Element $\mathbf{X}$ can react with $\text{AgF}$ under strictly dry and low-pressure conditions to obtain an isomer $\mathbf{A}_1$ of compound $\mathbf{A}$ with $C_2$ symmetry. Another isomer $\mathbf{A}_2$ can be obtained by reacting $\text{KSO}_3\text{F}$ with $\mathbf{B}$ in liquid $\text{SO}_2$. $\mathbf{A}_2$ is thermodynamically unstable and can disproportionate into compound $\mathbf{C}$ and elemental $\mathbf{X}$ under $\text{HF}$ catalysis. $\mathbf{C}$ can form a 1:1 adduct $\mathbf{D}$ with $\rm CsF$. Chlorination of $\mathbf{B}$ in the presence of $\text{FeCl}_3$ yields compound $\mathbf{E}$. Reacting $\mathbf{E}$ with NaF in warm diethyl ether yields a mixture of $\mathbf{B}$ and $\mathbf{C}$. Both $\mathbf{B}$ and $\mathbf{E}$ can react with $\text{NH}_4\text{Cl}$ to produce a cyclic compound $\mathbf{F}$. $\mathbf{F}$ is insoluble in water but soluble in concentrated $\text{NaOH}$ solution to form two sodium salts $\mathbf{G}$ and $\mathbf{H}$. The mass fractions of $\text{Na}$ in $\mathbf{G}$ and $\mathbf{H}$ are 29.09 % and 36.48 %, respectively. It is known that all substances represented by letters in the question contain $\mathbf{X}$. From the following options, choose the one that best fits the meaning of the question.
元素$\mathbf{X}$可与$\text{AgF}$在严格干燥且低压环境下反应得到化合物$\mathbf{A}$的一种具有$C_2$对称性的异构体$\mathbf{A}_1$，其另一种异构体$\mathbf{A}_2$可由$\text{KSO}_3\text{F}$与$\mathbf{B}$在液态$\text{SO}_2$中反应得到。$\mathbf{A}_2$在热力学上不稳定，在$\text{HF}$催化下可歧化为化合物$\mathbf{C}$及$\mathbf{X}$单质，$\mathbf{C}$可与$\rm CsF$形成1:1加合物$\mathbf{D}$。在$\text{FeCl}_3$存在下对$\mathbf{B}$进行氯化可得化合物$\mathbf{E}$，将$\mathbf{E}$在温热的乙醚中与NaF反应可得$\mathbf{B}$与$\mathbf{C}$的混合物，$\mathbf{B}$或$\mathbf{E}$均可与$\text{NH}_4\text{Cl}$反应得到一种环状化合物$\mathbf{F}$，$\mathbf{F}$不溶于水，但可溶于浓$\text{NaOH}$溶液生成两种钠盐$\mathbf{G}$与$\mathbf{H}$，$\mathbf{G}$与$\mathbf{H}$中$\text{Na}$的质量分数分别为29.09 %和36.48 %。已知题中所有以字母代表的物质均含$\mathbf{X}$。从下列选项中，选择最贴合题意的一项。
All other options are incorrect
$\mathbf{A}_1$ and $\mathbf{A}_2$ have different structures, but both possess a diad axis.
$\mathbf{A}_2$ and $\mathbf{B}$ have different structures, but their symmetry elements are completely identical.
The structure of $\mathbf{C}$ and the corresponding structure of the substance in $\mathbf{D}$ that forms a 1:1 adduct, although slightly different, share the same hybridization mode of the central atom.
The structure of $\mathbf{C}$ and the corresponding structure of the substance in $\mathbf{D}$ that forms a 1:1 adduct, although slightly different in the hybridization mode of the central atom, share the same symmetry elements.
The valence states of $\mathbf{C, E}$ are consistent with those of the cyclic compound $\mathbf{F}$ formed.
The valence states of $\mathbf{C}$ and $\mathbf{E}$, as well as the valence state of the formed cyclic compound $\mathbf{F}$, are all distinct from each other.
其他选项均不正确
$\mathbf{A}_1$和$\mathbf{A}_2$虽然结构不同，但是都拥有二重轴
$\mathbf{A}_2,\mathbf{B}$虽然结构不同，但是对称元素完全相同
$\mathbf{C}$的结构和形成1:1加合物的$\mathbf{D}$中物质对应的结构虽然略有差别，但是中心原子的杂化方式是相同的。
$\mathbf{C}$的结构和形成1:1加合物的$\mathbf{D}$中物质对应的结构虽然略有差别，中心原子的杂化方式不同，但是对称元素都一样。
$\mathbf{C,E}$的价态和形成的环状化合物$\mathbf{F}$的价态是一致的
$\mathbf{C}$,$\mathbf{E}$的价态和形成的环状化合物$\mathbf{F}$的价态两两之间均不相同。
The breakthrough point of the problem lies in the information of compounds $\mathbf{G}$ and $\mathbf{H}$.

