Atomic Structure and Periodic Law

Question 1

Question 1: 1. The chemical formula of gold mica is $\mathrm { KMg } _ { 3 } \mathrm { AlSi } _ { 3 } \mathrm { ...

1. The chemical formula of gold mica is $\mathrm { KMg } _ { 3 } \mathrm { AlSi } _ { 3 } \mathrm { O } _ { 10 } \mathrm {~F} _ { x } ( \mathrm { OH } ) _ { ( 2 - x ) }$ The following statement is correct $\mathrm { HF } > \mathrm { H } _ { 2 } \mathrm { O }$

A. A. Radius size: ${ } ^ { r ( F ) > r ( \mathrm { Al } ) }$
B. B. Alkalinity : $\mathrm { Al } ( \mathrm { OH } ) _ { 3 } > \mathrm { Mg } ( \mathrm { OH } ) _ { 2 }$
C. C. Stability strength :
D. D. Ionization energy magnitude: $\mathrm { I } _ { 1 } ( \mathrm { Si } ) > \mathrm { I } _ { 1 } ( \mathrm { O } )$

Answer

Answer: C

Solution: A. The more electron layers the larger the radius, when the number of electron layers is the same, the larger the nuclear charge, the smaller the radius; atomic radius size : $\mathrm { r } ( \mathrm { F } ) < \mathrm { r } ( \mathrm { Al } )$, so A error; B. The stronger the metallicity, the stronger the basicity of the corresponding hydrate of the highest valence oxide, the stronger the basicity: $\mathrm { Al } ( \mathrm { OH } ) _ { 3 } < \mathrm { Mg } ( \mathrm { OH } ) _ { 2 }$, so B is wrong; C. The stronger the nonmetallicity, the stronger the stability of the simple hydride, then stability: $\mathrm { HF } > \mathrm { H } _ { 2 } \mathrm { O }$, so C is correct; D. the same main group from top to bottom of the first ionization energy decreases, the first ionization energy: $\mathrm { O } > \mathrm { S }$, the same period from left to right the first ionization energy tends to increase, the first ionization energy $: \mathrm { S } > \mathrm { Si }$, so the ionization energy magnitude $: \mathrm { I } _ { 1 } ( \mathrm { Si } ) < \mathrm { I } _ { 1 } ( \mathrm { O } )$, the D is wrong; Therefore, choose C.

Question 2

Question 2: 3. Under certain conditions, the conversion of graphite into diamond requires the absorption of ener...

3. Under certain conditions, the conversion of graphite into diamond requires the absorption of energy. The following statements are wrong (1) graphite and diamond are isotopes of each other (2) diamond is more stable than graphite (3) the conversion of diamond into graphite is an exothermic reaction (4) the total energy of equal amounts of C (diamond) is higher than that of C (graphite) (5) Diamond and graphite of equal mass are fully burned in a sufficient amount of oxygen, and graphite releases more energy.

A. A. (2) (3) (4)
B. B. (2) (4) (5)
C. C. (1) (2) (5)
D. D. (1) (2) (4)

Answer

Answer: C

Solution: (1) Graphite and diamond are different monomers composed of carbon elements and are isomers of each other, (1) is wrong; (2) The conversion of graphite to diamond absorbs heat, indicating that graphite is lower in energy and more stable, (2) is incorrect; (3) The inverse of a heat-absorbing reaction (diamond $\rightarrow$ graphite) is exothermic, (3) is correct; (4) The conversion of graphite into diamond requires the absorption of energy, which means that an equal mass of graphite has less energy than diamond, and an equal amount of C (diamond) has a higher total energy than C (graphite), (4) is correct; (5) equal amount of C (diamond) than the total energy of C (graphite) is high, then equal mass of diamond and graphite are in a sufficient amount of oxygen in the full combustion, diamond released more energy, (5) error; In summary, the wrong combination of statements is (1)(2)(5), corresponding to option C.

Question 3

Question 3: 4. $\mathrm { R } ^ { 2 + }$ is known to have b neutrons and a electron outside the nucleus, the cor...

4. $\mathrm { R } ^ { 2 + }$ is known to have b neutrons and a electron outside the nucleus, the correct symbol for the R atom is ( )

A. A. ${ } _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { R }$
B. B. $\begin{array} { r } \mathrm { a } + \mathrm { b } - 2 \\ \mathrm { a } ^ { - 2 } \end{array} \mathrm { R }$
C. C. $\begin{array} { r } \mathrm { a } + \mathrm { b } + 2 \\ \mathrm { a } + 2 \end{array}$ R
D. D. ${ } ^ { a + b + 2 } { } _ { a - 2 } R$

Answer

Answer: C

Solution: The ground state of the element $R$ has a number of electrons outside the nucleus of $a + 2$, which gives a proton number of $a + 2$ and a relative atomic mass of $a + b + 2$, so C is correct.

Question 4

Question 4: 5. The conclusions that can be drawn directly from studying the information in the table below are. ...

5. The conclusions that can be drawn directly from studying the information in the table below are. | Elements | F | Cl | Br | I | | :--- | :--- | :--- | :--- | :--- | :--- | | Hydrogen Composition | HF | HCl | HBr | HI | | Ease of Hydrogen Compound Formation | $\mathrm { H } _ { 2 } , \mathrm {~F} _ { 2 }$ Mixed, occurs with intense chemicalization in cold darkness | $\mathrm { H } _ { 2 } , \mathrm { Cl } _ { 2 }$ Mixed, occurs with light or ignition | $\mathrm { H } _ { 2 }$ Mixed with $\mathrm { Br } _ { 2 }$, occurs with heat | [[[]]] Mixed, occurs with heat | [[]]] Mixed, occurs with heat | [[]]] Mixed, occurs with heat INLINE_FORMULA_4]], $\mathrm { I } _ { 2 }$ mixed, chemically synthesized by heat, and at the same time | | | Explosion | Explosion | Biochemical Synthesis | Decomposition | | :--- | :--- | :--- | :--- | :--- | :--- | | Hydrogen compounds are thermally stable Qualitative | Very stable | Stable | More stable | Less stable |

A. A. Elemental properties vary periodically with increasing atomic number
B. B. The nonmetallicity of elements of the same period increases with increasing atomic number
C. C. Metallicity of group VII A elements decreases with increasing number of electron layers
D. D. The nonmetallicity of group VIIA elements decreases with the number of electron layers

Answer

Answer: D

Solution: A. The four elements F, Cl, Br and I belong to Group VIIA, so the periodic changes in the properties of the elements cannot be seen, so A is not chosen; B. The four elements F, Cl, Br, and I belong to Group VIIA and do not belong to the same cycle, so it is not possible to see the variation of nonmetallic properties of the elements in the same cycle, so B is not chosen; C. F, Cl, Br and I are all non-metallic elements with no metallic properties, so C is not chosen; D. F, Cl, Br, I four elements with the increase in the number of electron layers, hydride formation difficulty gradually increased, hydride stability gradually decreased, can see the non-metallic with the increase in the number of electron layers and weakened, so D selected;

Question 5

Question 5: 6. Human understanding of the structure of the atom has gone through a long historical stage. One of...

6. Human understanding of the structure of the atom has gone through a long historical stage. One of the first scientists to propose that "the electrons outside the nucleus of the atom are moving in atomic orbitals and follow certain laws. "The scientist is

A. A. Dalton (name)
B. B. Thomson (name)
C. C. Rutherford (Earnest, an early nuclear physicist from New Zealand)
D. D. Bohr (name)

Answer

Answer: D

Solution: Dalton proposed that "atoms constitute the most basic ions of matter, they are solid, indivisible solid ball"; Thomson proposed that "the atom is an average distribution of positively charged ball, this ball is embedded with many electrons, the negative charge of the electrons and the positive charge in the small ball neutralization Rutherford proposed that "the atom is composed of the nucleus and the electrons outside the nucleus"; Pohl proposed that "the electrons outside the nucleus of the atom are moving in the atomic orbitals and follow certain laws"; In summary, the answer is D.

Question 6

Question 6: 7. The following statements are correct

7. The following statements are correct

A. A. Particles with the same number of protons must belong to the same element
B. B. The mass numbers of atoms of different elements must not be equal
C. C. The number of protons determines the type of element, and the number of protons and neutrons determines the type of atom
D. D. Particles with the same number of protons must have the same number of electrons outside the nucleus, but not necessarily the same chemical properties

Answer

Answer: C

Solution: A. The object of study of the elements is the atom, and particles with the same number of protons do not necessarily belong to the same element, such as $\mathrm { CH } _ { 4 }$ and $\mathrm { H } _ { 2 } \mathrm { O }$, if they are atoms, then they belong to the same element, so A is wrong; B. Atoms of different elements may have the same mass number, such as ${ } ^ { 3 } \mathrm { H }$ and ${ } ^ { 3 } \mathrm { He }$, so B is wrong; C. The type of element is determined by the number of protons, and the type of atom is determined by the number of protons and neutrons, so C is correct; D. The number of electrons outside the nucleus is not necessarily the same for particles with the same number of protons, such as sodium ions and sodium atoms, so D is wrong.

