A reactive metastable diatomic particle $\mathbf{X}$ detected in interstellar gas can react with a reddish-brown gas $\mathbf{A}$ in a 1:1 ratio to yield various substances. Computational chemistry indicates that the reaction first involves the combination of $\mathbf{X}$ and $\mathbf{A}$ to produce two isomeric molecules $\mathbf{B_1}$ and $\mathbf{B_2}$. $\mathbf{B_1}$ can directly decompose into $\mathbf{C}$ and $\mathbf{D}$, and $\mathbf{D}$ can be oxidized to $\mathbf{A}$ in air; while $\mathbf{B_2}$ cannot decompose directly, but first undergoes isomerization to obtain $\mathbf{B_3}$, and $\mathbf{B_3}$ then decomposes into $\mathbf{E}$ and hydroxyl radicals. $\mathbf{X}$ is oxidized in oxygenated water to yield equal amounts of sulfur dioxide and $\mathbf{F}$. $\mathbf{F}$ does not contain sulfur and is a highly reactive monoprotic acid. $\mathbf{X}$ can ionize in deoxygenated water to produce a negative ion $\mathbf{G}$, and $\mathbf{G}$ can catalyze the cis-trans isomerization of double bonds in lipid compounds.

Please provide the chemical formula or structural formula of each substance and select the matching options.
在星际气体中探测到了一种具有反应活性的亚稳态双原子粒子 $\mathbf{X}$ 能与红棕色气体 $\mathbf{A}$ 按 1 :1 的比例反应可以得到多种物质。计算化学表明, 该反应首先由 $\mathbf{X}$ 与 $\mathbf{A}$ 结合, 得到两种互为同分异构体的分子 $\mathbf{B_1}$ 和 $\mathbf{B_2}$，$\mathbf{B_1}$ 可以直接分解为 $\mathbf{C}$ 和 $\mathbf{D}$, $\mathbf{D}$ 在空气中可被氧化为 $\mathbf{A}$; 而$\mathbf{B_2}$ 不能直接分解, 而是先经异构化得到 $\mathbf{B_3}$, $\mathbf{B_3}$再分解成 $\mathbf{E}$ 和羟基自由基。$\mathbf{X}$ 在含氧的水中被氧化, 得到等量的二氧化硫和 $\mathbf{F}$, $\mathbf{F}$ 不含硫, 是一种活性很高的一元酸。$\mathbf{X}$ 在不含氧的水中可电离出负离子 $\mathbf{G}$, $\mathbf{G}$ 能催化脂类化合物中双键的顺反异构化。

请给出各个物质的化学式或者结构式，并且选择匹配的选项。
All other options are incorrect
The chemical formula of $\mathbf{X}$ is $\mathrm{SO}$.
The chemical formula of $\mathbf{A}$ is $\mathrm{N_2O_4}$
The structural formula of $\mathbf{B_1}$ is
<MultiModal>![OSON=O](/media/uploads/ae8f3373-9f46-4c76-a612-2f6b70f22ab2.png)</MultiModal>
The structural formula of $\mathbf{B_2}$ is
<MultiModal>![O1N=[S-][O+]=N1](/media/uploads/22905c6e-825f-459e-94e2-a815d4d31535.png)</MultiModal>
The structural formula of $\mathbf{B_3}$ is
<MultiModal>![O=[N+](O)[S-]](/media/uploads/ad7cb368-1aa1-44a9-8311-e96eaa7c89e5.png)</MultiModal>
$\mathbf{C}$'s chemical formula is $\mathrm{SO}$
The chemical formula of $\mathbf{D}$ is $\mathrm{NOH}$.
The structural formula of $\mathbf{E}$ is
<MultiModal>![SN=O](/media/uploads/00941b57-c9a4-4066-a0fe-4b31e92e94c5.png)</MultiModal>
The chemical formula of $\mathbf{F}$ is $\mathrm{H_2O_2}$
$\mathbf{G}$ catalyzed cis-trans isomerization of double bonds proceeds via cycloaddition reactions.
其他选项均不正确
$\mathbf{X}$的化学式为$\mathrm{SO}$
$\mathbf{A}$的化学式为$\mathrm{N_2O_4}$
$\mathbf{B_1}$的结构式为
<MultiModal>![OSON=O](/media/uploads/ae8f3373-9f46-4c76-a612-2f6b70f22ab2.png)</MultiModal>
$\mathbf{B_2}$的结构式为
<MultiModal>![O1N=[S-][O+]=N1](/media/uploads/22905c6e-825f-459e-94e2-a815d4d31535.png)</MultiModal>
$\mathbf{B_3}$的结构式为
<MultiModal>![O=[N+](O)[S-]](/media/uploads/ad7cb368-1aa1-44a9-8311-e96eaa7c89e5.png)</MultiModal>
$\mathbf{C}$的化学式为$\mathrm{SO}$
$\mathbf{D}$的化学式为$\mathrm{NOH}$
$\mathbf{E}$的结构式为
<MultiModal>![SN=O](/media/uploads/00941b57-c9a4-4066-a0fe-4b31e92e94c5.png)</MultiModal>
$\mathbf{F}$的化学式为$\mathrm{H_2O_2}$
$\mathbf{G}$催化双键顺反异构是通过环加成反应进行的
(The structural representation uses SMILES coding, and chemical formulas use ordinary representation methods.)