Calculating the molar mass of $\mathbf{G}$ and $\mathbf{H}$:

For $\mathbf{G}$: The mass fraction of sodium is 29.09%.
The molar mass of $\text{G}$ satisfies:
If $n=1$, $M(\mathbf{G})\approx79.03\text{g/mol}$, the mass of the anion part $\approx79.03-22.99=56.04\text{g/mol}$.
If $n=2$, $M(\mathbf{G})\approx158.06\text{g/mol}$, the mass of the anion part $\approx158.06-45.98=112.08\text{g/mol}$.
If $n=3$, $M(\mathbf{G})\approx237.09\text{g/mol}$, the mass of the anion part $\approx237.09-68.97=168.12\text{g/mol}$.
Because it is treated in concentrated $\text{NaOH}$ solution, and the anion is composed of elements $\mathbf{X}$ and $\text{O}$, $\mathbf{G}$ is $\text{Na}_2\text{S}_2\text{O}_3$.

<Checkpoint>[$\mathbf{G}$ is $\text{Na}_2\text{S}_2\text{O}_3$](1)</Checkpoint>

Similarly, for $\mathbf{H}$: The mass fraction of sodium is 36.48%, calculated to be $\mathbf{H}$ is $\text{Na}_2\text{SO}_3$.

<Checkpoint>[$\mathbf{H}$ is $\text{Na}_2\text{SO}_3$](1)</Checkpoint>

Therefore, element $\mathbf{X}$ is $\text{S}$.

$\mathbf{F}$ is hydrolyzed in $\text{NaOH}$ to obtain $\text{Na}_2\text{SO}_3$ and $\text{Na}_2\text{S}_2\text{O}_3$. We know that $\mathbf{F}$ is a cyclic compound, so we can deduce that $\mathbf{F}$ is cyclic $\text{S}_4\text{N}_4$.

<Checkpoint>[$\mathbf{F}$ is cyclic $\text{S}_4\text{N}_4$](1)</Checkpoint>

$\mathbf{B}$ or $\mathbf{E}$ can react with $\text{NH}_4\text{Cl}$ to obtain $\text{S}_4\text{N}_4$; $\mathbf{B}$ is chlorinated in the presence of $\text{FeCl}_3$ to obtain $\mathbf{E}$, one of which is $\text{S}_2\text{Cl}_2$, and the other should be $\text{SCl}_2$. The latter has a higher valence, so $\mathbf{B}=\text{S}_2\text{Cl}_2$, $\mathbf{E}=\text{SCl}_2$.

<Checkpoint>[$\mathbf{B}=\text{S}_2\text{Cl}_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{E}=\text{SCl}_2$](1)</Checkpoint>

Elemental sulfur reacts with $\text{AgF}$ to obtain $\mathbf{A}$, whose isomers $\mathbf{A}_1$ belong to the $\text{C}_2$ point group and $\mathbf{A}_2$ belong to the $\text{C}_s$ point group, that is, $\mathbf{A}_1=\text{F-S-S-F}$, $\mathbf{A}_2=\text{S=SF}_2$. It can be known that the symmetry elements of $\mathbf{A}_2$ and $\mathbf{A}_1,\mathbf{B}$ are different.

<Checkpoint>[$\mathbf{A}$ is $\text{S}_2\text{F}_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}_1$ is $\text{F-S-S-F}$, belonging to the $\text{C}_2$ point group](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}_2=\text{S=SF}_2$, belonging to the $\text{C}_s$ point group](1)</Checkpoint>

<Checkpoint>[The symmetry elements of $\mathbf{A}_2$ and $\mathbf{A}_1,\mathbf{B}$ are different](1)</Checkpoint>

$\mathbf{A}_2$ disproportionates to produce $\mathbf{C}$ and $\text{S}$, so $\mathbf{C}=\text{SF}_4$, and the hybridization is $sp^3d$ hybridization.

<Checkpoint>[$\mathbf{C}=\text{SF}_4$](1)</Checkpoint>

According to the above information, we can know that the valences of $\mathbf{C}$, $\mathbf{E}$ and the valence of the cyclic compound $\mathbf{F}$ formed are all different from each other.

<Checkpoint>[The valences of $\mathbf{C}$, $\mathbf{E}$ and the valence of the cyclic compound $\mathbf{F}$ formed are all different from each other.](1)</Checkpoint>

$\mathbf{C}$ reacts with $\text{CsF}$ to form a $1:1$ adduct $\mathbf{D}$, that is, $\mathbf{D}=\text{CsSF}_5$, and the hybridization is $sp^3d^2$ hybridization.

