$\mathbf{X}$ is a metal-organic compound of element $\mathbf{M}$, containing only the four elements $\mathrm{C}$, $\mathrm{H}$, $\mathrm{O}$, and $\mathbf{M}$, with no crystalline water. When $\mathrm{3.414~g}$ of compound $\mathbf{X}$ is fully calcined in air, the residual solid weighs $\mathrm{3.318~g}$. Mixing $\mathbf{X}$ with $\mathrm{H}_{3} \mathrm{PO}_{2}$ solution reduces $\mathbf{X}$ to compound $\mathbf{Y}$. The mass fraction of $\mathbf{M}$ in $\mathbf{X}$ is $74.71\%$, and in $\mathbf{Y}$, it is $79.83\%$. Determine the identities of these compounds.

$\mathbf{X}$ further reacts with an equivalent amount of ethanedithiol (abbreviated as $\mathrm{edtH}_{2}$) to yield $\mathbf{B}$. $\mathbf{B}$ then reacts with an equivalent amount of ethanedithiol to form $\mathbf{C}$, and $\mathbf{C}$ undergoes a one-step reduction to yield $\mathbf{D}$, which has the same molecular formula as $\mathbf{C}$. During the reactions from $\mathbf{X}$ to $\mathbf{B}$ and $\mathbf{B}$ to $\mathbf{C}$, equivalent amounts of water are produced. Deduce the structure of $\mathbf{D}$.

Researchers found that when $\mathbf{X}$ reacts with $\mathrm{mtpH}_{2}$ (structure shown in Figure 1), a dinuclear complex $\mathbf{E}$ forms in a $1:2$ molar ratio.

<MultiModal>![Fig. 1, Structure of $\mathrm{mtpH}_{2}$, represented by SMILES: SC1=CC=CC=C1CS](/media/uploads/ef60494e-f313-4b02-af21-3dbdecce00ca.png)</MultiModal>

$\mathbf{E}$ reacts with pyridine oxide (py-O, structure shown in Figure 2) to regenerate $\mathbf{X}$. In $\mathbf{E}$, the coordination number of metal $\mathbf{M}$ is 5, and the mass fraction of $\mathbf{M}$ is $50.12\%$.

<MultiModal>![Fig. 2, SMILES representation: [O-][N+]1=CC=CC=C1](/media/uploads/ab8f5829-ebe9-46d5-9415-99ba6e1eff45.png)</MultiModal>

Deduce the structure of the dinuclear complex $\mathbf{E}$.

1. $\mathbf{X}$ contains 7 chemical bonds (multiple bonds counted as one bond).
2. $\mathbf{X}$ contains 4 chemical bonds (multiple bonds counted as one bond).
3. In one $\mathbf{D}$ molecule, there exists an element with three atoms having the same bond count.
4. In one $\mathbf{D}$ molecule, there is no element where three atoms share the same bond count.
5. In one $\mathbf{E}$ molecule, there exists an element with two atoms each having a bond count of 3.
6. In one $\mathbf{E}$ molecule, there exists an element where all atoms have a bond count of 2.

Which of the following statements is correct?
$\mathbf{X}$ 是元素 $\mathbf{M}$ 的金属有机化合物, 只含有 $\mathrm{C} 、 \mathrm{H} 、 \mathrm{O}$ 以及 $\mathbf{M}$ 四种元素, 不含有结晶水。将 $\mathrm{3.414
~g}$ 的化合物 $\mathbf{X}$ 置于空气中充分煅烧, 残余固体称重后质量为 $\mathrm{3.318~g}$ 。将 $\mathbf{X}$ 与 $\mathrm{H}_{3} \mathrm{PO}_{2}$ 溶液混合,  $\mathbf{X}$ 会被还原为化合物  $\mathbf{Y}$。 $\mathbf{X}$ 中  $\mathbf{M}$ 的质量分数为 $74.71 \%$,  $\mathbf{Y}$ 中  $\mathbf{M}$ 的质量分数为 $79.83 \%$ 。推断各化合物分别是什么。

