A preparation method of a common metal complex is as follows: $1.8 \ \mathrm { g } \ \mathrm {\mathbf{ M} } _ { 2 } \mathrm { O }$, $50\ \rm{mL}\ CH_2Cl_2$, $3\ \rm{mL}$ 3-hexyne (approximately $26.3\rm{\ mmol}$), $1.1 \ \mathrm { g }$ oxalic acid, stirred for several hours, and the product was obtained by recrystallization with $\mathrm { C H_2Cl_2}$ at low temperature. It is known that the metal is a dinuclear complex, does not contain metal-metal bonds, and the metal content is $33.50\%$.

Thermogravimetric analysis of the complex was performed under inert conditions, and the results are shown in the figure below. Which of the following statements is incorrect?

<MultiModal>![This is a line graph. The abscissa is "Temperature($\degree C$)", the scale division is 200, and the range is 0-600; the ordinate is "Weight(X)", the scale division is 40, and the range is 20-100. There is only one broken line in the figure, and there are the following inflection points (including the starting point): (70,100)(80,100)(100,75)(120,75)(130,55)(250,55)(280,30)(600,30)](/media/uploads/ccdc86d4-9ee6-49c0-bedd-9f039d0dcf44.png)</MultiModal>
某常见金属的配合物制备方法如下， $1 . 8 \ \mathrm { g } \ \mathrm {\mathbf{ M} } _ { 2 } \mathrm { O }$，$50\ \rm{mL}\ CH_2Cl_2$，$3\ \rm{mL}$ 3-己炔(约$26.3\rm{\ mmol}$)，$1 . 1 \ \mathrm { g } $草酸，搅拌数小时后，用 $\mathrm { C H_2Cl_2}$在低温下重结晶得到产物。已知该金属为双核配合物，不含有金属-金属键且金属含量为 $3 3 . 5 0 \%$。

在惰性条件下对该配合物进行热重分析，结果如下图所示。下列说法中错误的是

<MultiModal>![这是一张折线图。横坐标为“Temperature($\degree C$)”，刻度的分度值为200，范围为0-600；纵坐标为“Weight(X)”，刻度的分度值为40，范围为20-100。图中仅有一条折线，存在以下拐点（包含起始点）：(70,100)(80,100)(100,75)(120,75)(130,55)(250,55)(280,30)(600,30)](/media/uploads/ccdc86d4-9ee6-49c0-bedd-9f039d0dcf44.png)</MultiModal>
All other options are incorrect
The product has the molecular formula $\mathrm{\mathbf{M}_2C_{14}H_{20}O_4}$
The product contains two five-membered rings.
The common salt solution of $\mathbf{M}$ is blue.
The final product of thermogravimetry is the element $\mathbf{M}$.
The first two thermogravimetric losses involve the same molecule.
其他选项均不正确
产物的分子式为$\mathrm{\mathbf{M}_2C_{14}H_{20}O_4}$
产物中含有两个五元环
$\mathbf{M}$的常见盐溶液呈蓝色
热重的最终产物为$\mathbf{M}$单质
前两次热重失去的分子相同
According to the description in the question stem, "the metal is a dinuclear complex, does not contain metal-metal bonds, and the metal content is $33.50\%$", it is not difficult to guess that oxalate is a bridging ligand, coordinating with two $\bf{M}$. According to the preparation process, it can be seen that 3-hexyne participates in the reaction. It may be assumed that one $\bf{M}$ coordinates with one 3-hexyne. Therefore, the chemical formula of the product can be described as $\rm{\mathbf{M}_2(C_2O_4)(C_6H_{10})_2}$. The molecular formula of the product is $\mathrm{\mathbf{M}_2C_{14}H_{20}O_4}$, B is correct.

<Checkpoint>[The chemical formula of the product can be described as $\rm{\mathbf{M}_2(C_2O_4)(C_6H_{10})_2}$](1)</Checkpoint>

In the product, the oxalate radical acts as a tetradentate ligand, coordinating with two $\rm{Cu}$ respectively, and two 3-hexynes coordinate with $\rm{Cu}$, forming two five-membered rings, C is correct.

