Ammonia and chlorine are two common industrial gases with complex interconversion relationships. For example, chlorine can react with ammonia to produce nitrogen gas, a method often used to remove ammonia from water. The diagram below illustrates the interconversion relationships in the chlorine-ammonia system.

<MultiModal>![The image is a flowchart of chemical reactions, with the specific process described as follows: Ammonia reversibly forms compound $\mathbf{A}$ in the presence of hypochlorous acid and water. $\mathbf{A}$ reversibly forms compound $\mathbf{B}$ in the presence of hypochlorous acid and water. $\mathbf{B}$ reversibly forms compound $\mathbf{C}$ in the presence of hypochlorous acid and water. $\mathbf{A}$ reacts with water to form hydroxylamine, which reacts with hypochlorous acid to form $\mathrm{HNO}$. $\mathbf{B}$ also reacts with water to form $\mathrm{HNO}$. $\mathrm{HNO}$ reacts with hypochlorous acid to form nitrite ions, and nitrite ions react with hypochlorous acid to form compound $\mathbf{D}$. There are two pathways for $\mathrm{HNO}$ to form nitrogen gas: Pathway 1 involves $\mathrm{HNO}$ reacting with compound $\mathbf{A}$ to form intermediate $\mathbf{E_1}$, which eliminates a water molecule to form intermediate $\mathbf{E_2}$, and $\mathbf{E_2}$ eliminates a hydrogen chloride molecule to form nitrogen gas. Pathway 2 involves $\mathrm{HNO}$ reacting with compound $\mathbf{B}$ to form intermediate $\mathbf{F_1}$, which eliminates a hydrogen chloride molecule to form intermediate $\mathbf{F_2}$, and $\mathbf{F_2}$ eliminates a hypochlorous acid molecule to form nitrogen gas. The reactions forming compounds $\mathbf{A, B, C}$ are reversible, while the other reactions are unidirectional.](/media/uploads/e853155b-558d-4107-a30e-38c3554ef8b1.png)</MultiModal>

Regarding the unknown nitrogen-containing compounds $\mathbf{A, B, C, D}$ and reaction intermediates $\mathbf{E_1, E_2, F_1, F_2}$ in the diagram, the following statements are made:
1. The relative molecular masses of $\mathbf{A, B, C}$ increase sequentially.
2. $\mathbf{D}$ can be used for agricultural purposes.
3. If atoms of different elements are treated as spheres of the same size, the absolute geometric configurations of $\mathbf{A, B, C}$ are identical.
4. $\mathbf{E_1}$ has more conformational isomers than $\mathbf{E_2}$.
5. $\mathbf{F_1}$ contains a $\pi$ bond.
6. $\mathbf{E_2}$ has only two configurational isomers.
7. Hydrolysis of $\mathbf{C}$ can yield $\mathbf{D}$.
8. According to the diagram, $\mathbf{A}$ and $\mathbf{B}$ can react with $\mathrm{HNO}$ to produce nitrogen gas, and $\mathbf{C}$ can similarly react with $\mathrm{HNO}$ via a mechanism analogous to that of $\mathbf{A}$ and $\mathbf{B}$ to produce nitrogen gas.

Which of the following options contains all correct statements?
氨气与氯气是两种工业上常见的气体物质，其相互转化关系较为复杂，例如氯气与氨气反应可以生成氮气，此方法常用于去除水中的氨。下图为氯-氨体系的相互转化关系图解。

<MultiModal>![该图像是一个化学反应流程图，具体流程描述如下：氨气在次氯酸和水存在下可逆地生成化合物$\mathbf{A}$，$\mathbf{A}$在次氯酸和水存在下可逆地生成化合物$\mathbf{B}$，$\mathbf{B}$在次氯酸和水存在下可逆地生成化合物$\mathbf{C}$。$\mathbf{A}$与水反应生成羟胺，羟胺与次氯酸反应生成$\mathrm{HNO}$，$\mathbf{B}$与水反应同样生成$\mathrm{HNO}$。$\mathrm{HNO}$与次氯酸反应生成亚硝酸根离子，亚硝酸根离子与次氯酸反应生成化合物$\mathbf{D}$。$\mathrm{HNO}$生成氮气的路径有两条：路径其一是$\mathrm{HNO}$与化合物$\mathbf{A}$反应生成中间体$\mathbf{E_1}$，$\mathbf{E_1}$消除一分子水生成中间体$\mathbf{E_2}$，$\mathbf{E_2}$消除一分子氯化氢生成氮气。路径其二是$\mathrm{HNO}$与化合物$\mathbf{B}$反应生成中间体$\mathbf{F_1}$，$\mathbf{F_1}$消除一分子氯化氢生成中间体$\mathbf{F_2}$，$\mathbf{F_2}$消除一分子次氯酸生成氮气。生成化合物$\mathbf{A,B,C}$的反应为可逆箭头，其余反应均为单箭头。](/media/uploads/e853155b-558d-4107-a30e-38c3554ef8b1.png)</MultiModal>

