The element $\mathbf{X}$ is an indispensable element for the human body and a versatile life-sustaining nutrient, holding significant importance.

I. Purification of $\mathbf{X}$: $\mathbf{X}$ is often associated with another heavy element $\mathbf{Y}$ from the same group in some minerals, making the separation of $\mathbf{X}$ from $\mathbf{Y}$ crucial during purification.

Method 1 involves dissolving $\mathbf{X}$ containing trace amounts of $\mathbf{Y}$ in concentrated $\mathrm{HNO}_{3}$ solution, evaporating the $\mathrm{HNO}_{3}$, and then adding hydroiodic acid to the solution.

Method 2 involves dissolving $\mathbf{X}$ containing trace amounts of $\mathbf{Y}$ in concentrated $\mathrm{Na}_{2}\mathrm{SO}_{3}$ solution at $\mathrm{363 K}$.

II. Content Analysis of $\mathbf{X}$: Reacting $3$,$3'$-diaminobenzidine with $\mathbf{X}$ in a stable but non-highest oxidation state in $\mathrm{0.1} \mathrm{mol/L}^{-1}$ hydrochloric acid produces a bright yellow compound $\mathbf{M}$. After adjusting to $\mathrm{pH=6}$, toluene is used for extraction, followed by spectrophotometric determination, which is a sensitive and characteristic analytical method for $\mathbf{X}$.

Using the above method to determine the $\mathbf{X}$ content in a sample, which may contain trace metal impurities, high-purity $\mathbf{X}$ with a mass of $m_{1}$ is used as the starting material. After a series of treatments, a $25 \mathrm{~mL}$ aqueous solution of $\mathbf{M}$ is obtained. Extraction with $1 \mathrm{~mL}$ toluene and measurement at $z \mathrm{~nm}$ using a $1 \mathrm{~cm}$ cuvette with toluene as the reference yields absorbance $\mathrm{A}$. Repeated experiments provide the following data:

| $m_1/μg$ | 0.5   | 1.0   | 3.0   | 5.0   | 10.0  | 0.0 |
| :----- | :---- | :---- | :---- | :---- | :---- | :----- |
| $\mathrm{A}$      | 0.060 | 0.090 | 0.210 | 0.335 | 0.631 | 0.034  |

After preprocessing $1.0 \mathrm{~g}$ of the sample, a solution containing $\mathbf{X}$ with a mass of $m_{2}$ is added, where the oxidation state of $\mathbf{X}$ is the same as that used in the spectrophotometric method above.

Using the same method, absorbance $\mathrm{A}$ is measured at $z \mathrm{~nm}$ with a $1 \mathrm{~cm}$ cuvette and toluene as the reference. Repeated experiments provide the following data:

| $m_2/μg$ | 0     | 0.5   | 1.0   | 3.0   | 5.0   | 10.0  |
| :----- | :---- | :---- | :---- | :---- | :---- | :---- |
| $\mathrm{A}$      | 0.046 | 0.072 | 0.091 | 0.211 | 0.359 | 0.657 |

The partition coefficient of $\mathbf{M}$ in the toluene-water system is $\mathrm{k_D=25.4}$.

Regarding the above analytical process, the following statements are made:

1. In Method 1, when hydroiodic acid is added, the two main reactions occurring in the solution involve changes in the oxidation states of both $\mathbf{X}$ and $\mathbf{Y}$.

2. Both Method 1 and Method 2 require extraction and filtration to obtain purified $\mathbf{X}$.

3. The analytical method mentioned in "Content Analysis of $\mathbf{X}$" requires the addition of a small amount of disodium $\mathrm{EDTA}$ salt.

4. Compound $\mathbf{M}$ can possess a twofold rotation axis, two mutually perpendicular mirror planes, and a center of symmetry through the rotation of covalent bonds.

5. The $\mathbf{X}$ content in the sample is $\mathrm{0.100ppm}$.

This problem specifies that all data points must be included when fitting the curve.

