$\bf{M}$ is a gray-white metallic element, soluble in acids and bases, and highly toxic. It exists in nature in the form of silicate minerals with a $+2$ oxidation state. $\bf{M}$ reacts with chlorine gas to yield a white chloride $\bf{A}$; $\bf{A}$ reacts with tetraphenylphosphonium chloride in a $1:2$ molar ratio to yield $\bf{B}$, while reacting in a $1:1$ molar ratio yields $\bf{C}$. $\bf{B}$ and $\bf{C}$ are both $2:1$ type salts. When $\bf{A}$ is dissolved in a mixture of dinitrogen tetroxide and ethyl acetate at low temperature, $\bf{D}$ precipitates, containing only $+1$ valent cations and $-2$ valent anions, and no isolated solvent molecules; when $\bf{D}$ is heated to $50\degree C$, a reddish-brown gas $\bf{E}$ is released, and $\bf{F}$ is obtained; when $\bf{F}$ is further heated, a mixture of gases $\bf{E}$ and $\bf{G}$ is released, and compound $\bf{H}$ is obtained; $\bf{H}$ is a tetranuclear complex with the same symmetry as a methane molecule.
$\bf{M}$为灰白色金属单质,能溶于酸碱,有剧毒, 在自然界中以$+2$ 氧化态的硅酸盐矿物形式存在。$\bf{M}$ 与氯气反应,得到白色氯化物 $\bf{A}$;$\bf{A}$ 与四苯基氯化鏻按摩尔比 $1: 2$ 反应得到 $\bf{B}$ 而按摩尔比 $1: 1$ 反应则得到 $\bf{C}$, $\bf{B}$ 和 $\bf{C}$ 均为 $2: 1$ 型盐。低温下将 $\bf{A}$ 溶于四氧化二氮-乙酸乙酯混合溶剂中,析出仅含有$+1$ 价阳离子和$-2$ 价阴离子而不含孤立的溶剂分子的 $\bf{D}$;将 $\bf{D}$ 加热至 $50\degree C$, 放出红棕色气体 $\bf{E}$ 并得到 $\bf{F}$;将 $\bf{F}$ 继续加热,放出气体 $\bf{E}$ 和 $\bf{G}$ 的混合物,并 得到化合物 $\bf{H}$;$\bf{H}$ 为四核配合物,与甲烷分子具有相同的对称性。
All other options are incorrect.
Metal $\bf{M}$ is the metal with the weakest **X**-ray penetration ability.
The cations and anions in $\bf{B}$ possess different symmetries.
$\bf{B}$ and $\bf{C}$ have different coordination numbers for the metal $\bf{M}$.
$\bf{D}$, $\bf{F}$, and $\bf{H}$ contain exactly the same elements.
$\bf{H}$ contains only one kind of bidentate ligand.
$\bf{E}$ and $\bf{G}$ are diamagnetic at room temperature.
其他选项均不正确
金属 $\bf{M}$ 是透过**X**射线能力最弱的金属。
$\bf{B}$ 中阴阳离子具有不同的对称性。
$\bf{B}$ 、$\bf{C}$ 中金属 $\bf{M}$ 的配位数不同。
$\bf{D}$、$\bf{F}$ 、$\bf{H}$ 含有完全相同的元素。
$\bf{H}$ 中仅含一种双齿配体。
$\bf{E}$、$\bf{G}$ 在室温下为抗磁性。
$1$. Identification of Metal $\bf{M}$

$\bf{M}$ is a gray-white metallic element, soluble in both acids and bases (amphoteric), and highly toxic. It exists in nature in the $+2$ oxidation state.

Among common amphoteric metals, Aluminum $(\mathrm{Al})$ does not meet the oxidation state condition, Zinc $(\mathrm{Zn})$ is not highly toxic, and Tin $(\mathrm{Sn})$ exists in nature mostly in the $+4$ oxidation state, which does not meet the common natural oxidation state condition. Lead $(\mathrm{Pb})$ meets the natural oxidation state condition, but it mainly exists as the sulfide $\mathrm{PbS}$. None of these meet the requirements of the question.

At this point, consider the diagonal relationship. Due to the $+2$ valence state, consider Beryllium $\mathrm{Be}$, the diagonal of $\mathrm{Al}$, which is gray-white, amphoteric (soluble in acids and bases), highly toxic, and exists in nature in the +2 oxidation state (such as in beryl).

Therefore, $\bf{M}$ is beryllium ($\mathrm{Be}$).

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

$2$. Identification of Compound $\bf{A}$

$\mathrm{M}(\mathrm{Be})$ reacts with chlorine $\left(\mathrm{Cl}_{2}\right)$: $\mathrm{Be}+\mathrm{Cl}_{2} \rightarrow \mathrm{BeCl}_{2}$ .