<Checkpoint>[$\mathbf{D}=\text{CsSF}_5$](1)</Checkpoint>

<Checkpoint>[The hybridization modes of S atoms in $\mathbf{C}$ and $\mathbf{D}$ are different, which are $sp^3d$ hybridization and $sp^3d^2$ hybridization, respectively](1)</Checkpoint>

<Checkpoint>[The anion symmetry of $\mathbf{C}$ and $\mathbf{D}$ are different](1)</Checkpoint>
题目的突破口在于化合物$\mathbf{G}$和$\mathbf{H}$的信息。

计算$\mathbf{G}$和$\mathbf{H}$的摩尔质量：

对于$\mathbf{G}$：钠的质量分数为29.09%。
$\text{G}$的摩尔质量满足：
若$n=1$，$M(\mathbf{G})\approx79.03\text{g/mol}$，阴离子部分质量$\approx79.03-22.99=56.04\text{g/mol}$。
若$n=2$，$M(\mathbf{G})\approx158.06\text{g/mol}$，阴离子部分质量$\approx158.06-45.98=112.08\text{g/mol}$。
若$n=3$，$M(\mathbf{G})\approx237.09\text{g/mol}$，阴离子部分质量$\approx237.09-68.97=168.12\text{g/mol}$。
因为是在浓$\text{NaOH}$溶液中处理，阴离子由元素$\mathbf{X}$和$\text{O}$组成，故$\mathbf{G}$为$\text{Na}_2\text{S}_2\text{O}_3$。

<Checkpoint>[$\mathbf{G}$为$\text{Na}_2\text{S}_2\text{O}_3$](1)</Checkpoint>

类似地，对于$\mathbf{H}$：钠的质量分数为36.48%，计算得$\mathbf{H}$为$\text{Na}_2\text{SO}_3$。

<Checkpoint>[$\mathbf{H}$为$\text{Na}_2\text{SO}_3$](1)</Checkpoint>

因此元素$\mathbf{X}$为$\text{S}$。

$\mathbf{F}$在$\text{NaOH}$中水解得$\text{Na}_2\text{SO}_3$与$\text{Na}_2\text{S}_2\text{O}_3$，我们知道$\mathbf{F}$为环状化合物，可以推出$\mathbf{F}$为环状$\text{S}_4\text{N}_4$。

<Checkpoint>[$\mathbf{F}$为环状$\text{S}_4\text{N}_4$](1)</Checkpoint>

$\mathbf{B}$或$\mathbf{E}$均可与$\text{NH}_4\text{Cl}$反应得$\text{S}_4\text{N}_4$；$\mathbf{B}$在$\text{FeCl}_3$存在下氯化得$\mathbf{E}$，其中有一个是$\text{S}_2\text{Cl}_2$，另一个应该是$\text{SCl}_2$，后者化合价更高，故$\mathbf{B}=\text{S}_2\text{Cl}_2$，$\mathbf{E}=\text{SCl}_2$。

<Checkpoint>[$\mathbf{B}=\text{S}_2\text{Cl}_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{E}=\text{SCl}_2$](1)</Checkpoint>

硫单质与$\text{AgF}$反应得$\mathbf{A}$，其异构体$\mathbf{A}_1$属$\text{C}_2$点群、$\mathbf{A}_2$属$\text{C}_s$点群，即$\mathbf{A}_1=\text{F-S-S-F}$，$\mathbf{A}_2=\text{S=SF}_2$，可以知道$\mathbf{A}_2$和$\mathbf{A}_1,\mathbf{B}$的对称元素都不同。

<Checkpoint>[$\mathbf{A}$ 是 $\text{S}_2\text{F}_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}_1$是$\text{F-S-S-F}$，属于$\text{C}_2$点群](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}_2=\text{S=SF}_2$，属于$\text{C}_s$点群](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}_2$和$\mathbf{A}_1,\mathbf{B}$的对称元素都不同](1)</Checkpoint>

$\mathbf{A}_2$歧化生成$\mathbf{C}$和$\text{S}$，故$\mathbf{C}=\text{SF}_4$，杂化是$sp^3d$杂化。

<Checkpoint>[$\mathbf{C}=\text{SF}_4$](1)</Checkpoint>

根据上面的信息，我们可以知道$\mathbf{C}$,$\mathbf{E}$的价态和形成的环状化合物$\mathbf{F}$的价态两两之间均不相同。

<Checkpoint>[$\mathbf{C}$,$\mathbf{E}$的价态和形成的环状化合物$\mathbf{F}$的价态两两之间均不相同。](1)</Checkpoint>

$\mathbf{C}$与$\text{CsF}$形成$1:1$加合物$\mathbf{D}$，即$\mathbf{D}=\text{CsSF}_5$，杂化是$sp^3d^2$杂化。

<Checkpoint>[$\mathbf{D}=\text{CsSF}_5$](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$、$\mathbf{D}$中S原子杂化方式不同，分别为$sp^3d$杂化，$sp^3d^2$杂化](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$和$\mathbf{D}$阴离子对称性不同](1)</Checkpoint>