The non-radioactive element **A** can form a series of binary polymers with chain-like structures when combined with various halogens. The schematic structures are as follows (hollow/white spheres represent element **A**, solid/black spheres represent halogens), where the mass fractions of element **A** in **X1** to **X4** are 84.37%, 62.67%, 76.16%, and 50.14%, respectively:  

- Image for **X1**  

<MultiModal>![In the image, hollow/white spheres are linked to form a one-dimensional long chain similar to the carbon chain in polyethylene. If the one-dimensional chains are numbered 1, 2, 3..., then every hollow/white sphere numbered 3N is connected to two solid/black spheres as terminal groups.](/media/uploads/d9bd83ac-9c65-438c-8da8-fdc1c37d748e.png)</MultiModal>  

- Image for **X2**  

<MultiModal>![In the image, hollow/white spheres and solid/black spheres are alternately linked to form a one-dimensional long chain. Each hollow/white sphere is connected to five solid/black spheres, three of which are terminal groups and two are bridging.](/media/uploads/b9749f76-c268-4268-a987-f4b6a8baa37f.png)</MultiModal>  

- Image for **X3**  

<MultiModal>![In the image, hollow/white spheres form long chains of fused hexagon rings sharing edges, resembling the fusion of benzene rings in polyacenes. For every two hollow/white spheres on the same side of a hexagon that are not involved in edge-sharing, an additional solid/black sphere bridges them, forming an extra four-membered ring structure. The hollow/white spheres not involved in edge-sharing hexagons are connected to two hollow/white spheres that are part of edge-sharing hexagons and to two bridging solid/black spheres forming the four-membered ring.](/media/uploads/003606ea-df92-48f6-8c9b-fca03e8e9e9d.png)</MultiModal>  

- Image for **X4**  

<MultiModal>![In the image, hollow/white spheres form the same one-dimensional long chain as in **X1**. If the chain is numbered 1, 2, 3..., adjacent odd-numbered hollow/white spheres are bridged by an additional solid/black sphere to form a four-membered ring. Every odd-numbered hollow/white sphere has two bridging solid/black spheres, while even-numbered hollow/white spheres are connected to one terminal solid/black sphere.](/media/uploads/0338e0e4-b6d4-4b40-ac79-5fca66b9a432.png)</MultiModal>  

Based on the information about the inorganic chain-like polymers in the images, deduce the stoichiometric ratio of **A** atoms to halogen atoms in **X1-X4**, and calculate to determine which element **A** is and the chemical formulas of **X1-X4**. Then, identify the number of correct propositions from the following:  

1. Element **A** belongs to Group 15.  
2. Element **A** is located in the fifth period.  
3. When the halogen atoms in **X1-4** are ordered by their periodic table positions from low to high, the sequence is **X1**, **X2**, **X4**, **X3**.  
4. The ratio of **A** atoms to halogen atoms in **X1-4**, ordered from smallest to largest, is **X2**, **X4**, **X3**, **X1**.  
5. According to hybrid orbital theory, **A** atoms in **X1** exhibit two hybridization modes, and their ratio is 1:1.  
6. According to hybrid orbital theory, **A** atoms in **X2** are $sp^3d^2$ hybridized.  
7. **X3** contains two types of **A** atoms with different chemical situations, and their numbers are equal.
非放射性的元素 **A** 可以与各种卤素形成一系列具有链状结构的二元聚合物，其结构示意图如下（空心球/白球为元素 **A**，实心球/黑球为卤素），其中 **A** 元素的质量分数从**X1**到**X4** 依次为84.37%、62.67%、76.16%、50.14%：

- **X1** 的图片

<MultiModal>![图片中空心球/白球彼此链接形成类似聚乙烯中的碳链的一维长链，若将一维长链编号为1、2、3...，则编号为3N的空心球/白球有2个实心球/黑球作为端基与之相连](/media/uploads/d9bd83ac-9c65-438c-8da8-fdc1c37d748e.png)</MultiModal>

- **X2** 的图片

<MultiModal>![图片中空心球/白球与实心球/黑球间隔链接形成一维长链，每个空心球/白球与5个实心球/黑球相连，其中3个为端基、2个为桥联](/media/uploads/b9749f76-c268-4268-a987-f4b6a8baa37f.png)</MultiModal>