Question 7

Question 7: 8. The following statements about the periodic table of elements are not correct

8. The following statements about the periodic table of elements are not correct

A. A. The periodic table of the elements has seven horizontal rows, i.e., seven periods, eighteen vertical rows, and sixteen groups
B. B. Group IA, also known as the alkali metals, increase the density of the monomers in order as the number of nuclear charges increases
C. C. Group VIIA, also known as halogens, have monomers that become darker in color as their nuclear charge increases
D. D. The third period main group element with the smallest simple ionic radius is $\mathrm { Al } ^ { 3 + }$

Answer

Answer: B

Solution: A. The periodic table has seven horizontal rows, i.e., seven cycles, and eighteen vertical rows, of which columns 8, 9, and 10 are listed as group VIII, which includes sixteen groups, including the main group, the subgroups, group VIII, and group 0. A is correct; B. Group IA, also known as alkali metals, with the increase in the number of nuclear charges, the density of the monomers tends to increase, but the density of K is smaller than the density of Na, B error; C. Group VIIA, also known as halogen elements, with the increase in the number of nuclear charges, the color of the monomers from yellowish green to purple-black, more and more dark, C is correct; D. The larger the number of electron layers, the larger the radius of simple ions, and when the number of electron layers is the same, the smaller the number of nuclear charges, the larger the radius of simple ions, then the third period of the main group of elements, the smallest radius of simple ions is $\mathrm { Al } ^ { 3 + }$, D is correct;

Question 8

Question 8: 10. The following description of ${ } ^ { 60 } \mathrm { Co }$ is correct ( )

10. The following description of ${ } ^ { 60 } \mathrm { Co }$ is correct ( )

A. A. The number of protons is 33.
B. B. The number of electrons is 27.
C. C. The mass number is 27.
D. D. The neutron count is 60.

Answer

Answer: B

Solution: In ${ } ^ { 60 } \mathrm { Co }$, 60 represents the number of masses, the number of protons $=$ the number of extra-nuclear electrons $= 27$, and the number of neutrons $= 60 - 27 = 33$.

Question 9

Question 9: 11. The following chemical terms are expressed correctly

11. The following chemical terms are expressed correctly

A. A. Three nuclides of the element H: ${ } ^ { 1 } \mathrm { H } _ { 2 } , { } ^ { 2 } \mathrm { H } _ { 2 } , { } ^ { 3 } \mathrm { H } _ { 2 }$
B. B. The spatial structure of $\mathrm { H } _ { 2 } \mathrm { O }$ is rectilinear
C. C. Express the formation of ${ } _ { \mathrm { NaCl } }$ in electronic form: $\mathrm { Na } \widetilde { + } \cdot \underset { \bullet } { \bullet } \mathrm { l } : \longrightarrow \mathrm { Na } ^ { + } [ \stackrel { \bullet } { \mathrm { C } } \mathrm { l } : ] ^ { - }$
D. D. The space-filling model ( ) can represent both methane and carbon tetrachloride molecules

Answer

Answer: C

Solution: A. The three nuclides of the element H are denoted as ${ } ^ { 1 } \mathrm { H } , { } ^ { 2 } \mathrm { H } , { } ^ { 3 } \mathrm { H }$, and A is wrong; B. The space structure of $\mathrm { H } _ { 2 } \mathrm { O }$ is in the shape of $\mathrm { V } ^ { - }$, which is wrong; C. NaCl is an ionic compound, and the formation of NaCl is represented by the electronic formula: ![](/images/questions/atomic-structure/image-001.jpg) D. The atomic radius $\mathrm { Cl } > \mathrm { C } > \mathrm { H }$, then the space-filling model ( )) can represent the methane molecule, but not the carbon tetrachloride molecule, D is wrong;

Question 10

Question 10: 12. The following ions have the smallest radius

12. The following ions have the smallest radius

A. A. $\mathrm { F } ^ { - }$
B. B. $\mathrm { Cl } ^ { - }$
C. C. $\mathrm { Br } ^ { - }$
D. D. $\mathrm { I } ^ { - }$

Answer

Answer: A

Solution: The more layers of electrons outside the nucleus, the larger the ion radius, and the number of electron layers $\mathrm { I } ^ { - } > \mathrm { Br } ^ { - } > \mathrm { Cl } ^ { - } > \mathrm { F } ^ { - }$ results in an ion radius of $\mathrm { I } ^ { - } > \mathrm { Br } ^ { - } > \mathrm { Cl } ^ { - } > \mathrm { F } ^ { - }$, so the smallest radius of the ion is $\mathrm { F } ^ { - }$; Therefore, the answer is A.

Question 11

Question 11: 13. Tin is a metal element, as early as in ancient times, people have discovered and use of tin, acc...

13. Tin is a metal element, as early as in ancient times, people have discovered and use of tin, according to evidence, China's Zhou Dynasty, the use of tin has been very common. The following statement about ${ } ^ { 50 } \mathrm { Sn }$ is correct

A. A. ${ } ^ { 119 } \mathrm { Sn }$ and ${ } ^ { 122 } \mathrm { Sn }$ both have 50 extra-nuclear electrons.
B. B. ${ } ^ { 119 } \mathrm { Sn }$ Contains 50 neutrons.
C. C. Sn is a fourth-period element
D. D. Sn has a smaller atomic radius than Ge

Answer

Answer: A

Solution: A. ${ } ^ { 119 } \mathrm { Sn }$ and ${ } ^ { 122 } \mathrm { Sn }$ have the same number of protons, and the number of electrons outside the nucleus is equal to the number of protons, which is 50, so A is correct; B. ${ } ^ { 119 } \mathrm { Sn }$ has a mass number of 119 and a proton number of 50, and contains neutrons = mass number - proton number $= 119 - 50 = 69$, so B is wrong; C. The atomic number of Sn is 50, and the arrangement of electrons outside the nucleus is $2 , 8 , 18 , 18 , 4$, which is located in the fifth period, so C is wrong; D. Sn and Ge are both the IVA elements, and Sn is located below Ge, then the atomic radius of Sn is larger than that of Ge, so D is wrong; The answer is A. $14 . \mathrm { C }$ [Knowledge Points]Atomic Structure Schematic, Ionic Structure Schematic, Structure and Formation of Covalent Compounds, Ionization Equations, Elements, Nuclides, Isotopes [Detailed Explanation A. Be is the 4th element, the atomic structure schematic diagram is: (+4) $\frac { 1 } { 2 } \frac { 1 } { 2 }$/, so A is wrong; B. The electronic formula of HClO is H : $\underset { . } { \ddot { 0 } }$ : $\underset { . . } { \ddot { C } } 1 :$, so B is wrong; C. The hydrogen sulfite ion is an acid radical ion of a weak acid and cannot be split; the ionization equation of $\mathrm { NaHSO } _ { 3 }$ is: $\mathrm { NaHSO } _ { 3 } = \mathrm { Na } ^ { + } + \mathrm { HSO } _ { 3 } ^ { - }$, so C is correct; D. Isotopes are elements with the same number of protons but different numbers of neutrons, ${ } ^ { 14 } \mathrm { C }$ and ${ } ^ { 14 } \mathrm {~N}$ belong to different elements, D is wrong.

Question 12

Question 12: 14. The following chemical terms are correctly stated ![](/images/questions/atomic-structure/image-...

14. The following chemical terms are correctly stated ![](/images/questions/atomic-structure/image-002.jpg)

A. A. Schematic of the atomic structure of Be
B. B. The electronic formula for HClO is $\mathrm { H } : \stackrel { \bullet } { \mathrm { C } } _ { \bullet \bullet } \mathrm { l } : \stackrel { \circ } { \mathrm { O } } _ { \bullet }$:
C. C. The ionization equations $\mathrm { NaHSO } _ { 3 }$ for $\mathrm { NaHSO } _ { 3 } = { } ^ { \mathrm { HSO } _ { 3 } ^ { - } } + \mathrm { Na } ^ { + } \mathrm { D }$. are isotopes of each other : ${ } ^ { 14 } \mathrm { C }$ and ${ } ^ { 14 } \mathrm {~N}$
D. D. The ionization equations $\mathrm { NaHSO } _ { 3 }$ for $\mathrm { NaHSO } _ { 3 } = { } ^ { \mathrm { HSO } _ { 3 } ^ { - } } + \mathrm { Na } ^ { + } \mathrm { D }$. are isotopes of each other : ${ } ^ { 14 } \mathrm { C }$ and ${ } ^ { 14 } \mathrm {~N}$

Answer

Answer: C

Question 13

Question 13: 15. Tick-tock, tick-tock, where has the time gone! The history of timepieces is engraved with the co...

15. Tick-tock, tick-tock, where has the time gone! The history of timepieces is engraved with the contributions of chemistry. The following statements are correct

A. A. The stone used to make the sundial disk contains the element silicon, which is an element of group VIA of the third cycle.
B. B. The hairspring in a mechanical watch is made of molybdenum, cobalt and nickel, with cobalt and nickel in the p-zone.
C. C. A quartz watch with $\mathrm { SiO } _ { 2 } , \mathrm { SiO } _ { 2 }$ is a covalent crystal based on the oscillation characteristics of quartz crystals for timekeeping.
D. D. Currently, "Beijing Time" is based on a cesium atomic clock, and the neutron count of ${ } ^ { 135 } \mathrm { Cs }$ is 135.

Answer

Answer: C

Solution: A. Silicon is element No. 14, and silicon is an element of group IV A in the third cycle; B. Cobalt and chromium are located in the d-region in the winding of a mechanical watch composed of molybdenum, nickel and chromium, which is incorrect; C. The oscillating properties of quartz crystals for timekeeping contain $\mathrm { SiO } _ { 2 } , \mathrm { SiO } _ { 2 }$ which is formed by covalent bonding and belongs to covalent crystals, C is correct; D. The mass number of ${ } ^ { 135 } \mathrm { Cs }$ is 135, the number of protons is 55, and the number of neutrons is $135 - 55 = 80$, which is D wrong;

Question 14

Question 14: 16. W, X, Y and Z are short-period elements with increasing atomic numbers. $\mathrm { X } + \mathrm...