$\mathbf{X}$ 进一步与等当量的乙二硫醇(简写为 $\mathrm{edtH}_{2}$ )反应得到 $\mathbf{B}$,
$\mathbf{B}$ 接着与等当量的乙二硫醇反应得到 $\mathbf{C}$, $\mathbf{C}$ 经过一步还原得到分子式与 $\mathbf{C}$ 相同的 $\mathbf{D}$。在 $\mathbf{X}$
到 $\mathbf{B}$ 以及 $\mathbf{B}$ 到 $\mathbf{C}$ 的反应过程中均有等当量的水生成。推断$\mathbf{D}$的结构。

研究人员发现, 若 $\mathbf{X}$ 与 $\mathrm{mtpH}_{2}$ (结构如图1所示)发生反应, 则会以 $1: 2$ 的摩尔比反应形成双核配合物 $\mathbf{E}$。

<MultiModal>![Fig.1，$\mathrm{mtpH}_{2}$的结构，以SMILES表示为：SC1=CC=CC=C1CS](/media/uploads/ef60494e-f313-4b02-af21-3dbdecce00ca.png)</MultiModal>

 $\mathbf{E}$ 与吡啶氧化物 py-O(结构如图2所示)反应将重新生成 $\mathbf{X}$。 $\mathbf{E}$ 中金属 $\mathbf{M}$ 的配位数均为 5 , 且 $\mathbf{M}$ 的质量分数为 $50.12 \%$。

<MultiModal>![Fig. 2，SMILES表示为：[O-][N+]1=CC=CC=C1](/media/uploads/ab8f5829-ebe9-46d5-9415-99ba6e1eff45.png)</MultiModal>

推断双核配合物$\mathbf{E}$的结构。

1. $\mathbf{X}$中包含7根化学键（多重键按一根化学键计）
2. $\mathbf{X}$中包含4根化学键（多重键按一根化学键计）
3. 一个$\mathbf{D}$分子中存在一种元素，该元素中有三个原子的成键数相同
4. 一个$\mathbf{D}$分子中不存在可以使该元素中有三个原子的成键数相同的元素
5. 一个$\mathbf{E}$分子中存在一种元素，该元素中有两个原子的成键数均为3
6. 一个$\mathbf{E}$分子中存在一种元素，该元素的所有原子成键数均为2

下列选项中说法正确的是？
1,3,5
1,3,6
1,4,5
1,4,6
2,3,5
2,3,6
2,4,5
2,4,6
1,4
4,5
5,6
1,5,6
1,3,5,6
1,3,5
1,3,6
1,4,5
1,4,6
2,3,5
2,3,6
2,4,5
2,4,6
1,4
4,5
5,6
1,5,6
1,3,5,6
The problem does not provide extensive elemental properties, so the element can only be inferred based on mathematical relationships. Metal-organic compounds often lose all $\mathrm{C}$ and $\mathrm{H}$ elements when calcined in air, forming stable higher oxidation state metal oxides in air. The problem gives the mass fraction of the $\mathbf{M}$ element in $\mathbf{X}$, allowing the calculation of the total mass of $\mathbf{M}$ in $\mathbf{X}$ and thus the molar amount of $\mathrm{O}$ in the metal oxide obtained by calcining $\mathbf{X}$. This further enables the estimation of the possible molar mass of the $\mathbf{M}$ element.

The total mass of $\mathbf{M}$ in $\mathbf{X}$:  

$\mathrm{3.414g \times 74.71\% = 2.551g}$  

The molar amount of $\mathrm{O}$ in the metal oxide obtained by calcining $\mathbf{X}$:  

$\mathrm{(3.318g - 2.551g) \div 16.00 \, g \cdot mol^{-1} = 0.0479 \, mol}$  

Assuming the molecular formula of the calcined metal oxide is $\mathrm{MO_x}$, the molar mass of element $\mathbf{M}$ is:  

$\mathrm{(3.318g \cdot x) \div 0.0479 \, mol - 16 \, g \cdot mol^{-1} = 53.27 \cdot x \, g \cdot mol^{-1}}$  

Based on common metal oxide forms, for  

$\mathrm{x = 0.333, 0.5, 1, 1.333, 1.5, 2, 2.5, 3, 3.5, 4}$  

tabulation shows that when $\mathrm{x = 3.5}$, the molar mass of $\mathbf{M}$ is $\mathrm{186.4 \, g/mol}$, matching the molar mass of $\mathrm{Re}$. Moreover, the oxide $\mathrm{Re_2O_7}$ aligns with the elemental properties, so $\mathbf{M}$ is $\mathrm{Re}$.  

