , i.e. the iron atom has an unfilled 3d-sublevel.
Determine the atoms of which of the elements indicated in the series belong to s-elements.
Write down the numbers of the selected elements in the answer field.
Answer: 15
Explanation:
Helium is an element of the main subgroup of the second group and the first period of the Periodic system of D. I. Mendeleev, the electronic configuration of the helium atom is 1 s 2 , i.e. The valence electrons of a helium atom are located only on 1s-sublevel, therefore, helium can be attributed to s-elements.
Phosphorus is an element of the main subgroup of the fifth group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the phosphorus atom is 3 s 2 3p 3, therefore, phosphorus refers to p-elements.
s 2 3p 1, therefore, aluminum belongs to p-elements.
Chlorine is an element of the main subgroup of the seventh group and the third period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the chlorine atom is 3s 2 3p 5, therefore, chlorine belongs to p-elements.
Lithium is an element of the main subgroup of the first group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the lithium atom is 2 s 1, therefore, lithium belongs to s-elements.
Determine the atoms of which of the elements indicated in the row in the excited state have the electronic configuration of the external energy level ns 1 np 2.
Write down the numbers of the selected elements in the answer field.
Answer: 12
Explanation:
Boron is an element of the main subgroup of the third group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the boron atom in the ground state is 2 s 2 2p 1 . When the boron atom goes into an excited state, the electronic configuration becomes 2 s 1 2p 2 due to electron hopping from 2 s- on 2 p- orbital.
Aluminum is an element of the main subgroup of the third group and the third period of the Periodic Table, the electronic configuration of the outer layer of the aluminum atom is 3 s 2 3p one . When an aluminum atom goes into an excited state, the electronic configuration becomes 3 s 1 3 p 2 due to electron hopping from 3 s- for 3 p- orbital.
Fluorine is an element of the main subgroup of the seventh group and the second period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the fluorine atom is 3 s 2 3p 5 . In this case, in the excited state it is impossible to obtain the electronic configuration of the external electronic level n s 1n p 2 .
Iron is an element of a side subgroup of the eighth group and the fourth period of the Periodic Table of D. I. Mendeleev, the electronic configuration of the outer layer of the iron atom is 4 s 2 3d 6 . In this case, in the excited state, it is also impossible to obtain the electronic configuration of the external electronic level n s 1n p 2 .
Nitrogen is an element of the main subgroup of the fifth group and the second period of the Periodic system, and the electronic configuration of the outer layer of the nitrogen atom is 2 s 2 2p 3. In this case, in the excited state, it is also impossible to obtain the electronic configuration of the external electronic level n s 1n p 2 .
Determine for which atoms of the elements indicated in the series a transition to an excited state is possible.
Write down the numbers of the selected elements in the answer field.
Answer: 23
Explanation:
Rubidium and cesium - elements of the main subgroup of the first group of the Periodic Table of D. I. Mendeleev, are alkali metals, the atoms of which have one electron at the external energy level. Because the s-orbital for the atoms of these elements is external, it is impossible for an electron to jump from s- on the p-orbital, and therefore, the transition of an atom to an excited state is not characteristic.
The nitrogen atom is not able to go into an excited state, because he has the 2nd energy level filled in and there are no free orbitals at this energy level.
Aluminum is an element of the main subgroup of the third group of the Periodic Table of Chemical Elements, the electronic configuration of the outer layer of the aluminum atom is 3 s 2 3p one . When an aluminum atom passes into an excited state, an electron jumps from 3 s- for 3 p- orbital, and the electronic configuration of the aluminum atom becomes 3 s 1 3 p 2 .
Carbon is an element of the main subgroup of the fourth group of the Periodic Table, the electronic configuration of the outer layer of the carbon atom is 2 s 2 2p2. When a carbon atom passes into an excited state, an electron jumps from 2 s- on 2 p- orbital, and the electron configuration of the carbon atom becomes 2s 1 2p 3 .
Determine the atoms of which of the elements indicated in the series correspond to the electronic configuration of the outer electronic layer ns 2 np 3 .
Write down the numbers of the selected elements in the answer field.
Answer: 23
Explanation:
Electronic configuration of the outer electron layer ns 2 np 3 says that the element to be filled in is p sublevel, i.e. this is p-elements. All p-elements are located in the last 6 cells of each period in a group whose number is equal to the sum of electrons on s and p sublevels of the outer layer, i.e. 2 + 3 \u003d 5. Thus, the required elements are nitrogen and phosphorus.
Determine the atoms of which of the elements indicated in the series have a similar configuration of the external energy level.
Write down the numbers of the selected elements in the answer field.
Answer: 34
Among these elements, bromine and fluorine have a similar electronic configuration. The electronic configuration of the outer layer has the form ns 2 np 5
Determine the atoms of which of the elements indicated in the series have a completely completed second electronic level.
Write down the numbers of the selected elements in the answer field.
Answer: 13
Explanation:
The filled 2nd electronic level has the noble gas neon, as well as any chemical element located in the periodic table after it.
Determine the atoms of which of the elements indicated in the series lack 2 electrons to complete the external energy level.
Write down the numbers of the selected elements in the answer field.
Answer: 34
Before the completion of the external electronic level, 2 electrons are missing p-elements of the sixth group. Recall that all p-elements are located in the last 6 cells of each period.
