Complex $\mathbf{E}$ can be synthesized according to the following route:

<MultiModal>![Substance $\mathbf{A}$ and $CH_3COOH$ generate $\mathbf{B}$ under heating conditions. Subsequently, $\mathbf{B}$ reacts with [Na]OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1 and $CH_3ONa$ in $CH_2Cl_2$ to generate $\mathbf{C}$. Next, $\mathbf{C}$ reacts with $\mathrm{PdCl_2(PhCN)_2}$ in $CH_2Cl_2$ to generate $\mathbf{D}$. Finally, $\mathbf{D}$ reacts with $\mathrm{Et_4NBH_4}$ to generate $\mathbf{E}$](/media/uploads/82c7b9e5-350e-429b-8afa-ba1616d8e61a.png)</MultiModal>

$\mathbf{A}$ is a mononuclear carbonyl complex of molybdenum that satisfies the 18-electron rule. $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$, and $\mathbf{E}$ are all compounds containing $\mathrm{Mo-Mo}$ bonds, and all metal atoms have 16 electrons around them. Their molar mass, $\mathrm{Mo-Mo}$ distance, non-metal atoms coordinated with the metal, and the highest order symmetry axis are shown in the following table:

|    Compound    | M/ ($\mathrm{g\cdot mol^{-1}}$) | d(Mo-Mo) / pm |  Coordinating Atom   | Highest Order Symmetry Axis |
| :----------: | :-----------------------------: | :-----------: | :---------: | :----------: |
| $\mathbf{B}$ |              428.1              |      210      |      O      |    $C_4$     |
| $\mathbf{C}$ |             1389.9              |      210      |    N、O     |    $I_4$     |
| $\mathbf{D}$ |             2169.2              |      210      | N、O、P、Cl |    $I_4$     |
| $\mathbf{E}$ |             2098.3              |      212      | N、O、P、Cl |    $I_4$     |

Which of the following statements is incorrect (choose F if all are correct)?
配合物$\mathbf{E}$可按如下路线合成：

<MultiModal>![物质$\mathbf{A}$和$CH_3COOH$在加热条件下生成$\mathbf{B}$。随后，$\mathbf{B}$会和[Na]OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1和$CH_3ONa$在$CH_2Cl_2$下反应，生成$\mathbf{C}$。接下来，$\mathbf{C}$和$\mathrm{PdCl_2(PhCN)_2}$在$CH_2Cl_2$中反应，生成$\mathbf{D}$。最后，$\mathbf{D}$和$\mathrm{Et_4NBH_4}$反应生成$\mathbf{E}$](/media/uploads/82c7b9e5-350e-429b-8afa-ba1616d8e61a.png)</MultiModal>

$\mathbf{A}$为钼的的单核羰基配合物，满足18电子规则。$\mathbf{B}$、$\mathbf{C}$、$\mathbf{D}$、$\mathbf{E}$均为含有$\mathrm{Mo-Mo}$键的化合物，所有的金属原子周围均有16个电子，其摩尔质量、$\mathrm{Mo-Mo}$间距、与金属配位的非金属原子和最高次对称轴如下表所示：

|    化合物    | M/ ($\mathrm{g\cdot mol^{-1}}$) | d(Mo-Mo) / pm |  配位原子   | 最高次对称轴 |
| :----------: | :-----------------------------: | :-----------: | :---------: | :----------: |
| $\mathbf{B}$ |              428.1              |      210      |      O      |    $C_4$     |
| $\mathbf{C}$ |             1389.9              |      210      |    N、O     |    $I_4$     |
| $\mathbf{D}$ |             2169.2              |      210      | N、O、P、Cl |    $I_4$     |
| $\mathbf{E}$ |             2098.3              |      212      | N、O、P、Cl |    $I_4$     |

下列说法中，错误的是(若均正确，选F)
$\mathbf{A}$ contains 6 CO
The bond order of the $\mathrm{Mo-Mo}$ bond in $\mathbf{B}$ is 4
The bond order of the $\mathrm{Mo-Mo}$ bond in $\mathbf{C}$ is 4
The bond order of the $\mathbf{D}$ Mo-Mo bond is 4.
The bond order of the $\mathbf{E}$ in $\mathrm{Mo-Mo}$ bond is 4.
All of the above options are correct.
$\mathbf{A}$中包含6个CO
$\mathbf{B}$中$\mathrm{Mo-Mo}$键键级为4
$\mathbf{C}$中$\mathrm{Mo-Mo}$键键级为4
$\mathbf{D}$中$\mathrm{Mo-Mo}$键键级为4
$\mathbf{E}$中$\mathrm{Mo-Mo}$键键级为4
以上选项都正确
From the prompt "$\mathbf{A}$ is a mononuclear carbonyl complex of molybdenum that satisfies the 18-electron rule," it can be determined that $\mathbf{A}$ is $\mathrm{Mo(CO)_6}$.