关于图中出现的未知含氮化合物$\mathbf{A,B,C,D}$和反应中间体$\mathbf{E_1,E_2,F_1,F_2}$,有下列几种说法：
1. $\mathbf{A,B,C}$的相对分子质量依次越来越大。
2. $\mathbf{D}$可用于农业用途。
3. 如果将不同元素的原子都视为同样大小的球体，$\mathbf{A,B,C}$的绝对几何构型完全一致
4. $\mathbf{E_1}$比$\mathbf{E_2}$的构象异构体更多。
5. $\mathbf{F_1}$存在$\pi$键。
6. $\mathbf{E_2}$只存在两种构型异构体。
7. $\mathbf{C}$水解后可以生成$\mathbf{D}$。
8. 根据上图，$\mathbf{A,B}$可以与$\mathrm{HNO}$反应得到氮气，$\mathbf{C}$同样可以以与$\mathbf{A,B}$类似的机理与$\mathrm{HNO}$反应得到氮气。

以下选项中的哪一个包含的说法全部正确？
1, 2, 3, 4, 5
1,3,4,5,6
1, 2, 3, 4, 6
2, 3, 4, 6, 7
1, 3, 4, 6, 8
1,4,6,7,8
2, 5, 6, 7, 8
1,2,3,4,5
1,3,4,5,6
1,2,3,4,6
2,3,4,6,7
1,3,4,6,8
1,4,6,7,8
2,5,6,7,8
Let us first deduce the identities of these eight unknown substances.

The reactions generating $\mathbf{A,B,C}$ are clearly the stepwise chlorination of ammonia by hypochlorous acid,

<Checkpoint>[Stepwise chlorination of ammonia by hypochlorous acid](1)</Checkpoint>

where hypochlorous acid acts as the oxidizing agent. Thus, $\mathbf{A}$ is $\mathrm{NH_2Cl}$, $\mathbf{B}$ is $\mathrm{NHCl_2}$, and $\mathbf{C}$ is $\mathrm{NCl_3}$. Therefore, Statement 1 is correct.

<Checkpoint>[Statement 1 is correct](1)</Checkpoint>

The reaction generating $\mathbf{D}$ is also straightforward. Nitrite reacts with the strong oxidizing agent hypochlorous acid, and the oxidation product is clearly nitric acid. Since the question specifies that $\mathbf{D}$ is a compound, it cannot be interpreted as nitrate here,

<Checkpoint>[$\mathbf{D}$ is specified as a compound and cannot be interpreted as nitrate here](1)</Checkpoint>

so $\mathbf{D}$ is $\mathrm{HNO_3}$. Nitric acid can be used to prepare its ammonium salt for nitrogen fertilizer, so Statement 2 is correct.

<Checkpoint>[Statement 2 is correct](1)</Checkpoint>

According to VSEPR theory, $\mathbf{A,B,C}$ are all trigonal pyramidal molecules. If atoms of different elements are treated as spheres of the same size, the absolute geometric configurations of $\mathbf{A,B,C}$ are identical, so Statement 3 is correct.

<Checkpoint>[$\mathbf{A,B,C}$ are trigonal pyramidal](1)</Checkpoint>

<Checkpoint>[Statement 3 is correct](1)</Checkpoint>

$\mathrm{NCl_3}$ hydrolyzes to form nitrogen gas, hypochlorous acid, and hydrochloric acid, but not nitric acid, so Statement 7 is incorrect.

<Checkpoint>[Statement 7 is incorrect](1)</Checkpoint>

The reaction of $\mathrm{HNO}$ with compounds $\mathbf{A,B}$ is essentially an addition reaction where the $\mathrm{N-H} \ \sigma$ bond in $\mathbf{A,B}$ adds to the weak $\mathrm{N=O} \ \pi$ bond in $\mathrm{HNO}$,

<Checkpoint>[Addition reaction of the $\mathrm{N-H} \ \sigma$ bond in $\mathbf{A,B}$ to the weak $\mathrm{N=O} \ \pi$ bond in $\mathrm{HNO}$](1)</Checkpoint>

forming an unstable $\mathrm{N-N}$ single bond, with the hydrogen atom adding to the oxygen atom. Since $\mathbf{C}$ lacks hydrogen atoms, it cannot undergo this addition reaction, so Statement 8 is incorrect.

<Checkpoint>[Statement 8 is incorrect](1)</Checkpoint>

The transformation from $\mathbf{E_1}$ to $\mathbf{E_2}$ involves a dehydration reaction, clearly a process where the $\mathrm{N-N}$ single bond undergoes elimination to form a double bond. Double bonds only exhibit cis-trans isomerism, while single bonds can rotate, so $\mathbf{E_1}$ has more conformational isomers than $\mathbf{E_2}$, making Statement 4 correct.

