The chloride $\bf{A}$ of a first-row transition metal $\bf{M}$ is a red paramagnetic liquid with a tetrahedral molecular geometry and a magnetic moment of $\mu = 1.7 B.M.$ When $\bf{A}$ is heated in an inert atmosphere, it releases chlorine gas and forms compound $\bf{B}$; in $\bf{B}$, chlorine forms a hexagonal close-packed structure, and metal atoms fill one-third of the octahedral voids. Under a high-pressure carbon monoxide atmosphere, in the presence of bis(2-methoxyethyl) ether (which can be abbreviated as $\bf{L}$), $\bf{B}$ is reduced by metallic $\rm{Na}$ to obtain a 1:1 salt $\bf{C}$; the cation of $\bf{C}$ is an octahedral coordination ion of $\rm{Na}$, and the anion is an octahedral coordination ion of $\bf{M}$. $\bf{C}$ reacts with a large amount of dilute phosphoric acid, releasing hydrogen gas and obtaining a green solid $\bf{D}$; $\bf{D}$ is paramagnetic, sublimable, and easily ignites. $\bf{D}$ undergoes disproportionation in tetrahydrofuran (which can be abbreviated as $\rm{THF}$), and the disproportionation product $\bf{E}$ can be recrystallized from dichloromethane; $\bf{E}$ is a 1:2 salt, and its cation is connected to 2 anions through the electrostatic interaction of $\bf{M} \cdots \rm{O}$, so that all $\bf{M}$ in its structure are octahedrally coordinated. Based on the above information, select the correct option below.
第一过渡系金属$\bf{M}$的氯化物$\bf{A}$为红色顺磁性液体，其分子为四面体构型，磁矩 $\mu = 1.7 B.M.$。$\bf{A}$在惰性气氛中加热时，放出氯气，生成化合物$\bf{B}$；$\bf{B}$中氯做六方最密堆积，金属原子填入三分之一的八面体空隙。在高压的一氧化碳气氛下，若有二乙二醇二甲醚，以下可简记为$\bf{L}$)存在，$\bf{B}$被金属$\rm{Na}$ 还原，得到1:1型盐$\bf{C}$;$\bf{C}$ 的阳离子为$\rm{Na}$的八面体配离子，阴离子为$\bf{M}$的八面体配离子。$\bf{C}$与大量稀磷酸反应，放出氢气并得到绿色固体$\bf{D}$；$\bf{D}$具有顺磁性，可升华，易着火。$\bf{D}$在四氢呋喃(可简记为$\rm{THF}$)中发生歧化，用二氯甲烷可重结晶出歧化产物$\bf{E}$；$\bf{E}$为$1:2$型盐，其阳离子通过 $\bf{M} \cdots \rm{O}$ 的静电作用和2个阴离子相连接，使得在其结构中所有的$\bf{M}$均为八面体配位。根据以上信息，选出下列正确选项。
All other options are incorrect
$\bf{A}$ contains 2 unpaired electrons
$\bf{M}$'s common oxides do not include $\bf{M_2}\rm{O_5}$
$\bf{B}$ contains 2 chlorine atoms.
$\bf{C}$'s cation contains three diglyme molecules.
The anion of $\bf{C}$ possesses strong oxidizing properties.
The structure of $\bf{D}$ is octahedral and contains four elements.
The anion in $\bf{E}$ is not the same as the anion in $\bf{C}$.
In $\bf{E}$, for the $\bf{M}$ with the higher oxidation state, the ligands attached to it have two different coordination environments
其他选项均不正确
$\bf{A}$中有2个未成对电子
$\bf{M}$的常见氧化物中不存在$\bf{M_2}\rm{O_5}$
$\bf{B}$中含有2个氯
$\bf{C}$的阳离子中含有三个二乙二醇二甲醚
$\bf{C}$的阴离子具有较强的氧化性
$\bf{D}$的结构是八面体构型且含有四种元素
$\bf{E}$中的阴离子与$\bf{C}$中的阴离子不相同
$\bf{E}$中价态更高的$\bf{M}$的周围配体中配位原子有两种不同的化学环境
$\bf{A}$ being a tetrahedral configuration chloride indicates that the chemical formula of $\bf{A}$ is $\rm{MCl_4}$. According to the magnetic moment calculation formula $\mu = \sqrt{n(n + 2)}$, the number of unpaired electrons is 1, so option $\bf{B}$ is incorrect.

