$1$ equivalent of azide **A** and $5 ~\mathrm{mol}\%$ of compound **B** react with $3 ~\mathrm{mol}\%$ of a specific catalyst **C** in toluene at $22\degree \mathrm{C}$ for 2 hours to form polymer **P**.

The structure of azide **A** is:

<MultiModal>![CCCCCCCCCCCCC(CCCCCCCCCCCC)N1C2=C(C=CC(N=[N+]=[N-])=C2)C3=C1C=C(N=[N+]=[N-])C=C3](/media/uploads/79d39621-eec7-4317-8dd2-ccf76795ae11.png)</MultiModal>

The structure of compound **B** is:

<MultiModal>![[N-]=[N+]=NC1=CC=CC=C1C(F)(F)F](/media/uploads/adbc3002-3ff7-4c00-a469-62928deb5ab7.png)</MultiModal>

The structure of catalyst **C** is:

<MultiModal>![The image describes the structure of a complex: C1=CC(C(C)C)=C(N2[Ni]3[N]4=C5N6[Ni]3[N(=C(C)C6=CC=C5C=CC4=C2C)C2C(C(C)C)=CC=CC=2C(C)C)C(C(C)C)=C1, also a neutral benzene molecule is bonded to the two nickel atoms, with dashed lines from two para-carbons to the two nickel atoms respectively.](/media/uploads/058be38a-0b75-47bb-b0e6-6b3db1f4d086.png)</MultiModal>

Read the following options and select the correct one.
$1$ 当量的叠氮化合物 **A** 和 $5 ~\mathrm{mol}\%$化合物 **B** 与 $3 ~\mathrm{mol}\%$的特殊催化剂 **C** 的作用，在甲苯中 $22\degree \mathrm{C}$ 下反应 2 小时可以形成高分子 **P**。

叠氮化合物 **A** 的结构为：

<MultiModal>![CCCCCCCCCCCCC(CCCCCCCCCCCC)N1C2=C(C=CC(N=[N+]=[N-])=C2)C3=C1C=C(N=[N+]=[N-])C=C3](/media/uploads/79d39621-eec7-4317-8dd2-ccf76795ae11.png)</MultiModal>

化合物 **B** 的结构为：

<MultiModal>![[N-]=[N+]=NC1=CC=CC=C1C(F)(F)F](/media/uploads/adbc3002-3ff7-4c00-a469-62928deb5ab7.png)</MultiModal>

催化剂 **C** 的结构为：

<MultiModal>![图片描述了一个配合物的结构：C1=CC(C(C)C)=C(N2[Ni]3[N]4=C5N6[Ni]3[N(=C(C)C6=CC=C5C=CC4=C2C)C2C(C(C)C)=CC=CC=2C(C)C)C(C(C)C)=C1, 另外还有一个中性苯分子与两个镍原子结合, 对位的两个碳原子各有一条虚线分别指向两个镍原子](/media/uploads/058be38a-0b75-47bb-b0e6-6b3db1f4d086.png)</MultiModal>

阅读以下选项，选择其中正确的一选。
The oxidation state of nickel in catalyst **C** is $+2$.
The role of compound **B** is similar to that of trimethylchlorosilane introduced in siloxane polymerization.
All nickel atoms in catalyst **C** obey the EAN rule.
The polymerization reaction is a chain polymerization.
The polymerization reaction is an addition polymerization.
The ratio of the number of fluorine atoms to the number of nitrogen atoms in the end groups of polymer **P** is $\frac{3}{2}$.
The number of fluorine atoms in the end group of polymer **P** is twice the number of nitrogen atoms.
Ignoring side reactions, the theoretical average degree of polymerization is $20$.
Ignoring side reactions, the theoretical average degree of polymerization is $80$.
Ignoring side reactions, the theoretical average degree of polymerization is $10$.
All other options are incorrect.
催化剂 **C** 中镍的化合价为 $+2$。
化合物 **B** 的作用与硅氧烷聚合中引入的三甲基氯硅烷的作用有相似之处。
催化剂 **C** 中镍均符合EAN规则。
该聚合反应是链式聚合
该聚合反应是加聚反应
聚合物 **P** 的端基中包含的氟原子个数与氮原子个数比为 $\frac{3}{2}$ 。
聚合物 **P** 的端基中包含的氟原子个数是氮原子个数的两倍。
不考虑副反应，理论平均聚合度为 $20$。
不考虑副反应，理论平均聚合度为 $80$。
不考虑副反应，理论平均聚合度为 $10$。
其他选项均不正确
The complex in catalyst **C** is a neutral molecule, in which one ligand benzene is a neutral molecule, and the other ligand is an anion with two negative charges. Therefore, the two nickel atoms should have a total of two positive charges. The complex structure is symmetrical, and the valence of the two nickel atoms should be the same. Therefore, the oxidation state of nickel is $+1$. The oxidation state of nickel described in option A is $+2$, which is incorrect.