- **X3** 的图片

<MultiModal>![图片中空心球/白球形成共边相连的正六元环并环结构的长链，其并环方式类似于苯共边并环形成聚并苯，其中每两个未参与相邻共边六元环的同侧空心球/白球，都有一个额外的实心球/黑球进行桥联，形成额外的形成四元环结构，未参与相邻共边六元环的空心球/白球与两个参与相邻共边六元环的空心球/白球相连、与两个参与桥联组成四元环的实心球/黑球相连](/media/uploads/003606ea-df92-48f6-8c9b-fca03e8e9e9d.png)</MultiModal>

- **X4** 的图片

<MultiModal>![图片中空心球/白球形成与X1中相同的一维长链， 若将一维长链编号为1、2、3...，则相邻的奇数编号空心球/白球之间与一个额外的桥联的实心球/黑球形成四元环，每隔奇数编号的空心球/白球有两个桥联的实心球/黑球，而偶数编号空心球/白球与一个端基实心球/黑球相连](/media/uploads/0338e0e4-b6d4-4b40-ac79-5fca66b9a432.png)</MultiModal>

请结合图片中的无机链状聚合物的信息，推导**X1-X4** 的 **A** 原子与卤素原子的计量关系，并通过计算，推断 **A** 是何种元素以及 **X1-X4** 四个物种的化学式。然后从以下命题中找出正确命题的个数。

1.  **A** 元素位于第五主族
2.  **A** 元素位于第五周期
3.  将 **X1-4** 中的卤素原子按照周期高低排序，由低到高的顺序依次为**X1**、**X2**、**X4**、**X3**
4.  计算 **X1-4** 中 **A** 原子:卤素原子数目的比值，由小到大的顺序依次为**X2**、**X4**、**X3**、**X1**
5.  根据杂化轨道理论，**X1** 中的 **A** 原子有两种杂化方式，且原子数目比为 1:1
6.  根据杂化轨道理论，**X2** 中的 **A** 原子为$sp^3d^2$杂化
7.  **X3** 中含有两种不同化学环境的 **A** 原子，且数目相同
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First, based on the problem description and structural diagrams, determine the simplest integer ratio of **A** atoms to halogen atoms (**X**) for the four polymers (**X1**, **X2**, **X3**, **X4**).  

* **X1**: The structure consists of a long chain of **A** atoms, where every third **A** atom (after every two **A** atoms) is bonded to two terminal halogen atoms. That is, $\mathrm{...-A-A-A(X_2)-A-A-A(X_2)-...}$. Thus, the repeating unit is $\mathrm{A_3X_2}$.  
    * **A:X ratio = 3:2**  

<Checkpoint>[The simplest formula for **X1** is $\mathrm{A_3X_2}$](1)</Checkpoint>  

* **X2**: Each **A** atom is bonded to five halogen atoms, three of which are terminal and two are bridging. The bridging halogen atoms are shared between two **A** atoms. Therefore, the effective number of halogen atoms per **A** atom is 3 + 2 * (1/2) = 4. The repeating unit is $\mathrm{AX_4}$.  
    * **A:X ratio = 1:4**  

<Checkpoint>[The simplest formula for **X2** is $\mathrm{AX_4}$](1)</Checkpoint>  

* **X3**: The structure is a ladder-like chain composed of fused six-membered rings. From the diagram, the smallest repeating unit contains four **A** atoms (two on the inner side of the "ladder" and two on the outer side) and two bridging halogen atoms. Thus, the repeating unit is $\mathrm{A_4X_2}$, and the simplest formula is $\mathrm{A_2X}$.  
    * **A:X ratio = 2:1**  

<Checkpoint>[The simplest formula for **X3** is $\mathrm{A_2X}$](1)</Checkpoint>  

* **X4**: The structure is a long chain of **A** atoms, where **A** atoms at even positions are bonded to one terminal halogen atom, and **A** atoms at odd positions are bridged by one halogen atom. Considering an A(odd)-A(even) repeating unit: A(even) has one terminal halogen; A(odd) is connected to two bridging halogen atoms, each contributing 1/2, totaling 2 * (1/2) = 1 halogen atom. Thus, two **A** atoms correspond to two halogen atoms. The repeating unit is $\mathrm{A_2X_2}$, and the simplest formula is $\mathrm{AX}$.  
    * **A:X ratio = 1:1**  

