Chemical Bonding and Intermolecular Forces

Question 1

Question 1: 1. The reaction $^ { \mathrm { PbS } + 4 \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } = \mathrm { PbS...

1. The reaction $^ { \mathrm { PbS } + 4 \mathrm { H } _ { 2 } \mathrm { O } _ { 2 } = \mathrm { PbSO } _ { 4 } + 4 \mathrm { H } _ { 2 } \mathrm { O } }$ can be used for mural restoration. The following statement is correct

A. A. The structure of $\mathrm { S } ^ { 2 - }$ is schematically shown as (+16) $\begin{aligned} & \text { 春 } \\ & \text {//)} \end{aligned}$
B. B. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ contains both ionic and covalent bonds
C. C. The valence of the element S in $\mathrm { SO } _ { 4 } ^ { 2 - }$ is: + 6
D. D. Pb is in group IV of the periodic table.

Answer

Answer: C

Solution: A. There are 18 electrons outside the nucleus, and its structure is shown as + 16 //), A is wrong; B. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is a covalent compound, which contains only covalent bonds; B. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ is a covalent compound, which contains only covalent bonds, B is wrong. C. The element O in $\mathrm { SO } _ { 4 } ^ { 2 - }$ has a valence of $- 2 , \mathrm {~S}$ and the element $+ 6 , \mathrm { C }$ is correct; D. Pb is in Group IV A of the Sixth Period of the Periodic Table;

Question 2

Question 2: 2. The following processes do not break chemical bonds ()

2. The following processes do not break chemical bonds ()

A. A. Potassium sulfide melting
B. B. Acetic acid dissolves in water
C. C. Ammonium chloride heating
D. D. Sucrose dissolves in water

Answer

Answer: D

Solution: A. Potassium sulfide breaks ionic bonds at high temperatures when melted, so A is wrong; B. Acetic acid dissolved in water will be ionized, destroying covalent bonds, so B is wrong; C. Ammonium chloride will decompose when heated, destroying ionic bonds, so C is wrong; D. Sucrose is a non-electrolyte, dissolved in water does not ionize, did not destroy the chemical bond, so D is correct; Therefore, this question is D.

Question 3

Question 3: 3. The following statements are correct

3. The following statements are correct

A. A. A chemical reaction in which a single substance participates or in which a single substance is produced is not necessarily a redox reaction
B. B. A change in which a chemical bond is broken or a chemical bond is formed must be a chemical change
C. C. If the metallicity of the element A is stronger than that of the element B, then A must displace B in the solution of the salt B.
D. D. It is impossible to make a strong acid from a weak acid, nor is it possible to make a strong acid salt from a weak acid salt

Answer

Answer: A

Solution: A. There are elemental valence changes in the reaction for redox reactions, replacement reactions must be redox reactions, isomers in the conversion of no valence changes, does not belong to the redox reaction, so the A option is correct; B. The essence of the chemical reaction for the old bond breaking, new bond formation, both chemical bond breaking and chemical bond formation changes must be chemical changes, so B error; C. Na is more metallic than Fe, but it can't displace Fe from the salt solution, and the active metal reacts with the water in the salt solution first, so C is wrong; D. Hydrogen sulfuric acid and copper sulfate reaction to generate sulfuric acid, the weak acid may be strong acid; sodium carbonate and sulfuric acid reaction to generate sodium sulfate, the weak acid salt may be strong acid salt, so D error; To sum up, the correct option A.

Question 4

Question 4: 4. The following substances contain both ionic and polar bonding

4. The following substances contain both ionic and polar bonding

A. A. NaOH
B. B. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$
C. C. $\mathrm { MgCl } _ { 2 }$
D. D. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$

Answer

Answer: A

Solution: A. Sodium hydroxide contains both ionic and polar bonds, so A is correct; B. Sodium peroxide contains both ionic and nonpolar covalent bonds, so B is wrong; C. $\mathrm { MgCl } _ { 2 }$ contains only ionic bonds, so C is wrong; D. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ contains only polar and non-polar covalent bonds, so D is wrong; D. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ contains only polar covalent bonds and non-polar covalent bonds.

Question 5

Question 5: 5. In each of the following molecules, all atoms satisfy the 8-electron structure of the outermost l...

5. In each of the following molecules, all atoms satisfy the 8-electron structure of the outermost layer.

A. A. HCl
B. B. $\mathrm { PCl } _ { 3 }$
C. C. $\mathrm { PCl } _ { 5 }$
D. D. $\mathrm { CH } _ { 4 }$

Answer

Answer: B

Solution: A. The hydrogen atom in the HCl molecule has only two electrons in its outermost layer; B. P has 5 valence electrons in the outermost layer and Cl has 7 valence electrons in the outermost layer. P and 3 Cl form three pairs of shared electrons, so that both P and Cl have 8 electrons in the outermost layer, so B is chosen; C. P has 5 valence electrons in the outermost layer and 7 valence electrons in the outermost layer of Cl. P and 5 Cl form $\mathrm { PCl } _ { 5 }$, and it is impossible to reach an 8-electron structure, so C is not chosen; D. The hydrogen atom in the $\mathrm { CH } _ { 4 }$ molecule has only 2 electrons in the outermost layer, so D is not chosen;

Question 6

Question 6: 6. The short-period main group elements $X , Y , Z , W , Q$ have increasing atomic numbers; the oute...

6. The short-period main group elements $X , Y , Z , W , Q$ have increasing atomic numbers; the outermost electrons of the elemental atom of $X$ are two times as many as those of the inner electrons; Y and W are in the same main group, and the algebraic sum of the highest positive and lowest negative valences of W is 4; the periodic number of $Z$ is 1 greater than the group number; the following statements are incorrect. The number of cycles is 1 greater than the number of groups, and the following statements are false.

A. A. Atomic radius: $r ( \mathrm { Y } ) < r ( \mathrm {~W} ) < r ( \mathrm { Z } )$
B. B. Stability of simple gaseous hydrides: $\mathrm { W } < \mathrm { Y }$
C. C. The compounds $\mathrm { XW } _ { 2 } , \mathrm { ZQ } _ { 2 }$ all contain mainly covalent bonds
D. D. The acidity of the hydrides of the highest valence oxides of W and Q: W < Q

Answer

Answer: C

Solution: A. The atomic radii of elements in the same periodic main group decrease from left to right, and increase from top to bottom in the same main group, so the atomic radii $\mathrm { O } < \mathrm { S } < \mathrm { Mg }$, that is $r ( \mathrm { Y } ) < r ( \mathrm {~W} ) < r ( \mathrm { Z } )$, so A is correct; B. The simple gaseous hydrides of W and Y are $\mathrm { H } _ { 2 } \mathrm {~S} , \mathrm { H } _ { 2 } \mathrm { O }$ and non-metallic $\mathrm { S } < \mathrm { O }$, so they are stable $\mathrm { H } _ { 2 } \mathrm {~S} < \mathrm { H } _ { 2 } \mathrm { O }$, so B is correct; C. The compound $\mathrm { XW } _ { 2 } , \mathrm { ZQ } _ { 2 }$ decomposes into $\mathrm { CS } _ { 2 } , \mathrm { MgCl } _ { 2 } , \mathrm { MgCl } _ { 2 }$ as an ionic compound, so C is wrong; D. The highest valence oxides of W and Q are $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } , \mathrm { HClO } _ { 4 }$ and are acidic $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } < \mathrm { HClO } _ { 4 }$ because of the nonmetallicity $\mathrm { S } < \mathrm { Cl }$, so D is correct;

Question 7

Question 7: 7. The following values indicate the atomic numbers of the elements concerned, and the groups of ato...

7. The following values indicate the atomic numbers of the elements concerned, and the groups of atoms indicated can be bonded to each other by ionic bonds to form stable compounds.

A. A. 10 and 12
B. B. 8 vs. 17
C. C. 19 vs. 17
D. D. 6 vs. 14

Answer

Answer: C

Solution: A. Element 10 is Ne and element 12 is magnesium. Ne is very stable and cannot form compounds with magnesium, so A is wrong; B. Element 8 is the element O and element 17 is the element chlorine, and only covalent bonds exist in the covalent compound $\mathrm { ClO } _ { 2 } , \mathrm { ClO } _ { 2 }$ formed by the two, so B is wrong; C. Element 19 is the K element and element 17 is the Cl element, and they form the ionic compound KCl by gaining and losing electrons, and there are only ionic bonds in KCl, so C is correct; D. Element 6 is the C element and element 14 is the Si element, and they form the covalent compound $\mathrm { SiC } , \mathrm { SiC }$ in which only covalent bonds exist, so D is wrong;

Question 8

Question 8: 8. The chemical equation for the explosive reaction of black powder is $3 \mathrm { C } + \mathrm { ...