16. W, X, Y and Z are short-period elements with increasing atomic numbers. $\mathrm { X } + \mathrm { Y } = \mathrm { W } + \mathrm { Z }$; the compound $\mathrm { XW } _ { 3 }$ produces white smoke when it meets WZ. The following statement is correct

A. A. Electronegativity: $\mathrm { W } > \mathrm { X } > \mathrm { Y } > \mathrm { Z }$
B. B. Atomic radius: $Z > Y > X > W$
C. C. The oxygenated acids of Z are all strong acids
D. D. The hydrates of the oxides of Y are strong bases

Answer

Answer: D

Solution: A. The electronegativity of the atoms of the elements of the same period increases from left to right, and the electronegativity of the atoms of the elements of the same main group decreases from top to bottom, so the electronegativities of the four elements are $\mathrm { Cl } > \mathrm { N } > \mathrm { Na } > \mathrm { H }$ from largest to smallest, and A is wrong; B. The atomic radii of the elements of the same period decrease from left to right, and the atomic radii of the elements of the same main group increase from top to bottom, so the atomic radii are $\mathrm { Na } > \mathrm { Cl } > \mathrm { N } > \mathrm { H }$, and B is wrong; C. The oxygen-containing acids of the element N are not necessarily all strong acids, such as $\mathrm { HNO } _ { 2 }$ is a weak acid, C is wrong; D. The oxide hydrate of Y is NaOH, which is a strong base; D. The hydrate of the oxide of Y is NaOH, which is a strong base.

Question 15

Question 15: 17. The following statements are correct

17. The following statements are correct

A. A. Chlorine atom with neutron number 20: ${ } _ { 17 } ^ { 20 } \mathrm { Cl }$
B. B. $\mathrm { H } _ { 2 } , \mathrm {~T} _ { 2 }$ is an allotrope of hydrogen
C. C. Hypochlorous acid has the structural formula $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$
D. D. $\mathrm { NaHCO } _ { 3 }$ Ionization equation in aqueous solution: $\mathrm { NaHCO } _ { 3 } = \mathrm { Na } ^ { + } + \mathrm { H } ^ { + } + \mathrm { CO } ^ { 3 }$

Answer

Answer: C

Solution: A. Mass number = proton number + neutron number, chlorine atom with neutron number 20: ${ } ^ { 37 } \mathrm { Cl } , \mathrm { A }$ is wrong; B. $\mathrm { H } _ { 2 } , \mathrm {~T} _ { 2 }$ are all hydrogen, which is the same substance; C.Oxygen is the central atom in hypochlorous acid, and the structural formula is $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$, C is correct; D. The ionization equation of $\mathrm { NaHCO } _ { 3 }$ in aqueous solution is $\mathrm { NaHCO } _ { 3 } = \mathrm { Na } ^ { + } + \mathrm { HCO } ^ { - }$, which is wrong;

Question 16

Question 16: 18. A substance used to synthesize drugs for the treatment of immune diseases has the structure show...

