Transition metal $\bf{M}$ is difficult to extract due to its dispersed presence in nature. The highest-valent oxide $\bf{A}$ of $\bf{M}$ reacts with coke and chlorine gas at high temperature to produce $\bf{B}$. $\bf{B}$ reacts with $\bf{M}$ to produce $\bf{C}$. $\bf{C}$ is often used to catalyze ethylene polymerization. Pyrolysis of $\bf{C}$ can yield $\bf{B}$ and $\bf{D}$. The crystal of $\bf{D}$ has a layered structure, in which all $\bf{M}$ are octahedrally coordinated, and all octahedra are connected to six adjacent octahedra by edge-sharing. Heating $\bf{D}$ can also yield $\bf{M}$ and $\bf{B}$. $\bf{B}$ reacts with excess ethanol to produce $\bf{F}$. $\bf{F}$ is a tetramer, and the ratio of oxygen atoms with different coordination numbers is 1:2:5. $\bf{B}$ reacts with excess $\mathrm{NaCp}$ ( $\mathrm{Cp=C_5H_5}$ ) to produce $\bf{G}$ or $\bf{H}$. Heating $\bf{G}$ can also produce $\bf{H}$. $\bf{G}$ satisfies the 16-electron rule, while $\bf{H}$ contains two bridging hydrogen atoms.

The following statements are made:

1. The minimum mass fraction of $\bf{M}$ in $\bf{A-D}$ is less than 25%.

2. Ignoring carbon and hydrogen atoms, the point group of $\bf{F}$ is $D_{2h}$.

3. There are four chemically distinct types of hydrogen in $\bf{G}$.

4. $\bf{M}$ in $\bf{H}$ satisfies the 18-electron rule.

Then the following options contain all the correct statements:
过渡金属 $\bf{M}$ 由于在自然界中存在分散而较难提取。$\bf{M}$ 的最高价氧化物 $\bf{A}$ 在高温下与焦炭和氯气反应生成 $\bf{B}$，$\bf{B}$ 与 $\bf{M}$ 反应生成 $\bf{C}$，$\bf{C}$ 常用于催化乙烯聚合。$\bf{C}$ 热解可以得到 $\bf{B}$ 和 $\bf{D}$，$\bf{D}$ 的晶体为层状结构，其中 $\bf{M}$ 均为八面体配位，所有八面体均与相邻的六个八面体共棱连接。$\bf{D}$ 加热又可得到 $\bf{M}$ 和 $\bf{B}$。$\bf{B}$ 与足量乙醇反应生成 $\bf{F}$，$\bf{F}$ 为四聚体，不同配位数的氧原子的比例为 $1:2:5$。$\bf{B}$ 与足量 $\mathrm{NaCp}$（ $\mathrm{Cp=C_ 5H_5}$ ）反应，生成 $\bf{G}$ 或 $\bf{H}$，$\bf{G}$ 受热分解也可以产生 $\bf{H}$。$\bf{G}$ 满足16电子规则，而 $\bf{H}$ 中存在两个氢桥键。

有下列说法：

1. $\bf{A-D}$ 中 $\bf{M}$ 的质量分数的最小值小于25%。

2. 忽略碳原子与氢原子，则 $\bf{F}$ 所属点群为 $D_{2h}$。

3. $\bf{G}$ 中有四种化学环境的氢。

4. $\bf{H}$ 中 $\bf{M}$ 满足18电子规则。

则下列选项中包含所有正确说法的是：
All other options are incorrect
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
其他选项均不正确
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
Based on the fact that $\bf{C}$ is often used to catalyze ethylene polymerization, it can be inferred that the core element of this question is $\mathrm{Ti}$.

<Checkpoint>[$\bf{M}$ is $\mathrm{Ti}$](1)</Checkpoint>

Therefore, it can be immediately obtained that $\bf{A}$ is $\mathrm{TiO_2}$, $\bf{B}$ is $\mathrm{TiCl_4}$, $\bf{C}$ is $\mathrm{TiCl_3}$, and $\bf{D}$ is $\mathrm{TiCl_2}$. Among them, the mass fraction of $\mathrm{Ti}$ is the smallest in $\mathrm{TiCl_4}$, which is 25.23%.

<Checkpoint>[The minimum mass fraction of $\bf{M}$ in $\bf{A-D}$ is 25.23%](1)</Checkpoint>

$\bf{B}$ reacts with excess ethanol to generate $\bf{F}$. $\bf{F}$ is a tetramer. Based on the oxygen atom ratio, it can be deduced that $\bf{F}$ is a tetramer of $\mathrm{Ti(C_2H_5O)_4}$. From the oxygen atom ratio, it can be deduced that the structure of $\bf{F}$ is (carbon and hydrogen atoms are omitted in the figure):

<MultiModal>![CCO1[Ti]2(OCC)(OCC)(O([Ti]3(OCC)(O4CC)(O2([Ti]15(OCC)(O([Ti]4(OCC)(OCC)(O53CC)OCC)CC)OCC)CC)OCC)CC)OCC](/media/uploads/b63f55d9-0feb-4d8f-8f2a-14a67af36ac8.png)</MultiModal>

This structure has a $C_2$ axis and a mirror plane perpendicular to it, with $C_{2h}$ symmetry.

<Checkpoint>[Ignoring carbon and hydrogen atoms, the point group to which $\bf{F}$ belongs is $C_{2h}$](1)</Checkpoint>

$\bf{B}$ reacts with excess $\mathrm{NaCp}$ to generate $\bf{G}$ or $\bf{H}$. $\bf{G}$ should be a mononuclear complex, so it is speculated that $\bf{G}$ is $\rm{TiCp_4}$. Since $\bf{G}$ satisfies the 16-electron rule, it can be obtained that two $\rm{Cp}$ provide 6 electrons and two $\rm{Cp}$ provide 2 electrons. Its structure is easily obtained as:

<MultiModal>![The figure shows the structure of $\rm{TiCp_4}$, where two $\rm{Cp}$ ligands provide 6-electron coordination and two $\rm{Cp}$ ligands provide 2-electron coordination.](/media/uploads/08b6c6be-5245-4adc-bde2-f5f5c4f86135.png)</MultiModal>

There are 4 kinds of hydrogen with different chemical environments.