$\bf{A}$ is a white chloride, consistent with the properties of $\mathrm{BeCl}_{2}$ (white solid).

Therefore, $\bf{A}$ is $\mathrm{BeCl}_{2}$ .

<Checkpoint>[Obtained the chemical formula of **A**](1)</Checkpoint>

3. Identification of Compounds $\bf{B}$ and $\bf{C}$

$\mathrm{A}\left(\mathrm{BeCl}_{2}\right)$ reacts with tetraphenylphosphonium chloride $\left(\mathrm{PPh}_{4} \mathrm{Cl}\right)$.

$\mathrm{PPh}_{4} \mathrm{Cl}$ provides $\mathrm{PPh}_{4}{ }^{+}$ and $\mathrm{Cl}^{-}, \mathrm{BeCl}_{2}$ is a Lewis acid and can form chloro complexes.

A $1:2$ molar ratio reaction yields $\bf{B}$:

$\mathrm{BeCl}_{2}+2 \mathrm{PPh}_{4} \mathrm{Cl} \rightarrow\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{BeCl}_{4}\right]$.

The anion $\left[\mathrm{BeCl}_{4}\right]^{2-}$ has a $-2$ charge, and the cation $\mathrm{PPh}_{4}^{+}$ has a $+1$ charge, forming a $2:1$ type salt (cation number:anion number $=2:1$).

Therefore, $\bf{B}$ is $\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{BeCl}_{4}\right]$.

<Checkpoint>[Obtained the chemical formula of $\bf{B}$](1)</Checkpoint>

A $1:1$ molar ratio reaction yields $\bf{C}$:

$2 \mathrm{BeCl}_{2}+2 \mathrm{PPh}_{4} \mathrm{Cl} \rightarrow\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]$.

$\mathrm{BeCl}_{2}$ can dimerize in solution to form the $\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]^{2-}$ anion (containing a chlorine-bridged structure).

The anion $\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]^{2-}$ has a $-2$ charge, and the cation $\mathrm{PPh}_{4}{ }^{+}$ has a $+1$ charge, forming a $2:1$ type salt (cation number:anion number $=2:1$).

Therefore, $\bf{C}$ is $\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]$.

<Checkpoint>[Obtained the chemical formula of $\bf{C}$](1)</Checkpoint>

4. Identification of Compound $\bf{D}$

$\mathrm{A}\left(\mathrm{BeCl}_{2}\right)$ is dissolved in a mixture of dinitrogen tetroxide $\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)$ and ethyl acetate, and $\bf{D}$ precipitates out at low temperature.

$\mathrm{N}_{2} \mathrm{O}_{4}$ dissociation: $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{NO}^{+}+ \mathrm{NO}_{3}^{-}$.

Ethyl acetate $\left(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right)$ may participate in the reaction or hydrolysis to provide acetate $\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)$, but at low temperature, the main reaction is the
displacement of chloride by nitrate.

$\bf{D}$ contains only $+1$ valent cations and $-2$ valent anions, with no isolated solvent molecules.

The cation is $\mathrm{NO}^{+}$ (from $\mathrm{N}_{2} \mathrm{O}_{4}$), and the anion is a -2 valent anion containing $\mathrm{Be}$.

$\mathrm{BeCl}_{2}$ reacts with $\mathrm{NO}_{3}{ }^{-}$ to form $\left[\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{4}\right]^{2-}$

Therefore, $\bf{D}$ is $\mathrm{(NO)_{2}[Be(NO_{3})_{4}]}$, consistent with a $2:1$ type salt.

<Checkpoint>[Obtained the chemical formula of $\bf{D}$](1)</Checkpoint>

5. Identification of Compounds $\bf{E}$, $\bf{F}$, $\bf{G}$, and $\bf{H}$

$\bf{D}$ is heated to $50\degree C$ to obtain $\bf{E}$ and $\bf{F}$:

$\mathrm{(NO)_{2}[Be(NO_{3})_{4}]}$ decomposes upon heating, releasing the reddish-brown gas $\mathrm{NO}_{2}$.

Therefore, $\bf{E}$ is $\mathrm{NO_{2}}$, and $\bf{F}$ is $\mathrm{Be(NO_{3})_{2}}$.