<Checkpoint>[$\mathbf{A}$ is $\mathrm{Mo(CO)_6}$](1)</Checkpoint>

The process of $\mathbf{A}\to\mathbf{B}$ should be the replacement of $\mathrm{CO}$ by acetic acid. Since the coordinating atom is only $\mathrm{O}$, it indicates that complete substitution has occurred. According to the prompt "$\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$, and $\mathbf{E}$ are all compounds containing $\mathrm{Mo-Mo}$ bonds," let's assume there are 2 $\mathrm{Mo}$ first. Then the amount of $\mathrm{Mo}$ is $191.92 \ \mathrm{g\cdot mol^{-1}}$. Based on the molecular weight of $\mathbf{B}$, the remaining four acetate groups can be deduced. Considering that the highest-order symmetry axis of $\mathbf{B}$ is $C_4$, all four acetate groups should be bridging ligands. The chemical formula of $\mathbf{B}$ should be $\mathrm{Mo_2(CH_3COO)_4}$. It should be noted that $\mathrm{Mo}$ is in the $+2$ oxidation state. From the prompt "all metal atoms have 16 electrons around them," it can be determined that the bond order of $\mathrm{Mo}$ is $\frac{16\times 2-4\times 2-4\times 4}{2}=4$.

<Checkpoint>[The chemical formula of $\mathbf{B}$ should be $\mathrm{Mo_2(CH_3COO)_4}$, the bond order of $\mathrm{Mo}$ is $4$](1)</Checkpoint>

During the formation of $\mathbf{C}$, according to the data in the question, the $\mathrm{Mo-Mo}$ bond distance does not change, and the coordinating atoms are $\mathrm{N}$ and $\mathrm{O}$, indicating that ligand substitution has occurred. Due to $I_4$ symmetry, let's assume that all four ligands are substituted, i.e., the chemical formula of $\mathbf{C}$ should be $\mathrm{Mo_2(X)_4}$, where $\mathbf{X}$ is OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1. Calculating the molecular weight, it coincides with the prompt. Combined with $I_4$ symmetry, adjacent ligand molecules should adopt opposite orientations. The bond order of $\mathrm{Mo}$ should be the same as that of $\mathbf{A}$, which is $4$.

<Checkpoint>[The chemical formula of $\mathbf{C}$ should be $\mathrm{Mo_2(X)_4}$, where $\mathbf{X}$ is OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1, the bond order of $\mathrm{Mo}$ is $4$](1)</Checkpoint>

During the formation of $\mathbf{D}$, the coordinating atoms increase by $\mathrm{P}$ and $\mathrm{Cl}$. It can be seen that $\mathrm{PdCl_2(PhCN)_2}$ is coordinated to $\mathrm{P}$ in OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1, and the ligand $\mathrm{PhCN}$ is replaced by $\mathrm{P}$. The product is $\mathrm{Mo_2(X)_4(PdCl_2)_2}$, and the calculated molecular weight matches. The bond order of $\mathrm{Mo}$ should be the same as that of $\mathbf{A}$, which is $4$.

<Checkpoint>[The chemical formula of $\mathbf{D}$ should be $\mathrm{Mo_2(X)_4(PdCl_2)_2}$, the bond order of $\mathrm{Mo}$ is $4$](1)</Checkpoint>

During the formation of $\mathbf{E}$, the molecular weight decreases, and the $\mathrm{Mo-Mo}$ bond distance increases. It indicates that ligand departure may have occurred, and at the same time, $\mathrm{Pd}$ and $\mathrm{Mo}$ are connected. Calculating the decreased molecular weight, it can be determined that two $\mathrm{Cl}$ have departed, and the product is $\mathrm{Mo_2(X)_4(PdCl)_2}$. Due to the formation of the $\mathrm{Pd-Mo}$ bond, the bond order of $\mathrm{Mo}$ changes. Two $\mathrm{Pd-Mo}$ bonds will reduce the bond order by 1, and the bond order of $\mathrm{Mo}$ is 3.