Determine the atoms of which of the elements indicated in the row in the excited state have the electronic formula of the external energy level n s 1n p 3 .
Write down the numbers of the selected elements in the answer field.
Answer: 24
Explanation:
s 1n p 3 tells us that there are 4 electrons (1 + 3) in the outer energy level (electronic layer). Among these elements, only silicon and carbon atoms have 4 electrons at the outer level.
The electronic configuration of the external energy level of these elements in the ground state has the form n s 2n p 2 , and in excited n s 1n p 3 (when carbon and silicon atoms are excited, the s-orbital electrons are depaired and one electron enters the free p-orbital).
Determine the atoms of which of the elements indicated in the row in the ground state have the electronic formula of the external energy level n s 2n p 4 .
Write down the numbers of the selected elements in the answer field.
Answer: 25
Explanation:
Formula of external energy level n s 2n p 4 tells us that there are 6 electrons (2+4) in the outer energy level (electronic layer). The number of electrons in the outer electronic level for the elements of the main subgroups is always equal to the group number. Thus, the electronic configuration n s 2n p 4 among these elements have selenium and sulfur atoms, since these elements are located in the VIA group.
Determine the atoms of which of the elements indicated in the series in the ground state have only one unpaired electron.
Write down the numbers of the selected elements in the answer field.
Answer: 25
Determine the atoms of which of the elements has the configuration of the external electronic level n s 2n p 3 .
Answer: 45
Determine the atoms of which of the elements indicated in the series in the ground state do not contain unpaired electrons.
Write down the numbers of the selected elements in the answer field.
For the correct answer to each of the tasks 1-8, 12-16, 20, 21, 27-29, 1 point is given.
Tasks 9–11, 17–19, 22–26 are considered completed correctly if the sequence of numbers is correctly indicated. For a complete correct answer in tasks 9–11, 17–19, 22–26, 2 points are given; if one mistake is made - 1 point; for an incorrect answer (more than one mistake) or its absence - 0 points.
Theory on assignment:
1) F 2) S 3) I 4) Na 5) Mg
Determine the atoms of which of the indicated elements in the ground state, before the completion of the outer electron layer, one electron is missing.
1
The eight-electron shell corresponds to the shell of an inert gas. For each of the substances in the period in which they are located, they correspond to an inert gas, neon for fluorine, argon for sulfur, xenon for iodine, argon for sodium and magnesium, but of the listed elements, only fluorine and iodine lack one electron to an eight-electron shell, since they are in the seventh group.
To complete the task, use the following row of chemical elements. The answer in the task is a sequence of three numbers, under which the chemical elements in this row are indicated.
1) Be 2) H 3) N 4) K 5) C
Determine the atoms of which of the indicated elements in the ground state contain the same number of unpaired electrons.
1
4 Be Beryllium: 1s 2 2s 2
7 N Nitrogen: 1s 2 2s 2 2p 3
Number of unpaired electrons - 1
6 C Carbon: 1s 2 2s 2 2p 2
1s2 | 2s 2 | 2p 3 | ||
↓ | ↓ |
Number of unpaired electrons - 2
Hence it is obvious that for hydrogen and for potassium the number of unpaired electrons is the same.
To complete the task, use the following row of chemical elements. The answer in the task is a sequence of three numbers, under which the chemical elements in this row are indicated.
1) Ge 2) Fe 3) Sn 4) Pb 5) Mn
Determine the atoms of which of the elements indicated in the series have valence electrons both on the s- and on the d-sublevels.
1
To solve this task, it is necessary to paint the upper electronic level of the elements:
- 32 Ge Germanium: 3d 10 4s 2 4p 2
- 26Fe Iron: 3d 6 4s 2
- 50 Sn Tin: 4d 10 5s 2 5p 2
- 82 Pb Lead: 4f 14 5d 10 6s 2 6p 2
- 25 Mn Manganese: 3d 5 4s 2
Iron and manganese have valence electrons in the s- and d-sublevels.
To complete the task, use the following row of chemical elements. The answer in the task is a sequence of three numbers, under which the chemical elements in this row are indicated.
1) Br 2) Si 3) Mg 4) C 5) Al
Determine the atoms of which of the elements indicated in the row in the excited state have the electronic formula of the external energy level ns 1 np 3
1
For an unexcited state electronic formula ns 1 np 3 will represent ns 2 np 2, it is the elements of this configuration that we need. Let's write the upper electronic level of the elements (or simply find the elements of the fourth group):
- 35 Br Bromine: 3d 10 4s 2 4p 5
- 14Si Silicon: 3s 2 3p 2
- 12 Mg Magnesium: 3s 2
- 6 C Carbon: 1s 2 2s 2 2p 2
- 13 Al Aluminum: 3s 2 3p 1
For silicon and carbon, the upper energy level coincides with the desired
To complete the task, use the following row of chemical elements. The answer in the task is a sequence of three numbers, under which the chemical elements in this row are indicated.
1) Si 2) F 3) Al 4) S 5) Li
- The displacement is called the vector connecting the start and end points of the trajectory The vector connecting the beginning and end of the path is called
- Trajectory, path length, displacement vector Vector connecting the initial position
- Calculating the area of a polygon from the coordinates of its vertices The area of a triangle from the coordinates of the vertices formula
- Acceptable Value Range (ODZ), theory, examples, solutions