<Checkpoint>[Obtained the chemical formulas of $\bf{E}$ and $\bf{F}$](1)</Checkpoint>

Continued heating leads to the loss of $\mathrm{NO_{2}}$. Some of the nitrate ligands in the complex are converted to oxide ions. Furthermore, according to the question, this complex is similar to methane and has $T_{d}$ symmetry. Therefore, the $4$ $\mathrm{Be}$ atoms it contains are located at the vertices of a tetrahedron. From symmetry, the nitrate groups are distributed along the edges of the tetrahedron, with a total of 6. Valence calculation reveals that one $\mathrm{O^{2-}}$ ion can only be located at the center, coordinated to the $4$ $\mathrm{Be}$ atoms. In this case, each $\mathrm{Be}$ atom is in a tetrahedral coordination environment, which is more reasonable.

Therefore, $\bf{H}$ is $\mathrm{Be_{4}O(NO_{3})_{6}}$.

<Checkpoint>[Obtained the chemical formula of $\bf{H}$](1)</Checkpoint>

The metal $\mathrm{Be}$ has the strongest X-ray transmission ability.

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

The cation and anion in $\bf{B}$ are both tetrahedral and have the same symmetry.

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

The metal $\mathrm{Be}$ in $\bf{B}$ and $\bf{C}$ are both tetracoordinate, with the same coordination number.

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

$\bf{D}$, $\bf{F}$, and $\bf{H}$ contain exactly the same elements, namely $\mathrm{Be}$, $\mathrm{N}$, and $\mathrm{O}$.

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

$\bf{H}$ contains not only one type of bidentate ligand, but also an $\mathrm{O^{2-}}$ ion that can only be located at the center. The coordination number of this $\mathrm{O^{2-}}$ ion is $4$.

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

$\bf{E}$ and $\bf{G}$ both contain unpaired electrons, so they are both paramagnetic.

<Checkpoint>[$\bf{G}$ is incorrect.](1)</Checkpoint>
$1$. 金属 $\bf{M}$ 的鉴定

$\bf{M}$ 为灰白色金属单质,能溶于酸和碱 (两性),有剧毒，在自然界中以 $+2$ 氧化态存在。

常见两性金属中， 铝 $(\mathrm{Al})$ 不满足氧化态条件， 锌 $(\mathrm{Zn})$ 无剧毒， 锡 $(\mathrm{Sn})$在自然界中的氧化态绝大部分为 $+4$ 价,不符合常见自然界氧化态条件 、铅 $(\mathrm{Pb})$ 符合自然界氧化态条件，但其主要以硫化物 $\mathrm{PbS}$ 存在。均不满足题意。

此时考虑对角线原则，由于$+2$价态，考虑$\mathrm{Al}$的对角线铍$\mathrm{Be}$，，灰白色，两性(溶于酸和碱)，剧毒，自然界中以+2氧化态存在(如绿柱石)。

因此，$\bf{M}$为铍($\mathrm{Be}$)。

<Checkpoint>[推理得到$\bf{M}$为$\mathrm{Be}$](1)</Checkpoint>

$2$.化合物$\bf{A}$的鉴定

$\mathrm{M}(\mathrm{Be})$ 与氯气 $\left(\mathrm{Cl}_{2}\right)$ 反应: $\mathrm{Be}+\mathrm{Cl}_{2} \rightarrow \mathrm{BeCl}_{2}$ 。

$\bf{A}$ 为白色氯化物, 符合 $\mathrm{BeCl}_{2}$ 的性质(白色固体)。

因此, $\bf{A}$ 为 $\mathrm{BeCl}_{2}$ 。

<Checkpoint>[得到 **A** 的化学式 。](1)</Checkpoint>

3.化合物$\bf{B}$和$\bf{C}$的鉴定

$\mathrm{A}\left(\mathrm{BeCl}_{2}\right)$ 与四苯基氯化鳞 $\left(\mathrm{PPh}_{4} \mathrm{Cl}\right)$ 反应。

$\mathrm{PPh}_{4} \mathrm{Cl}$ 提供 $\mathrm{PPh}_{4}{ }^{+}$和 $\mathrm{Cl}^{-}, \mathrm{BeCl}_{2}$ 为路易斯酸,可形成氯配合物。

摩尔比$1:2$反应得$\bf{B}$

$\mathrm{BeCl}_{2}+2 \mathrm{PPh}_{4} \mathrm{Cl} \rightarrow\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{BeCl}_{4}\right]$ 。

阴离子 $\left[\mathrm{BeCl}_{4}\right]^{2-}$ 为 $-2$价, 阳离子 $\mathrm{PPh}_{4}^{+}$为 $+1$ 价, 形成 $2: 1$ 型盐 (阳离子数:阴离子数 $=2: 1$ )。

因此, $\bf{B}$ 为 $\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{BeCl}_{4}\right]$ 。

