In nature, $\bf {A}$ often exists in trace amounts dispersed in ores such as bauxite and sphalerite. Bauxite is placed in an autoclave and treated with sodium hydroxide solution to transfer A into the solution; $\mathrm {CO} _ {2}$ is bubbled into it and filtered to remove a large amount of aluminum; $\mathrm {CO} _ {2}$ is continuously bubbled in until $\mathrm {pH} < 9.7$, precipitating compound $\boldsymbol { \bf {B} }$; $\bf {B}$ is placed under high temperature for dehydration, and then treated with thionyl chloride to obtain a white powder $\bf {C}$; $\bf {C}$ is directly mixed with triethylsilane at low temperature to generate $\bf {D}$, which is a liquid dimer at room temperature with a melting point of $29^{ \circ } \mathrm {C}$; $\bf {D}$ is heated to $150 ^ { \circ } \mathrm {C}$ for complete decomposition, with a weight loss of $0.71\%$, releasing flammable gas $\bf{E}$ and converting into colorless solid crystal $\bf{F}$, which is a $1:1$ type ionic compound. Triethylphosphine is added to an ether solution of $\bf {D}$, cooled, and a precipitate $\bf {G}$ is precipitated, which is a $1:1$ type $Lewis$ acid-base adduct. Triethylphosphine is slowly added to a toluene solution of $\boldsymbol { \bf { F } }$ at $ -78 ^ { \circ } \mathrm { C }$, the solution turns red; it is raised to room temperature and cooled again, the red color fades, and a colorless solid $\bf {H}$ is formed, which contains $\bf {A}$ $26.9\%$, and phosphorus has only one chemical environment.
自然界中，$\bf {A}$ 常以微量分散于铝土矿、闪锌矿等矿石中。将铝土矿置于高压釜中，用氢氧化钠溶液处理使A转移到溶液中；向其中通入 $\mathrm {CO} _ {2}$ 并过滤以除去大量的铝；继续通入 $\mathrm {CO} _ {2}$ 至$\mathrm {pH} < 9.7$ ，析出化合物 $\boldsymbol { \bf {B} }$ ；将$\bf {B}$置于高温下脱水，然后用亚硫酰氯处理，可制得白色粉末$\bf {C}$；将 $\bf {C}$ 与三乙基甲硅烷在低温下直接混合，生成 $\bf {D}$ ， $\bf{D}$在室温下为液态二聚体，熔点$29^{ \circ } \mathrm {C}$ ；将  $\bf {D}$ 加热至 $150 ^ { \circ } \mathrm {C}$ 完全分解，失重 $0.71 \%$ ，放出可燃气体$\bf{E}$并转化为无色固体结晶$\bf{F}$，$\bf {F}$ 为$1:1$型离子化合物。将三乙基麟加入到  $\bf {D}$ 的乙醚溶液中，冷却，析出沉淀$\bf {G}$，$\bf {G}$为$1:1$型$Lewis$酸碱加合物。在$ -78 ^ { \circ } \mathrm { C }$ 下将三乙基膦缓慢加入到 $\boldsymbol { \bf { F } }$ 的甲苯溶液中，溶液变红;升高至室温并再次冷却，红色褪去，生成无色固体$\bf {H}$，$\bf {H}$中含 $\bf {A}$ $26.9 \%$ ，磷只有一种化学环境。
All other options are incorrect.
$\bf{B}$ and $\bf{F}$ contain the same number of elements.
$\bf{D}$ and $\bf{H}$ molecules possess similar structures.
Compound $\bf{C}$ contains $\bf{A}$ $20. 2 \%$.
The most stable conformation of the $\bf{H}$ molecule contains only one center of symmetry.
$\bf{F}$ contains only one valence state of $\bf{A}$.
其他选项均不正确
$\bf{B}$ 和 $\bf{F}$ 所含元素种数相同。
$\bf{D}$ 和 $\bf{H}$分子具有相似的结构。
化合物 $\bf{C}$ 中含 $\bf{A}$ $20. 2 \%$.
$\bf{H}$分子的最稳定构象中仅含有一个对称中心。
$\bf{F}$ 中仅含有一种价态的 $\bf{A}$。
$\bf{A}$ is present in trace amounts dispersed in bauxite (mainly containing $\mathrm { Al } $) and sphalerite (mainly containing $\mathrm { Zn } $). Since the coexistence of elements in the same group in ore crystals is very common in nature, it can be inferred that it is an element in the same group as $\mathrm { Al } $ or $\mathrm { Zn } $.