18. A substance used to synthesize drugs for the treatment of immune diseases has the structure shown in the figure, in which $\mathrm { X } , \mathrm { Y } , \mathrm { Z } , \mathrm { Q } , \mathrm {~W}$ is an element of the main group of $1 \sim 20$ and the atomic number increases sequentially, Z and Q are of the same main group, and the simple ions of Q and W have the same electronic layer structure, and the following statements are not correct <img class="imgSvg" id = "mi1n0j6gxqdoa6anxl" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW4wajZneHFkb2E2YW54bCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTc0IDEwNC45OTk5OTk5OTk5NzQ2IiBzdHlsZT0id2lkdGg6IDE3NC4yNzk4MTQzNTY0ODA2cHg7IGhlaWdodDogMTA0Ljk5OTk5OTk5OTk3NDZweDsgb3ZlcmZsb3c6IHZpc2libGU7Ij48ZGVmcz48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbjBqNmd4cWRvYTZhbnhsLTEiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTA0Ljk5OTk5OTk5OTk3NDYyIiB5MT0iNTIuNTAwMDI4Mjc0NjAxNjYiIHgyPSIxMjAuNzQ5OTc1NTEzNDMzOTUiIHkyPSI3OS43Nzk4NDI2MzExMDc2OCI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbjBqNmd4cWRvYTZhbnhsLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTA0Ljk5OTk5OTk5OTk3NDYyIiB5MT0iNTIuNTAwMDI4Mjc0NjAxNjYiIHgyPSIxMDUuMDAwMDI4Mjc0NTg4OTciIHkyPSIyMS4wMDAwMjgyNzQ2MTQzNjUiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4wajZneHFkb2E2YW54bC01IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjEwNC45OTk5OTk5OTk5NzQ2MiIgeTE9IjUyLjUwMDAyODI3NDYwMTY2IiB4Mj0iMTMyLjI3OTgxNDM1NjQ4MDYiIHkyPSIzNi43NTAwNTI3NjExNDIzMyI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbjBqNmd4cWRvYTZhbnhsLTciIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNzMuNDk5OTk5OTk5OTg3MyIgeTE9IjUyLjQ5OTk5OTk5OTk4NzMxIiB4Mj0iMTA0Ljk5OTk5OTk5OTk3NDYyIiB5Mj0iNTIuNTAwMDI4Mjc0NjAxNjYiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4wajZneHFkb2E2YW54bC05IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjQyIiB5MT0iNTIuNDk5OTcxNzI1MzcyOTYiIHgyPSI3My40OTk5OTk5OTk5ODczIiB5Mj0iNTIuNDk5OTk5OTk5OTg3MzEiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4wajZneHFkb2E2YW54bC0xMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI3My40OTk5OTk5OTk5ODczIiB5MT0iNTIuNDk5OTk5OTk5OTg3MzEiIHgyPSI3My41MDAwMjgyNzQ2MDE2OCIgeTI9IjIxIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFuMGo2Z3hxZG9hNmFueGwtMTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNzMuNDk5OTcxNzI1MzcyOTQiIHkxPSI4My45OTk5OTk5OTk5NzQ2MiIgeDI9IjczLjQ5OTk5OTk5OTk4NzMiIHkyPSI1Mi40OTk5OTk5OTk5ODczMSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+PC9saW5lYXJHcmFkaWVudD48L2RlZnM+ PG1hc2sgaWQ9InRleHQtbWFzay1taTFuMGo2Z3hxZG9hNmFueGwiPjxyZWN0IHg9IjAiIHk9IjAiIHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIGZpbGw9IndoaXRlIj48L3JlY3Q +PGNpcmNsZSBjeD0iMTIwLjc0OTk3NTUxMzQzMzk1IiBjeT0iNzkuNzc5ODQyNjMxMTA3NjgiIHI9IjcuODc1IiBmaWxsPSJibGFjayI+PC9jaXJjbGU+ PGNpcmNsZSBjeD0iMTA0Ljk5OTk5OTk5OTk3NDYyIiBjeT0iNTIuNTAwMDI4Mjc0NjAxNjYiIHI9IjcuODc1IiBmaWxsPSJibGFjayI+PC9jaXJjbGU+ PGNpcmNsZSBjeD0iMTA1LjAwMDAyODI3NDU4ODk3IiBjeT0iMjEuMDAwMDI4Mjc0NjE0MzY1IiByPSI3Ljg3NSIgZmlsbD0iYmxhY2siPjwvY2lyY2xlPjxjaXJjbGUgY3g9IjEzMi4yNzk4MTQzNTY0ODA2IiBjeT0iMzYuNzUwMDUyNzYxMTQyMzMiIHI9IjcuODc1IiBmaWxsPSJibGFjayI +PC9jaXJjbGU+ PGNpcmNsZSBjeD0iNzMuNDk5OTk5OTk5OTg3MyIgY3k9IjUyLjQ5OTk5OTk5OTk4NzMxIiByPSI3Ljg3NSIgZmlsbD0iYmxhY2siPjwvY2lyY2xlPjxjaXJjbGUgY3g9IjQyIiBjeT0iNTIuNDk5OTcxNzI1MzcyOTYiIHI9IjcuODc1IiBmaWxsPSJibGFjayI +PC9jaXJjbGU+ PGNpcmNsZSBjeD0iNzMuNTAwMDI4Mjc0NjAxNjgiIGN5PSIyMSIgcj0iNy44NzUiIGZpbGw9ImJsYWNrIj48L2NpcmNsZT48Y2lyY2xlIGN4PSI3My40OTk5NzE3MjUzNzI5NCIgY3k9IjgzLjk5OTk5OTk5OTk3NDYyIiByPSI3Ljg3NSIgZmlsbD0iYmxhY2siPjwvY2lyY2xlPjwvbWFzaz48c3R5bGU + CiAgICAgICAgICAgICAgICAuZWxlbWVudC1taTFuMGo2Z3hxZG9hNmFueGwgewogICAgICAgICAgICAgICAgICAgIGZvbnQ6IDE0cHggSGVsdmV0aWNhLCBBcmlhbCwgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgICAgICBhbGlnbm1lbnQtYmFzZWxpbmU6ICdtaWRkbGUnOwogICAgICAgICAgICAgICAgfQogICAgICAgICAgICAgICAgLnN1Yi1taTFuMGo2Z3hxZG9hNmFueGwgewogICAgICAgICAgICAgICAgICAgIGZvbnQ6IDguNHB4IEhlbHZldGljYSwgQXJpYWwsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICAgICB9CiAgICAgICAgICAgIDwvc3R5bGU +PGcgbWFzaz0idXJsKCN0ZXh0LW1hc2stbWkxbjBqNmd4cWRvYTZhbnhsKSI+ PGxpbmUgeDE9IjEwNC45OTk5OTk5OTk5NzQ2MiIgeTE9IjUyLjUwMDAyODI3NDYwMTY2IiB4Mj0iMTIwLjc0OTk3NTUxMzQzMzk1IiB5Mj0iNzkuNzc5ODQyNjMxMTA3NjgiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFuMGo2Z3hxZG9hNmFueGwtMScpIj48L2xpbmU + PGxpbmUgeDE9IjEwNC45OTk5OTk5OTk5NzQ2MiIgeTE9IjUyLjUwMDAyODI3NDYwMTY2IiB4Mj0iMTA1LjAwMDAyODI3NDU4ODk3IiB5Mj0iMjEuMDAwMDI4Mjc0NjE0MzY1IiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbjBqNmd4cWRvYTZhbnhsLTMnKSI + PC9saW5lPjxsaW5lIHgxPSIxMDQuOTk5OTk5OTk5OTc0NjIiIHkxPSI1Mi41MDAwMjgyNzQ2MDE2NiIgeDI9IjEzMi4yNzk4MTQzNTY0ODA2IiB5Mj0iMzYuNzUwMDUyNzYxMTQyMzMiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFuMGo2Z3hxZG9hNmFueGwtNScpIj48L2xpbmU + PGxpbmUgeDE9IjczLjQ5OTk5OTk5OTk4NzMiIHkxPSI1Mi40OTk5OTk5OTk5ODczMSIgeDI9IjEwNC45OTk5OTk5OTk5NzQ2MiIgeTI9IjUyLjUwMDAyODI3NDYwMTY2IiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbjBqNmd4cWRvYTZhbnhsLTcnKSI + PC9saW5lPjxsaW5lIHgxPSI0MiIgeTE9IjUyLjQ5OTk3MTcyNTM3Mjk2IiB4Mj0iNzMuNDk5OTk5OTk5OTg3MyIgeTI9IjUyLjQ5OTk5OTk5OTk4NzMxIiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbjBqNmd4cWRvYTZhbnhsLTknKSI + 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 + PC9nPjxnPjx0ZXh0IHg9IjExNS40OTk5NzU1MTM0MzM5NSIgeT0iODUuMDI5ODQyNjMxMTA3NjgiIGNsYXNzPSJlbGVtZW50LW1pMW4wajZneHFkb2E2YW54bCIgZmlsbD0iY3VycmVudENvbG9yIiBzdHlsZT0idGV4dC1hbmNob3I6IHN0YXJ0OyB3cml0aW5nLW1vZGU6IGhvcml6b250YWwtdGI7IHRleHQtb3JpZW50YXRpb246IG1peGVkOyBsZXR0ZXItc3BhY2luZzogbm9ybWFsOyBkaXJlY3Rpb246IGx0cjsiPjx0c3Bhbj5ZPC90c3Bhbj48L3RleHQ + PHRleHQgeD0iMTIwLjc0OTk3NTUxMzQzMzk1IiB5PSI3OS43Nzk4NDI2MzExMDc2OCIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iMTA0Ljk5OTk5OTk5OTk3NDYyIiB5PSI0NC42MjUwMjgyNzQ2MDE2NiIgY2xhc3M9ImVsZW1lbnQtbWkxbjBqNmd4cWRvYTZhbnhsIiBmaWxsPSJjdXJyZW50Q29sb3IiIHN0eWxlPSJ0ZXh0LWFuY2hvcjogc3RhcnQ7IGdseXBoLW9yaWVudGF0aW9uLXZlcnRpY2FsOiAwOyB3cml0aW5nLW1vZGU6IHZlcnRpY2FsLXJsOyB0ZXh0LW9yaWVudGF0aW9uOiB1cHJpZ2h0OyBsZXR0ZXItc3BhY2luZzogLTFweDsgZGlyZWN0aW9uOiBsdHI7Ij48dHNwYW4 +WTwvdHNwYW4+ PC90ZXh0Pjx0ZXh0IHg9IjEwNC45OTk5OTk5OTk5NzQ2MiIgeT0iNTIuNTAwMDI4Mjc0NjAxNjYiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZAm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9Ijk5Ljc1MDAyODI3NDU4ODk3IiB5PSIyNi4yNTAwMjgyNzQ2MTQzNjUiIGNsYXNzPSJlbGVtZW50LW1pMW4wajZneHFkb2E2YW54bCIgZmlsbD0iY3VycmVudENvbG9yIiBzdHlsZT0idGV4dC1hbmNob3I6IHN0YXJ0OyB3cml0aW5nLW1vZGU6IGhvcml6b250YWwtdGI7IHRleHQtb3JpZW50YXRpb246IG1peGVkOyBsZXR0ZXItc3BhY2luZzogbm9ybWFsOyBkaXJlY3Rpb246IGx0cjsiPjx0c3Bhbj5ZPC90c3Bhbj48L3RleHQ + PHRleHQgeD0iMTA1LjAwMDAyODI3NDU4ODk3IiB5PSIyMS4wMDAwMjgyNzQ2MTQzNjUiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjEyOC4zNDIzMTQzNTY0ODA2IiB5PSI0Mi4wMDAwNTI3NjExNDIzMyIgY2xhc3M9ImVsZW1lbnQtbWkxbjBqNmd4cWRvYTZhbnhsIiBmaWxsPSJjdXJyZW50Q29sb3IiIHN0eWxlPSJ0ZXh0LWFuY2hvcjogc3RhcnQ7IHdyaXRpbmctbW9kZTogaG9yaXpvbnRhbC10YjsgdGV4dC1vcmllbnRhdGlvbjogbWl4ZWQ7IGxldHRlci1zcGFjaW5nOiBub3JtYWw7IGRpcmVjdGlvbjogbHRyOyI + PHRzcGFuPlk8L3RzcGFuPjwvdGV4dD48dGV4dCB4PSIxMzIuMjc5ODE0MzU2NDgwNiIgeT0iMzYuNzUwMDUyNzYxMTQyMzMiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjczLjQ5OTk5OTk5OTk4NzMiIHk9IjYwLjM3NDk5OTk5OTk4NzMxIiBjbGFzcz0iZWxlbWVudC1taTFuMGo2Z3hxZG9hNmFueGwiIGZpbGw9ImN1cnJlbnRDb2xvciIgc3R5bGU9InRleHQtYW5jaG9yOiBzdGFydDsgZ2x5cGgtb3JpZW50YXRpb24tdmVydGljYWw6IDA7IHdyaXRpbmctbW9kZTogdmVydGljYWwtcmw7IHRleHQtb3JpZW50YXRpb246IHVwcmlnaHQ7IGxldHRlci1zcGFjaW5nOiAtMXB4OyBkaXJlY3Rpb246IHJ0bDsgdW5pY29kZS1iaWRpOiBiaWRpLW92ZXJyaWRlOyI + PHRzcGFuPlk8L3RzcGFuPjwvdGV4dD48dGV4dCB4PSI3My40OTk5OTk5OTk5ODk5ODczIiB5PSI1Mi40OTk5OTk5OTk5ODk5ODczMSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iNDcuMjUiIHk9IjU3Ljc0OTk3MTcyNTM3Mjk2IiBjbGFzcz0iZWxlbWVudC1taTFuMGo2Z3hxZG9hNmFueGwiIGZpbGw9ImN1cnJlbnRDb2xvciIgc3R5bGU9InRleHQtYW5jaG9yOiBzdGFydDsgd3JpdGluZy1tb2RlOiBob3Jpem9udGFsLXRiOyB0ZXh0LW9yaWVudGF0aW9uOiBtaXhlZDsgbGV0dGVyLXNwYWNpbmc6IG5vcm1hbDsgZGlyZWN0aW9uOiBydGw7IHVuaWNvZGUtYmlkaTogYmlkaS1vdmVycmlkZTsiPjx0c3Bhbj5aPC90c3Bhbj48L3RleHQ + 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 + 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 + PC90ZXh0Pjx0ZXh0IHg9IjczLjQ5OTk3MTcyNTM3Mjk0IiB5PSI4My45OTk5OTk5OTk5NzQ2MiIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICIj48L3RleHQ +PC9nPjwvc3ZnPg== "/> $A$. Simple Ionic Radius : $W < Q$

A. A. Simple ion radius : $W < Q$
B. B. Boiling point of simplest hydride: $\mathrm { Q } < \mathrm { Z }$
C. C. Compound $\mathrm { YZ } _ { 2 }$ is similar to YZQ molecular structure.
D. D. An aqueous solution of a compound formed by X, Y and Z must be acidic.

Answer

Answer: D

Solution: A. $\mathrm { K } ^ { + }$ and $\mathrm { S } ^ { 2 - }$ have the same electron layer structure, $\mathrm { S } ^ { 2 - }$ has a smaller nuclear charge and a larger radius; B. $\mathrm { H } _ { 2 } \mathrm { O }$ can form hydrogen bonds between molecules and has a higher boiling point than $\mathrm { H } _ { 2 } \mathrm {~S}$; C. The compounds $\mathrm { YZ } _ { 2 }$ and YZQ are $\mathrm { CO } _ { 2 }$ and COS, respectively. Both are triatomic, linear molecules, and the C-atom is located in the center, so their structures are similar; D. The compounds formed by H, C and O can be carboxylic acids, aldehydes and alcohols, and the aqueous solutions of aldehydes and alcohols are neutral; The answer is D.

Question 17

Question 17: 19. It is known that astatine (${ } _ { 85 } \mathrm { At }$) is located in Group VII A of the Sixth...

19. It is known that astatine (${ } _ { 85 } \mathrm { At }$) is located in Group VII A of the Sixth Cycle, the correct statement about astatine (${ } _ { 85 } \mathrm { At }$) and its compounds is

A. A. Stability: HAt>HCl
B. B. Oxidizing properties: $\mathrm { At } _ { 2 } > \mathrm { I } _ { 2 }$
C. C. Reduction: HAt>HCl
D. D. $\mathrm { At } _ { 2 }$ is a white, non-toxic crystal.