<Checkpoint>[In the product, the oxalate radical acts as a tetradentate ligand, coordinating with two $\rm{Cu}$ respectively, and two 3-hexynes coordinate with $\rm{Cu}$](1)</Checkpoint>

According to the metal content, it can be judged that the metal is $\rm{Cu}$, and the molecular weight of the product is $379.42\ \rm{g/mol}$. The common salt solution of $\mathbf{M}$ is blue, D is correct.

<Checkpoint>[The metal is $\rm{Cu}$](1)</Checkpoint>

The thermogravimetric image is discussed below. The first weight loss is about $25\%$, corresponding to about $95\ \rm{g/mol}$, corresponding to one 3-hexyne. Since it is reading the image, the actual weight loss is $21\%$. The second weight loss is about $25\%$, corresponding to about $95\ \rm{g/mol}$, corresponding to one 3-hexyne. The actual weight loss is $21\%$. The third weight loss is about $20\%$, corresponding to about $76\ \rm{g/mol}$, corresponding to 2 $\rm{CO_2}$, and the actual weight loss is $23\%$. It should be pointed out that the molecular weights of one 3-hexyne and two $\rm{CO_2}$ are relatively close. The judgment of the three weight losses takes into account the thermal stability of the coordination: $\pi$ ligands are easier to leave than $\sigma$ ligands. Therefore, E and F are both correct. Choose A.

<Checkpoint>[The first weight loss corresponds to 3-hexyne, the second weight loss corresponds to 3-hexyne, the third weight loss corresponds to 2 $\rm{CO_2}$, and the final residue is $\rm{Cu}$](1)</Checkpoint>
根据题干描述“该金属为双核配合物，不含有金属-金属键且金属含量为 $3 3 . 5 0 \%$”，不难猜测草酸为桥连配体，和两个$\bf{M}$配位。根据制备过程，可以看出3-己炔参与了反应，不妨假设一个$\bf{M}$与一个3-己炔配位。因此，产物的化学式可以描述为$\rm{\mathbf{M}_2(C_2O_4)(C_6H_{10})_2}$。产物的分子式为$\mathrm{\mathbf{M}_2C_{14}H_{20}O_4}$，B正确。

<Checkpoint>[产物的化学式可以描述为$\rm{\mathbf{M}_2(C_2O_4)(C_6H_{10})_2}$](1)</Checkpoint>

产物中草酸根作四齿配体，分别和两个$\rm{Cu}$配位，两个3-己炔和$\rm{Cu}$配位，存在两个五元环，C正确。

<Checkpoint>[产物中草酸根作四齿配体，分别和两个$\rm{Cu}$配位，两个3-己炔和$\rm{Cu}$配位](1)</Checkpoint>

依据金属含量，可以判断金属为$\rm{Cu}$，产物的分子量为$379.42\ \rm{g/mol}$。$\mathbf{M}$的常见盐溶液呈蓝色，D正确。

<Checkpoint>[金属为$\rm{Cu}$](1)</Checkpoint>

下讨论热重图像。第一次失重约$25\%$，对应约$95\ \rm{g/mol}$，对应一个3-己炔，由于是读图，实际失重为$21\%$。第二次失重约$25\%$，对应约$95\ \rm{g/mol}$，对应一个3-己炔，实际失重为$21\%$。第二次失重约$20\%$，对应约$76\ \rm{g/mol}$，对应2个$\rm{CO_2}$，实际失重为$23\%$。需要指出的是，一个3-己炔和2个$\rm{CO_2}$分子量较接近，三次失重的前后判断考虑了配位的热稳定性：$\pi$配体相比$\sigma$配体更易离去。因此，E、F均正确。选择A。

<Checkpoint>[第一次失重对应3-己炔，第二次失重对应3-己炔第三次失重对应2个$\rm{CO_2}$，最终剩余$\rm{Cu}$](1)</Checkpoint>