<Checkpoint>[The number of unpaired electrons is calculated to be 1 based on the magnetic moment](1)</Checkpoint>

Tetrahedral fields are usually high-spin, and since the number of unpaired electrons is 1, the metal $\bf{M}$ is in the +4 valence state. Therefore, it is inferred that the metal itself has 5 valence layer electrons, so $\bf{M}$ is $\rm{V}$. $\rm{V}$ has a common oxide $\rm{V_2O_5}$, so option $\bf{C}$ is incorrect.

<Checkpoint>[It is inferred that $\bf{M}$ is $\rm{V}$](1)</Checkpoint>

$\bf{B}$ is obtained from $\bf{A}$ by releasing chlorine gas, so it should also be a chloride of $\rm{V}$. At the same time, $\rm{V}$ occupies 1/3 of the octahedral voids formed by $\rm{Cl}$, so $\bf{B}$ is $\rm{VCl_3}$, and option $\bf{D}$ is incorrect.

<Checkpoint>[It is inferred that $\bf{B}$ is $\rm{VCl_3}$](1)</Checkpoint>

In $\bf{C}$, the cations and anions are octahedrally coordinated and formed in a high-pressure $\rm{CO}$ atmosphere. It is speculated that the anion is the $\rm{CO}$ hexa-coordinated $\rm{V}$. The cation is $\rm{Na}$ coordinated by the ligand $\bf{L}$. The ligand $\bf{L}$ is a tridentate ligand, so the cation chemical formula is $\rm{NaL_2}^+$, and the chemical formula of $\bf{C}$ is $\rm{[NaL_2][V(CO)_6]}$.

<Checkpoint>[It is inferred that $\bf{C}$ is $\rm{[NaL_2][V(CO)_6]}$](2)</Checkpoint>

$\bf{C}$ reacts with acid to release hydrogen gas to generate $\bf{D}$, indicating that $\bf{C}$ is oxidized. It should be $\rm{V(CO)_6}^-$ losing electrons to generate $\rm{V(CO)_6}$. In option $\bf{G}$, it is correct that $\bf{D}$ has an octahedral configuration, but it only contains three elements. $\bf{G}$ is incorrect.

<Checkpoint>[It is inferred that $\bf{D}$ is $\rm{V(CO)_6}$](1)</Checkpoint>

$\bf{D}$ undergoes a disproportionation reaction to generate $\bf{E}$. The anion should be $\rm{V(CO)_6}^-$. Since the ratio of cations to anions is 2:1, the cation center is $\rm{V^{2+}}$. Also, because the cation is connected to the $\rm{O}$ of two anions, there are four remaining coordination sites, which should be 4 $\rm{THF}$. Therefore, $\bf{E}$ is $\rm{[V(THF)_4][V(CO)_6]_2}$. So, option $\bf{E}$ states that the cation only contains two diethylene glycol dimethyl ethers, which is incorrect. In option $\bf{H}$, $\bf{E}$ and $\bf{C}$ have the same anion, which is incorrect.

In option $\bf{F}$, $\rm{V(CO)_6}^-$ has a negative charge and has strong reducing properties rather than oxidizing properties, so $\bf{F}$ is incorrect.

The structure of $\bf{E}$ is linear, with the cation coordinated by 4 tetrahydrofuran molecules sandwiched between two anionic complexes. One $\rm{CO}$ from each of the two $\rm{V(CO)_6}^-$ participates in the coordination of the central cation. There are two types of $\rm{O}$ environments, $\rm{THF}$ and $\rm{CO}$, participating in the cation coordination, so option $\bf{I}$ is correct.