Answer

Answer: C

Solution: A. In the halogen group of elements, the more nonmetallic the element, the more stable the hydride, i.e., stability: $\mathrm { HAt } < \mathrm { HCl } , \mathrm { A }$ is incorrect; B. From top to bottom of the same main group, the metallicity increases and the nonmetallicity becomes weaker; that is, monatomic oxidizability: $\mathrm { At } _ { 2 } < \mathrm { I } _ { 2 }$, B is wrong; C. monomer oxidation: $\mathrm { At } _ { 2 } < \mathrm { Cl } _ { 2 }$, then hydride reduction: $\mathrm { HAt } > \mathrm { HCl }$, C is correct; D. As the halogen group goes from top to bottom, the monomers deepen in color, and $\mathrm { At } _ { 2 }$ will not be white crystals, D is wrong;

Question 18

Question 18: 20. If element 116 has been discovered and is in the same group as sulfur, the following inferences ...

20. If element 116 has been discovered and is in the same group as sulfur, the following inferences are correct: (1) $R$ is a metallic element (2) $R$ is a nonmetallic element (3) the chemical formula of its sodium salt is $\mathrm { Na } _ { 2 } \mathrm { R }$ (4) the acidity of its highest-valent oxide is stronger than that of sulfuric acid (5) the outermost electron level has six electrons (5) the acidity of its hydride is stronger than that of sulfuric acid (6) the outermost electron level has six electrons (6) the chemical formula of the sulfur salt is the same as the sulfur salt. hydrate is more acidic than sulfuric acid (5) It has 6 electrons in the outermost electron shell ([INLINE_FORMULA_2]).

A. A. (1) (3) (4)
B. B. (1) (3) (5)
C. C. (2) (4) (5)
D. D. (1) (5)

Answer

Answer: B

Solution: If the seventh cycle is full, the last element of the element number 118, for the zero group elements, the 116th element R is located in the seventh cycle of the periodic table VI A group, for the oxygen group elements. (1) Metallicity increases and nonmetallicity decreases from top to bottom of the same main group, and element 116 is a metallic element, (1) is correct; (2) The metallicity of elements in the same main group gradually increases from top to bottom, and the sixth periodic element is Po, which is a metallic element, so element 116 is definitely a metallic element, not a nonmetallic element, and item (2) is incorrect; (3) According to the above analysis, $R$ is located in Group VI A, with a maximum valence of + 6 valences and a minimum valence of $6 - 8 = - 2$ valences, and the chemical formula of its sodium salt is $\mathrm { Na } _ { 2 } \mathrm { R }$, (3) is correct; (4) The nonmetallicity of elements of the same main group gradually decreases and the metallicity gradually increases from the top to the bottom element, and the hydrides of their highest valence oxides should be alkaline, (4) is incorrect; (5) According to the above analysis, R is located in Group VI A and is an element of the oxygen group, and the number of electrons in the outermost layer of the atom is 6, (5) is correct. In summary, (1) (3) (5) is correct, and (B) is correct. The answer is B. INLINE_FORMULA_5]]) by comparing the size of the fixed period, which can save time and improve the accuracy of the questions.

Question 19

Question 19: 21. The following statements about the periodic table and the periodic law of the elements are false...

21. The following statements about the periodic table and the periodic law of the elements are false.

A. A. The nonmetallicity of the second-period elements increases gradually from Li to F.
B. B. From left to right, the atomic radii of elements in the same periodic main group gradually increase
C. C. Elements of the third period: The highest positive valence of $\mathrm { P } , \mathrm {~S} , \mathrm { Cl }$ increases in order.
D. D. Elements of the main group present the highest positive valence, the value of which is equal to the number of electrons in the outermost shell of the atom of that element

Answer

Answer: B

Solution: A. According to the periodic law of the elements: the atoms of the elements of the same period from left to right, the nonmetallicity of the elements gradually increase can be deduced that the second period elements from Li to F nonmetallicity gradually increase, so A is correct; B. The main group elements of the same period increase in nuclear charge and decrease in atomic radius from left to right, so B is wrong; As the atomic number of $\mathrm { C } , \mathrm { P } , \mathrm {~S} , \mathrm { Cl }$ increases gradually, the highest positive value of $\mathrm { P } , \mathrm {~S} , \mathrm { Cl }$ increases gradually, so C is correct; D. The atoms of nonmetallic elements have several electrons in their outermost layers and form at most a few shared electron pairs, so the highest valence presented is equal to the number of electrons in the outermost layer of the atom, so D is correct. The answer is B.

Question 20

Question 20: 22. It is known that $A , B , C , D$ is one of the first three elements in the periodic table with i...

22. It is known that $A , B , C , D$ is one of the first three elements in the periodic table with increasing atomic numbers, $A , B , C$ is a non-metallic element in the same period and $\mathrm { B } , \mathrm { C }$ has two unpaired electrons, and that the D element has the same number of electrons in the s and p energy levels. The following statements are correct

A. A. Atomic radius: $A > B > C > D$
B. B. First ionization energy: $\mathrm { A } < \mathrm { B } < \mathrm { C } < \mathrm { D }$
C. C. D and C can form compounds with high melting points
D. D. Boiling point of lowest valent gaseous hydride: $A < C < B$

Answer

Answer: C

Solution: A. The greater the number of electron layers the greater the radius, and for the same number of electron layers, the greater the nuclear charge, the smaller the radius; Atomic radius: $\mathrm { Mg } > \mathrm { B } > \mathrm { C } > \mathrm { O }$ , A is wrong; B. the same main group with the atomic number becomes larger, the atomic radius becomes larger, the first ionization energy becomes smaller; the same period with the atomic number becomes larger, the first ionization energy becomes larger, the first ionization energy: $\mathrm { Mg } < \mathrm { B } < \mathrm { C } < \mathrm { O }$, B error; C. D and C can form the high melting point compound magnesium oxide, C is correct; D. Simple hydride water can form hydrogen bonds leading to a higher boiling point, so the boiling point of simple lowest valence gaseous hydride water is greater than that of methane, D error ;

Question 21

Question 21: 23. The following law of progression is correct for the halogen elements as the atomic number increa...

23. The following law of progression is correct for the halogen elements as the atomic number increases ( )

A. A. Gradual decrease in melting and boiling points of monomers
B. B. The oxidizing properties of the monomers gradually increase
C. C. Gradual increase in atomic radius
D. D. Gradual increase in the stability of gaseous hydrides

Answer

Answer: C

Solution: A. With the increase of atomic number, the relative molecular mass of the halogen group elements increases gradually, van der Waals force increases gradually, and the melting and boiling points of the monomers increase gradually, so A is wrong; B. From top to bottom of the same main group, the non-metallic nature of the element is gradually weakened, and the oxidizing property of the monomers is gradually weakened, so B is wrong; C. The atom of the same main group of elements from top to bottom, the number of electron layers gradually increased, the atomic radius gradually increased, so C is correct; D. with the main group from top to bottom, the non-metallic nature of the elements gradually weakened, the stability of gaseous hydrides gradually weakened, the D error; answer C.

Question 22

Question 22: 24. According to the periodic law of the elements, the following statements are incorrect

24. According to the periodic law of the elements, the following statements are incorrect

A. A. Be atoms have a weaker ability to lose electrons than Ca atoms
B. B. K reacts more violently with water than Mg does with water.
C. C. HCl is more stable than HBr
D. D. $\mathrm { H } _ { 2 } \mathrm { SeO } _ { 4 }$ is more acidic than $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$.

Answer

Answer: D

Solution: A, Be and Ca are elements of the same main group, due to the metallicity: $\mathrm { Ca } > \mathrm { Be }$, so the Be atoms are more capable of losing electrons than A is correct; B. Because of metallicity: $\mathrm { K } > \mathrm { Mg }$, K loses electrons more easily than Mg, so the reaction between K and water is more violent than that between Mg and water, so B is correct; C. $\mathrm { Cl } , \mathrm { Br }$ is an element of the same main group, and the nonmetallicity of the element: $\mathrm { Cl } > \mathrm { Br }$, so the stability: $\mathrm { HCl } > \mathrm { HBr }$, so C is correct; D. Because of nonmetallicity: $\mathrm { Se } < \mathrm { S }$, the acidity of the hydride corresponding to the highest valence oxide: $\mathrm { H } _ { 2 } \mathrm { SeO } _ { 4 } < \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$, so D is wrong. The answer is D. [点睛]本题考查元素周期律的应用,注意掌握同主族,同周期元素性质的递变规律. 25 . [Knowledge Points] understanding and application of the law of variation of the properties of the elements of the same period, understanding and application of the law of variation of the properties of the elements of the same main group, understanding and application of the law of variation of the metallicity and non-metallicity of the elements. [Detailed Explanation]A. Metallicity gradually decreases from left to right in the same period, then metallicity: $\mathrm { Na } > \mathrm { Mg } > \mathrm { Al }$, A error; B.The atomic radius gradually increases from top to bottom of the same main group, then the atomic radius: $\mathrm { I } > \mathrm { Br } > \mathrm { Cl } , \mathrm { B }$ is correct; C. The more nonmetallic, the more stable the hydride. The non-metallicity gradually decreases from top to bottom in the same main group, then the hydride stability: $\mathrm { H } _ { 2 } \mathrm { O } > \mathrm { H } _ { 2 } \mathrm {~S} > \mathrm { H } _ { 2 } \mathrm { Se }$ , C is correct; D. The stronger the nonmetallicity, the stronger the acidity of the highest-valued oxygen-containing acid, nonmetallicity $\mathrm { Cl } > \mathrm { S } > \mathrm { P }$, then acidity: $\mathrm { HClO } _ { 4 } > \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } > \mathrm { H } _ { 3 } \mathrm { PO } _ { 4 }$ , D is correct; Answer choice A.

Question 23

Question 23: 25. The following law of progression of the properties of an element is not correct.