8. The chemical equation for the explosive reaction of black powder is $3 \mathrm { C } + \mathrm { S } + 2 \mathrm { KNO } _ { 3 } \xlongequal { \text { high temperature } } \mathrm { K } _ { 2 } \mathrm {~S} + 3 \mathrm { CO } _ { 2 } \uparrow + \mathrm { N } _ { 2 } \uparrow$. The following statement is correct ( )

A. A. $\mathrm { K } _ { 2 } \mathrm {~S}$ contains covalent bonds
B. B. $\mathrm { CO } _ { 2 }$ is a covalent compound
C. C. The only oxidizer in this reaction is $\mathrm { KNO } _ { 3 }$
D. D. ${ } ^ { \mathrm { K } _ { 2 } \mathrm {~S} }$ is an oxidation product

Answer

Answer: B

Solution: A. $\mathrm { K } _ { 2 } \mathrm {~S}$ contains only ionic bonds, A is wrong; B. $\mathrm { CO } _ { 2 }$ contains only covalent bonds and is a covalent compound; B. $\mathrm { CO } _ { 2 }$ contains only covalent bonds, so it is a covalent compound, B is correct; C. The elemental valence of $\mathrm { S } , \mathrm {~N}$ decreases in this reaction, so the oxidizing agents are S and $\mathrm { KNO } _ { 3 }$, C is wrong; D. In this reaction, the element $S$ gains electrons from 0-valent to -2-valent, so ${ } ^ { K _ { 2 } S }$ is a product of reduction, D is wrong;

Question 9

Question 9: 9. Chemical terms can express chemical processes, the following chemical terms are expressed incorre...

9. Chemical terms can express chemical processes, the following chemical terms are expressed incorrectly () ![](/images/questions/chem-bonding/image-001.jpg) ![](/images/questions/chem-bonding/image-002.jpg)

A. A. VSEPR modeling of $\mathrm { NH } _ { 3 }$ molecules:
B. B. Oxygen with a neutron number of 10 originates from: ${ } _ { 8 } ^ { 18 } \mathrm { O }$
C. C. Schematic representation of $\mathrm { H } - \mathrm { H }$ of $\mathrm { S } - \mathrm { S } \sigma$ key formation using an electron cloud contour map:
D. D. Formation of $\mathrm { K } _ { 2 } \mathrm {~S}$ in electronic form: $\mathrm { K } \stackrel { ( } { \times } + \cdot \underset { . \cdot + } { \ddot { S } } \times \mathrm { K } \longrightarrow \mathrm { K } ^ { + } [ \underset { \cdot } { \ddot { x } } \times ] ^ { 2 - } \mathrm { K } ^ { + }$

Answer

Answer: A

Solution: A. The lone electron pair of $\mathrm { NH } _ { 3 }$ is $\frac { 5 - 3 \times 1 } { 2 } = 1$, so its VSEPR model is a tetrahedral structure; B. The upper-left corner of the nuclide symbol represents the mass number and the lower-left corner represents the number of protons, the number of protons of O is 8, the mass number $=$ the number of protons + the number of neutrons $= 8 + 10 = 18$, B is correct; C. The electrons in the 1 s energy level outside the nucleus of each H-atom combine to form $\mathrm { s } - \mathrm { s } \sigma$ bonds in a "head-to-head" manner, C is correct; D. In $\mathrm { K } _ { 2 } \mathrm {~S}$, potassium atoms lose electrons and sulfur atoms gain electrons to form ionic bonds, and the electronic formula of the formation process is correct, D is correct;

Question 10

Question 10: 10. The following information about $\mathrm { CH } _ { 4 } , \mathrm { CH } _ { 3 } \mathrm { CH } ...

10. The following information about $\mathrm { CH } _ { 4 } , \mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 }$ and <img class="imgSvg" id = "mi1mzzezvodqt64gre" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16emV6dm9kcXQ2NGdyZSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTM5IDg5LjI1MDAwNjEyMTYzMjM4IiBzdHlsZT0id2lkdGg6IDEzOC41NTk2MDA0Mzg0MTgyN3B4OyBoZWlnaHQ6IDg5LjI1MDAwNjEyMTYzMjM4cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16emV6dm9kcXQ2NGdyZS0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjY5LjI3OTc5NjY4NDg4MjM0IiB5MT0iMzYuNzUwMDA2MTIxNjMzMTgiIHgyPSI5Ni41NTk2MDA0Mzg0MTgyNyIgeTI9IjIxLjAwMDAxMjI0MzI2NzE1NyI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXp6ZXp2b2RxdDY0Z3JlLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNjkuMjc5Nzg5NjE2MjI4NzQiIHkxPSI2OC4yNTAwMDYxMjE2MzIzOCIgeDI9IjY5LjI3OTc5NjY4NDg4MjM0IiB5Mj0iMzYuNzUwMDA2MTIxNjMzMTgiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16emV6dm9kcXQ2NGdyZS01IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjQyIiB5MT0iMjEiIHgyPSI2OS4yNzk3OTY2ODQ4ODIzNCIgeTI9IjM2Ljc1MDAwNjEyMTYzMzE4Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjwvZGVmcz48bWFzayBpZD0idGV4dC1tYXNrLW1pMW16emV6dm9kcXQ2NGdyZSI +PHJlY3QgeD0iMCIgeT0iMCIgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0id2hpdGUiPjwvcmVjdD48L21hc2s+ 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 + PGxpbmUgeDE9IjY5LjI3OTc4OTYxNjIyODc0IiB5MT0iNjguMjUwMDA2MTIxNjMyMzgiIHgyPSI2OS4yNzk3OTY2ODQ4ODIzNCIgeTI9IjM2Ljc1MDAwNjEyMTYzMzE4IiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbXp6ZXp2b2RxdDY0Z3JlLTMnKSI + PC9saW5lPjxsaW5lIHgxPSI0MiIgeTE9IjIxIiB4Mj0iNjkuMjc5Nzk2Njg0ODgyMzQiIHkyPSIzNi43NTAwMDYxMjE2MzMxOCIgc3R5bGU9InN0cm9rZS1saW5lY2FwOnJvdW5kO3N0cm9rZS1kYXNoYXJyYXk6bm9uZTtzdHJva2Utd2lkdGg6MS4yNiIgc3Ryb2tlPSJ1cmwoJyNsaW5lLW1pMW16emV6dm9kcXQ2NGdyZS01JykiPjwvbGluZT48L2c +PGc+ PHRleHQgeD0iOTYuNTU5NjAwNDM4NDE4MjciIHk9IjIxLjAwMDAxMjI0MzI2NzE1NyIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iNjkuMjc5Nzk2Njg0ODgyMzQiIHk9IjM2Ljc1MDAwNjEyMTYzMzE4IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI2OS4yNzk3ODk2MTYyMjg3NCIgeT0iNjguMjUwMDA2MTIxNjMyMzgiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjQyIiB5PSIyMSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgICIj48L3RleHQ +PC9nPjwvc3ZnPg== "/> The narrative of the correct

A. A. can be expressed by the generalized formula $\mathrm { C } _ { \mathrm { n } } \mathrm { H } _ { 2 \mathrm { n } + 2 }$.
B. B. Isomorphic with all alkanes
C. C. They are both alkanes and have both polar and non-polar covalent bonds in their structures
D. D. They are identical in nature.