Among the above statements, the correct options are (for calculation-related options, an error margin of less than $5$% is considered correct).
某元素$\mathbf{X}$是人体不可或缺的一种元素,也是一种多功能的生命营养素,具有重要意义。

一、$\mathbf{X}$的提纯：$\mathbf{X}$与和它同族的另一种重元素$\mathbf{Y}$在一些矿物中常常伴生, 因此提纯$\mathbf{X}$时将其与$\mathbf{Y}$元素分离至关重要。

 方法$1$是将含有少量$\mathbf{Y}$的$\mathbf{X}$溶于浓$\mathrm{HNO}_{3}$ 溶液中, 蒸除 $\mathrm{HNO}_{3}$ 后, 向溶液中加入氢碘酸。

方法$2$是将含有少量$\mathbf{Y}$的$\mathbf{X}$在 $\mathrm{363 K}$ 下溶于浓 $\mathrm{Na}_{2}\mathrm{SO}_{3}$ 溶液。

二、$\mathbf{X}$的含量分析：使用$3$,$3'$-二氨基联苯胺与处于某一稳定价态但并非最高价的$\mathbf{X}$在$\mathrm{0.1} \mathrm{mol/L}^{-1}$盐酸中反应可以生成一种鲜黄色化合物$\mathbf{M}$,调节$\mathrm{pH=6}$后用甲苯进行萃取,对其进行分光光度测定,是对$\mathbf{X}$灵敏且特征的分析方法。

用上述分析方法测定某样品中的$\mathbf{X}$含量，样品中可能含有微量杂质金属。以质量为 $m_{1}$ 的高纯$\mathbf{X}$单质为起始物, 经过一系列处理后得到 $25 \mathrm{~mL} \mathbf{~M}$ 的水溶液, 用 $1 \mathrm{~mL}$ 甲苯萃取后, 于 $z \mathrm{~nm}$ 处用 $1 \mathrm{~cm}$ 比色皿
以甲苯作参比测得吸光度 $\mathrm{A}$ 。重复多次实验, 得到以下数据:

| $m_1/μg$ | 0.5   | 1.0   | 3.0   | 5.0   | 10.0  | 0.0 |
| :----- | :---- | :---- | :---- | :---- | :---- | :----- |
| $\mathrm{A}$      | 0.060 | 0.090 | 0.210 | 0.335 | 0.631 | 0.034  |

将 $1.0 \mathrm{~g}$ 该样品预处理后，加入含$\mathbf{X}$元素的质量为 $m_{2}$ 的溶液，且该溶液中$\mathbf{X}$元素的价态与上一步分光光度法中所用的$\mathbf{X}$相同。

采用相同的方法, 于 $z \mathrm{~nm}$
处用 $1 \mathrm{~cm}$ 比色皿以甲苯为参比测得吸光度 $\mathrm{A}$ 。重复多次实验, 得到以下数据:

| $m_2/μg$ | 0     | 0.5   | 1.0   | 3.0   | 5.0   | 10.0  |
| :----- | :---- | :---- | :---- | :---- | :---- | :---- |
| $\mathrm{A}$      | 0.046 | 0.072 | 0.091 | 0.211 | 0.359 | 0.657 |