25. The following law of progression of the properties of an element is not correct.

A. A. Metal properties: Na>Al>Mg
B. B. Atomic radius: $\mathrm { I } > \mathrm { Br } > \mathrm { Cl }$
C. C. Hydride stability: $\mathrm { H } _ { 2 } \mathrm { O } > \mathrm { H } _ { 2 } \mathrm {~S} > \mathrm { H } _ { 2 } \mathrm { Se }$
D. D. Acidity: $\mathrm { HClO } _ { 4 } > \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } > \mathrm { H } _ { 3 } \mathrm { PO } _ { 4 }$

Answer

Answer: A

Question 24

Question 24: 26. All atoms in the following molecules satisfy the structure with 8 electrons in the outermost lay...

26. All atoms in the following molecules satisfy the structure with 8 electrons in the outermost layer ( ) (1) $\mathrm { BeCl } _ { 2 }$ (2) $\mathrm { PCl } _ { 5 }$ (3) $\mathrm { SF } _ { 6 }$ (4) $\mathrm { CS } _ { 2 }$ (5) $\mathrm { CH } _ { 4 }$ (6) $\mathrm { SiO } _ { 2 } ( 7 ) \mathrm { CF } _ { 2 } \mathrm { Cl } _ { 2 }$ (7) $\mathrm { SiO } _ { 2 } ( 7 ) \mathrm { CF } _ { 2 } \mathrm { Cl } _ { 2 }$ (7) $\mathrm { SiO } _ { 2 } ( 7 ) \mathrm { CF } _ { 2 } \mathrm { Cl } _ { 2 }$ (8) FORMULA_5]]

A. A. (1) (2) (5)
B. B. (4) (6) (7)
C. C. (3) (4) (7)
D. D. (3) (4) (6)

Answer

Answer: B

Solution: (1) The element beryllium in $\mathrm { BeCl } _ { 2 }$ has 2 electrons in the outermost layer with a valence of + 2, so the total number of electrons in the outermost layer is $2 + 2 = 4$, so it is wrong; (2) The number of electrons in the outermost layer of the atom phosphorus in $\mathrm { PCl } _ { 5 }$ is 5, and the valence of the element is + 5, so the total number of electrons in the outermost layer is [[[]]]. INLINE_FORMULA_3]] electrons, so it is wrong; (3) The sulfur atom in $\mathrm { SF } _ { 6 }$ has 6 electrons in the outermost layer, and the valence is + 6, so the total number of $6 + 6 = 12$ electrons in the outermost layer is wrong; (4) The number of electrons in the outermost layer of $\mathrm { CS } _ { 2 }$ is 5, and the valence is + 5, so there are a total of $6 + 6 = 12$ electrons in the outermost layer. In $\mathrm { CS } _ { 2 }$, the carbon atom has $4 + 4 = 8$ and the sulfur atom has $6 + 2 = 8$ electrons in its outermost layer, so it is correct; (5) In $\mathrm { CH } _ { 4 }$, the hydrogen atom has 2 electrons in its outermost layer, so it is wrong; (6) In $\mathrm { SiO } _ { 2 }$, the silicon atom has $\mathrm { SiO } _ { 2 }$ electrons in its outermost layer. In $\mathrm { SiO } _ { 2 }$, the silicon atom has $4 + 4 = 8$ and the oxygen atom has $6 + 2 = 8$, so it is correct; (7) The carbon atom in $\mathrm { CF } _ { 2 } \mathrm { Cl } _ { 2 }$ has $4 + 4 = 8$ and $7 + 1 = 8$ in $7 + 1 = 8$. $4 + 4 = 8$, the number of outermost electrons of fluorine atom is $7 + 1 = 8$, and the number of outermost electrons of chlorine atom is $7 + 1 = 8$, so it is correct. Therefore, choose B. [Eyesight] in determining whether all atoms in the molecule to meet the outermost layer of the 8-electron structure, can be based on the following steps to compare the simple, first of all, to see whether the molecule contains hydrogen or rare gas elements, these elements do not meet the 8-electron stabilization of the structure, and secondly, to see whether to write the electronic formulas, if they have been written to write the electronic formulas of the 8-electron stabilization of the structure can be met, and finally through the calculation of the judgment, the atom according to the outermost layer of the Finally, by calculation, according to the outermost layer of the atom, the absolute value of electrons + valence is calculated, and if the result is 8, then 8 electrons are satisfied, otherwise it is not satisfied. This 3-step process can reduce the amount of calculations and increase the rate of doing the problem.

Question 25

Question 25: 27. An experiment using the following setup and reagents can achieve the purpose of the experiment ...

27. An experiment using the following setup and reagents can achieve the purpose of the experiment | ![](/images/questions/atomic-structure/image-003.jpg) | ![](/images/questions/atomic-structure/image-004.jpg) | ![](/images/questions/atomic-structure/image-005.jpg) | ![](/images/questions/atomic-structure/image-006.jpg) | ![](/images/questions/atomic-structure/image-005.jpg) | ![](/images/questions/atomic-structure/image-005.jpg) | ![](/images/questions/atomic-structure/image-006.jpg) | ![](/images/questions/atomic-structure/image-006.jpg) | :--- | :--- | :--- | :--- | | A. Prepare a small amount of $\mathrm { O } _ { 2 }$ | B. Verify nonmetallicity: $\mathrm { Cl } > \mathrm { C } > \mathrm { Si }$ | C. Verify that hydrogen precipitation corrosion occurs in iron nails | D. Prepare a small amount of $\mathrm { NH } _ { 3 }$ | | :--- | :--- | :--- | :--- | :--- | .

A. A. A
B. B. B
C. C. C
D. D. D

Answer

Answer: C

Solution: A. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ is a powdered solid and it is not suitable to use a simple kip generator to achieve the purpose of the experiment, so A does not fit the question; B. To verify the nonmetallicity of $\mathrm { Cl } > \mathrm { C } > \mathrm { Si }$, we should compare the acidity of $\mathrm { HClO } _ { 4 } > \mathrm { H } _ { 2 } \mathrm { CO } _ { 3 } > \mathrm { H } _ { 2 } \mathrm { SiO } _ { 3 }$ with that of $\mathrm { HClO } _ { 4 } > \mathrm { H } _ { 2 } \mathrm { CO } _ { 3 } > \mathrm { H } _ { 2 } \mathrm { SiO } _ { 3 }$, and should not use hydrochloric acid, so we cannot achieve the purpose of the experiment, therefore B does not meet the meaning of the question; C. ammonium chloride ammonium hydrolysis acidic, according to the red ink left low right high verification of iron nails hydrogen corrosion, can achieve the purpose of the experiment, so C meets the meaning of the question; D. When heating the solid, the test tube should be placed downward diagonally, so it can not achieve the purpose of the experiment, so D does not meet the meaning of the question. D. When heating the solid, the test tube should be placed at a downward angle, so the purpose of the experiment cannot be achieved.

Question 26

Question 26: 28. Chemistry is closely related to production, life and society. The following statements are wrong

28. Chemistry is closely related to production, life and society. The following statements are wrong

A. A. Spiral carbon nanotubes TEM and graphene are isomers of each other
B. B. Cement is made from clay and limestone, calcined in a rotary kiln.
C. C. Actively promoting the use of new energy sources, such as solar, wind and geothermal energy, to reduce the use of fossil fuels, practicing the concept of "green mountains are golden mountains".
D. D. Chinese scientists in recent years developed a new type of "continuous fiber toughened" aerospace materials (the main component is made of silicon carbide ceramics and carbon fiber composite), which is lighter than steel, hard, but brittle!

Answer

Answer: D

Solution: A. spiral carbon nanotubes TEM and graphene are carbon monomers, each other isotopes, so A is correct; B. cement is clay and limestone as the main raw material, calcined in the rotary kiln in the cement, and then add gypsum produced, so B is correct; C. Actively promote the use of solar energy, wind energy, geothermal energy and other new energy sources to reduce the use of fossil fuels, in line with the requirements of sustainable development, so C is correct; D. The new "continuous fiber toughened" aerospace materials made of silicon carbide ceramics and carbon fiber composite, this material can increase the toughness, so the material will not have brittle, so D error;

Question 27

Question 27: 29. The following statements about $\mathrm { Na } , \mathrm { Al }$ are correct.

29. The following statements about $\mathrm { Na } , \mathrm { Al }$ are correct.

A. A. Have the same number of outermost electrons
B. B. Both react vigorously with water
C. C. NaOH is less basic than $\mathrm { Al } ( \mathrm { OH } ) _ { 3 }$.
D. D. They are all third-period elements.

Answer

Answer: D

Solution: A. The number of electrons in the outermost layers of Na and Al are 1 and 3, respectively, and A is incorrect; $B$. Sodium reacts violently with water, but aluminum does not react with water, $B$ is incorrect; C. Because sodium is more metallic than aluminum, NaOH is more basic than $\mathrm { Al } ( \mathrm { OH } ) _ { 3 }$, C is incorrect; D. The atomic structures of sodium and aluminum are shown schematically as (-111) ///// $\left. ( - 13 ) \frac { 281 } { 28 } \right)$, and they both have three electron layers outside the nucleus, and they are both third-period elements, D is correct;

Question 28

Question 28: 30. The figure shows a part of the periodic table of the elements, the following description is not ...