Answer

Answer: A

Solution: A. The molecular formula of $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 }$ is $\mathrm { C } _ { 2 } \mathrm { H } _ { 6 }$. <img class="imgSvg" id = "mi1mzzf79s0ne0isvs" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16emY3OXMwbmUwaXN2cyIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTM5IDg5LjI1MDAwNjEyMTYzMjM4IiBzdHlsZT0id2lkdGg6IDEzOC41NTk2MDA0Mzg0MTgyN3B4OyBoZWlnaHQ6IDg5LjI1MDAwNjEyMTYzMjM4cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16emY3OXMwbmUwaXN2cy0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjY5LjI3OTc5NjY4NDg4MjM0IiB5MT0iMzYuNzUwMDA2MTIxNjMzMTgiIHgyPSI5Ni41NTk2MDA0Mzg0MTgyNyIgeTI9IjIxLjAwMDAxMjI0MzI2NzE1NyI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXp6Zjc5czBuZTBpc3ZzLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNjkuMjc5Nzg5NjE2MjI4NzQiIHkxPSI2OC4yNTAwMDYxMjE2MzIzOCIgeDI9IjY5LjI3OTc5NjY4NDg4MjM0IiB5Mj0iMzYuNzUwMDA2MTIxNjMzMTgiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16emY3OXMwbmUwaXN2cy01IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjQyIiB5MT0iMjEiIHgyPSI2OS4yNzk3OTY2ODQ4ODIzNCIgeTI9IjM2Ljc1MDAwNjEyMTYzMzE4Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjwvZGVmcz48bWFzayBpZD0idGV4dC1tYXNrLW1pMW16emY3OXMwbmUwaXN2cyI +PHJlY3QgeD0iMCIgeT0iMCIgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0id2hpdGUiPjwvcmVjdD48L21hc2s+ 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 + PGxpbmUgeDE9IjY5LjI3OTc4OTYxNjIyODc0IiB5MT0iNjguMjUwMDA2MTIxNjMyMzgiIHgyPSI2OS4yNzk3OTY2ODQ4ODIzNCIgeTI9IjM2Ljc1MDAwNjEyMTYzMzE4IiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbXp6Zjc5czBuZTBpc3ZzLTMnKSI + PC9saW5lPjxsaW5lIHgxPSI0MiIgeTE9IjIxIiB4Mj0iNjkuMjc5Nzk2Njg0ODgyMzQiIHkyPSIzNi43NTAwMDYxMjE2MzMxOCIgc3R5bGU9InN0cm9rZS1saW5lY2FwOnJvdW5kO3N0cm9rZS1kYXNoYXJyYXk6bm9uZTtzdHJva2Utd2lkdGg6MS4yNiIgc3Ryb2tlPSJ1cmwoJyNsaW5lLW1pMW16emY3OXMwbmUwaXN2cy01JykiPjwvbGluZT48L2c +PGc+ PHRleHQgeD0iOTYuNTU5NjAwNDM4NDE4MjciIHk9IjIxLjAwMDAxMjI0MzI2NzE1NyIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iNjkuMjc5Nzk2Njg0ODgyMzQiIHk9IjM2Ljc1MDAwNjEyMTYzMzE4IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI2OS4yNzk3ODk2MTYyMjg3NCIgeT0iNjguMjUwMDA2MTIxNjMyMzgiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjQyIiB5PSIyMSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgICIj48L3RleHQ +PC9nPjwvc3ZnPg== "/> The molecular formula of [[INL]] is [[INL]]. The molecular formula of $\mathrm { C } _ { 4 } \mathrm { H } _ { 10 }$ is $\mathrm { C } _ { 4 } \mathrm { H } _ { 10 }$ and all three can be represented by the general formula $\mathrm { C } _ { \mathrm { n } } \mathrm { H } _ { 2 \mathrm { n } + 2 }$, A is correct; B. Different monomers composed of the same element are isomers of each other, all three are compounds, and the other alkanes are also chemical compounds. The other alkanes are also chemical compounds and are not isomers of each other; C. All three are alkanes, but methane contains only C-H polar covalent bonds, not non-polar covalent bonds, C error; D. All three are alkanes with similar chemical properties but different physical properties such as melting and boiling points; The answer is A.

Question 11

Question 11: 12. The following substances contain both ionic and covalent bonding

12. The following substances contain both ionic and covalent bonding

A. A. $\mathrm { NH } _ { 4 } \mathrm { Cl }$
B. B. $\mathrm { Na } _ { 2 } \mathrm {~S}$
C. C. $\mathrm { H } _ { 2 } \mathrm { O }$
D. D. KCl

Answer

Answer: A

Solution: A. In $\mathrm { NH } _ { 4 } \mathrm { Cl }$, there is an ionic bond between ${ } ^ { \mathrm { NH } _ { 4 } ^ { + } }$ and $\mathrm { Cl } ^ { - }$, and in $\mathrm { NH } _ { 4 } ^ { + }$, there is a covalent bond between N and H. Therefore, A is correct; B. There is only ionic bonding in $\mathrm { Na } _ { 2 } \mathrm {~S}$, so B is wrong; C. In $\mathrm { H } _ { 2 } \mathrm { O }$, there is only covalent bonding, so C is wrong; D. Only ionic bonds exist in KCl, so D is wrong;

Question 12

Question 12: 13. The following descriptions of chemical terms are incorrect. <img class="imgSvg" id = "mi1mzzf1b...

13. The following descriptions of chemical terms are incorrect. <img class="imgSvg" id = "mi1mzzf1b178ghq1ha" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16emYxYjE3OGdocTFoYSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgODQgNDIiIHN0eWxlPSJ3aWR0aDogODRweDsgaGVpZ2h0OiA0MnB4OyBvdmVyZmxvdzogdmlzaWJsZTsiPjxkZWZzPjwvZGVmcz48bWFzayBpZD0idGV4dC1tYXNrLW1pMW16emYxYjE3OGdocTFoYSI + PHJlY3QgeD0iMCIgeT0iMCIgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0id2hpdGUiPjwvcmVjdD48Y2lyY2xlIGN4PSI0MiIgY3k9IjIxIiByPSI3Ljg3NSIgZmlsbD0iYmxhY2siPjwvY2lyY2xlPjwvbWFzaz48c3R5bGU + CiAgICAgICAgICAgICAgICAuZWxlbWVudC1taTFtenpmMWIxNzhnaHExaGEgewogICAgICAgICAgICAgICAgICAgIGZvbnQ6IDE0cHggSGVsdmV0aWNhLCBBcmlhbCwgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgICAgICBhbGlnbm1lbnQtYmFzZWxpbmU6ICdtaWRkbGUnOwogICAgICAgICAgICAgICAgfQogICAgICAgICAgICAgICAgLnN1Yi1taTFtenpmMWIxNzhnaHExaGEgewogICAgICAgICAgICAgICAgICAgIGZvbnQ6IDguNHB4IEhlbHZldGljYSwgQXJpYWwsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICAgICB9CiAgICAgICAgICAgIDwvc3R5bGU +PGcgbWFzaz0idXJsKCN0ZXh0LW1hc2stbWkxbXp6ZjFiMTc4Z2hxMWhhKSI+ PC9nPjxnPjx0ZXh0IHg9IjQyIiB5PSIxMy4xMjUiIGNsYXNzPSJlbGVtZW50LW1pMW16emYxYjE3OGdocTFoYSIgZmlsbD0iY3VycmVudENvbG9yIiBzdHlsZT0idGV4dC1hbmNob3I6IHN0YXJ0OyBnbHlwaC1vcmllbnRhdGlvbi12ZXJ0aWNhbDogMDsgd3JpdGluZy1tb2RlOiB2ZXJ0aWNhbC1ybDsgdGV4dC1vcmllbnRhdGlvbjogdXByaWdodDsgbGV0dGVyLXNwYWNpbmc6IC0xcHg7IGRpcmVjdGlvbjogbHRyOyI + PHRzcGFuPkM8L3RzcGFuPjx0c3BhbiBzdHlsZT0idW5pY29kZS1iaWRpOiBwbGFpbnRleHQ7Ij5IPHRzcGFuIGJhc2VsaW5lLXNoaWZ0PSJzdWIiIGNsYXNzPSJzdWItbWkxbXp6ZjFiMTc4Z2hxMWhhIj40PC90c3Bhbj48L3RzcGFuPjwvdGV4dD48dGV4dCB4PSI0MiIgeT0iMjEiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI +PC90ZXh0PjwvZz48L3N2Zz4="/>

A. A. Structural formula of methane:
B. B. Diagram of the atomic structure of phosphorus : (+15) $\begin{aligned} & \text { )} \\ & \text {///)} \end{aligned}$
C. C. ${ } ^ { 209 } \mathrm { Bi }$ and ${ } ^ { 210 } \mathrm { Bi }$ are isotopes of each other.
D. D. Formation of NaCl from Na and Cl $: ~ \mathrm { Na } : + \underset { . : } { \mathrm { Cl } } : \rightarrow \mathrm { Na } ^ { + } [ : \underset { . : } { \mathrm { Cl } } : ] ^ { - }$