By treating bauxite with sodium hydroxide and $\mathrm { CO_ {2}} $, $\bf{A}$ enters the solution and forms compound $\bf{B}$. It can be seen that $\bf{A}$ has similar properties to $\mathrm { Al } $, that is, it can be dissolved in alkaline solutions. It is speculated that it may be a group IIIA element. Since $\mathrm { In } $ and $\mathrm { Tl } $ are not particularly close to $\mathrm { Al } $ in chemical properties, it is especially likely to be $\mathrm { Ga } $.

<Checkpoint>[So element $\bf{A}$ may be $\mathrm { Ga } $.] (1)</Checkpoint>

Therefore, carbon dioxide is passed into the solution dissolved in sodium hydroxide to adjust the $\bf{pH}$. Since the hydroxides of $\mathrm { Al } $ and $\mathrm { Ga } $ are both amphoteric, and the $\bf{pH}$ at which the two precipitate are different, $\mathrm { Ga(OH)_{3} } $ can be separated out.

<Checkpoint>[So $\bf{B}$ is $\mathrm { Ga(OH)_{3} } $.] (1)</Checkpoint>

Dehydrating $\bf{B}$ at high temperature yields its oxide $\mathrm { Ga_{2} O_{3}} $.

Treatment with thionyl chloride is a one-step chlorination reaction, producing $\bf{C}$:
$\mathrm { G a _ { 2 } O _ { 3 } + 3 \ S O C l _ { 2 } \rightarrow 2 \ G a C l _ { 3 } + 3 \ S O _ { 2 } }$

<Checkpoint>[So $\bf{C}$ is $\mathrm { GaCl_{3} } $.] (1)</Checkpoint>

$\bf{C}$ $\mathrm { GaCl_{3} } $ reacts with triethylsilane, and the result is to replace $\mathrm { Cl } $ in $\mathrm { GaCl_{3} } $ with $\mathrm {H} $ to generate $\bf{D}$.

Furthermore, based on the small decomposition weight loss of $\bf{D}$ and the flammability of the generated gas, it is speculated that $\mathrm {H} $ in $\bf{D}$ may leave in the form of hydrogen gas. Using data for calculation, it can be obtained that: $\bf{D}$ is $\mathrm { GaHCl_{2} } $, and $\bf{F}$ is $\mathrm { GaCl_{2} } $. So the generated gas $\bf{E}$ is $\mathrm { H_{2} } $.

Then, substitute the data given in the question for verification:

$\frac{1.008}{1.008+69.72+35.45\times2}=0.7117 \%$, which is consistent with the meaning of the question, verifying the assumption.

<Checkpoint>[$\bf{D}$ is $\mathrm { GaHCl_{2} } $, $\bf{F}$ is $\mathrm { GaCl_{2} } $, $\bf{E}$ is $\mathrm { H_{2} } $.] (3)</Checkpoint>

$\bf{C}$ to $\bf{D}$ in one step:

$$
2 \mathrm{GaCl}_{3}+2\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{SiH} \rightarrow\left(\mathrm{GaHCl}_{2}\right)_{2}+2\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{SiCl}
$$

 $\bf{D}$ decomposition in one step:
$\bf{F}$ is a $1:1$ ionic compound. Since $\mathrm { Ga } $ mainly exists in monovalent or trivalent forms, $\mathrm { Ga } $ in $\bf{F}$ exists in monovalent and trivalent forms in a $1:1$ ratio, that is, this step of the reaction is:
$$
\left(\mathrm{GaHCl}_{2}\right)_{2} \rightarrow[\mathrm{Ga}]\left[\mathrm{GaCl}_{4}\right]+\mathrm{H}_{2}
$$

$\bf{D}$ reacts with triethylphosphine to generate the $Lewis$ adduct $\bf{G}$. Triethylphosphine is a $Lewis$ base, so it should be that the monomer after dissociation of $\bf{D}$ undergoes addition as a $Lewis$ acid.

<Checkpoint>[So $\bf{G}$ is $\mathrm{HCl}_{2} \mathrm{Ga}-\mathrm{PEt}_{3}$.] (1)</Checkpoint>

$\bf{F}$ reacts with triethylphosphine, presumably triethylphosphine first coordinates with monovalent $\mathrm{Ga}$, and after raising the temperature, $\bf{H}$ is generated. Therefore, it is speculated that $\bf{H}$ should contain triethylphosphine, $\mathrm{Ga}$, and $\mathrm { Cl } $.

Calculate according to the given data:

Suppose one $\mathrm{Ga}$ is coordinated with $x$ triethylphosphines and $y$ $\mathrm { Cl } $s, then there is the equation:

$\frac{69.72}{69.72+x\times118.15+y\times35.45}=26.9\%$

Solving for: $x=1，y=2$.

The simplest formula for $\bf{H}$ is $\mathrm{GaCl_{2}PEt_{3}}$.