30. The figure shows a part of the periodic table of the elements, the following description is not correct | B | C | N | O | F | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | Al | Si | P | S | Cl | Al | Si | P | S | Cl | Ga | Ge | S | S | S | S | S | Ga | Ge | As | Se | Br | In | Sn | S | S | S | S | S | S | S | S | In | Sn | Sb | Te | I | | Tl | Pb | Bi | Po | At |

A. A. The highest positive value for Pb is + 4
B. B. Atomic radius comparison: $\mathrm { Al } > \mathrm { F }$
C. C. Acid strength: $\mathrm { H } _ { 3 } \mathrm { AsO } _ { 4 } < \mathrm { H } _ { 2 } \mathrm { SeO } _ { 4 }$
D. D. The chemical formula of the hydride of the highest valence oxide of Br is $\mathrm { HBrO } _ { 3 }$

Answer

Answer: D

Question 29

Question 29: 31. The following statements are correct

31. The following statements are correct

A. A. Dalton proposed the doctrine of atoms and molecules
B. B. ${ } ^ { 18 } \mathrm { O }$ has an approximate relative atomic mass of 18
C. C. The number of electrons outside the nucleus of an atom is 2 or 8 or 18 in the second outer layer.
D. D. Archaeology often uses ${ } ^ { 13 } \mathrm { C }$ to date artifacts, ${ } ^ { 14 } \mathrm { C }$ to analyze the food structure of ancient humans

Answer

Answer: B

Solution: A.Dalton proposed the atomic theory and Avogadro proposed the molecular theory, so A is wrong; B. ${ } ^ { 18 } \mathrm { O }$ is an oxygen atom with 8 protons and 10 neutrons, and the mass number, 18 protons + neutrons, is the approximate relative atomic mass of the atom, so B is correct; C. The number of electrons in the second outer layer can also be $9 \sim 17$, such as iron's atomic electron arrangement outside the nucleus is $2 , 8 , 14 , 2$, so C is wrong; D.${ } ^ { 13 }$ C is used to analyze the food structure of ancient human beings, ${ } ^ { 14 } \mathrm { C }$ is used to determine the age of cultural relics, so D is wrong;

Question 30

Question 30: 32. W, X, Y, Z for the atomic number of not more than 20 and increase in order of the elements, by t...

32. W, X, Y, Z for the atomic number of not more than 20 and increase in order of the elements, by their composition of the relevant substances can be produced by the reaction of bleach solution, the following is correct

A. A. Non-metallic: $\mathrm { W } > \mathrm { X } > \mathrm { Y }$
B. B. The oxygenated acids of Y are all weak acids
C. C. Atomic radius: $Z > Y > W$
D. D. $\mathrm { ZW } _ { 2 }$ contains covalent bonds

Answer

Answer: C

Solution: A. Among the three nonmetallic elements $\mathrm { W } ( \mathrm { H } ) , \mathrm { X } ( \mathrm { O } ) , \mathrm { Y } ( \mathrm { Cl } )$, $\mathrm { X } ( \mathrm { O } )$ is the most nonmetallic and $\mathrm { W } ( \mathrm { H } )$ is the least nonmetallic, so A is wrong; B. Among the oxygen-containing acids in $\mathrm { Y } ( \mathrm { Cl } )$, HClO is a weak acid and $\mathrm { HClO } _ { 3 }$ is a strong acid; C. The ionic radius $\mathrm { W } ( \mathrm { H } )$ is the smallest, $\mathrm { Z } ( \mathrm { Ca } ) > \mathrm { X } ( \mathrm { O } ) > \mathrm { Y } ( \mathrm { Cl } )$, and C is correct; D. $\mathrm { ZW } _ { 2 }$, that is, $\mathrm { CaH } _ { 2 }$, contains only ionic bonding (between $\mathrm { Ca } ^ { 2 + }$ and $\mathrm { H } ^ { - }$), and does not contain covalent bonding, so option D is wrong;

Question 31

Question 31: 33. The following narratives are correct

33. The following narratives are correct

A. A. Water solubility: $\mathrm { SO } _ { 2 } > \mathrm { NH } _ { 3 } > \mathrm { Cl } _ { 2 }$
B. B. Thermal stability: $\mathrm { HF } > \mathrm { H } _ { 2 } \mathrm { O } > \mathrm { NH } _ { 3 } > \mathrm { PH } _ { 3 }$
C. C. 18 g of heavy water $\left( \mathrm { D } _ { 2 } \mathrm { O } \right)$ contains $10 \mathrm {~N} _ { \mathrm { A } }$ of electrons
D. D. Under standard conditions, 2.24 L of pentane contains the atomic number $1.7 \mathrm {~N} _ { \mathrm { A } }$

Answer

Answer: B

Solution: A. $\mathrm { NH } _ { 3 }$ can form hydrogen bonds with water molecules, and it is very soluble in water, and its solubility in water is greater than that of $\mathrm { SO } _ { 2 }$, which is wrong; B. Non-metallicity: $\mathrm { F } > \mathrm { O } > \mathrm { N } > \mathrm { P }$, covalent bond energy: $\mathrm { H } - \mathrm { F } > \mathrm { H } - \mathrm { O } > \mathrm { H } - \mathrm { N } > \mathrm { H } - \mathrm { P }$, then thermal stability: $\mathrm { HF } > \mathrm { H } _ { 2 } \mathrm { O } > \mathrm { NH } _ { 3 } > \mathrm { PH } _ { 3 }$ , B is correct ; C. The $\mathrm { D } _ { 2 } \mathrm { O }$ molecule contains 10 electrons, and the amount of heavy water $\left( \mathrm { D } _ { 2 } \mathrm { O } \right)$ substance in 18 g of water is $\frac { 18 \mathrm {~g} } { 20 \mathrm {~g} / \mathrm { mol } } = 0.9 \mathrm {~mol}$, the number of electrons contained in it is $9 N _ { A }$ , C is wrong; D. Pentane is not a gas at standard conditions and cannot be calculated by molar volume of gas, D is wrong;

Question 32

Question 32: 34. The following setup and principles do not accomplish the purpose of the experiment ![](/images/...

34. The following setup and principles do not accomplish the purpose of the experiment ![](/images/questions/atomic-structure/image-007.jpg) (1) (2) Saturated $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 }$ solution ![](/images/questions/atomic-structure/image-008.jpg) (3) ![](/images/questions/atomic-structure/image-009.jpg) (4)

A. A. (1) Laboratory preparation of $\mathrm { Cl } _ { 2 }$
B. B. (2) Remove a small amount of HCl gas from $\mathrm { CO } _ { 2 }$.
C. C. (3) Verification of non-metallicity: S>C>Si
D. D. (4) Iodine extraction by sublimation

Answer

Answer: B

Question 33

Question 33: 36. The International Commission on Inorganic Chemical Nomenclature (ICIN) decided in 1989 to abolis...

36. The International Commission on Inorganic Chemical Nomenclature (ICIN) decided in 1989 to abolish the main and subgroups of the long form of the periodic table as well as the number of the group, and to replace them with columns $1 \sim 18$ from left to right, e.g., column 1 for group IA and column 18 for group Zero. According to this rule, the following statements are not correct

A. A. There are definitely no non-metallic elements in column 2.
B. B. The hydrides corresponding to the highest valence oxides of the elements in column 15 are $\mathrm { HRO } _ { 3 }$
C. C. The first element of column 17 has no supreme oxygen-containing acid
D. D. Column 3 has the largest number of elemental species and column 14 has the largest number of compounds formed by elements

Answer

Answer: B

Solution: A. Column 2 is in group IIA of the periodic table, and there are no nonmetallic elements; B. Column 15 is a group VA element in the periodic table, in which the hydride corresponding to the highest valence oxide of phosphorus is $\mathrm { H } _ { 3 } \mathrm { RO } _ { 4 } , \mathrm {~B}$ incorrect; C. The 17th element is an element of group VIIA in the periodic table, and the first element is fluorine, which has no highest-valent oxygenated acid, C is correct; D. The elements in column 3 contain the largest number of hook and hook system species, and the elements in column 14 contain the largest number of organic compounds formed by carbon, D is correct; The answer is B D. Column 3 contains the largest number of lanthanides and steel systems, and column 14 contains the largest number of organic compounds formed from carbon.

Question 34

Question 34: 37 . The following statements are incorrect ![](/images/questions/atomic-structure/image-010.jpg) ...

37 . The following statements are incorrect ![](/images/questions/atomic-structure/image-010.jpg) ) has only one monochlorine substituent

A. A. Lactic acid $\left( \mathrm { C } _ { 3 } \mathrm { H } _ { 6 } \mathrm { O } _ { 3 } \right)$ and glucose $\left( \mathrm { C } _ { 6 } \mathrm { H } _ { 12 } \mathrm { O } _ { 6 } \right)$ have the same simplest formula and are homologues of each other
B. B. Neopentane (
C. C. Glycine and nitroethane are isomers of each other
D. D. Different monomers of the same element formed due to differences in the number of atoms and the arrangement of atoms are known as isoforms