<Checkpoint>[得到 $\bf{B}$ 的化学式 。](1)</Checkpoint>

摩尔比 $1:1$ 反应得 $\bf{C}$:

$2 \mathrm{BeCl}_{2}+2 \mathrm{PPh}_{4} \mathrm{Cl} \rightarrow\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]$ 。

$\mathrm{BeCl}_{2}$ 在溶液中可二聚形成 $\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]^{2-}$阴离子(含氯桥连结构)。

阴离子 $\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]^{2-}$ 为 $-2$ 价, 阳离子 $\mathrm{PPh}_{4}{ }^{+}$为 $+1$ 价，形成$ 2:1 $型盐 (阳离子数:阴离子数 $=2: 1$ )。

因此， $\bf{C}$ 为 $\left(\mathrm{PPh}_{4}\right)_{2}\left[\mathrm{Be}_{2} \mathrm{Cl}_{6}\right]$ 。

<Checkpoint>[得到 $\bf{C}$ 的化学式 。](1)</Checkpoint>

4. 化合物 $\bf{D}$ 的鉴定

$\mathrm{A}\left(\mathrm{BeCl}_{2}\right)$ 溶于四氧化二氮 $\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)$ -乙酸乙酯混合溶剂中，低温析出 $\bf{D}$ 。

$\mathrm{N}_{2} \mathrm{O}_{4}$ 解离: $\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{NO}^{+}+ \mathrm{NO}_{3}^{-}$。

乙酸乙酯 $\left(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right)$ 可能参与反应或水解提供乙酸根 $\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)$，但在低温下， 主要反应为
氯与硝酸根的置换。

$\bf{D}$ 仅含 $+1$ 价阳离子和 $-2$ 价阴离子，无孤立溶剂分子。

阳离子为 $\mathrm{NO}^{+}$(来自 $\mathrm{N}_{2} \mathrm{O}_{4}$ ), 阴离子为含 $\mathrm{Be}$ 的 -2 价阴离子。

$\mathrm{BeCl}_{2}$ 与 $\mathrm{NO}_{3}{ }^{-}$反应形成 $\left[\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{4}\right]^{2-}$

因此，$\bf{D}$ 为 $\mathrm{(NO)_{2}[Be(NO_{3})_{4}]}$ , 符合 $2:1$ 型盐。

<Checkpoint>[得到 $\bf{D}$ 的化学式 。](1)</Checkpoint>

5.化合物$\bf{E}$、$\bf{F}$、$\bf{G}$和$\bf{H}$的鉴定

$\bf{D}$ 加热至 $50\degree C$ 得$\bf{E}$和$\bf{F}$:

$\mathrm{(NO)_{2}[Be(NO_{3})_{4}]}$加热分解, 释放红棕色气体 $\mathrm{NO}_{2}$ 。

因此，$\bf{E}$ 为 $\mathrm{NO_{2}}$，$\bf{F}$ 为 $\mathrm{Be(NO_{3})_{2}}$。

<Checkpoint>[得到 $\bf{E}$、$\bf{F}$ 的化学式 。](1)</Checkpoint>

继续加热失去$\mathrm{NO_{2}}$，配合物的部分硝酸根配体转化为氧离子，又由题意可知，该配合物与甲烷类似，具有 $T_{d}$ 对称性，故其含有的 $4$ 个 $\mathrm{Be}$ 位于四面体顶点，由对称性可知硝酸根分布于四面体的边上，共6个，并通过化合价计算得到该化合物有一个$\mathrm{O^{2-}}$离子只能处于中心，与$4$ 个 $\mathrm{Be}$ 配位，此时每个 $\mathrm{Be}$都处于四面体配位形式，较为合理。

故 $\bf{H}$ 为$\mathrm{Be_{4}O(NO_{3})_{6}}$ 。

<Checkpoint>[得到 $\bf{H}$ 的化学式 。](1)</Checkpoint>

金属  $\mathrm{Be}$ 是透过X射线能力最强的金属。

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

$\bf{B}$ 中阴阳离子均为正四面体，具有相同的对称性。

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

$\bf{B}$ 、$\bf{C}$ 中金属  $\mathrm{Be}$ 均为四配位，配位数相同。

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

$\bf{D}$、$\bf{F}$ 、$\bf{H}$ 含有完全相同的元素，即 $\mathrm{Be}$、 $\mathrm{N}$、 $\mathrm{O}$。

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

$\bf{H}$ 中不仅含一种双齿配体，还含有$\mathrm{O^{2-}}$离子只能处于中心，此$\mathrm{O^{2-}}$离子配位数为$4$。

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

$\bf{E}$、$\bf{G}$ 都含有未成对电子，故均为顺磁性。

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