<Checkpoint>[The simplest formula for **X4** is $\mathrm{AX}$](1)</Checkpoint>  

### 2. Inferring Element **A** and the Chemical Formulas of the Species  

Let the atomic weight of element **A** be $M_A$, and the atomic weight of the halogen be $M_X$. Using the mass fraction formula $w_A = \frac{n_A \cdot M_A}{n_A \cdot M_A + n_X \cdot M_X}$, we can derive $M_A = \frac{w_A \cdot n_X \cdot M_X}{n_A \cdot (1 - w_A)}$.  

We match the four stoichiometric ratios and mass fractions, testing possible halogens (F=19.0, Cl=35.5, Br=79.9, I=126.9) to solve for a consistent $M_A$.  

* **For the formula AX₄ (A:X = 1:4), mass fraction $w_A = 62.67\%$**:  
    $M_A = \frac{0.6267 \cdot 4 \cdot M_X}{1 \cdot (1 - 0.6267)} \approx 6.716 \cdot M_X$.  
    If **X** is **F** (fluorine), $M_A \approx 6.716 \cdot 19.0 = 127.6$.  

* **For the formula A₃X₂ (A:X = 3:2), mass fraction $w_A = 84.37\%$**:  
    $M_A = \frac{0.8437 \cdot 2 \cdot M_X}{3 \cdot (1 - 0.8437)} \approx 3.601 \cdot M_X$.  
    If **X** is **Cl** (chlorine), $M_A \approx 3.601 \cdot 35.5 = 127.8$.  

* **For the formula A₂X (A:X = 2:1), mass fraction $w_A = 76.16\%$**:  
    $M_A = \frac{0.7616 \cdot 1 \cdot M_X}{2 \cdot (1 - 0.7616)} \approx 1.597 \cdot M_X$.  
    If **X** is **Br** (bromine), $M_A \approx 1.597 \cdot 79.9 = 127.6$.  

* **For the formula AX (A:X = 1:1), mass fraction $w_A = 50.14\%$**:  
    $M_A = \frac{0.5014 \cdot 1 \cdot M_X}{1 \cdot (1 - 0.5014)} \approx 1.0056 \cdot M_X$.  
    If **X** is **I** (iodine), $M_A \approx 1.0056 \cdot 126.9 = 127.6$.  

All calculations point to $M_A \approx 127.6$, identifying the element as **tellurium (Te)**. It is a non-radioactive element, consistent with the problem statement.  

<Checkpoint>[Element **A** is **Te**](2)</Checkpoint>  

Thus, the chemical formulas of the four species are:  
* **X1**: $\mathrm{Te_3Cl_2}$  
* **X2**: $\mathrm{TeF_4}$  
* **X3**: $\mathrm{Te_2Br}$  
* **X4**: $\mathrm{TeI}$  

<Checkpoint>[X1-X4 are $\mathrm{Te_3Cl_2}$, $\mathrm{TeF_4}$, $\mathrm{Te_2Br}$, $\mathrm{TeI}$ respectively](1)</Checkpoint>  
---  

### 3. Evaluating the Truth of Propositions  

Now, we analyze the eight given propositions one by one:  

1.  **Element **A** belongs to Group V (15)**  
    Tellurium (Te) is located in Group VI (16, the chalcogen group) of the periodic table, not Group V (15, the pnictogen group). Thus, this proposition is **false**.  

2.  **Element **A** is in the fifth period**  
    The valence electron shell of tellurium (Te) is the fifth shell (n=5), placing it in the fifth period. Thus, this proposition is **true**.  

<Checkpoint>[Tellurium (Te) is in Group VI, Period 5. Proposition 1 is false, 2 is true](1)</Checkpoint>  

3.  **When ordering the halogen atoms in X1-4 by increasing period number, the sequence is X1, X2, X4, X3**  
    * X1: Cl (chlorine, Period 3)  
    * X2: F (fluorine, Period 2)  
    * X3: Br (bromine, Period 4)  
    * X4: I (iodine, Period 5)  
    The correct order from low to high period is: F (X2) < Cl (X1) < Br (X3) < I (X4). The given sequence is X1, X2, X4, X3. Thus, this proposition is **false**.  