Answer

Answer: B

Solution: A. Methane is an orthotetrahedral structure containing 4 C-H bonds and the structural formula is <img class="imgSvg" id = "mi1mzzf82hincaq5p1" src="data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16emY4MmhpbmNhcTVwMSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTY2IDU3Ljc1MDA1NTA5NDY5ODY1IiBzdHlsZT0id2lkdGg6IDE2NS44MzkzOTAwNTQ2MzA1NHB4OyBoZWlnaHQ6IDU3Ljc1MDA1NTA5NDY5ODY1cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16emY4MmhpbmNhcTVwMS0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9Ijk2LjU1OTYwMDQzODQwNzI3IiB5MT0iMjEuMDAwMDM2NzI5ODAxNDgiIHgyPSIxMjMuODM5MzkwMDU0NjMwNTQiIHkyPSIzNi43NTAwNTUwOTQ2OTg2NSI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXp6ZjgyaGluY2FxNXAxLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNjkuMjc5Nzg5NjE2MjIzMjUiIHkxPSIzNi43NTAwMTgzNjQ4OTcxNyIgeDI9Ijk2LjU1OTYwMDQzODQwNzI3IiB5Mj0iMjEuMDAwMDM2NzI5ODAxNDgiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16emY4MmhpbmNhcTVwMS01IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjQyIiB5MT0iMjEiIHgyPSI2OS4yNzk3ODk2MTYyMjMyNSIgeTI9IjM2Ljc1MDAxODM2NDg5NzE3Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjwvZGVmcz48bWFzayBpZD0idGV4dC1tYXNrLW1pMW16emY4MmhpbmNhcTVwMSI +PHJlY3QgeD0iMCIgeT0iMCIgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0id2hpdGUiPjwvcmVjdD48L21hc2s+ 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 + PGxpbmUgeDE9IjY5LjI3OTc4OTYxNjIyMzI1IiB5MT0iMzYuNzUwMDE4MzY0ODk3MTciIHgyPSI5Ni41NTk2MDA0Mzg0MDcyNyIgeTI9IjIxLjAwMDAzNjcyOTgwMTQ4IiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbXp6ZjgyaGluY2FxNXAxLTMnKSI + PC9saW5lPjxsaW5lIHgxPSI0MiIgeTE9IjIxIiB4Mj0iNjkuMjc5Nzg5NjE2MjIzMjUiIHkyPSIzNi43NTAwMTgzNjQ4OTcxNyIgc3R5bGU9InN0cm9rZS1saW5lY2FwOnJvdW5kO3N0cm9rZS1kYXNoYXJyYXk6bm9uZTtzdHJva2Utd2lkdGg6MS4yNiIgc3Ryb2tlPSJ1cmwoJyNsaW5lLW1pMW16emY4MmhpbmNhcTVwMS01JykiPjwvbGluZT48L2c +PGc+ PHRleHQgeD0iMTIzLjgzOTM5MDA1NDYzMDU0IiB5PSIzNi43NTAwNTUwOTQ2OTg2NSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ + PHRleHQgeD0iOTYuNTU5NjAwNDM4NDA3MjciIHk9IjIxLjAwMDAzNjcyOTgwMTQ4IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI2OS4yNzk3ODk2MTYyMjMyNSIgeT0iMzYuNzUwMDE4MzY0ODk3MTciIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjQyIiB5PSIyMSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgICAgIj48L3RleHQ +PC9nPjwvc3ZnPg=="/> Therefore, A is correct; B. The number of protons in the diagram is 15, and the number of electrons outside the nucleus is 18, which is the structure of $\mathrm { P } ^ { 3 }$, so B is wrong; C. Isotope refers to the same number of protons, neutron number of different nuclides of the same element, ${ } ^ { 209 } \mathrm { Bi }$ and ${ } ^ { 210 } \mathrm { Bi }$ is an isotope of Bi, so C is correct; D. Na and Cl form NaCl in which Na loses electrons to $\mathrm { Na } ^ { + } , \mathrm { Cl }$ and gains electrons to $\mathrm { Cl } ^ { - }$, which is shown as $\mathrm { Na } \overparen { + } . \underset { . } { \mathrm { C } } \mathrm { i } : \longrightarrow \mathrm { Na } ^ { + } [ : \stackrel { \ddot { \mathrm { C } } \mathrm { i } } { : } : ] ^ { - }$, so D is correct; Therefore, the answer is B. $14 . \mathrm { C }$ [Knowledge Points]the interconversion of quantities of substances, the conceptual judgment of redox reactions, the understanding and application of the law of metallicity and nonmetallicity of elements, covalent bonding [Detailed Explanation]A.Elements of the main group of the same cycle gradually increase in nonmetallicity from left to right, and nonmetallicity $\mathrm { C } < \mathrm { O } , \mathrm { A }$ is wrong; B. 0.1 mol of ${ } ^ { \mathrm { C } _ { 24 } \mathrm { O } _ { 6 } }$ contains $19.2 { } ^ { \mathrm { N } _ { \mathrm { A } } }$ electrons, and 0.1 mol of ${ } ^ { \mathrm { C } _ { 22 } \mathrm { O } _ { 4 } }$ contains $15.6 { } ^ { \mathrm { N } _ { \mathrm { A } } }$ electrons. INLINE_FORMULA_12]] contains $3.6 { } ^ { \mathrm { N } _ { \mathrm { A } } }$ more electrons than 0.1 mol of ${ } ^ { \mathrm { C } _ { 22 } \mathrm { O } _ { 4 } }$, so 0.1 mol of ${ } ^ { \mathrm { C } _ { 22 } \mathrm { O } _ { 4 } }$ contains $3.6 { } ^ { \mathrm { N } _ { \mathrm { A } } }$ more electrons; C. The diagram shows that the $\mathrm { C } _ { 22 } \mathrm { O } _ { 4 }$ molecule contains $\mathrm { C } = \mathrm { O }$ polar bonds and $\mathrm { C } - \mathrm { C } , \mathrm { C } = \mathrm { C } , \mathrm { C } \equiv \mathrm { C }$ nonpolar bonds, C is correct; D. When $\mathrm { C } _ { 24 } \mathrm { O } _ { 6 }$ goes to $\mathrm { C } _ { 22 } \mathrm { O } _ { 4 }$, the elemental valence of carbon decreases and a reduction reaction occurs, D is wrong; The answer is C.

Question 13

Question 13: 14. In recent years, scientists have synthesized cyclic ${ } ^ { \mathrm { C } } { } ^ { 18 }$ molec...

14. In recent years, scientists have synthesized cyclic ${ } ^ { \mathrm { C } } { } ^ { 18 }$ molecules with semiconducting properties, and the method of synthesis is shown below: ![](/images/questions/chem-bonding/image-003.jpg) The following statement is correct

A. A. C is a second-period element with O. Non-metallic C $> \mathrm { O }$
B. B. 0.1 mol of ${ } ^ { \mathrm { C } _ { 24 } \mathrm { O } _ { 6 } }$ has more $0.4 { } ^ { \mathrm { N } _ { \mathrm { A } } }$ electrons than 0.1 mol of ${ } ^ { \mathrm { C } _ { 22 } \mathrm { O } _ { 4 } }$.
C. C. $\mathrm { C } _ { 22 } \mathrm { O } _ { 4 }$ molecule contains both polar and non-polar bonds
D. D. $\mathrm { C } _ { 24 } \mathrm { O } _ { 6 }$ oxidizes to form $\mathrm { C } _ { 22 } \mathrm { O } _ { 4 }$

Answer

Answer: C

Question 14

Question 14: 15. All of the following compounds are covalently bonded ( )

15. All of the following compounds are covalently bonded ( )

A. A. ammonia (gas)
B. B. potassium hydroxide
C. C. barium chloride
D. D. ammonium chloride

Answer

Answer: A

Solution: A. There are covalent bonds between nitrogen and hydrogen atoms in the ammonia molecule; B. In potassium hydroxide, there are ionic bonds between potassium ions and hydroxide ions, and covalent bonds between oxygen atoms and hydrogen atoms. B. B is not selected; C. There are ionic bonds between chlorine atoms and barium atoms in barium chloride, so C is not chosen; D. Ammonium chloride ammonium ions and chloride ions between the existence of ionic bonds, ammonium ions in the N atoms and the existence of covalent bonds between H atoms, so D is not selected.