From the fact that $\mathbf{X}$ can be ionized as an acid in water under anaerobic conditions, it can be known that it contains hydrogen, and that it reacts to produce sulfur dioxide under oxygen-containing conditions, it can be known that another atom in it is a sulfur atom, and then it can be known that the chemical formula of $\mathbf{X}$ is $\mathrm{HS·}$.

<Checkpoint>[The chemical formula of $\mathbf{X}$ is $\mathrm{HS·}$](1.5)</Checkpoint>

Next, since $\mathbf{X}$ reacts with $\mathbf{A}$ to produce two isomers, it should have two binding sites. Combined with the reddish-brown color information, it can be known that the chemical formula of $\mathbf{A}$ is $\mathrm{NO_2}$

<Checkpoint>[The chemical formula of $\mathbf{A}$ is $\mathrm{NO_2}$](1)</Checkpoint>

Next, since $\mathbf{D}$ can be oxidized to $\mathbf{A}$ in the air and is obtained by the decomposition of $\mathbf{B_1}$, it should be $\mathrm{NO}$

<Checkpoint>[The chemical formula of $\mathbf{D}$ is $\mathrm{NO}$](1)</Checkpoint>

Since $\mathbf{B_1}$ can be directly decomposed to obtain $\mathbf{D}$ and another molecule, it indicates that the fragment corresponding to $\mathrm{NO}$ should be relatively independent. When $\mathbf{B_1}$ is generated, the oxygen in $\mathrm{NO_2}$ should be connected to the sulfur atom, so its structure should be
<MultiModal>![O=NOS](/media/uploads/23e5ad3e-82fb-4ece-83ab-0c579aba65e0.png)</MultiModal>

<Checkpoint>[The structure of $\mathbf{B_1}$ is O=NOS](1.5)</Checkpoint>

Correspondingly, the structure of the isomer $\mathbf{B_2}$ is
<MultiModal>![O=[N+](S)[O-]](/media/uploads/d918e45e-1547-47ea-8756-0705513ebadf.png)</MultiModal>

<Checkpoint>[The structure of $\mathbf{B_2}$ is O=[N+](S)[O-]](1.5)</Checkpoint>

Next, $\mathbf{C}$ is the remaining part of $\mathbf{B_1}$ after removing $\mathrm{NO}$, so it can be known that it is
<MultiModal>![[H][S]=O](/media/uploads/90b3b6c4-2912-45ba-8f47-2ef9b396dd86.png)</MultiModal>

<Checkpoint>[The structure of $\mathbf{C}$ is [H][S]=O](1)</Checkpoint>

Since $\mathbf{B_3}$ converted from $\mathbf{B_2}$ decomposes to generate hydroxyl radicals, it indicates that the oxygen atom is connected to the hydrogen atom, and it can be known that its structure is
<MultiModal>![O=[N+]([S-])O](/media/uploads/7f4981e4-3cae-43ca-a08f-969b80057bb6.png)</MultiModal>

<Checkpoint>[The structure of $\mathbf{B_3}$ is O=[N+]([S-])O](1.5)</Checkpoint>

$\mathbf{E}$ is the part of $\mathbf{B_3}$ after removing the hydroxyl radical, so its structure is
<MultiModal>![O=N[S]](/media/uploads/290011e0-cfeb-443c-a9b0-91ba5c4f0d1b.png)</MultiModal>


<Checkpoint>[The structure of $\mathbf{E}$ is O=N[S]](1)</Checkpoint>

Next, sulfur dioxide is also generated during the generation of $\mathbf{F}$. The system contains water and oxygen. At this time, it is found that considering the reaction involving water, it is impossible to produce an equivalent amount of highly active monoprotic acid. Therefore, considering the reaction that only occurs with oxygen, it can be known that the generated product should be $\mathbf{HO_2}$

<Checkpoint>[The chemical formula of $\mathbf{F}$ is $\mathbf{HO_2}$](1)</Checkpoint>

Finally, $\mathbf{G}$ is the product of simple ionization of $\mathbf{X}$, so it can be known that it is $\mathrm{·S^-}$

<Checkpoint>[The chemical formula of $\mathbf{G}$ is $\mathrm{·S^-}$](0.5)</Checkpoint>

The cis-trans isomerization of double bonds catalyzed by $\mathbf{G}$ should proceed through a radical addition-elimination process. After addition and before elimination, the double bond becomes a single bond and can rotate freely.

