When sodium carbonate is added to a mineral containing element $\mathbf{A}$ and burned with oxygen, a sodium salt $\mathbf{B}$ containing $\mathbf{A}$ is obtained. After acidification and evaporative crystallization of $\mathbf{B}$, colorless crystals $\mathbf{C}$ are formed. $\mathbf{C}$ readily dehydrates to yield a white solid $\mathbf{D}$. When $\mathbf{D}$ reacts with $\mathrm{SF_{4}}$ above $110^{\circ}\mathrm{C}$, only gas $\mathbf{E}$ containing $\mathbf{A}$ with a mass fraction of $50.96\%$ and another gas $\mathbf{F}$ are produced. $\mathbf{F}$ reacts with an aqueous solution of $\mathbf{C}$ to precipitate red crystals $\mathbf{G}$. When $\mathbf{G}$ reacts with $\mathrm{SbF_{5}}$ at $-23^{\circ}\mathrm{C}$ in $\mathrm{SO_{2}}$ in a 1:5 ratio, an ionic compound $\mathbf{H}$ and $\mathrm{SbF_{3}}$ are formed in a 1:1 product ratio. Which of the following statements is correct:
在含元素 $\mathbf { A }$ 的矿物中加入碳酸钠并通入氧气灼烧，得到含 $\mathbf { A }$ 的钠盐$\mathbf { B }$ ，$\mathbf { B }$酸化后蒸发结晶得到无色晶体$\mathbf { C }$，$\mathbf { C }$易脱水得到白色固体 $\mathbf { D }$ ， $\mathbf { D }$ 与 $\mathrm { S F _ { 4 } }$ 在 $1 1 0 ^ { \circ } \mathrm { C }$ 以上反应，仅得到含$\mathbf{A}$质量分数为 $5 0 . 9 6 \%$ 的气体$\mathbf { E }$和另一种气体$\mathbf { F }$，$\mathbf { F }$与 $\mathbf { C }$ 的水溶液反应可沉淀出红色晶体$\mathbf { G }$， $\mathbf { G }$ 与 ${ \mathrm { S b F } } _ { 5 }$ 在 $- 2 3 ^ { \circ } \mathrm { C }$ 下于 $\mathrm { S O } _ { 2 }$ 中1:5反应，生成一种离子化合物 $\mathbf { H }$ 和 ${ \mathrm { S b F } } _ { 3 }$ ，产物比例为1:1。下列说法正确的是：
The element $\mathbf { A }$ in $\mathbf { B }$ is in its highest oxidation state
In the infinite chain structure of $\mathbf { D }$, there are three chemically distinct oxygen atoms.
The point group of molecule $\mathbf{E}$ is the same as that of $\mathrm{CCl_{4}}$.
The element $\mathbf { A }$ in $\mathbf { G }$ is in its lowest valence state
In the cation of $\mathbf{H}$, the $\mathbf{A}$ atom has two coordination numbers.
There is a fourfold rotation axis for the anions in $\mathbf { H }$
$\mathbf { B }$中的元素$\mathbf { A }$处于其最高价态
无限链状结构的$\mathbf { D }$中有三种化学环境的氧原子
分子$\mathbf { E }$的点群与$\mathrm { CCl _ {4} }$相同
$\mathbf { G }$中的元素$\mathbf { A }$处于其最低价态
$\mathbf { H }$的阳离子中，$\mathbf { A }$原子有两种配位数
$\mathbf { H }$的阴离子存在四重旋转轴
The key breakthrough point of the question lies in the gas $\mathbf { E }$. $\mathbf { D }$ reacts with $\mathrm{SF_4}$ to produce gas $\mathbf { E }$ and gas $\mathbf { F }$. $\mathrm{SF_4}$ is a commonly used fluorinating agent that can replace oxygen atoms in oxides with fluorine atoms. Therefore, $\mathbf { D }$ is an oxide of $\mathbf { A }$, and $\mathbf { E }$ is a fluoride of $\mathbf { A }$. The mass fraction of $\mathbf { A }$ in $\mathbf { E }$ is 50.96%, the atomic weight of $\mathbf { A }$ is M, and the chemical formula of $\mathbf { E }$ is $\mathrm{A_xF_y}$ (the atomic weight of F is 19). Then we have: $\frac{xM}{xM+19y}=0.5096$. By trial and error, we can obtain the most suitable result: element $\mathbf { A }$ is Se, and $\mathbf { E }$ is $\mathrm{SeF_4}$.

