Element $\bf{X}$ is one of the essential non-metallic elements for life. Its simple substance reacts with $Cl_2$ to obtain a compound $\bf{A}$ with a structure similar to $H_2O_2$. $\bf{A}$ hydrolyzes and disproportionates into the simple substance and $\bf{B}$. $\bf{A}$ reacts with $O_3$ to obtain $\bf{B}$ and $\bf{C}$. The simple substance reacts with CIF to obtain $\bf{D}$. The oxidation states of $\bf{X}$ in $\bf{B}$, $\bf{C}$, and $\bf{D}$ are all the same, all are binary compounds, and $\bf{B}$ mainly exhibits oxidizing properties in chemical reactions. $\bf{D}$ and the highest-valent oxide $\bf{E}$ of $\bf{X}$ react in a 1:1 ratio to obtain $\bf{F}$ and $\bf{G}$ in a 1:1 ratio. $\bf{F}$ and $\bf{G}$ are both ternary compounds, where $\bf{F}$ can also be obtained by reacting the hydrate $\bf{H}$ of $\bf{G}$ with fluorosulfuric acid. $\bf{H}$ is a strong acid.

Please select the correct options from the following statements.
元素 $\bf{X}$ 是生命必要的非金属元素之一，其单质与 $Cl_2$ 发生化合反应得到结构类似于 $H_2O_2$ 的化合物 $\bf{A}$，$\bf{A}$ 会水解歧化分解为单质与 $\bf{B}$，$\bf{A}$ 与 $O_3$ 反应则可以得到 $\bf{B}$ 和 $\bf{C}$。单质与 $ClF$ 作用可得到 $\bf{D}$，$\bf{B}$、$\bf{C}$、$\bf{D}$ 中 $\bf{X}$ 的氧化态都相同，都是二元化合物，且 $\bf{B}$ 在化学反应中主要体现氧化性。$\bf{D}$ 与 $\bf{X}$ 的最高价氧化物 $\bf{E}$ $1:1$地作用可以$1:1$地得到 $\bf{F}$ 和 $\bf{G}$， $\bf{F}$ 和 $\bf{G}$ 都是三元化合物，其中 $\bf{F}$ 也可以通过 $\bf{G}$ 的水化物 $\bf{H}$ 与氟磺酸作用而来，$\bf{H}$ 是强酸。

请选出下列说法正确的选项。
From the information in the question, we know that $\bf{X}$ is a non-metallic element and has multiple variable valences, so X is arsenic.
The molecular structure of $\bf{A}$ possesses a center of symmetry.
$\bf{B}$, $\bf{C}$, and $\bf{X}$ all have the same oxidation state for $\bf{X}$, which is +4, so $\bf{X}$ is silicon.
$\bf{A}$ reacts with $BCl_3$ at low temperature to produce an ionic compound with a relative molecular mass of 345.98.
The valence of $\bf{X}$ in $\bf{F}$ and $\bf{G}$ are equal, both being +6.
The products of the reaction between $\bf{A}$ and $O_3$ are only $\bf{B}$ and $\bf{C}$, with no other substances.
All of the above options are incorrect.
由题目中信息可知，$\bf{X}$为非金属元素，且具有多种变价，所以X为砷。
$\bf{A}$的分子结构中具有对称中心。
$\bf{B}$、$\bf{C}$、$\bf{X}$ 中 $\bf{X}$ 的氧化态都相同，都为+4价，所以$\bf{X}$为硅。
$\bf{A}$ 与 $BCl_3$ 在低温下作用，生成离子化合物，相对分子质量为345.98。
$\bf{F}$ 和 $\bf{G}$ 中 $\bf{X}$ 的价态相等，都为+6价。
$\bf{A}$ 与 $O_3$ 反应的产物只有 $\bf{B}$ 和 $\bf{C}$，没有别的物质。
上述选项均不正确
Element $\bf{X}$ is one of the essential non-metal elements for life, first narrowing the scope to C, N, O, P, S, Se, etc.

$X_2Cl_2$ is structurally similar to $H_2O_2$, so $\bf{X}$ should be a chalcogen element.

$\bf{A}$ disproportionates upon hydrolysis to form an element and $\bf{B}$, $\bf{B}$ should be $XO_2$. It is also mentioned that it often exhibits oxidizing properties in chemical reactions. $SO_2$ mostly exhibits reducing properties, and $SeO_2$ is often used as an oxidant, so $\bf{X}$ is selenium.