<Checkpoint>[The oxidation state of nickel is $+1$, and the oxidation state of nickel described in option A is $+2$, which is incorrect.](1)</Checkpoint>

The function of compound **B** is to regulate the degree of polymerization or polymer molecular weight as an end group, which is similar to the role of trimethylchlorosilane introduced in siloxane polymerization. Therefore, option B is correct.

<Checkpoint>[The function of compound **B** is to regulate the degree of polymerization or polymer molecular weight as an end group, which is similar to the role of trimethylchlorosilane introduced in siloxane polymerization. Option B is correct.](1)</Checkpoint>

In addition to benzene ring coordination, each nickel itself has $10$ valence electrons, coordinated by $2$ nitrogens and one lone pair of electrons, and one negative ion is coordinated. The $2$ nitrogens provide a total of $3$ electrons, and $1$ metal bond provides $1$ electron. There are $14$ electrons around it. To satisfy the EAN rule, benzene needs to provide $4$ electrons to each nickel, which cannot be satisfied. Therefore, option C is incorrect.

<Checkpoint>[To satisfy the EAN rule, benzene needs to provide $4$ electrons to each nickel, which cannot be satisfied. Option C is incorrect](1)</Checkpoint>

This polymerization reaction has no clear initiation, growth, and termination stages, no specific active center, and the main monomer contains two obvious functional groups with condensation reaction performance, which is in line with the characteristics of step-growth reaction. Therefore, this polymerization reaction is a step-growth reaction rather than a chain reaction, and option D is incorrect.

<Checkpoint>[This polymerization reaction is a step-growth reaction rather than a chain reaction, and option D is incorrect.](1)</Checkpoint>

This polymerization reaction has small molecule generation, which is a condensation polymerization reaction rather than an addition polymerization reaction, and option E is incorrect.

<Checkpoint>[This polymerization reaction is a condensation polymerization reaction rather than an addition polymerization reaction, and option E is incorrect.](1)</Checkpoint>

According to the characteristics of the azido dimerization reaction, the structure of **P** can be deduced:

<MultiModal>![Image describes a polymer: [N]=NC1=CC=CC=C1C(F)(F)F is one end group, FC(C1=CC=CC=[C]1)(F)F is the other end group, CCCCCCCCCCCCC(CCCCCCCCCCCC)N1C2=C(C=C[C]=C2)C3=C1C=C(N=[N])C=C3 is the repeating segment](/media/uploads/d928d4c5-1996-4d3c-8060-faa6583496d3.png)</MultiModal>

The end groups on both sides contain a total of 6 fluorine atoms and 2 nitrogen atoms. The number of fluorine atoms contained in the end group of polymer **P** is three times the number of nitrogen atoms. Therefore, options F and G are incorrect.

<Checkpoint>[The end groups on both sides contain a total of 6 fluorine atoms and 2 nitrogen atoms. The number of fluorine atoms contained in the end group of polymer **P** is 3 times the number of nitrogen atoms. Therefore, options F and G are incorrect.](1)</Checkpoint>

The azido dimerization reaction proceeds to a large extent and can be regarded as a complete reaction. Without considering side reactions, calculate the theoretical average degree of polymerization:

$\overline{X_{\mathrm{n}}}=\frac{n_{\mathrm{monomer}}}{n_{\mathrm{polymer\,chain}}}=\frac{n_{1}}{\frac{1}{2} \times n_{2}}=\frac{1}{\frac{1}{2} \times 0.05}=40$

The degree of polymerization is 40, so options H, I, and J are all incorrect.