Since there is only one chemical environment for phosphorus, it is speculated to be a polymer. The $\mathrm{Ga}$ in this compound also exhibits a $+2$ valence and has a single electron, so it may take a dimeric form.

<Checkpoint>[$\bf{H}$ is $\mathrm{(GaCl_{2}PEt_{3})_{2}}$.] (1)</Checkpoint>

$\bf{B}$ is $\mathrm { Ga(OH)_{3} } $, containing 3 kinds of elements, $\bf{F}$ is $\mathrm { Ga_{2}Cl_{4} } $, containing 2 kinds of elements, which are different.

<Checkpoint>[$\bf{B}$ is incorrect.] (1)</Checkpoint>

$\bf{D}$ is a dimer of $\mathrm { GaHCl_{2} } $, and its structure is similar to that of aluminum trichloride, that is, chlorine atoms act as bridges.

The $\bf{H}$ molecule is due to the fact that $+2$ valent $\mathrm { Ga}$ has a single electron, and the dimer contains a $\mathrm { Ga-Ga}$ bond. The overall bonding mode is similar to that of ethane. There are no bridging ligands, and the two are different.

<Checkpoint>[$\bf{C}$ is incorrect.] (1)</Checkpoint>

The compound $\bf{C}$ is calculated to contain $\bf{A}$ :

$\frac{69.72}{69.72+3\times35.45}=39.6\%$.

<Checkpoint>[$\bf{D}$ is incorrect.] (1)</Checkpoint>

The most stable conformation of the $\bf{H}$ molecule is staggered, containing a center of symmetry and a $\bf{C_{2}}$ symmetry axis, as shown in the figure:

<MultiModal>![分子SMILES表达式：Cl[Ga-]([P+](CC)(CC)CC)(Cl)[Ga-](Cl)(Cl)[P+](CC)(CC)CC](/media/uploads/1a5e8fdb-ba69-4c5c-ba65-e2a8547287f8.png)</MultiModal>

<Checkpoint>[$\bf{E}$ is incorrect.] (1)</Checkpoint>

It can be seen from the above reasoning that $\bf{F}$ contains $2$ kinds of valence states of $\bf{A}$.

<Checkpoint>[$\bf{F}$ is incorrect.] (1)</Checkpoint>

All other answers are incorrect.

<Checkpoint>[$\bf{A}$ is correct.] (1)</Checkpoint>
$\bf{A}$ 微量分散于铝土矿（主要含$\mathrm { Al } $）和闪锌矿（主要含$\mathrm { Zn } $）中。由于在自然界中，同族元素在矿石晶体中共生非常常见，可推测为$\mathrm { Al } $或$\mathrm { Zn } $的同族元素。

通过氢氧化钠和$\mathrm { CO_ {2}} $处理铝土矿，$\bf{A}$进入溶液并形成化合物$\bf{B}$。可见$\bf{A}$具有与$\mathrm { Al } $相似的性质，即可溶解在碱溶液中，推测其可能为第三主族元素，由于$\mathrm { In } $和$\mathrm { Tl } $在化学性质上与$\mathrm { Al } $不算特别接近，故尤其可能是$\mathrm { Ga } $。

<Checkpoint>[所以元素$\bf{A}$有可能是$\mathrm { Ga } $。](1)</Checkpoint>

故向溶解于氢氧化钠的溶液中通入二氧化碳调节$\bf{pH}$，由于$\mathrm { Al } $ 和 $\mathrm { Ga } $的氢氧化物均具有两性，二者沉淀$\bf{pH}$不同，即可分离出$\mathrm { Ga(OH)_{3} } $。

<Checkpoint>[所以 $\bf{B}$ 为$\mathrm { Ga(OH)_{3} } $。](1)</Checkpoint>

将 $\bf{B}$ 在高温下脱水，即可得到其氧化物 $\mathrm { Ga_{2} O_{3}} $。

使用亚硫酰氯处理，是一步氯化反应，生成$\bf{C}$：
$\mathrm { G a _ { 2 } O _ { 3 } + 3 \ S O C l _ { 2 } \rightarrow 2 \ G a C l _ { 3 } + 3 \ S O _ { 2 } }$

<Checkpoint>[所以 $\bf{C}$ 为$\mathrm { GaCl_{3} } $。](1)</Checkpoint>

$\bf{C}$ $\mathrm { GaCl_{3} } $与三乙基甲硅烷反应，从结果上是将$\mathrm { GaCl_{3} } $中的$\mathrm { Cl } $替换为$\mathrm {H} $生成$\bf{D}$。