<Checkpoint>[$\bf{X}$ is selenium](1)</Checkpoint>

Therefore, the substances are as follows:

$\bf{A}$: $Se_2Cl_2$

<Checkpoint>[$\bf{A}$: $Se_2Cl_2$](1)</Checkpoint>

$\bf{B}$: $SeO_2$

<Checkpoint>[$\bf{B}$: $SeO_2$](1)</Checkpoint>

$\bf{C}$: $SeCl_4$

<Checkpoint>[$\bf{C}$: $SeCl_4$](1)</Checkpoint>

$\bf{D}$: $SeF_4$

<Checkpoint>[$\bf{D}$: $SeF_4$](1)</Checkpoint>

$\bf{E}$: $SeO_3$

<Checkpoint>[$\bf{E}$: $SeO_3$](1)</Checkpoint>

$\bf{F}$: $SeO_2F_2$

<Checkpoint>[$\bf{F}$: $SeO_2F_2$](1)</Checkpoint>

$\bf{G}$: $SeOF_2$

<Checkpoint>[$\bf{G}$: $SeOF_2$](1)</Checkpoint>

$\bf{H}$: $H_2SeO_4$

<Checkpoint>[$\bf{H}$: $H_2SeO_4$](1)</Checkpoint>

Based on this, analyze the options:

A, The element C is incorrect;

B is incorrect, $\bf{A}$ has no center of symmetry;

<Checkpoint>[$\bf{A}$ has no center of symmetry](1)</Checkpoint>

E is incorrect, $\bf{G}$ has a valence of +4;

<Checkpoint>[$\bf{G}$ has a valence of +4](1)</Checkpoint>

D is correct, $\bf{A}$($Se_2Cl_2$) reacts with $BCl_3$ at low temperature, which can be regarded as the transfer of $Cl^-$ ions from $Se_2Cl_2$ to $BCl_3$ to generate $[Se_2Cl]^+[BCl_4]^-$ ionic compounds, and the relative molecular mass is correct;

F is incorrect, the reaction formula is $3Se_2Cl_2 + 4O_3 = 4SeO_2 + 2SeCl_4 + O_2$. In addition to $\bf{B}$ and $\bf{C}$, oxygen is also produced;

<Checkpoint>[In addition to $\bf{B}$ and $\bf{C}$, oxygen is also produced](1)</Checkpoint>

G is incorrect because option D is correct.
元素 $\bf{X}$ 是生命必要的非金属元素之一，首先缩小范围到C、N、O、P、S、Se等。

$X_2Cl_2$与$H_2O_2$结构类似，所以 $\bf{X}$ 应该为氧族元素。

$\bf{A}$ 水解歧化得到单质与 $\bf{B}$，$\bf{B}$应该为 $XO_2$，又提到在化学反应中常体现氧化性，$SO_2$多体现还原性，$SeO_2$常作为氧化剂，故$\bf{X}$ 为硒。

<Checkpoint>[$\bf{X}$ 为硒](1)</Checkpoint>

因此，各物质如下：

$\bf{A}$：$Se_2Cl_2$

<Checkpoint>[$\bf{A}$：$Se_2Cl_2$](1)</Checkpoint>

$\bf{B}$：$SeO_2$

<Checkpoint>[$\bf{B}$：$SeO_2$](1)</Checkpoint>

$\bf{C}$：$SeCl_4$

<Checkpoint>[$\bf{C}$：$SeCl_4$](1)</Checkpoint>

$\bf{D}$：$SeF_4$

<Checkpoint>[$\bf{D}$：$SeF_4$](1)</Checkpoint>

$\bf{E}$：$SeO_3$

<Checkpoint>[$\bf{E}$：$SeO_3$](1)</Checkpoint>

$\bf{F}$：$SeO_2F_2$

<Checkpoint>[$\bf{F}$：$SeO_2F_2$](1)</Checkpoint>

$\bf{G}$：$SeOF_2$

<Checkpoint>[$\bf{G}$：$SeOF_2$](1)</Checkpoint>

$\bf{H}$：$H_2SeO_4$

<Checkpoint>[$\bf{H}$：$H_2SeO_4$](1)</Checkpoint>

据此分析选项：

A，C元素错误；

B错误，$\bf{A}$ 没有对称中心；

<Checkpoint>[$\bf{A}$ 没有对称中心](1)</Checkpoint>

E错误，$\bf{G}$为+4价；

<Checkpoint>[$\bf{G}$为+4价](1)</Checkpoint>

D正确，$\bf{A}$($Se_2Cl_2$)与$BCl_3$在低温下发生反应，可看作$Cl^-$离子从$Se_2Cl_2$转移至$BCl_3$，生成$[Se_2Cl]^+[BCl_4]^-$离子化合物，相对分子质量正确；

F错误，反应式为 $3Se_2Cl_2 + 4O_3 = 4SeO_2 + 2SeCl_4 + O_2$，除了 $\bf{B}$ 和 $\bf{C}$，还生成了氧气；

<Checkpoint>[除了 $\bf{B}$ 和 $\bf{C}$，还生成了氧气](1)</Checkpoint>

G错误，因为选项D是正确的。