<Checkpoint>[F (X2) < Cl (X1) < Br (X3) < I (X4). Proposition 3 is false](1)</Checkpoint>  

4.  **When ordering the A:halogen atom ratio in X1-4 from smallest to largest, the sequence is X2, X4, X3, X1**  
    * X1 ($\mathrm{Te_3Cl_2}$): A:X = 3:2 = 1.5  
    * X2 ($\mathrm{TeF_4}$): A:X = 1:4 = 0.25  
    * X3 ($\mathrm{Te_2Br}$): A:X = 2:1 = 2.0  
    * X4 ($\mathrm{TeI}$): A:X = 1:1 = 1.0  
    The correct order from smallest to largest is: 0.25 (X2) < 1.0 (X4) < 1.5 (X1) < 2.0 (X3). The given sequence is X2, X4, X3, X1. Thus, this proposition is **false**.  

<Checkpoint>[Halogen atom ratios: 0.25 (X2) < 1.0 (X4) < 1.5 (X1) < 2.0 (X3). Proposition 4 is false](1)</Checkpoint>  

5.  **According to hybridization theory, **X1** has two types of hybridized **A** atoms, with a 1:1 ratio**  
    In the **X1** ($\mathrm{Te_3Cl_2}$) chain, there are two distinct tellurium environments:  
    * Te bonded to two other Te atoms: 2 bonding pairs and 2 lone pairs, $\mathrm{sp^3}$ hybridization.  
    * Te bonded to two Te atoms and two Cl atoms: Te(IV) with 4 bonding pairs and 1 lone pair (total 5 valence electron pairs), $\mathrm{sp^3d}$ hybridization.  
    The ratio of these Te atoms is 2:1, not 1:1. Thus, this proposition is **false**.  

<Checkpoint>[**X1** has two distinct Te environments with a 2:1 ratio. Proposition 5 is false](1)</Checkpoint>  

6.  **According to hybridization theory, **X2** has **A** atoms with $\mathrm{sp^3d^2}$ hybridization**  
    In **X2** ($\mathrm{TeF_4}$), each Te atom bonds with five F atoms, and Te is in the +4 oxidation state with 1 lone pair. The total valence electron pairs are 5 (bonding) + 1 (lone) = 6. Six electron pairs correspond to $\mathrm{sp^3d^2}$ hybridization. Thus, this proposition is **true**.  

<Checkpoint>[**X2** Te atoms are $\mathrm{sp^3d^2}$ hybridized. Proposition 6 is true](1)</Checkpoint>  

7.  **X3 contains two chemically distinct **A** atoms in equal numbers**  
    In **X3**'s ladder structure, there are two distinct Te environments:  
    * Inner Te: bonded to 3 Te atoms.  
    * Outer Te: bonded to 2 Te atoms and 1 Br atom.  
    Their counts are equal in the repeating unit (2 each). Thus, this proposition is **true**.  

<Checkpoint>[**X3** has two chemically distinct Te atoms in equal numbers. Proposition 7 is true](1)</Checkpoint>  

In summary, the correct propositions are 2, 6, and 7, totaling **3**.  

**The correct answer is D**.
首先，根据题目描述和结构示意图，确定四种聚合物（**X1**、**X2**、**X3**、**X4**）中 **A** 原子与卤素原子（**X**）的最简整数比。

* **X1**: 结构为 **A** 原子组成的长链，每隔2个 **A** 原子，第3个 **A** 原子上连接2个端基卤素原子。即 $\mathrm{...-A-A-A(X_2)-A-A-A(X_2)-...}$。因此，重复单元为 $\mathrm{A_3X_2}$。
    * **A:X 比例 = 3:2**

<Checkpoint>[**X1**的最简式为$\mathrm{A_3X_2}$](1)</Checkpoint>

* **X2**: 每个 **A** 原子与5个卤素原子相连，其中3个为端基，2个为桥联。桥联卤素原子被两个 **A** 原子共享。因此，每个 **A** 原子实际占有的卤素原子数为3 + 2 * (1/2) = 4。重复单元为$\mathrm{AX_4}$。
    * **A:X 比例 = 1:4**