(Alternatively, assuming $\mathbf{X}$ is reduced to $\mathbf{Y}$ by losing only N $\mathrm{O}$ atoms, the compound can also be inferred.)  

<Checkpoint>[Based on the mass fraction, element $\mathbf{M}$ is inferred to be $\mathrm{Re}$](1)</Checkpoint>  

From the mass fractions of $\mathrm{Re}$ in $\mathbf{X}$ and $\mathbf{Y}$, their molar masses can be calculated, revealing that the molar masses of $\mathrm{C, H, O}$ in the two compounds are $\mathrm{63.03 \, g \cdot mol^{-1}}$ and $\mathrm{47.05 \, g \cdot mol^{-1}}$, respectively, differing by exactly one $\mathrm{O}$ atom's molar mass. It is thus deduced that the molecular formula of $\mathbf{X}$ is $\mathrm{ReO_3(CH_3)}$ and that of $\mathbf{Y}$ is $\mathrm{ReO_2(CH_3)}$. Since all atoms in $\mathbf{X}$ are in a single environment, the $\mathrm{O}$ atoms are all directly bonded to $\mathrm{Re}$. Given that the methyl group contains three C-H bonds, the number of chemical bonds in $\mathbf{X}$ is 7, making Statement 1 correct and Statement 2 incorrect.  

<Checkpoint>[Based on the mass fractions, the molecular formula of $\mathbf{X}$ is deduced as $\mathrm{ReO_3(CH_3)}$ and $\mathbf{Y}$ as $\mathrm{ReO_2(CH_3)}$. The number of chemical bonds in $\mathbf{X}$ is 7, making Statement 1 correct and Statement 2 incorrect.](1)</Checkpoint>  

$\mathbf{X}$ reacts with $\mathrm{edtH_2}$ to form water molecules, a common ligand substitution reaction. Thus, the products $\mathbf{B}$ and $\mathbf{C}$ have molecular formulas $\mathrm{ReO_2(edt)CH_3}$ and $\mathrm{ReO(edt)_2CH_3}$, respectively. The molecular formula of $\mathbf{C}$ remains unchanged after reduction, indicating intramolecular reductive elimination. The least stable ligand is the methyl group, which likely couples with a sulfur atom. The most probable structure of compound $\mathbf{D}$ is shown in Figure 3, where sulfur has three atoms each with a bond order of 2, making Statement 3 correct and Statement 4 incorrect.  

<MultiModal>![Fig. 3, structure described in SMILES as: C[S]1CCS[Re]12(SCCS2)=O](/media/uploads/44d2193b-e338-4706-ae11-015759ec0140.png)</MultiModal>  

<Checkpoint>[Based on the unchanged molecular formula of $\mathbf{C}$ to $\mathbf{D}$, the methyl group is inferred to migrate to a sulfur atom. In compound $\mathbf{D}$, sulfur has three atoms each with a bond order of 2, making Statement 3 correct and Statement 4 incorrect.](1)</Checkpoint>  

$\mathbf{X}$ most likely undergoes a ligand substitution reaction with $\mathrm{mtpH_2}$. However, the product $\mathbf{E}$ is reoxidized by $\mathrm{py-O}$ to regenerate $\mathbf{X}$, indicating that $\mathbf{X}$ undergoes reduction when forming $\mathbf{E}$, likely losing an oxygen atom. Based on the mass fraction of $\mathbf{M}$, a dinuclear complex contains two $\mathrm{mtpH_2}$ ligands and two $\mathrm{O}$ atoms. The reaction ratio of $\mathbf{X}$ to $\mathrm{mtpH_2}$ is $\mathrm{1:2}$, suggesting $\mathrm{mtpH_2}$ acts as a common reducing agent and is oxidized to form a disulfide bond, while $\mathbf{C}$ is reduced to $\mathbf{D}$. The reaction equation is balanced to verify this conclusion. According to the problem, the coordination number of $\mathrm{Re}$ is 5, implying one ligand bridges two $\mathrm{Re}$ atoms. Given the transition metal properties of $\mathrm{Re}$, the $\mathrm{Re=O}$ double bond is stable, and the sulfur atom, with its larger electron cloud, is most likely to form the bridge. Considering coordination capacity, the thiol $\mathrm{S}$ atom in $\mathrm{mtpH_2}$ is more likely to coordinate than the thiophenol $\mathrm{S}$ atom. The structure of $\mathbf{E}$ is shown in Figure 4.  