再由$\bf{D}$的分解失重较少且生成气体可燃，推测可能是$\bf{D}$中$\mathrm {H} $以氢气的形式离去，使用数据进行计算可得：$\bf{D}$为$\mathrm { GaHCl_{2} } $，$\bf{F}$为$\mathrm { GaCl_{2} } $。所以生成的气体$\bf{E}$为$\mathrm { H_{2} } $。

再代入题目中所给数据，进行验证：

$\frac{1.008}{1.008+69.72+35.45\times2}=0.7117 \%$，与题意符合，验证假设。

<Checkpoint>[$\bf{D}$为$\mathrm { GaHCl_{2} } $，$\bf{F}$为$\mathrm { GaCl_{2} } $，$\bf{E}$为$\mathrm { H_{2} } $。](3)</Checkpoint>

$\bf{C}$ 到 $\bf{D}$ 一步：

$$
2 \mathrm{GaCl}_{3}+2\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{SiH} \rightarrow\left(\mathrm{GaHCl}_{2}\right)_{2}+2\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{SiCl}
$$

 $\bf{D}$ 分解一步：
$\bf{F}$ 为$1:1$离子化合物，由于 $\mathrm { Ga } $ 基本以一价或三价存在，故$\bf{F}$中的$\mathrm { Ga } $ 以一价和三价$1:1$的形式存在，即此步反应为：
$$
\left(\mathrm{GaHCl}_{2}\right)_{2} \rightarrow[\mathrm{Ga}]\left[\mathrm{GaCl}_{4}\right]+\mathrm{H}_{2}
$$

$\bf{D}$与三乙基膦反应生成$Lewis$加合物$\bf{G}$，三乙基膦为$Lewis$碱，故应该是$\bf{D}$解离后单体作为$Lewis$酸进行加和。

<Checkpoint>[故$\bf{G}$为$\mathrm{HCl}_{2} \mathrm{Ga}-\mathrm{PEt}_{3}$。](1)</Checkpoint>

 $\bf{F}$与三乙基膦反应，应该是三乙基膦先与一价$\mathrm{Ga}$配位，升高温度后生成$\bf{H}$，故推测$\bf{H}$中应该含有三乙基膦、$\mathrm{Ga}$和$\mathrm { Cl } $。

根据给出数据计算：

设一个$\mathrm{Ga}$与$x$个三乙基膦配位，与$y$个$\mathrm { Cl } $ 配位，有方程：

$\frac{69.72}{69.72+x\times118.15+y\times35.45}=26.9\%$

解得：$x=1，y=2$。

$\bf{H}$最简式为$\mathrm{GaCl_{2}PEt_{3}}$。

由于磷只有一种化学环境，推测为多聚体，该化合物中的$\mathrm{Ga}$同样表现为$+2$价，有一个单电子，故可能采取二聚的形式。

<Checkpoint>[$\bf{H}$为$\mathrm{(GaCl_{2}PEt_{3})_{2}}$。](1)</Checkpoint>

$\bf{B}$ 为$\mathrm { Ga(OH)_{3} } $，所含元素种数为3， $\bf{F}$为$\mathrm { Ga_{2}Cl_{4} } $所含元素种数为2，不同。

<Checkpoint>[$\bf{B}$错误。](1)</Checkpoint>

$\bf{D}$为$\mathrm { GaHCl_{2} } $二聚体，其结构与三氯化铝类似，即氯原子作为桥连。

$\bf{H}$分子则是由于$+2$价的 $\mathrm { Ga}$具有单电子，二聚体含有$\mathrm { Ga-Ga}$键，总体成键模式与乙烷类似。无桥连配体，二者不同。

<Checkpoint>[$\bf{C}$错误。](1)</Checkpoint>

计算得化合物 $\bf{C}$ 中含 $\bf{A}$ ：

$\frac{69.72}{69.72+3\times35.45}=39.6\%$.

<Checkpoint>[$\bf{D}$错误。](1)</Checkpoint>

$\bf{H}$分子的最稳定构象为交叉式，含有一个对称中心和$\bf{C_{2}}$对称轴，如图所示：

<MultiModal>![分子SMILES表达式：Cl[Ga-]([P+](CC)(CC)CC)(Cl)[Ga-](Cl)(Cl)[P+](CC)(CC)CC](/media/uploads/1a5e8fdb-ba69-4c5c-ba65-e2a8547287f8.png)</MultiModal>

<Checkpoint>[$\bf{E}$错误。](1)</Checkpoint>

由上述推理可知，$\bf{F}$ 中含有$2$种价态的 $\bf{A}$。

<Checkpoint>[$\bf{F}$错误。](1)</Checkpoint>

其他答案均不正确。

<Checkpoint>[$\bf{A}$正确。](1)</Checkpoint>