<Checkpoint>[**X2**的最简式为 $\mathrm{AX_4}$](1)</Checkpoint>

* **X3**: 结构为并六元环构成的梯形长链。从图中可以看出，最小重复单元包含4个 **A** 原子（2个在“梯子”内侧，2个在外侧）和2个桥联卤素原子。因此，重复单元为$\mathrm{A_4X_2}$，最简式为 $\mathrm{A_2X}$。
    * **A:X 比例 = 2:1**

<Checkpoint>[**X3**的最简式为 $\mathrm{A_2X}$](1)</Checkpoint>

* **X4**: 结构为 **A** 原子长链，偶数位 **A** 原子连接1个端基卤素，奇数位 **A** 原子之间由1个卤素原子桥联。考虑一个A(奇)-A(偶)的重复单元：A(偶)拥有1个端基卤素; A(奇)连接了两个桥联卤素，每个贡献1/2，总计 2 * (1/2) = 1个卤素。因此，2个 **A** 原子对应2个卤素原子。重复单元为$\mathrm{A_2X_2}$，最简式为$\mathrm{AX}$。
    * **A:X 比例 = 1:1**

<Checkpoint>[**X4**的最简式为 $\mathrm{AX}$](1)</Checkpoint>

### 2. 推断元素 A 及各物种化学式

设元素 **A** 的原子量为 $M_A$，卤素的原子量为 $M_X$。根据质量分数公式 $w_A = \frac{n_A \cdot M_A}{n_A \cdot M_A + n_X \cdot M_X}$，可以推导出 $M_A = \frac{w_A \cdot n_X \cdot M_X}{n_A \cdot (1 - w_A)}$。

我们将四组化学计量比和质量分数进行匹配，尝试可能的卤素（F=19.0, Cl=35.5, Br=79.9, I=126.9）以求解一致的 $M_A$。

* **对于化学式 AX₄ (A:X = 1:4)，质量分数 $w_A = 62.67\%$**:
    $M_A = \frac{0.6267 \cdot 4 \cdot M_X}{1 \cdot (1 - 0.6267)} \approx 6.716 \cdot M_X$。
    若 **X** 为 **F** (氟), $M_A \approx 6.716 \cdot 19.0 = 127.6$。

* **对于化学式 A₃X₂ (A:X = 3:2)，质量分数 $w_A = 84.37\%$**:
    $M_A = \frac{0.8437 \cdot 2 \cdot M_X}{3 \cdot (1 - 0.8437)} \approx 3.601 \cdot M_X$。
    若 **X** 为 **Cl** (氯), $M_A \approx 3.601 \cdot 35.5 = 127.8$。

* **对于化学式 A₂X (A:X = 2:1)，质量分数 $w_A = 76.16\%$**:
    $M_A = \frac{0.7616 \cdot 1 \cdot M_X}{2 \cdot (1 - 0.7616)} \approx 1.597 \cdot M_X$。
    若 **X** 为 **Br** (溴), $M_A \approx 1.597 \cdot 79.9 = 127.6$。

* **对于化学式 AX (A:X = 1:1)，质量分数 $w_A = 50.14\%$**:
    $M_A = \frac{0.5014 \cdot 1 \cdot M_X}{1 \cdot (1 - 0.5014)} \approx 1.0056 \cdot M_X$。
    若 **X** 为 **I** (碘), $M_A \approx 1.0056 \cdot 126.9 = 127.6$。

所有计算都指向 $M_A \approx 127.6$，该元素为**碲 (Tellurium, Te)**。它是一种非放射性元素，符合题意。

<Checkpoint>[元素**A**为**Te**](2)</Checkpoint>

因此，四种物质的化学式分别为：
* **X1**: $\mathrm{Te_3Cl_2}$
* **X2**: $\mathrm{TeF_4}$
* **X3**: $\mathrm{Te_2Br}$
* **X4**: $\mathrm{TeI}$

<Checkpoint>[得到X1-X4分别为$\mathrm{Te_3Cl_2}$、$\mathrm{TeF_4}$、$\mathrm{Te_2Br}$、$\mathrm{TeI}$](1)</Checkpoint>
---