Question 15

Question 15: 16. Hydrogen fluoride gas is present in: $\mathrm { nHF } _ { ( \mathrm { g } ) } \square \psi ( \ma...

16. Hydrogen fluoride gas is present in: $\mathrm { nHF } _ { ( \mathrm { g } ) } \square \psi ( \mathrm { HF } ) _ { \mathrm { n } ( \mathrm { g } ) } \Delta \mathrm { H } < 0 \quad ( \mathrm { n } = 2 \sim 8 )$ and the following statements are correct.

A. A. HF is more stable than HCl because of the hydrogen bonding between HF molecules.
B. B. The vapor density method for measuring the relative molecular mass of HF is suitable for high temperature and low pressure.
C. C. The chemical bonds present in hydrogen fluoride crystals are ${ } ^ { \mathrm { S } - \mathrm { p } \sigma }$ bonds, hydrogen bonds
D. D. All atoms in $( H F ) _ { n }$ are on the same line.

Answer

Answer: B

Solution: A. As the nonmetallicity of elements of the same main group increases from top to bottom, the stability of their hydrides also becomes more stable. By nonmetallicity: $\mathrm { F } > \mathrm { Cl }$, the stability: $\mathrm { HF } > \mathrm { HCl }$ is not due to the effect of hydrogen bonding, A is wrong; B. $\mathrm { nHF } _ { ( \mathrm { g } ) }$ [in $\downarrow ( \mathrm { HF } ) _ { \mathrm { ng } ( \mathrm { g } ) } \quad \Delta \mathrm { H } < 0$ for the gas volume reduction, exothermic reaction, then reduce the pressure, increase the temperature can make the equilibrium reversed, then the vapor density method of measuring the relative molecular mass of HF is appropriate for high temperature and low pressure, B correct; C. Hydrogen bonding is an intermolecular force, not a chemical bond, C error; D. $( H F ) _ { n }$ is a non-linear hydrogen bond, all atoms are not in the same straight line, D error;

Question 16

Question 16: 17. Hydrogen chloride is incorrectly stated as ( )

17. Hydrogen chloride is incorrectly stated as ( )

A. A. Breaks covalent bonds when dissolved in water
B. B. Can be used for fountain experiments
C. C. Produces white smoke when exposed to ammonia
D. D. Can form crystals through covalent bonds

Answer

Answer: D

Solution: A: HCl is a covalent compound, and when dissolved in water the covalent bonds break and ionize into hydrogen ions and chloride ions; Item B: HCl gas is very soluble in water and can be used in fountain experiments; B: HCl gas is very soluble in water and can be used in fountain experiments, B is correct. C: HCl gas meets ammonia to produce solid particles of ammonium chloride (white smoke), C is correct; C: HCl gas meets ammonia gas to produce solid particles of ammonium chloride (white smoke). D: HCl molecules form molecular crystals by van der Waals forces.

Question 17

Question 17: 18. The following judgments about chemical bonding are correct ( )

18. The following judgments about chemical bonding are correct ( )

A. A. There are chemical bonds in any substance
B. B. Covalent compounds may contain ionic bonds
C. C. Compounds containing covalent bonds are covalent compounds
D. D. Compounds that contain ionic bonds are ionic compounds

Answer

Answer: D

Solution: A. Rare gases are monatomic molecules with no chemical bonding, so A is wrong; B. Compounds containing only covalent bonds are called covalent compounds, and covalent compounds must not contain ionic bonds, so B is wrong; C. Compounds containing only covalent bonds are covalent compounds, so C is wrong; D. A compound containing only ionic bonds is an ionic compound, so D is correct; The answer is D.

Question 18

Question 18: 19. The following substances contain non-polar covalent bonds ()

19. The following substances contain non-polar covalent bonds ()

A. A. NaOH
B. B. $\mathrm { Na } _ { 2 } \mathrm { O }$
C. C. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$
D. D. $\mathrm { CaCl } _ { 2 }$

Answer

Answer: C

Solution: A. NaOH contains ionic and polar bonds, A is wrong; B. $\mathrm { Na } _ { 2 } \mathrm { O }$ contains ionic bonds, B is wrong; C. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ contains ionic and nonpolar bonds, C is right; D. $\mathrm { CaCl } _ { 2 }$ contains ionic bonds, D is wrong, the answer chooses C. Point: understand the meaning of ionic bond and covalent bond, the formation of conditions is the key to answer, generally active metals and active nonmetals are easy to form ionic bonds, nonmetallic elements are easy to form covalent bonds between the atoms, in which the shared electron pairs do not shift is a nonpolar bond, shared electron pairs shifted is a polar bond.

Question 19

Question 19: 20. Each of the following groups has both ionic and covalent bonds ( )

20. Each of the following groups has both ionic and covalent bonds ( )

A. A. $\mathrm { NaOH } \quad \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } \quad \mathrm { NH } _ { 4 } \mathrm { Cl }$
B. B. $\mathrm { MgO } \quad \mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 } \quad \mathrm { HNO } _ { 3 }$
C. C. $\mathrm { HCl } \quad \mathrm { Al } _ { 2 } \mathrm { O } _ { 3 } \quad \mathrm { MgCl } _ { 2 }$
D. D. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 } \quad \mathrm { KOH } \quad \mathrm { Na } _ { 3 } \mathrm { PO } _ { 4 }$

Answer

Answer: D

Solution: A. NaOH contains both ionic and covalent bonds, $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$ contains only covalent bonds, and $\mathrm { NH } _ { 4 } \mathrm { Cl }$ contains both ionic and covalent bonds; B. MgO contains only ionic bonds, $\mathrm { Na } _ { 2 } \mathrm { SO } _ { 4 }$ contains both ionic and covalent bonds, and $\mathrm { HNO } _ { 3 }$ contains only covalent bonds; C. HCl contains only covalent bonds, $\mathrm { Al } _ { 2 } \mathrm { O } _ { 3 }$ contains only ionic bonds and $\mathrm { MgCl } _ { 2 }$ contains only ionic bonds; D. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 } , \mathrm { KOH } , \mathrm { Na } _ { 3 } \mathrm { PO } _ { 4 }$ contains both ionic and covalent bonds. D. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 } , \mathrm { KOH } , \mathrm { Na } _ { 3 } \mathrm { PO } _ { 4 }$ contains both ionic and covalent bonds. Ionic compounds must contain ionic bonds and may contain covalent bonds; covalent compounds must contain covalent bonds and must not contain ionic bonds.

Question 20

Question 20: 21. There is a correspondence between the structure and properties of the following substances or be...

21. There is a correspondence between the structure and properties of the following substances or between the properties and uses of a substance ().

A. A. $\mathrm { FeCl } _ { 3 }$ Brownish yellow solution for etching copper printed circuit boards.
B. B. $\mathrm { H } _ { 2 } \mathrm { O }$ is intermolecularly hydrogen bonded, $\mathrm { H } _ { 2 } \mathrm { O }$ is stable
C. C. HClO is weakly acidic and can be used for sterilization
D. D. $\mathrm { H } - \mathrm { Cl }$ has a bond energy greater than $\mathrm { H } - \mathrm { Br } , \mathrm { HCl }$ has a thermal stability greater than HBr

Answer

Answer: D

Solution: A. $\mathrm { FeCl } _ { 3 }$ solution is used to etch copper printed circuit boards using the oxidizing properties of $\mathrm { Fe } ^ { 3 + }$; B. The stability of $\mathrm { H } _ { 2 } \mathrm { O }$ is related to the bond energy of the $\mathrm { H } - \mathrm { O }$ bonds, but not to the existence of hydrogen bonds between ${ } ^ { \mathrm { H } _ { 2 } \mathrm { O } }$ molecules; C. HClO is used to sterilize and disinfect by using its strong oxidizing property, so C is not chosen; D. The larger the bond energy, the more stable it is. The bond energy of $\mathrm { H } - \mathrm { Cl } ^ { 2 }$ is larger than that of $\mathrm { H } - \mathrm { Br }$, so the thermal stability of HCl is larger than that of HBr, so D. The bond energy of $\mathrm { H } - \mathrm { Cl } ^ { 2 }$ is larger than that of $\mathrm { H } - \mathrm { Br }$; D.