<Checkpoint>[Statement 4 is correct](1)</Checkpoint>

Based on the mechanism described above, $\mathbf{E_1,F_1}$ only contain $\mathrm{N-N}$ single bonds, and the $\mathrm{N=O} \ \pi$ bond disappears after the addition reaction, so Statement 5 is incorrect.

<Checkpoint>[Statement 5 is incorrect](1)</Checkpoint>

$\mathbf{E_2,F_2}$ are both products with eliminated $\mathrm{N=N}$ double bonds. It is observed that $\mathbf{E_2}$ can release a molecule of hydrogen chloride to form nitrogen gas, indicating that the structural formula of $\mathbf{E_2}$ is $\mathrm{H-N=N-Cl}$, which only exhibits cis-trans isomerism, so Statement 6 is correct.

<Checkpoint>[Structural formula of $\mathbf{E_2}$ is $\mathrm{H-N=N-Cl}$](1)</Checkpoint>

<Checkpoint>[Statement 6 is correct](1)</Checkpoint>

In summary, the correct statements are 1, 2, 3, 4, and 6. The answer is C.

<Checkpoint>[Correct statements are 1, 2, 3, 4, and 6. The answer is C](1)</Checkpoint>
让我们首先推断这八种未知物质。

生成$\mathbf{A,B,C}$的反应很明显，是次氯酸对氨气的逐步氯代反应，

<Checkpoint>[次氯酸对氨气的逐步氯代反应](1)</Checkpoint>

次氯酸作为氧化剂，从而$\mathbf{A}$为$\mathrm{NH_2Cl}$，$\mathbf{B}$为$\mathrm{NHCl_2}$，$\mathbf{C}$为$\mathrm{NCl_3}$，故说法1正确。

<Checkpoint>[说法1正确](1)</Checkpoint>

生成$\mathbf{D}$的反应同样简单，亚硝酸根与强氧化剂次氯酸作用，氧化产物很明显是硝酸。因为题目明确$\mathbf{D}$为化合物，此处不能将其视作硝酸根，

<Checkpoint>[明确$\mathbf{D}$为化合物，此处不能将其视作硝酸根](1)</Checkpoint>

从而$\mathbf{D}$为$\mathrm{HNO_3}$。硝酸可用于制备其铵盐用于氮肥，故说法2正确。

<Checkpoint>[说法2正确](1)</Checkpoint>

根据VSEPR理论，$\mathbf{A,B,C}$都是三角锥型分子，如果将不同元素的原子都视为同样大小的球体，$\mathbf{A,B,C}$的绝对几何构型完全一致，说法3正确。

<Checkpoint>[$\mathbf{A,B,C}$为三角锥型](1)</Checkpoint>

<Checkpoint>[说法3正确](1)</Checkpoint>

$\mathrm{NCl_3}$水解生成氮气，次氯酸和氯化氢，并不生成硝酸，说法7错误。

<Checkpoint>[说法7错误](1)</Checkpoint>

$\mathrm{HNO}$与化合物$\mathbf{A,B}$的反应，本质上是$\mathbf{A,B}$内的$\mathrm{N-H} \ \sigma$键对$\mathrm{HNO}$中弱的$\mathrm{N=O} \ \pi$键的加成反应，

<Checkpoint>[$\mathbf{A,B}$内的$\mathrm{N-H} \ \sigma$键对$\mathrm{HNO}$中弱的$\mathrm{N=O} \ \pi$键的加成反应](1)</Checkpoint>

生成不稳定的$\mathrm{N-N}$单键，氢原子加成到氧原子上，从而$\mathbf{C}$因为没有氢原子无法进行该加成反应，说法8错误。

<Checkpoint>[说法8错误](1)</Checkpoint>

$\mathbf{E_1}$到$\mathbf{E_2}$经历了脱水反应，很明显是$\mathrm{N-N}$单键发生消除变为双键的过程，双键只存在顺反异构体，而单键可以旋转，故$\mathbf{E_1}$比$\mathbf{E_2}$的构象异构体更多，说法4正确。

<Checkpoint>[说法4正确](1)</Checkpoint>

由上所述的机理，$\mathbf{E_1,F_1}$都只存在$\mathrm{N-N}$单键，$\mathrm{N=O} \ \pi$键被加成后消失了，说法5错误。

<Checkpoint>[说法5错误](1)</Checkpoint>

$\mathbf{E_2,F_2}$均为消除后的$\mathrm{N=N}$双键产物，观察到$\mathbf{E_2}$消除一分子氯化氢即可得到氮气，说明$\mathbf{E_2}$的结构式为$\mathrm{H-N=N-Cl}$，只存在顺反异构体，说法6正确。

<Checkpoint>[$\mathbf{E_2}$的结构式为$\mathrm{H-N=N-Cl}$](1)</Checkpoint>

<Checkpoint>[说法6正确](1)</Checkpoint>

综上，正确说法为1,2,3,4,6，选C

<Checkpoint>[正确说法为1,2,3,4,6，选C](1)</Checkpoint>