Answer

Answer: A

Solution: A. Lactic acid ($\mathrm { C } _ { 3 } \mathrm { H } _ { 6 } \mathrm { O } _ { 3 }$) contains hydroxyl and carboxyl groups, and glucose ($\mathrm { C } _ { 6 } \mathrm { H } _ { 12 } \mathrm { O } _ { 6 }$) contains hydroxyl and aldehyde groups, which are not homologous because of their different functional groups, so A is wrong; B. Neopentane ( <img class="imgSvg" id = "mi1n0j6ntanv0l97xs" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW4wajZudGFudjBsOTd4cyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTM5IDU3Ljc1MDAxODM2NDg5NzE3IiBzdHlsZT0id2lkdGg6IDEzOC41NTk2MDA0Mzg0MDcyNHB4OyBoZWlnaHQ6IDU3Ljc1MDAxODM2NDg5NzE3cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW4wajZudGFudjBsOTd4cy0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjY5LjI3OTgxMDgyMjE4NDAyIiB5MT0iMjEiIHgyPSI5Ni41NTk2MDA0Mzg0MDcyNyIgeTI9IjM2Ljc1MDAxODM2NDg5NzE3Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFuMGo2bnRhbnYwbDk3eHMtMyIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI0MiIgeTE9IjM2Ljc0OTk4MTYzNTA5NTY5IiB4Mj0iNjkuMjc5ODEwODIyMTg0MDIiIHkyPSIyMSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+PC9saW5lYXJHcmFkaWVudD48L2RlZnM+ PG1hc2sgaWQ9InRleHQtbWFzay1taTFuMGo2bnRhbnYwbDk3eHMiPjxyZWN0IHg9IjAiIHk9IjAiIHdpZHRoPSIxMDAlIiBoZWlnaHQ9IjEwMCUiIGZpbGw9IndoaXRlIj48L3JlY3Q + 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 + PGxpbmUgeDE9IjQyIiB5MT0iMzYuNzQ5OTgxNjM1MDk1NjkiIHgyPSI2OS4yNzk4MTA4MjIxODQwMiIgeTI9IjIxIiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbjBqNm50YW52MGw5N3hzLTMnKSI + PC9saW5lPjwvZz48Zz48dGV4dCB4PSI5Ni41NTk2MDA0Mzg0MDcyNyIgeT0iMzYuNzUwMDE4MzY0ODk3MTciIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjY5LjI3OTgxMDgyMjE4NDAyIiB5PSIyMSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iNDIiIHk9IjM2Ljc0OTk4MTYzNTA5NTY5IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48L2c +PC9zdmc+"/> ) has only one equivalent hydrogen and only one monochlorine substituent, so B is correct; C. Glycine ( $\mathrm { H } _ { 2 } \mathrm {~N} - \mathrm { CH } _ { 2 } - \mathrm { COOH }$) and nitroethane ( $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } - \mathrm { NO } _ { 2 }$) have the same molecular formula ( $\left. \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { NO } _ { 2 } \right)$) and different structures and are isomers of each other, so C is correct; D. Different monomers of the same element are called isomers of each other, such as diamond and graphite, so D is correct.

Question 35

Question 35: 38. According to the following experimental operations, phenomena and the conclusions obtained corre...

38. According to the following experimental operations, phenomena and the conclusions obtained correct is | Select Item | Experimental Procedure and Phenomenon | Explanation or Conclusion | | :--- | :--- | :--- | | A | Industrial preparation of crude silicon with $\mathrm { SiO } _ { 2 }$ and coke at high temperatures, generating the gas CO , obtaining a black solid crude silicon | Non-metallicity: C > Si | B | put a piece of aluminum polished with sandpaper into a test tube, and then add $98 \%$ concentrated sulfuric acid 3 mL, the surface of the aluminum bar no obvious phenomenon | Aluminum and concentrated sulfuric acid at room temperature does not react | C | dilute nitric acid and coke to prepare crude silicon at high temperature conditions to produce gas CO C | Add dilute nitric acid to an excess of iron powder, react fully and then add drops of KSCN solution: gas is generated, the solution is not blood red | Dilute nitric acid can only oxidize Fe to $\mathrm { Fe } ^ { 2 + }$ | D | Add 3 mL of concentrated sulfuric acid to a beaker of $98 \%$. | D | add about 20 g of finely ground $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 } \cdot 8 \mathrm { H } _ { 2 } \mathrm { O }$ crystals and $10 \mathrm { gNH } _ { 4 } \mathrm { Cl }$ crystals to a beaker and place the beaker on a sheet of glass with water dripping on it, stirring rapidly with a glass rod; irritating smelling gas is produced and ice forms on the bottom of the beaker, sticking to the sheet of glass | the reaction of adsorption of heat does not have to be heated in order to take place | the reaction of absorption of heat does not have to be heated in order for it to proceed

A. A. A
B. B. B
C. C. C
D. D. D

Answer

Answer: D

Solution: A. In the reaction between coke and $\mathrm { SiO } _ { 2 }$ at high temperature $2 \mathrm { C } + \mathrm { SiO } _ { 2 } \xlongequal { \text { high temperature } } 2 \mathrm { CO } \uparrow + \mathrm { Si }$, the C elemental valence increases while the Si elemental valence decreases, and the reaction can't be concluded, so A is wrong; B. Aluminum and $98 \%$ concentrated sulfuric acid at room temperature passivation reaction, so B error; C. Nitric acid oxidizes iron to $\mathrm { Fe } ^ { 3 + }$, excess Fe and $\mathrm { Fe } ^ { 3 + }$ occurs $\mathrm { Fe } ^ { + } 2 \mathrm { Fe } ^ { 3 + } = 3 \mathrm { Fe } ^ { 2 + }$, drops of KSCN solution, the solvent is dissolved. KSCN solution, the solution does not show red color, the conclusion is not correct, so C is wrong; D. $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 } \cdot 8 \mathrm { H } _ { 2 } \mathrm { O }$ reacts with $\mathrm { NH } _ { 4 } \mathrm { Cl }$ at room temperature, and ice forms at the bottom of the beaker, which indicates that the reaction is a heat-absorbing reaction, and the reaction is carried out at room temperature, so it can be concluded that the heat-absorbing reaction doesn't necessarily need to be heated to be carried out, and therefore D is correct; The answer is D.

Question 36

Question 36: 39. The following experimental program design, phenomena and conclusions are correct. | | Experimen...

39. The following experimental program design, phenomena and conclusions are correct. | | Experimental Program | Experimental Phenomena | Experimental Conclusions | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | A | Pass $\mathrm { SO } _ { 2 }$ into purple litmus solution | Purple color fades | $\mathrm { SO } _ { 2 }$ has bleaching properties | | B | NaBr solution Starch KI soluble braid | Left cotton ball turns orange, right cotton ball turns blue | nonmetallic: $\mathrm { Cl } > \mathrm { Br } > \mathrm { I }$ | | :--- | :--- | :--- | :--- | ... | C | take $1 \mathrm {~mL} 0.1 \mathrm {~mol} / \mathrm { LKI }$ solution, add $5 \mathrm {~mL} 0.1 \mathrm {~mol} / \mathrm { LFeCl } _ { 3 }$ solution drop by drop, after full reaction, add KSCN solution drop by drop | solution turns red | $\mathrm { Fe } ^ { 3 + }$ reacts with $\mathrm { I } ^ { - }$ in a reversible way | | D | add $\mathrm { I } ^ { - }$ to the $\mathrm { CuSO } _ { 4 }$ solution. | D | a small piece of sodium is added to the $\mathrm { CuSO } _ { 4 }$ solution | colorless bubbles are produced and a blue precipitate is formed at the bottom of the solution | Na does not react directly with the $\mathrm { CuSO } _ { 4 }$ in the solution |

A. A. A
B. B. B
C. C. C
D. D. D

Answer

Answer: D

Solution: A. Pass $\mathrm { SO } _ { 2 }$ into the purple litmus test solution, $\mathrm { SO } _ { 2 }$ reacts with water to form sulfite, and the solution turns red, and the solution does not fade because $\mathrm { SO } _ { 2 }$ does not bleach acid-base indicators, A is wrong; B. Chlorine can oxidize bromide ions to bromine monomers, so the left cotton ball turns orange, oxidizing: $\mathrm { Cl } _ { 2 } > \mathrm { Br } _ { 2 }$, non-metallic: $\mathrm { Cl } > \mathrm { Br }$; chlorine, bromine monomers (volatile) can oxidize iodide ions to iodine monomers, so the right cotton ball turns blue, since it can't be determined whether it is chlorine oxidizing the iodide ions, The right cotton ball turns blue. Since it is impossible to determine whether chlorine gas oxidized the iodide ion, the bromine monomers generated on the left oxidized the iodide ion, or chlorine gas oxidized the iodide ion together with the bromine monomers, it is impossible to determine the size of the oxidizing properties of $\mathrm { Br } _ { 2 }$ and $\mathrm { I } _ { 2 }$ and the strength of non-metallic properties of Br and I. B is incorrect; C. If the given $\mathrm { FeCl } _ { 3 }$ solution is too much, there must be $\mathrm { Fe } ^ { 3 + }$ in the final solution, and when KSCN solution is added dropwise, $\mathrm { Fe } ^ { 3 + }$ combines with SCN to form the red $\mathrm { Fe } ( \mathrm { SCN } ) _ { 3 }$, and the solution must turn red. Therefore, the reddening of the solution does not indicate that the reaction between $\mathrm { Fe } ^ { 3 + }$ and $\mathrm { I } ^ { - }$ is reversible, C is wrong; D. When a small piece of sodium is added to the $\mathrm { CuSO } _ { 4 }$ solution, sodium reacts with water to form sodium hydroxide and hydrogen, resulting in colorless bubbles, and sodium hydroxide reacts with copper sulfate to form a precipitate of sodium sulfate and copper hydroxide, resulting in a blue precipitate at the bottom of the solution, which indicate that Na does not react with $\mathrm { CuSO } _ { 4 }$ in the solution directly. 14]] in the solution to produce copper, D is correct;

Question 37

Question 37: 40. The short-period main group elements $W , X , Y , Z$ have increasing atomic numbers, and the ato...

40. The short-period main group elements $W , X , Y , Z$ have increasing atomic numbers, and the atomic $W$ orbitals in the ground state $2 p$ are in the half-period. The $2 p$ orbitals of atoms in the ground state $W$ are in a half-filled state, atoms in the ground state X have only one pair of paired electrons in the 2p energy level, and atoms in the ground state Y have only one state of motion of the outermost electrons. The element Z is in the same main group as X. The following statements are incorrect High School Chemistry Assignment, October 31, 2025

A. A. First ionization energy: $W > X > Y$
B. B. Electronegativity: $\mathrm { Y } > \mathrm { W } > \mathrm { Z }$
C. C. Simple Ion Radius: $\mathrm { Z } > \mathrm { X } > \mathrm { Y }$
D. D. Boiling point of simplest hydride: $X > W$

Answer

Answer: B

Atomic Structure and Periodic Law

Topic Overview

Key Points

Study Tips

Practicing topics ≠ Passing the exam

Related Learning Resources