Question 21

Question 21: 22. The following statements are correct

22. The following statements are correct

A. A. The bond types in $\mathrm { CaCl } _ { 2 }$ and $\mathrm { Na } _ { 2 } \mathrm { O }$ are identical.
B. B. Each atom in $\mathrm { CH } _ { 4 }$ has an 8-electron stable structure in its outermost electron layer
C. C. HCl is more stable than ${ } ^ { \mathrm { H } _ { 2 } \mathrm {~S} }$ because the intermolecular forces are stronger in the former than in the latter.
D. D. NaOH is an ionic compound and the substance contains only ionic bonds

Answer

Answer: A

Solution: A. $\mathrm { CaCl } _ { 2 }$ and $\mathrm { Na } _ { 2 } \mathrm { O }$ are both ionic compounds containing only ionic bonds, A is correct; B. The H atom in methane has only two electrons in its outermost layer; C. The stability of HCl and ${ } ^ { \mathrm { H } _ { 2 } \mathrm {~S} }$ is related to the strength of intramolecular bonds, not to intermolecular forces; D. NaOH is an ionic compound consisting of $\mathrm { Na } ^ { + }$ and $\mathrm { OH } ^ { - }$ bonded by ionic bonding, and in the anion $\mathrm { OH } ^ { - }$, [INLINE_FORMULA_6]] atoms of $\mathrm { H } , \mathrm { O }$ are bonded together by ionic bonding, and $\mathrm { OH } ^ { - }$ atoms of $\mathrm { OH } ^ { - }$ and $\mathrm { H } , \mathrm { O }$ are bonded by ionic bonding. are bonded by $\mathrm { H } - \mathrm { O }$ covalent bonds, so the substance contains both ionic and covalent bonds, D is wrong;

Question 22

Question 22: 23. The following substances are ionic compounds containing polar bonds

23. The following substances are ionic compounds containing polar bonds

A. A. HCl
B. B. KOH
C. C. $\mathrm { CaCl } _ { 2 }$
D. D. $\mathrm { O } _ { 2 }$

Answer

Answer: B

Solution: A. HCl is a covalent compound, so A is wrong; B. KOH is an ionic compound, and the hydroxide root contains polar covalent bonds, so B is correct; C. $\mathrm { CaCl } _ { 2 }$ is an ionic compound, and it only contains ionic bonds, so C is wrong; D. $\mathrm { O } _ { 2 }$ is a monomers, and it contains nonpolar bonds, so D is wrong; therefore, B is chosen.

Question 23

Question 23: 24. An electrolyte that contains polar covalent bonds is the

24. An electrolyte that contains polar covalent bonds is the

A. A. $\mathrm { CaCl } _ { 2 }$
B. B. $\mathrm { H } _ { 2 } \mathrm { O }$
C. C. $\mathrm { NH } _ { 3 }$
D. D. $\mathrm { CH } _ { 4 }$

Answer

Answer: B

Solution: A. CaCl2 is an ionic compound containing only ionic bonds and no covalent bonds; B. $\mathrm { H } _ { 2 } \mathrm { O }$ is a covalent compound, $\mathrm { H } - \mathrm { O }$ is a polar covalent bond, and water is an electrolyte; C. $\mathrm { NH } _ { 3 }$ is a covalent compound and $\mathrm { N } - \mathrm { H }$ is a polar covalent bond, but ammonia is a non-electrolyte; D. $\mathrm { CH } _ { 4 }$ is a covalent compound, $\mathrm { H } - \mathrm { C }$ is a polar covalent bond, and $\mathrm { CH } _ { 4 }$ is a non-electrolyte, D is wrong; D is incorrect; B is correct.

Question 24

Question 24: 25. The following substances are covalent compounds

25. The following substances are covalent compounds

A. A. $F _ { 2 }$
B. B. $\mathrm { H } _ { 2 } \mathrm { O }$
C. C. KCl
D. D. $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 }$

Answer

Answer: B

Solution: A. $\mathrm { F } _ { 2 }$ contains only the element F, which is a nonmetallic monomer, not a compound; B. $\mathrm { H } _ { 2 } \mathrm { O }$ contains only covalent bonds and is a covalent compound; C. There are only ionic bonds in KCl molecules, which are ionic compounds; D. There are both ionic and covalent bonds in $\mathrm { Ba } ( \mathrm { OH } ) _ { 2 }$, which is an ionic compound; The answer is B.

Question 25

Question 25: 26. $\mathrm { NO } ^ { - }$ in drinking water is hazardous to human health. In order to reduce the ...

26. $\mathrm { NO } ^ { - }$ in drinking water is hazardous to human health. In order to reduce the concentration of $\mathrm { NO } _ { 3 } ^ { - }$ in drinking water, $\mathrm { NO } ^ { 3 }$ can be reduced to $\mathrm { N } _ { 2 }$ by using aluminum powder under alkaline conditions. The ionic equation is $$ 10 \mathrm { Al } + 6 \mathrm { NO } _ { 3 } ^ { - } + 4 \mathrm { OH } ^ { - } + 18 \mathrm { H } _ { 2 } \mathrm { O } = $$ $10 \left[ \mathrm { Al } ( \mathrm { OH } ) _ { 4 } \right] ^ { - } + 3 \mathrm {~N} _ { 2 } \uparrow$. Let $\mathrm { N } _ { \mathrm { A } }$ be the value of Avogadro's constant, the following statement is false

A. A. Consuming 1 molAl , the number of electrons transferred by the reaction is $3 \mathrm {~N} _ { \mathrm { A } }$
B. B. The number of valence electron pairs of N in $1 \mathrm { molNO } ^ { 3 }$ is $4 \mathrm {~N} _ { \mathrm { A } }$
C. C. In the standard condition, the number of keys in $11.2 \mathrm { LN } _ { 2 }$ containing $\pi$ is $\mathrm { N } _ { \mathrm { A } }$.
D. D. 2 mol of liquid water contains fewer hydrogen bonds than $4 \mathrm {~N} _ { \mathrm { A } }$

Answer

Answer: B

Solution: $\mathrm { A } . \mathrm { Al }$ is converted to ${ } ^ { \left[ \mathrm { Al } ( \mathrm { OH } ) _ { 4 } \right] ^ { - } }$ and the valence increases by 3, then 1 mol of Al is consumed and the number of electrons transferred in the reaction is $3 \mathrm {~N} _ { \mathrm { A } }$, so A is correct; B. The number of valence electron pairs of N in $\mathrm { NO } _ { 3 } ^ { - }$ is $3 + \frac { 5 + 1 - 3 \times 2 } { 2 } = 3$, so the number of valence electron pairs of N in $1 \mathrm { molNO } _ { 3 } ^ { - }$ is $3 \mathrm {~N} _ { \mathrm { A } }$, so B is wrong; C. 1 $\mathrm { N } _ { 2 }$ molecule contains 2 $\pi$ bonds, and the amount of $11.2 \mathrm { LN } _ { 2 }$ substance in the standard condition is $\frac { 11.2 \mathrm {~L} } { 22.4 \mathrm {~L} / \mathrm { mol } } = 0.5 \mathrm {~mol}$, and the number of $\pi$ bonds is $\pi$. The number of $\mathrm { N } _ { \mathrm { A } }$ is $\mathrm { N } _ { \mathrm { A } }$, so C is correct; D. Liquid water has fewer hydrogen bonds than ice, and 1 mol of ice contains 2 mol of hydrogen bonds, so 2 mol of liquid water contains fewer hydrogen bonds than $4 \mathrm {~N} _ { \mathrm { A } }$, so D is correct; The answer is B.