<Checkpoint>[There are four kinds of hydrogen with different chemical environments in $\bf{G}$.](1)</Checkpoint>

$\bf{G}$ can also generate $\bf{H}$ upon thermal decomposition. There are two hydrogen bridge bonds in $\bf{H}$, so it can be inferred that a dehydrogenative coupling reaction of $\rm{Cp}$ has occurred. Therefore, the chemical formula of $\bf{H}$ can be obtained as $\mathrm{Ti_2(Cp-Cp)Cp_2H_2}$, and its structure is:

<MultiModal>![The figure shows the structure of $\mathrm{Ti_2(Cp-Cp)Cp_2H_2}$, where two $\rm{Cp}$ rings in $\rm{Cp-Cp}$ provide 6-electron coordination to the two $\rm{Ti}$ respectively to form a bridge, two $\rm{Cp}$ provide 6-electron coordination to the two $\rm{Ti}$, and two $\rm{H}$ both connect the two $\rm{Ti}$ as bridging ligands.](/media/uploads/eaf5674d-007e-4205-89d1-79d7c72a1e0e.png)</MultiModal>

The number of electrons around $\rm{Ti}$ is 17, which does not satisfy the 18-electron rule.

<Checkpoint>[$\bf{M}$ in $\bf{H}$ does not satisfy the 18-electron rule.](1)</Checkpoint>
根据 $\bf{C}$ 常用于催化乙烯聚合可以推测该题核心元素为 $\mathrm{Ti}$。

<Checkpoint>[$\bf{M}$ 为 $\mathrm{Ti}$](1)</Checkpoint>

从而立即得到 $\bf{A}$ 为 $\mathrm{TiO_2}$，$\bf{B}$ 为 $\mathrm{TiCl_4}$，$\bf{C}$ 为 $\mathrm{TiCl_3}$，$\bf{D}$ 为 $\mathrm{TiCl_2}$。其中 $\mathrm{Ti}$ 质量分数最小的为 $\mathrm{TiCl_4}$，为25.23%。

<Checkpoint>[$\bf{A-D}$ 中 $\bf{M}$ 的质量分数的最小值25.23%](1)</Checkpoint>

$\bf{B}$ 与足量乙醇反应生成 $\bf{F}$，$\bf{F}$ 为四聚体，根据氧原子比例可以推出 $\bf{F}$ 为 $\mathrm{Ti(C_2H_5O)_4}$ 的四聚体，从氧原子比例可以推出 $\bf{F}$ 的结构为（图中省略碳氢原子）：

<MultiModal>![CCO1[Ti]2(OCC)(OCC)(O([Ti]3(OCC)(O4CC)(O2([Ti]15(OCC)(O([Ti]4(OCC)(OCC)(O53CC)OCC)CC)OCC)CC)OCC)CC)OCC](/media/uploads/b63f55d9-0feb-4d8f-8f2a-14a67af36ac8.png)</MultiModal>

该结构具有一根 $C_2$ 轴和一个与其垂直的镜面，具有 $C_{2h}$ 对称性。

<Checkpoint>[忽略碳原子与氢原子，则 $\bf{F}$ 所属点群为 $C_{2h}$](1)</Checkpoint>

$\bf{B}$ 与足量 $\mathrm{NaCp}$ 反应，生成 $\bf{G}$ 或 $\bf{H}$，$\bf{G}$ 应为单核配合物，因此推测 $\bf{G}$ 为 $\rm{TiCp_4}$，由于 $\bf{G}$ 满足16电子规则，因此可以得到两个 $\rm{Cp}$ 提供6电子，两个 $\rm{Cp}$ 提供2电子，易得其结构为：

<MultiModal>![图中展示了 $\rm{TiCp_4}$ 的结构，两个 $\rm{Cp}$ 提供6电子配位，两个 $\rm{Cp}$ 提供2电子配位。](/media/uploads/08b6c6be-5245-4adc-bde2-f5f5c4f86135.png)</MultiModal>

其中有4种化学环境的氢。

<Checkpoint>[$\bf{G}$ 中有四种化学环境的氢。](1)</Checkpoint>

$\bf{G}$ 受热分解也可以产生 $\bf{H}$，$\bf{H}$ 中存在两个氢桥键，因此可以推断发生了 $\rm{Cp}$ 的脱氢偶联反应，因此可以得 $\bf{H}$ 的化学式为 $\mathrm{Ti_2(Cp-Cp)Cp_2H_2}$，其结构为：

<MultiModal>![图中展示了 $\mathrm{Ti_2(Cp-Cp)Cp_2H_2}$ 的结构，$\rm{Cp-Cp}$ 中的两个 $\rm{Cp}$ 环分别对两个 $\rm{Ti}$ 提供6电子配位从而形成桥连，两个 $\rm{Cp}$ 对两个 $\rm{Ti}$ 提供6电子配位，两个 $\rm{H}$ 均作为桥连配体连接两个 $\rm{Ti}$。](/media/uploads/eaf5674d-007e-4205-89d1-79d7c72a1e0e.png)</MultiModal>

其中 $\rm{Ti}$ 周围的电子数为17，不满足18电子规则。

<Checkpoint>[$\bf{H}$ 中 $\bf{M}$ 不满足18电子规则。](1)</Checkpoint>