<Checkpoint>[$\mathbf { A }$ is $\mathrm{Se}$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf { E }$ is $\mathrm{SeF_4}$](0.5)</Checkpoint>

The point group of molecule $\mathbf { E }$ is $C_{2v}$. The two have the same point group. The point group of $\mathrm { CCl _ {4} }$ is $T_d$. The two are different. Option C is incorrect.

<Checkpoint>[Option C is incorrect](0.5)</Checkpoint>

$\mathbf { D }$ is an oxide of Se, which reacts with $\mathrm{SF_4}$ to produce $\mathrm{SeF_4}$, indicating that the valence state of Se in $\mathbf { D }$ is +4. So $\mathbf { D }$ is selenium dioxide ($\mathrm{SeO_2}$). $\mathrm{SeO_2}$ is a white solid, which matches the description. In the infinite chain structure of $\mathbf { D }$, there are two types of oxygen atoms: terminal and bridging. Option B is incorrect.

<Checkpoint>[$\mathbf { D }$ is $\mathrm{SeO_2}$](0.5)</Checkpoint>

<Checkpoint>[Option B is incorrect](0.5)</Checkpoint>

$\mathbf { C }$ is the hydrate of $\mathbf { D }$, which easily loses water to obtain $\mathbf { D }$. So $\mathbf { C }$ is selenous acid ($\mathrm{H_2SeO_3}$). $\mathrm{H_2SeO_3}$ is a colorless crystal, which matches the description.

<Checkpoint>[$\mathbf { C }$ is $\mathrm{H_2SeO_3}$](0.5)</Checkpoint>

$\mathbf { B }$ is acidified to obtain $\mathbf { C }$. This indicates that $\mathbf { B }$ is a selenite. Although the first step is burning in oxygen, which is a strong oxidation process, selenite was not obtained due to the strong oxidizing property of hexavalent selenium. $\mathbf { B }$ is $\mathrm{Na_2SeO_3}$, in which Se has a valence of +4, which is not the highest valence of +6. Option A is incorrect.

<Checkpoint>[$\mathbf { B }$ is $\mathrm{Na_2SeO_3}$](0.5)</Checkpoint>

<Checkpoint>[Option A is incorrect](0.5)</Checkpoint>

$\mathrm{SF_4}$, as a commonly used fluorinating agent, can replace oxygen atoms in oxides with fluorine atoms and obtain $\mathrm{SO_2}$ gas. $\mathbf { F }$ is $\mathrm{SO_2}$.

<Checkpoint>[$\mathbf { F }$ is $\mathrm{SO_2}$](0.5)</Checkpoint>

$\mathrm{SO_2}$ has reducing properties and reacts with $\mathrm{H_2SeO_3}$ to reduce it to red crystalline elemental Se. Analogous to the congener element S, $\mathbf { G }$ is $\mathrm{Se_8}$. The oxidation state of Se in it is 0, which is not the lowest oxidation state of -2. Option D is incorrect.

<Checkpoint>[$\mathbf { G }$ is $\mathrm{Se_8}$](0.5)</Checkpoint>

<Checkpoint>[Option D is incorrect](0.5)</Checkpoint>

$\mathrm{SbF_5}$ is a strong Lewis acid and a strong oxidizing agent. It will oxidize Se and form polymeric cations, commonly $\mathrm{Se_4^{2+}}$, $\mathrm{Se_8^{2+}}$. The corresponding anions may be $\mathrm{SbF_6^{-}}$, $\mathrm{Sb_2F_{11}^{-}}$. Combining the reaction ratio and the product ratio, we can obtain $\mathbf { H }$ as $\mathrm{Se_8(Sb_2F_{11})_2}$. In it, $\mathrm{Se_8^{2+}}$ has a five-membered ring fused ring structure, and there are two coordination numbers, 2 and 3. Option E is correct. The $\mathrm{Sb-F-Sb}$ bond angle of the anion $\mathrm{Sb_2F_{11}^{-}}$ is not $180^\circ$. The anion does not have a four-fold rotation axis. Option F is incorrect.