<Checkpoint>[The chemical formula of $\mathbf{E}$ should be $\mathrm{Mo_2(X)_4(PdCl)_2}$, the bond order of $\mathrm{Mo}$ is $3$](1)</Checkpoint>

<MultiModal>![Structures of $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$, and $\mathbf{E+}$. Only the coordinating atoms are shown in the figure.](/media/uploads/0a05f472-08b5-4c29-8d19-93534c92afda.png)</MultiModal>
由题干“$\mathbf{A}$为钼的的单核羰基配合物，满足18电子规则“，可以判断出$\mathbf{A}$为$\mathrm{Mo(CO)_6}$。

<Checkpoint>[$\mathbf{A}$为$\mathrm{Mo(CO)_6}$](1)</Checkpoint>

$\mathbf{A}\to\mathbf{B}$的过程应为$\mathrm{CO}$被乙酸取代。由于配位原子仅有$\mathrm{O}$，说明发生了完全取代。根据题干”$\mathbf{B}$、$\mathbf{C}$、$\mathbf{D}$、$\mathbf{E}$均为含有$\mathrm{Mo-Mo}$键的化合物“，不妨先假设有2个$\mathrm{Mo}$。则$\mathrm{Mo}$的量为$191.92 \ \mathrm{g\cdot mol^{-1}}$，根据$\mathbf{B}$的分子量，可以推算出剩余四个乙酸根。考虑到$\mathbf{B}$的最高次对称轴为$C_4$，四个乙酸根应当均为桥连配体。$\mathbf{B}$的化学式应为$\mathrm{Mo_2(CH_3COO)_4}$，需要注意$\mathrm{Mo}$为$+2$价。由题干”所有的金属原子周围均有16个电子“，可以判断$\mathrm{Mo}$的键级为$\frac{16\times 2-4\times 2-4\times 4}{2}=4$。

<Checkpoint>[$\mathbf{B}$的化学式应为$\mathrm{Mo_2(CH_3COO)_4}$，$\mathrm{Mo}$的键级为$4$](1)</Checkpoint>

生成$\mathbf{C}$过程中，根据题目数据，$\mathrm{Mo-Mo}$键距离未改变，配位原子有$\mathrm{N}$和$\mathrm{O}$，说明发生了配体取代。由于$I_4$对称性，不妨假设四个配体均发生取代，即$\mathbf{C}$的化学式应为$\mathrm{Mo_2(X)_4}$，$\mathbf{X}$即OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1。计算分子量，正好与题干吻合。结合$I_4$对称性，相邻的配体分子应当采用相反的取向。$\mathrm{Mo}$的键级应与$\mathbf{A}$相同，为$4$。

<Checkpoint>[$\mathbf{C}$的化学式应为$\mathrm{Mo_2(X)_4}$，$\mathbf{X}$即OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1，$\mathrm{Mo}$的键级为$4$](1)</Checkpoint>

生成$\mathbf{D}$过程中，配位原子增加了$\mathrm{P}$和$\mathrm{Cl}$。可以看出，$\mathrm{PdCl_2(PhCN)_2}$被OC1=NC(P(C2=CC=CC=C2)C3=CC=CC=C3)=CC=C1中的$\mathrm{P}$配位，同时配体$\mathrm{PhCN}$被$\mathrm{P}$取代，产物为$\mathrm{Mo_2(X)_4(PdCl_2)_2}$，计算分子量吻合。$\mathrm{Mo}$的键级应与$\mathbf{A}$相同，为$4$。

<Checkpoint>[$\mathbf{D}$的化学式应为$\mathrm{Mo_2(X)_4(PdCl_2)_2}$，$\mathrm{Mo}$的键级为$4$](1)</Checkpoint>

生成$\mathbf{E}$过程中，分子量降低，$\mathrm{Mo-Mo}$键距离增大。说明可能发生了配体的离去，同时$\mathrm{Pd}$和$\mathrm{Mo}$相连。计算降低的分子量，可以判断离去了两个$\mathrm{Cl}$，产物为$\mathrm{Mo_2(X)_4(PdCl)_2}$。由于$\mathrm{Pd-Mo}$键的形成，$\mathrm{Mo}$的键级发生变化。两根$\mathrm{Pd-Mo}$键会使键级降低1，$\mathrm{Mo}$的键级为3。

<Checkpoint>[$\mathbf{E}$的化学式应为$\mathrm{Mo_2(X)_4(PdCl)_2}$，$\mathrm{Mo}$的键级为$3$](1)</Checkpoint>

<MultiModal>![$\mathbf{B}$、$\mathbf{C}$、$\mathbf{D}$、$\mathbf{E+}$的结构。图中仅给出了配位原子。](/media/uploads/0a05f472-08b5-4c29-8d19-93534c92afda.png)</MultiModal>