In $\mathbf{E}$, sulfur has two atoms each with a bond order of 3, making Statement 5 correct. No element in the molecule forms only two bonds, making Statement 6 incorrect.  

<MultiModal>![Fig. 4, the complex structure in SMILES notation: C[Re]12([S]([Re]3(C)(SC4=C(C[S]32)C=CC=C4)=O)CC5=C(S1)C=CC=C5)=O](/media/uploads/304a9af1-6cff-4a25-99fa-baa211d3118f.png)</MultiModal>  

<Checkpoint>[Based on the reduction reaction from $\mathbf{X}$ to $\mathbf{E}$, the reaction ratio of $\mathbf{X}$ to $\mathrm{mtpH_2}$, and the mass fraction of $\mathbf{M}$, the molecular formula of $\mathbf{E}$ is deduced, with the thiol sulfur atom selected as the bridging ligand. In $\mathbf{E}$, sulfur has two atoms each with a bond order of 3, making Statement 5 correct. No element in the molecule forms only two bonds, making Statement 6 incorrect.](1)</Checkpoint>
题目中未给出较多的元素性质，因此只能根据数学关系推断元素。金属有机化合物在空气中煅烧往往会失去所有$\mathrm{C}$、$\mathrm{H}$元素，并形成空气中稳定的较高氧化态金属氧化物。题目给出了$\mathbf{X}$中$\mathbf{M}$元素的的质量分数，因此可以推算$\mathbf{X}$中$\mathbf{M}$元素的总质量，从而得到$\mathbf{X}$煅烧得到金属氧化物中$\mathrm{O}$元素的摩尔量。进而推算$\mathbf{M}$元素可能的摩尔质量。

$\mathbf{X}$中$\mathbf{M}$元素的总质量：

$\mathrm{3.414g\times74.71\%=2.551g}$

$\mathbf{X}$煅烧得到金属氧化物中$\mathrm{O}$元素的摩尔量：

$\mathrm{3.318g-2.551g\div16.00 g\cdot mol^{-1}=0.0479mol}$

设煅烧后金属氧化物的分子式为$\mathrm{MO_x}$，则元素$\mathbf{M}$的摩尔质量为：

$\mathrm{3.318g\cdot x\div 0.0479mol-16g\cdot mol^{-1}=53.27\cdot x ~g\cdot mol^{-1}}$

根据常见金属氧化物的形式，对

$\mathrm{x=0.333,0.5,1,1.333,1.5,2,2.5,3,3.5,4}$

打表，得出$\mathrm{x=3.5}$时元素$\mathbf{M}$的摩尔质量为$\mathrm{186.4 g/mol}$，符合$\mathrm{Re}$元素的摩尔质量，且氧化物$\mathrm{Re_2O_7}$符合元素性质，因此元素$\mathbf{M}$为$\mathrm{Re}$。

(假设$\mathbf{X}$还原为$\mathbf{Y}$仅失去N个$\mathrm{O}$原子，据此推断化合物也可以。)

<Checkpoint>[根据质量分数推断出元素$\mathbf{M}$为$\mathrm{Re}$](1)</Checkpoint>

根据$\mathbf{X}$和$\mathbf{Y}$中$\mathrm{Re}$的质量分数可以推算二者的摩尔质量，进而得到二者中的$\mathrm{C、H、O}$占的摩尔质量分别为$\mathrm{63.03~g\cdot mol^{-1}}$和$\mathrm{47.05~g\cdot mol^{-1}}$，恰好相差一个$\mathrm{O}$原子的摩尔质量，易推算出$\mathbf{X}$分子式为$\mathrm{ReO_3(CH_3)}$，$\mathbf{Y}$分子式为$\mathrm{ReO_2(CH_3)}$。鉴于$\mathbf{X}$中所有种类的原子都只有一种环境，可知$\mathrm{O}$原子均直接与$\mathrm{Re}$原子相连。考虑到甲基中含有三根碳-氢键，因此$\mathbf{X}$中化学键数为7，说法1正确，说法2错误。