<Checkpoint>[$\mathbf { H }$ is $\mathrm{Se_8(Sb_2F_{11})_2}$](0.5)</Checkpoint>

<Checkpoint>[Option E is correct](0.5)</Checkpoint>

<Checkpoint>[Option F is incorrect](0.5)</Checkpoint>
题目关键突破口在于气体$\mathbf { E }$。$\mathbf { D }$与$\mathrm{SF_4}$ 反应生成气体$\mathbf { E }$和气体$\mathbf { F }$。$\mathrm{SF_4}$是一种常用的氟化剂，能将氧化物中的氧原子替换为氟原子。因此，$\mathbf { D }$是$\mathbf { A }$的一种氧化物，$\mathbf { E }$是$\mathbf { A }$的一种氟化物。$\mathbf { E }$中$\mathbf { A }$的质量分数为50.96%，$\mathbf { A }$的原子量为M，$\mathbf { E }$的化学式为 $\mathrm{A_xF_y}$ （F的原子量为19）。则有：$\frac{xM}{xM+19y}=0.5096$，尝试即可得到最符合的结果，元素$\mathbf { A }$为Se，$\mathbf { E }$为$\mathrm{SeF_4}$。

<Checkpoint>[$\mathbf { A }$为$\mathrm{Se}$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf { E }$为$\mathrm{SeF_4}$](0.5)</Checkpoint>

分子$\mathbf { E }$点群为$C_{2v}$，二者点群相同，$\mathrm { CCl _ {4} }$点群为$T_d$，二者不同，选项C错误。

<Checkpoint>[选项C错误](0.5)</Checkpoint>

$\mathbf { D }$是Se的氧化物，与$\mathrm{SF_4}$反应生成$\mathrm{SeF_4}$，说明$\mathbf { D }$中Se的价态为+4。所以$\mathbf { D }$是二氧化硒 ($\mathrm{SeO_2}$) 。$\mathrm{SeO_2}$是白色固体，符合描述。无限链状结构的$\mathbf { D }$中有端接与桥连两种氧原子，选项B错误。

<Checkpoint>[$\mathbf { D }$是$\mathrm{SeO_2}$](0.5)</Checkpoint>

<Checkpoint>[选项B错误](0.5)</Checkpoint>

$\mathbf { C }$是$\mathbf { D }$的水合物，易脱水得到$\mathbf { D }$ 。所以$\mathbf { C }$是亚硒酸 ($\mathrm{H_2SeO_3}$)。$\mathrm{H_2SeO_3}$是无色晶体，符合描述。

<Checkpoint>[$\mathbf { C }$是$\mathrm{H_2SeO_3}$](0.5)</Checkpoint>

$\mathbf { B }$酸化后得到$\mathbf { C }$ 。说明$\mathbf { B }$是亚硒酸盐。 尽管第一步是在氧气中灼烧，这是一个强氧化过程，但由于六价硒的强氧化性，并未得到稀酸盐。$\mathbf { B }$为$\mathrm{Na_2SeO_3}$，其中Se为+4价，并非最高价+6，选项A错误。

<Checkpoint>[$\mathbf { B }$为$\mathrm{Na_2SeO_3}$](0.5)</Checkpoint>

<Checkpoint>[选项A错误](0.5)</Checkpoint>

$\mathrm{SF_4}$作为一种常用的氟化剂，能将氧化物中的氧原子替换为氟原子，并得到$\mathrm{SO_2}$气体，$\mathbf { F }$为$\mathrm{SO_2}$。

<Checkpoint>[$\mathbf { F }$为$\mathrm{SO_2}$](0.5)</Checkpoint>

$\mathrm{SO_2}$具有还原性，与$\mathrm{H_2SeO_3}$反应，可将其还原得到红色晶体单质Se，类比同族元素S，$\mathbf { G }$为$\mathrm{Se_8}$。其中Se的氧化态为0，并非最低氧化态-2，选项D错误。

<Checkpoint>[$\mathbf { G }$为$\mathrm{Se_8}$](0.5)</Checkpoint>

<Checkpoint>[选项D错误](0.5)</Checkpoint>

$\mathrm{SbF_5}$是强路易斯酸和强氧化剂。它会氧化Se并形成多聚阳离子，常见的有$\mathrm{Se_4^{2+}}$、$\mathrm{Se_8^{2+}}$，对应的阴离子可能为$\mathrm{SbF_6^{-}}$、$\mathrm{Sb_2F_{11}^{-}}$，结合反应比例与产物比例，可得到$\mathbf { H }$为$\mathrm{Se_8(Sb_2F_{11})_2}$。其中$\mathrm{Se_8^{2+}}$为五元环并环结构，存在2、3两种配位数，选项E正确。阴离子$\mathrm{Sb_2F_{11}^{-}}$的$\mathrm{Sb-F-Sb}$键角不为$180^\circ$，阴离子没有四重旋转轴，选项F错误。

<Checkpoint>[$\mathbf { H }$为$\mathrm{Se_8(Sb_2F_{11})_2}$](0.5)</Checkpoint>

<Checkpoint>[选项E正确](0.5)</Checkpoint>

<Checkpoint>[选项F错误](0.5)</Checkpoint>