已知$\mathbf{M}$在甲苯-水体系中的分配系数$\mathrm{k_D=25.4}$。

关于上述分析过程，有如下说法：

$1$ 方法$1$中加入氢碘酸时溶液中主要发生的$2$个反应，$\mathbf{X}$元素和$\mathbf{Y}$元素的价态均发生变化。

$2$ 方法$1$与方法$2$都需要通过萃取、过滤获得提纯后的$\mathbf{X}$元素。

$3$ “$\mathbf{X}$的含量分析”中提到的分析方法，需要加入少量$\mathrm{EDTA}$二钠盐。

$4$ 化合物$\mathbf{M}$通过共价键的旋转，可以同时拥有二重旋转轴、两个相互垂直的镜面、对称中心。

$5$ 该样品中$\mathbf{X}$元素含量为$\mathrm{0.100ppm}$。

本题规定：在进行拟合曲线时，所有的数据点都需要参与拟合。

以上说法中正确的选项有(涉及计算的选项，误差小于$5$%即正确)
$1$, $3$, $4$
$1$, $3$
$1$, $4$
$1$, $4$, $5$
$1$, $2$
$2$, $3$, $4$
$2$, $4$
$3$, $4$, $5$
$3$, $4$
$3$, $5$
$2$, $3$
$1$, $3$, $4$, $5$
$1$, $2$, $3$, $4$
$2$, $3$, $4$, $5$
$1$, $2$, $4$, $5$
$1$, $2$, $3$, $5$
$1$, $5$
$4$, $5$
$1$，$3$，$4$
$1$，$3$
$1$，$4$
$1$，$4$，$5$
$1$，$2$
$2$，$3$，$4$
$2$，$4$
$3$，$4$，$5$
$3$，$4$
$3$，$5$
$2$，$3$
$1$，$3$，$4$，$5$
$1$，$2$，$3$，$4$
$2$，$3$，$4$，$5$
$1$，$2$，$4$，$5$
$1$，$2$，$3$，$5$
$1$，$5$
$4$，$5$
From the question stem, $\mathbf{X}$ is an essential element for the human body and often coexists in nature with its heavier congener $\mathbf{Y}$ in mineral forms. It can be determined that:

<Checkpoint>[Element $\mathbf{X}$ is $\mathrm{Se}$](1)</Checkpoint>

<Checkpoint>[Element $\mathbf{Y}$ is $\mathrm{Te}$](1)</Checkpoint>

Method 1:

<Checkpoint>[$\mathrm{H}_{2} \mathrm{SeO}_{3}+4 \mathrm{HI} \rightarrow \mathrm{Se}+2 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{H}_{2} \mathrm{TeO}_{3}+6 \mathrm{HI} \rightarrow \mathrm{H}_{2} \mathrm{TeI}_{6}+3 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

Method 2:

<Checkpoint>[$\mathrm{Se}+\mathrm{Na}_{2} \mathrm{SO}_{3}\rightarrow \mathrm{Na}_{2}\mathrm{Se} \mathrm{SO}_{3}$](1)</Checkpoint>

In Method 1, the oxidation state of $\mathrm{Se}$ changes, and it precipitates, while the oxidation state of $\mathrm{Te}$ remains unchanged, staying in the solution. Filtration can directly separate $\mathrm{Se}$ in its elemental form.

In Method 2, the oxidation state of $\mathrm{Se}$ changes, remaining in the solution, while $\mathrm{Te}$ does not dissolve.

<Checkpoint>[In both Method 1 and Method 2, the oxidation state of $\mathrm{Se}$ changes, while in Method 1, the oxidation state of $\mathrm{Te}$ remains unchanged](1)</Checkpoint>

From the reaction equations, Statement 1 is incorrect, and Statement 2 is incorrect.

The disodium salt of $\mathrm{EDTA}$ can complex with transition metal ions in the sample, masking them to prevent oxidation of $3$,$3'$-diaminobenzidine or the formation of colored complexes with $3$,$3'$-diaminobenzidine, which would interfere with the measurement. Statement 3 is correct.

<Checkpoint>[The disodium salt of $\mathrm{EDTA}$ can mask transition metal ions to prevent interference in the measurement.](1)</Checkpoint>

Structure of compound $\mathbf{M}$:

<MultiModal>![C12=N[Se]N=C1C=CC(C3=CC4=N[Se]N=C4C=C3)=C2](/media/uploads/bfa0b5c2-2c15-4c50-b067-7e37b7a90592.png)</MultiModal>

<Checkpoint>[The structure of compound M is C12=N[Se]N=C1C=CC(C3=CC4=N[Se]N=C4C=C3)=C2](2)</Checkpoint>

From the structure of $\mathbf{M}$, it can be concluded that as the central carbon-carbon single bond rotates, it cannot simultaneously possess a twofold rotation axis, two mutually perpendicular mirror planes, and an inversion center. Statement 4 is incorrect.