Question 26

Question 26: 27. The following statements are correct

27. The following statements are correct

A. A. A molecule made up of polar bonds must be a polar molecule
B. B. Nonpolar bonds may exist in ionic compounds
C. C. $\mathrm { BCl } _ { 3 } , \mathrm { CH } _ { 4 }$ 8-electron stable structures are satisfied in each atom
D. D. $\mathrm { CH } _ { 4 } , \mathrm { NH } _ { 3 } , \mathrm { H } _ { 2 } \mathrm { O }$ mutual isoelectronics

Answer

Answer: B

Solution: A. A molecule composed of polar bonds, if the positive and negative charge centers overlap, it is a nonpolar molecule, such as carbon dioxide is a linear molecule, carbon and oxygen are polar bonds, but it is a nonpolar molecule, A error; B. Nonpolar bonds may exist in ionic compounds. Sodium peroxide is an ionic compound, but there are $\mathrm { O } - \mathrm { O }$ nonpolar bonds; C. B atoms cannot satisfy an 8-electron stable structure; D. The atomic number is equal, the total number of valence electrons is equal for isoelectronic body, therefore, not each other isoelectronic body, D is wrong; the answer is B.

Question 27

Question 27: 28. The structure of $\mathrm { CS } _ { 2 }$ is similar to $\mathrm { CO } _ { 2 }$, and the wrong ...

28. The structure of $\mathrm { CS } _ { 2 }$ is similar to $\mathrm { CO } _ { 2 }$, and the wrong statement about $\mathrm { CS } _ { 2 }$ is that

A. A. Can be used to remove sulfur adhering to the inside of test tubes
B. B. Contains nonpolar bonds
C. C. Structured as $\mathrm { S } = \mathrm { C } = \mathrm { S }$
D. D. The boiling point of $\mathrm { CS } _ { 2 }$ is higher than that of $\mathrm { CO } _ { 2 }$.

Answer

Answer: B

Solution: A. S is soluble in $\mathrm { CS } _ { 2 }$, and the sulfur adhering to the inner wall of the test tube can be removed with $\mathrm { CS } _ { 2 }$, A is correct; B. $\mathrm { CS } _ { 2 }$ has the structural formula $\mathrm { S } = \mathrm { C } = \mathrm { S }$ and contains polar covalent bonds; C. The structural formula of $\mathrm { CS } _ { 2 }$ is $\mathrm { S } = \mathrm { C } = \mathrm { S }$; D. The relative molecular mass of $\mathrm { CS } _ { 2 }$ is larger, so the boiling point of $\mathrm { CS } _ { 2 }$ is higher than that of $\mathrm { CO } _ { 2 }$. D is correct.

Question 28

Question 28: 29. The following groups of substances are all covalent compounds

29. The following groups of substances are all covalent compounds

A. A. $\mathrm { H } _ { 2 } \mathrm {~S}$ and $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$
B. B. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ and $\mathrm { CaF } _ { 2 }$
C. C. $\mathrm { NH } _ { 3 }$ and $\mathrm { N } _ { 2 }$
D. D. $\mathrm { HNO } _ { 3 }$ and $\mathrm { HClO } _ { 4 }$

Answer

Answer: D

Solution: A. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ is an ionic compound, so A is not chosen; B. $\mathrm { CaF } _ { 2 }$ is an ionic compound, so B is not chosen; C. $N _ { 2 }$ is a monomer, so C is not chosen; D. $\mathrm { HNO } _ { 3 }$ and $\mathrm { HClO } _ { 4 }$ are covalent compounds;

Question 29

Question 29: 31. Fabipiravir is a drug used in the treatment of neocoronaryngitis, the structural short form is s...

31. 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A. A. The polarity of the C-F bond in this molecule is greater than the polarity of the C-N bond
B. B. The highest energy level of the ground state F atom in this molecule has the symbol L
C. C. All C atoms in this molecule are $\mathrm { sp } ^ { 3 }$ heterozygous
D. D. The bond energy of the $\mathrm { C } = \mathrm { C }$ bond in this molecule is greater than the bond energy of the $\mathrm { C } - \mathrm { C }$ bond

Answer

Answer: C

Solution: A. Fluorine is more electronegative than nitrogen, the difference in electronegativity between C and F is greater than the difference in electronegativity between $\mathrm { C } , \mathrm {~N}$, and the polarity of the C-F bond is greater than the polarity of the C-N bond; B. F is located in the second period, there are two electron layers outside the nucleus, and the symbol of the highest energy layer is L. B is correct; C. All carbon atoms in the molecule valence layer electron pairs are 3, are taken $\mathrm { sp } ^ { 2 }$ hybridization, C error; D. The bond energy of a double bond is greater than that of a single bond, and the bond energy of a C=C bond is greater than that of a C-C bond, D is correct.

Question 30

Question 30: 32. Chinese chemists have successfully developed a spherical carbon salt ${ } ^ { \mathrm { C } _ { ...

32. Chinese chemists have successfully developed a spherical carbon salt ${ } ^ { \mathrm { C } _ { 60 } }$ formed by an alkali metal and $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$, which is experimentally measured to be an ionic compound with good superconductivity. The following analysis is correct

A. A. There are only ionic bonds in $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$.
B. B. No covalent bonds in $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$
C. C. The crystal conducts electricity in the molten state
D. D. ${ } ^ { 1 \mathrm { molK } _ { 3 } \mathrm { C } _ { 60 } }$ contains the number of ions as $63 \mathrm {~N} _ { \mathrm { A } }$

Answer

Answer: C

Solution: A. $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$ consists of $\mathrm { K } ^ { + }$ and $\mathrm { C } _ { 60 } ^ { 3 - }$, which are ionic compounds containing both ionic and covalent bonds; B. $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$ consists of $\mathrm { K } ^ { + }$ and $\mathrm { C } _ { 60 } ^ { 3 - }$, which is an ionic compound containing both ionic and covalent bonds; C. $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$ is an ionic compound. In the molten state, the ionic bond is broken and free-moving ions are ionized, so it can conduct electricity. is correct; D. $\mathrm { K } _ { 3 } \mathrm { C } _ { 60 }$ consists of $\mathrm { K } ^ { + }$ and $\mathrm { C } _ { 60 } ^ { 3 - }$, and the number of ions in $1 \mathrm { molK } _ { 3 } \mathrm { C } _ { 60 }$ is ${ } ^ { 4 \mathrm {~N} _ { \mathrm { A } } }$, which is wrong; The answer is C.

Question 31

Question 31: 33. Laboratory commonly used calcium carbide (commonly known as calcium carbide) and water reaction ...

33. Laboratory commonly used calcium carbide (commonly known as calcium carbide) and water reaction to prepare acetylene gas, calcium carbide cell shown in the figure. The following description of the error is ![](/images/questions/chem-bonding/image-004.jpg) $\mathrm { OCa } ^ { 2 + }$ \& $\mathrm { C } _ { 2 } ^ { 2 - }$

A. A. The mass of 1 calcium carbide cell is $\frac { 64 \times 4 } { \mathrm {~N} _ { \mathrm { A } } } \mathrm { g }$
B. B. Calcium carbide has both ionic and covalent bonds and is an ionic compound
C. C. The $\mathrm { C } - \mathrm { H }$ single key in acetylene is the $\mathrm { sp } - \mathrm { s } \sigma$ key.
D. D. In the calcium carbide cell, the nearest $\mathrm { Ca } ^ { 2 + }$ number to ${ } ^ { \mathrm { C } _ { 2 } ^ { 2 - } }$ is 8

Answer

Answer: D

Solution: A. According to the principle of equalization, the number of cells containing $\mathrm { C } _ { 2 } ^ { 2 - }$ is $8 \times \frac { 1 } { 8 } + 6 \times \frac { 1 } { 2 } = 4$ and the number of cells containing $\mathrm { Ca } ^ { 2 + }$ is $12 \times \frac { 1 } { 4 } + 1 = 4$, so the mass of one calcium carbide cell is $\frac { 64 \times 4 } { \mathrm {~N} _ { \mathrm { A } } } \mathrm { g }$, so A is correct. $12 \times \frac { 1 } { 4 } + 1 = 4$, so the mass of one calcium carbide cell is $\frac { 64 \times 4 } { \mathrm {~N} _ { \mathrm { A } } } \mathrm { g }$, so A is correct; B. Calcium carbide consists of $\mathrm { C } _ { 2 } ^ { 2 - } , \mathrm { Ca } ^ { 2 + }$, which has both ionic and covalent bonds, so it is an ionic compound, so B is correct; C. The hybridization of C atoms in acetylene is sp hybridization, the s orbitals of H atoms and the sp hybridized orbitals of C atoms form $\sigma$ bonds, so the $\mathrm { C } - \mathrm { H }$ single bond of acetylene is the $\mathrm { sp } - \mathrm { s } \sigma$ bond, so C is correct; D. According to the diagram, there are 6 $\mathrm { Ca } ^ { 2 + }$ numbers nearest to ${ } ^ { \mathrm { C } _ { 2 } ^ { 2 - } }$ in the calcium carbide cell, so D is wrong; D is incorrect.