<Checkpoint>[根据质量分数推算出$\mathbf{X}$分子式为$\mathrm{ReO_3(CH_3)}$，$\mathbf{Y}$分子式为$\mathrm{ReO_2(CH_3)}$，$\mathbf{X}$中化学键数为7，说法1正确，说法2错误。](1)</Checkpoint>

$\mathbf{X}$与$\mathrm{edtH_2}$反应生成水分子，是常见配体取代反应，因此生成的产物$\mathbf{B}$和$\mathbf{C}$分子式分别为$\mathrm{ReO_2(edt)CH_3}$和$\mathrm{ReO(edt)_2CH_3}$。$\mathbf{C}$被还原后分子式不变，说明还原消除在分子内发生。配体中最不稳定的为甲基，最可能是甲基与硫原子发生偶联，化合物$\mathbf{D}$最可能的结构如图3。其中硫元素有三个原子成键数均为2，说法3正确，说法4错误。

<MultiModal>![Fig. 3，结构以SMILES描述为：C[S]1CCS[Re]12(SCCS2)=O](/media/uploads/44d2193b-e338-4706-ae11-015759ec0140.png)</MultiModal>

<Checkpoint>[根据$\mathbf{C}$到$\mathbf{D}$分子式不变，推断出甲基迁移到硫原子上。化合物$\mathbf{D}$中硫元素有三个原子成键数均为2，说法3正确，说法4错误。](1)</Checkpoint>

$\mathbf{X}$与$\mathrm{mtpH_2}$最可能发生配体取代反应，然而产物$\mathbf{E}$被$\mathbf{py-O}$氧化重新生成$\mathbf{X}$，说明$\mathbf{X}$生成$\mathbf{E}$时发生了还原反应，很有可能失去氧原子。根据$\mathbf{M}$的质量分数可以计算得到一分子双核配合物中包含两个$\mathrm{mtpH_2}$与两个$\mathrm{O}$原子。$\mathbf{X}$与$\mathrm{mtpH_2}$反应比为$\mathrm{1:2}$，最可能是$\mathrm{mtpH_2}$作为常见还原剂被氧化形成二硫键，而化合物$\mathbf{C}$被还原为化合物$\mathbf{D}$，配平反应方程式验证以上结论。根据题目，$\mathrm{Re}$配位数是5，说明有一个配体同时配位了两个$\mathrm{Re}$原子形成了桥接。根据$\mathrm{Re}$的过渡金属性质，$\mathrm{Re=O}$双键较稳定，最容易形成桥接的是电子云范围较大的$\mathrm{S}$原子，考虑到配位能力，最可能发生配位的是$\mathrm{mtpH_2}$中的硫醇$\mathrm{S}$原子而非硫酚$\mathrm{S}$原子。$\mathbf{E}$的结构式如图4。
$\mathbf{E}$分子中的硫元素，该元素中有两个原子的成键数均为3，说法5正确。该分子内没有元素全部形成两根键，说法6错误。

<MultiModal>![Fig. 4，该配合物结构以SMILES表示为：C[Re]12([S]([Re]3(C)(SC4=C(C[S]32)C=CC=C4)=O)CC5=C(S1)C=CC=C5)=O](/media/uploads/304a9af1-6cff-4a25-99fa-baa211d3118f.png)</MultiModal>

<Checkpoint>[根据$\mathbf{X}$到$\mathbf{E}$发生还原反应，$\mathbf{X}$与$\mathrm{mtpH_2}$的反应比以及$\mathbf{M}$的质量分数，推断出$\mathbf{E}$的分子式，并选择硫醇硫原子作为桥接配体。$\mathbf{E}$分子中的硫元素，该元素中有两个原子的成键数均为3，说法5正确。该分子内没有元素全部形成两根键，说法6错误。](1)</Checkpoint>