### 3. 判断命题真伪

现在我们逐一分析给出的8个命题：

1.  **A 元素位于第五主族**
    碲 (Te) 位于元素周期表第六主族（VIA族，氧族），而非第五主族（VA族，氮族）。故此命题错误。

2.  **A 元素位于第五周期**
    碲 (Te) 的价电子层为第五层 (n=5)，位于第五周期。故此命题 **正确**。

<Checkpoint>[碲 (Te) 位于元素周期表第六主族第五周期，命题1错误，2正确](1)</Checkpoint>

3.  **将 X1-4 中的卤素原子按照周期高低排序，由低到高的顺序依次为X1、X2、X4、X3**
    * X1: Cl (氯，第三周期)
    * X2: F (氟，第二周期)
    * X3: Br (溴，第四周期)
    * X4: I (碘，第五周期)
    按周期从低到高排序应为：F (X2) < Cl (X1) < Br (X3) < I (X4)。题目给出的顺序是 X1, X2, X4, X3。故此命题 **错误**。

<Checkpoint>[F (X2) < Cl (X1) < Br (X3) < I (X4)，故命题3错误](1)</Checkpoint>

4.  **计算 X1-4 中 A 原子:卤素原子数目的比值，由小到大的顺序依次为X2、X4、X3、X1**
    * X1 ($\mathrm{Te_3Cl_2}$): A:X = 3:2 = 1.5
    * X2 ($\mathrm{TeF_4}$): A:X = 1:4 = 0.25
    * X3 ($\mathrm{Te_2Br}$): A:X = 2:1 = 2.0
    * X4 ($\mathrm{TeI}$): A:X = 1:1 = 1.0
    从小到大排序为：0.25 (X2) < 1.0 (X4) < 1.5 (X1) < 2.0 (X3)。题目给出的顺序是 X2, X4, X3, X1。故此命题 **错误**。

<Checkpoint>[卤素原子数目的比值0.25 (X2) < 1.0 (X4) < 1.5 (X1) < 2.0 (X3)，命题4错误](1)</Checkpoint>

5.  **根据杂化轨道理论，X1 中的 A 原子有两种杂化方式，且原子数目比为 1:1**
    在 X1 ($\mathrm{Te_3Cl_2}$) 链中，有两种不同环境的碲原子：
    * 只与两个碲原子相连的 Te：有2对成键电子和2对孤对电子，为 $\mathrm{sp^3}$ 杂化。
    * 与两个碲原子和两个氯原子相连的 Te：Te(IV) 价态，有4个成键电子对和1对孤对电子，共5个价层电子对，为 $\mathrm{sp^3d}$ 杂化。
    这两种 Te 原子的数量比为 2:1，而非 1:1。故此命题 **错误**。

<Checkpoint>[ **X1**有两种不同环境的碲原子，数量比为 2:1，命题5错误](1)</Checkpoint>

6.  **根据杂化轨道理论，X2 中的 A 原子为$\mathrm{sp^3d^2}$ 杂化**
    在 X2 ($\mathrm{TeF_4}$) 的聚合物结构中，每个 Te 原子与5个 F 原子成键，且 Te 为+4价，中心原子有1对孤对电子。总价层电子对数为 5 (成键) + 1 (孤对) = 6。6个价层电子对对应的杂化方式为 $\mathrm{sp^3d^2}$。故此命题 **正确**。

<Checkpoint>[**X2**中碲原子的为$\mathrm{sp^3d^2}$杂化，命题6正确](1)</Checkpoint>

7.  **X3 中含有两种不同化学环境的 A 原子，且数目相同**
    在 X3 的梯形结构中，有两种不同化学环境的 Te 原子：
    * 内侧 Te：只与3个 Te 原子相连。
    * 外侧 Te：与2个 Te 和1个 Br 相连。
它们在重复单元中数量相同（各2个）。故此命题 **正确**。

<Checkpoint>[**X3**中存在两种化学环境的 Te 原子，且在重复单元中数量相同，命题7正确](1)</Checkpoint>

综上所述，正确的命题共有3个，分别是命题2、命题6和命题7。

**正确答案为 D**。