<Checkpoint>[$\mathbf{G}$ catalyzes cis-trans isomerization of double bonds through an addition-elimination process](0.5)</Checkpoint>

To sum up, option F is correct.
（结构表述采用SMILES编码，化学式采用平常表述方法）

由$\mathbf{X}$可在无氧的条件下作为酸在水中电离可知其中含有氢，且含氧条件下反应生成二氧化硫，可以得知其中另一个原子为硫原子，进而可知$\mathbf{X}$的化学式为$\mathrm{HS·}$。

<Checkpoint>[$\mathbf{X}$的化学式为$\mathrm{HS·}$](1.5)</Checkpoint>

接下来，由于$\mathbf{X}$与$\mathbf{A}$反应生成两种同分异构体，故而其应有两种结合位点，结合红棕色的颜色信息，可知$\mathbf{A}$的化学式为$\mathrm{NO_2}$

<Checkpoint>[$\mathbf{A}$的化学式为$\mathrm{NO_2}$](1)</Checkpoint>

接下来，由于$\mathbf{D}$在空气中可以氧化为$\mathbf{A}$，且其通过$\mathbf{B_1}$分解得到，可知其应为$\mathrm{NO}$

<Checkpoint>[$\mathbf{D}$的化学式为$\mathrm{NO}$](1)</Checkpoint>

由于$\mathbf{B_1}$可以直接分解得到$\mathbf{D}$和另一分子，说明其中$\mathrm{NO}$对应的片段应相对独立，生成$\mathbf{B_1}$时应为$\mathrm{NO_2}$中的氧与硫原子相连，故而其结构应为

<MultiModal>![O=NOS](/media/uploads/23e5ad3e-82fb-4ece-83ab-0c579aba65e0.png)</MultiModal>

<Checkpoint>[$\mathbf{B_1}$的结构为O=NOS](1.5)</Checkpoint>

对应的，同分异构体$\mathbf{B_2}$的结构为
<MultiModal>![O=[N+](S)[O-]](/media/uploads/d918e45e-1547-47ea-8756-0705513ebadf.png)</MultiModal>

<Checkpoint>[$\mathbf{B_2}$的结构为O=[N+](S)[O-]](1.5)</Checkpoint>

接着，$\mathbf{C}$为$\mathbf{B_1}$去掉$\mathrm{NO}$后余下的部分，故而可知其为

<MultiModal>![[H][S]=O](/media/uploads/90b3b6c4-2912-45ba-8f47-2ef9b396dd86.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$的结构为[H][S]=O](1)</Checkpoint>

由于$\mathbf{B_2}$转化生成的$\mathbf{B_3}$分解生成羟基自由基，故而说明其中氧原子与氢原子相连，可知其结构为

<MultiModal>![O=[N+]([S-])O](/media/uploads/7f4981e4-3cae-43ca-a08f-969b80057bb6.png)</MultiModal>

<Checkpoint>[$\mathbf{B_3}$的结构为O=[N+]([S-])O](1.5)</Checkpoint>

$\mathbf{E}$为$\mathbf{B_3}$去掉羟基自由基的部分，故而其结构为

<MultiModal>![O=N[S]](/media/uploads/290011e0-cfeb-443c-a9b0-91ba5c4f0d1b.png)</MultiModal>

<Checkpoint>[$\mathbf{E}$的结构为O=N[S]](1)</Checkpoint>

接着，生成$\mathbf{F}$的过程中还生成了二氧化硫，体系中含有水和氧气，此时发现考虑水参与的反应，无法产生等当量的高活性一元酸，故而考虑仅和氧气发生的反应，可知其生成的应为$\mathbf{HO_2}$

<Checkpoint>[$\mathbf{F}$的化学式为$\mathbf{HO_2}$](1)</Checkpoint>

最后，$\mathbf{G}$为$\mathbf{X}$简单电离后的产物，所以可知其为$\mathrm{·S^-}$

<Checkpoint>[$\mathbf{G}$的化学式为$\mathrm{·S^-}$](0.5)</Checkpoint>

而$\mathbf{G}$催化双键顺反异构应通过自由基加成-消除过程。在加成后，消除前，双键变为单键，可以自由旋转。

<Checkpoint>[$\mathbf{G}$催化双键顺反异构通过加成消除过程](0.5)</Checkpoint>

综上，选项F正确。