Question 32

Question 32: 34. Studies have shown that ${ } ^ { \mathrm { N } _ { 2 } \mathrm { O } }$ reacts with CO under cer...

34. Studies have shown that ${ } ^ { \mathrm { N } _ { 2 } \mathrm { O } }$ reacts with CO under certain conditions with the energy changes and the reaction process shown in the figure, the following statements are correct ( ) ![](/images/questions/chem-bonding/image-005.jpg) Reaction process

A. A. During the reaction, there is breaking and creation of polar covalent and nonpolar bonds
B. B. $\mathrm { FeO } ^ { + }$ makes the reaction easier and is the catalyst for the reaction
C. C. The overall reaction equation for this reaction is $2 \mathrm {~N} _ { 2 } \mathrm { O } + \mathrm { CO } + \mathrm { Fe } ^ { + } = \mathrm { FeO } ^ { + } + 2 \mathrm {~N} _ { 2 } + \mathrm { CO } _ { 2 }$
D. D. $\mathrm { Fe } ^ { + } + \mathrm { N } _ { 2 } \mathrm { O } \rightarrow \mathrm { FeO } ^ { + } + \mathrm { N } _ { 2 } , \quad \mathrm { FeO } ^ { + } + \mathrm { CO } \rightarrow \mathrm { Fe } ^ { + } + \mathrm { CO } _ { 2 }$ Both steps are exothermic reactions

Answer

Answer: D

Solution: A. As can be seen from the diagram, the reaction involves the breaking and formation of polar carbon-oxygen bonds, but not the breaking and formation of nonpolar bonds; B. From the diagram, $\mathrm { Fe } ^ { + }$ is the catalyst for the reaction, and $\mathrm { FeO } ^ { + }$ is the intermediate product of the reaction, B error; C. From the diagram, the total reaction equation for the reaction is $\mathrm { N } _ { 2 } \mathrm { O } + \mathrm { CO } = \mathrm { N } _ { 2 } + \mathrm { CO } _ { 2 }$, C is wrong; D. From the diagram, $\mathrm { Fe } ^ { + } + \mathrm { N } _ { 2 } \mathrm { O } \rightarrow \mathrm { FeO } ^ { + } + \mathrm { N } _ { 2 } , \mathrm { FeO } ^ { + } + \mathrm { CO } \rightarrow \mathrm { Fe } ^ { + } + \mathrm { CO } _ { 2 }$ is a two-step reaction in which the reactants are high in energy and the products are low in energy, and both are exothermic reactions, D is correct;

Question 33

Question 33: 35. The following substances are ionic compounds ()

35. The following substances are ionic compounds ()

A. A. $\mathrm { CH } _ { 3 } \mathrm { COOH }$
B. B. HCl
C. C. $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$
D. D. NaCl

Answer

Answer: D

Solution: A. $\mathrm { CH } _ { 3 } \mathrm { COOH }$ contains only covalent bonds and is a covalent compound, so A is not chosen; B. HCl contains only covalent bonds and is a covalent compound, so B is not chosen; C. $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$ contains only covalent bonds and is a covalent compound, so C is not chosen; D. NaCl consists of $\mathrm { Na } ^ { + } , \mathrm { Cl }$, which contains ionic bonds and is an ionic compound; D.

Question 34

Question 34: 38. The following narratives are correct ( )

38. The following narratives are correct ( )

A. A. Both ionic and covalent bonds are present in $\mathrm { NaOH } , \mathrm { NaCl } , \mathrm { NaClO }$.
B. B. Compounds composed of nonmetallic elements contain only covalent bonds
C. C. All substances contain chemical bonds
D. D. $\mathrm { Na } _ { 2 } \mathrm { O }$ It is the ionic bonds that are broken when melting

Answer

Answer: D

Solution: A. NaCl contains only ionic bonds, not covalent bonds, A is wrong; B. Compounds composed of nonmetallic elements may contain only covalent bonds, such as $\mathrm { H } _ { 2 } \mathrm { O } , \mathrm { HCl }$, etc.; or they may contain both ionic and covalent bonds, such as $\mathrm { NH } _ { 4 } \mathrm { Cl }$, which is an ionic compound formed exclusively by nonmetallic elements, and contains both ionic and covalent bonds, B error; C.Noble gases are substances composed of molecules, an atom is a molecule, and there is no chemical bonding in the substance, C error; D. $\mathrm { Na } _ { 2 } \mathrm { O }$ is an ionic compound formed by $\mathrm { Na } ^ { + }$ and $\mathrm { O } ^ { 2 - }$ through ionic bonding, and the ionic bonding is broken when it is melted to free-moving $\mathrm { Na } ^ { + }$ and $\mathrm { O } ^ { 2 - }$. FORMULA_6]] , D is correct ; Therefore, the reasonable choice is D.

Question 35

Question 35: 39. The following statements are incorrect ( )

39. The following statements are incorrect ( )

A. A. $\mathrm { CH } _ { 4 } , \mathrm { NH } _ { 3 } , \mathrm { CO } _ { 2 }$ The bond angles in the molecule increase in sequence
B. B. $\mathrm { HCl } , \mathrm { HBr } , \mathrm { HI }$ The bond lengths in the molecule increase sequentially
C. C. $\mathrm { H } _ { 2 } \mathrm { O } , \mathrm { H } _ { 2 } \mathrm {~S} , \mathrm { H } _ { 2 } \mathrm { Se }$ The bond energies in the molecule decrease sequentially
D. D. $\mathrm { H } _ { 2 } \mathrm { O } , \mathrm { PH } _ { 3 } , \mathrm { SiH } _ { 4 }$ The molecules are sequentially less stable

Answer

Answer: A

Solution: A. $\mathrm { CH } _ { 4 } , \mathrm { NH } _ { 3 } , \mathrm { CO } _ { 2 }$ are orthotetrahedral $\left( 109 ^ { \circ } 28 ^ { \prime } \right)$, triangular-conical $\left( 107 ^ { \circ } \right)$, and rectilinear $\left( 180 ^ { \circ } \right)$, respectively, so A is wrong; B. The larger the atomic radius, the larger the bond length of the covalent bond formed. The atomic radius of $\mathrm { Cl } , \mathrm { Br } , \mathrm { I }$ increases sequentially, so the bond length of the covalent bond formed with H increases sequentially, so B is correct; C. The stronger the nonmetallicity, the more stable the covalent bond is, and the larger the bond energy of the covalent bond is, the bond energy in the molecule of $\mathrm { H } _ { 2 } \mathrm { O } , \mathrm { H } _ { 2 } \mathrm {~S} , \mathrm { H } _ { 2 } \mathrm { Se }$ decreases in that order, so C is correct; D. Non-metallicity $\mathrm { O } > \mathrm { P } > \mathrm { Si }$, then the stability of simple hydride: $\mathrm { H } _ { 2 } \mathrm { O } > \mathrm { PH } _ { 3 } > \mathrm { SiH } _ { 4 }$, so D is correct;

Question 36

Question 36: 40. The following statements are correct () High School Chemistry Assignment, October 31, 2025

40. The following statements are correct () High School Chemistry Assignment, October 31, 2025

A. A. Ionic compounds cannot contain covalent bonds
B. B. A compound containing an ionic bond must be an ionic compound
C. C. Covalent compounds may contain ionic bonds
D. D. A compound containing covalent bonds must be a covalent compound

Answer

Answer: B

Solution: A. Ionic compounds contain covalent bonds, such as KOH, so A is wrong; B. Compounds containing ionic bonds are ionic compounds, and ionic compounds may contain covalent bonds, such as NaOH, so B is correct; C. Compounds containing only covalent bonds are covalent compounds, so covalent compounds must not contain ionic bonds, so C is wrong; D. Compounds containing only covalent bonds are covalent compounds, and compounds containing covalent bonds may be ionic compounds, such as $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$, so D is wrong; so B is wrong. D. Compounds containing only covalent bonds are covalent compounds, and compounds containing covalent bonds may be ionic compounds, such as $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$, so D is wrong.

Chemical Bonding and Intermolecular Forces

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