<Checkpoint>[It is inferred that $\bf{E}$ is $\rm{[V(THF)_4][V(CO)_6]_2}$](2)</Checkpoint>


<Checkpoint>[The structure of $\bf{E}$ is linear, with the cation coordinated by 4 tetrahydrofuran molecules sandwiched between two anionic complexes. One $\rm{CO}$ from each of the two $\rm{V(CO)_6}^-$ participates in the coordination of the central cation](2)</Checkpoint>
$\bf{A}$为四面体构型氯化物说明$\bf{A}$的化学式为 $\rm{MCl_4}$，根据磁矩计算公式 $\mu = \sqrt{n(n + 2)}$，得出单电子数量为1，$\bf{B}$选项不对

<Checkpoint>[根据磁矩计算出单电子数量为1](1)</Checkpoint>

四面体场通常都为高自旋，又因为单电子数量为1，金属 $\bf{M}$为+4价，所以推测出金属本身有5个价层电子，因此$\bf{M}$为$\rm{V}$，$\rm{V}$存在常见氧化物$\rm{V_2O_5}$，$\bf{C}$选项不对

<Checkpoint>[推断出$\bf{M}$为$\rm{V}$](1)</Checkpoint>

$\bf{B}$由$\bf{A}$放出氯气得到，也应为$\rm{V}$的氯化物，同时$\rm{V}$占据$\rm{Cl}$构成的1/3八面体空隙，所以$\bf{B}$为$\rm{VCl_3}$，$\bf{D}$选项不对。

<Checkpoint>[推断出$\bf{B}$为$\rm{VCl_3}$](1)</Checkpoint>

$\bf{C}$中阴阳离子为八面体配位，在高压$\rm{CO}$气氛中生成，推测阴离子为$\rm{V}$的$\rm{CO}$六配位。阳离子为$\rm{Na}$被配体$\bf{L}$配位。配体$\bf{L}$为三齿配体，所以阳离子化学式为$\rm{NaL_2}^+$，$\bf{C}$化学式为$\rm{[NaL_2][V(CO)_6]}$

<Checkpoint>[推断出$\bf{C}$为$\rm{[NaL_2][V(CO)_6]}$](2)</Checkpoint>

$\bf{C}$与酸反应放出氢气生成$\bf{D}$，说明$\bf{C}$被氧化，应为$\rm{V(CO)_6}^-$失去电子，生成$\rm{V(CO)_6}$。$\bf{G}$选项中$\bf{D}$为八面体构型没错，但仅含有三种元素。
$\bf{G}$不对。

<Checkpoint>[推断出$\bf{D}$为$\rm{V(CO)_6}$](1)</Checkpoint>

$\bf{D}$发生歧化反应生成$\bf{E}$，阴离子应为$\rm{V(CO)_6}^-$，由于阴阳离子比例为2:1，所以阳离子中心为$\rm{V^{2+}}$，又因阳离子与两个阴离子的$\rm{O}$相连，所以剩余四个配位位点，应为4个$\rm{THF}$。所以$\bf{E}$为$\rm{[V(THF)_4][V(CO)_6]_2}$。所以$\bf{E}$选项阳离子中仅含两个二乙二醇二甲醚，$\bf{E}$不对。$\bf{H}$选项中$\bf{E}$和$\bf{C}$有相同阴离子，$\bf{H}$不对。

$\bf{F}$选项，$\rm{V(CO)_6}^-$具有负电荷，具有较强的还原性而非氧化性，$\bf{F}$不对。

$\bf{E}$的结构为直线型，两个阴配离子中间夹有被4个四氢呋喃配位的阳离子，两个$\rm{V(CO)_6}^-$各自的其中一个$\rm{CO}$参与中心阳离子配位，共有$\rm{THF}$和$\rm{CO}$两种环境的$\rm{O}$参与阳离子配位，$\bf{I}$选项正确

<Checkpoint>[推断出$\bf{E}$为$\rm{[V(THF)_4][V(CO)_6]_2}$](2)</Checkpoint>


<Checkpoint>[$\bf{E}$的结构为直线型，两个阴配离子中间夹有被4个四氢呋喃配位的阳离子，两个$\rm{V(CO)_6}^-$各自的其中一个$\rm{CO}$参与中心阳离子配位](2)</Checkpoint>