<Checkpoint>[The degree of polymerization is 40, so options H, I, and J are all incorrect.](1)</Checkpoint>
催化剂 **C** 中配合物为中性分子，其中一个配体苯为中性分子，另一个配体为带两个负电荷的阴离子，因此两个镍应当总共带有两个正电荷，配合物结构对称，两个镍的价态应当相同，因此镍的化合价为 $+1$，选项A描述的镍的化合价为 $+2$ 错误。

<Checkpoint>[镍的化合价为 $+1$，选项A描述的镍的化合价为 $+2$ 错误。](1)</Checkpoint>

化合物 **B** 的作用为作为端基调节聚合度或聚合物分子量，与硅氧烷聚合中引入的三甲基氯硅烷的作用相似，因此选项B正确。

<Checkpoint>[化合物 **B** 的作用为作为端基调节聚合度或聚合物分子量，与硅氧烷聚合中引入的三甲基氯硅烷的作用相似，选项B正确。](1)</Checkpoint>

除苯环配位以外，每个镍自身 $10$ 个价电子，周围 $2$ 个氮一个孤对电子配位，一个负离子配位，$2$ 个氮共提供 $3$ 个电子，$1$ 根金属键提供 $1$ 个电子，周围 $14$ 个电子，要满足 EAN 规则需要苯对每个镍提供 $4$ 个电子，是无法满足的。因此选项C错误。

<Checkpoint>[满足 EAN 规则需要苯对每个镍提供 $4$ 个电子，无法满足。选项C错误](1)</Checkpoint>

该聚合反应无明确的引发、增长、终止阶段，无特定活性中心，主要单体包含了两个明显的具有缩合反应性能的官能团，符合逐步反应的特点。因此该聚合反应是逐步反应而不是链式反应，选项D错误。

<Checkpoint>[该聚合反应是逐步反应而不是链式反应，选项D错误。](1)</Checkpoint>

该聚合反应有小分子生成，是缩聚反应而不是加聚反应，选项E错误。

<Checkpoint>[该聚合反应是缩聚反应而不是加聚反应，选项E错误。](1)</Checkpoint>

根据叠氮二聚反应的特点，可以推出 **P** 的结构：

<MultiModal>![图片描述了一种聚合物：[N]=NC1=CC=CC=C1C(F)(F)F为一侧端基，FC(C1=CC=CC=[C]1)(F)F为另一侧端基，CCCCCCCCCCCCC(CCCCCCCCCCCC)N1C2=C(C=C[C]=C2)C3=C1C=C(N=[N])C=C3为重复的链段](/media/uploads/d928d4c5-1996-4d3c-8060-faa6583496d3.png)</MultiModal>

两侧的端基中共包含 6 个氟原子和 2 个氮原子，聚合物 **P** 的端基中包含的氟原子个数是氮原子个数的三倍。因此选项F和G错误。

<Checkpoint>[两侧的端基中共包含 6 个氟原子和 2 个氮原子，聚合物 **P** 的端基中包含的氟原子个数是氮原子个数的 3 倍。因此选项F和G错误。](1)</Checkpoint>

叠氮二聚反应进行程度很大，可以当做完全反应，在不考虑副反应情况下，计算理论平均聚合度：

$\overline{X_{\mathrm{n}}}=\frac{n_{\mathrm{单体}}}{n_{\mathrm{高分子链}}}=\frac{n_{1}}{\frac{1}{2} \times n_{2}}=\frac{1}{\frac{1}{2} \times 0.05}=40$

聚合度为 40，因此选项H，I，J均错误。

<Checkpoint>[聚合度为 40，选项H，I，J均错误。](1)</Checkpoint>