<Checkpoint>[a twofold rotation axis, an inversion center and two mutually perpendicular mirror planes cannot coexist](1)</Checkpoint>

Sample analysis calculation process:

First, calculate the extraction efficiency $\mathrm{E}$ of using $\mathrm{1 ~mL}$ toluene to extract $\mathrm{25 ~mL}$ of $\mathbf{M}$ solution:

$\mathrm{E=k_D/(k_D+V_{water}/V_{sample})=25.4/(25.4+25)=0.504}$

<Checkpoint>[The extraction efficiency $\mathrm{E}$ of using $\mathrm{1 ~mL}$ toluene to extract $\mathrm{25 ~mL}$ of $\mathbf{M}$ solution is $\mathrm{E=0.504}$](2)</Checkpoint>

Concentration of selenium in the toluene solution: $\mathrm{c}_{S e}=\frac{\mathrm{n}_{S e, o}}{V}=\frac{E \times \mathrm{n}_{S e, w}}{V}=\frac{\mathrm{E} m_{1}}{M_{S e} V}$ 

Since one molecule of $\mathbf{M}$ contains 2 $\mathrm{Se}$ atoms, the concentration of $\mathbf{M}$: $c_{\mathbf{M}}=\frac{1}{2} c_{S e}=\frac{E m_{1}}{2 M_{S e} V}$

Absorbance $\mathrm{A}=\varepsilon l c_{\mathrm{M}}+\mathrm{A}_{0}=\frac{\varepsilon l E m_{1}}{2 M_{s e} V}+\mathrm{A}_{0}$, indicating a linear relationship between $A$ and $m_{1}$ with a slope of $\frac{\varepsilon l E}{2 M_{s e} V}$.

<Checkpoint>[$\mathrm{A}=\frac{\varepsilon l E m_{1}}{2 M_{s e} V}+\mathrm{A}_{0}$](1)</Checkpoint>

Fitting the data from Table 1: (yields $y=a_{1} x+b_{1}, a_{1}=0.060035\left(\mu \mathrm{g}^{-1}\right), b_{1}=0.031554, r=0.99995$)

<Checkpoint>[Table 1 fitting result: $y=0.060035 x+0.031554$](1)</Checkpoint>

$\varepsilon=\frac{2 M_{S e} V}{l E} \times a=1.88 \times 10^{4} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{cm}^{-1}$ 

Selenium content in the sample:

For the sample, $\mathrm{A}=\varepsilon l c_{\mathrm{M}}+\mathrm{A}_{0}=\frac{\varepsilon l E\left(m+m_{2}\right)}{2 M_{s e} V}+\mathrm{A}_{0}=\frac{\varepsilon l E}{2 M_{s e} V} \times m_{2}+\frac{\varepsilon l E m}{2 M_{s e} V}+\mathrm{A}_{0}$,
showing a linear relationship between $A$ and $m2$;

<Checkpoint>[$\mathrm{A}=\frac{\varepsilon l E}{2 M_{s e} V} \times m_{2}+\frac{\varepsilon l E m}{2 M_{s e} V}+\mathrm{A}_{0}$](1)</Checkpoint>

Fitting the data from the table: yields $y=a_{2} x+b_{2}, a_{2}=0.062024\left(\mu \mathrm{g}^{-1}\right), b_{2}=0.037754, \quad r=0.99920$.

<Checkpoint>[Table 2 fitting result: $y=0.062024 x+0.037754$](1)</Checkpoint>

From $b_{1}=\mathrm{A}_{0}, b_{2}=\frac{\varepsilon l E m}{2 M_{s e} V}+\mathrm{A}_{0}$, it follows that $a_{2} = \frac{k^{\prime} \varepsilon l E}{2 M_{s e} V}=\frac{b_{2}-b_{1}}{\mathrm{n}}$,

$\quad m=\frac{b_{2}-b_{1}}{a_{2}}=0.09996 \mu \mathrm{g}$

$\mathrm{w}=\frac{m}{m_{0}} \times 1 \times 10^{6} \mathrm{ppm}=0.100 \mathrm{ppm}$. Statement 5 is correct.

If calculated using $a_{1}$, the result is $0.103 \mathrm{ppm}$, which is within the error margin and also correct.

<Checkpoint>[ $\mathrm{w}=0.100 \mathrm{~ppm}$ or $0.103 \mathrm{~ppm}$](2)</Checkpoint>

Thus, Statements 3 and 5 are correct, and the answer is J.
由题干，$\mathbf{X}$是人体必要元素，且在自然界中经常以矿物形式和同族重元素$\mathbf{Y}$伴生，可以确定：

<Checkpoint>[$\mathbf{X}$元素是$\mathrm{Se}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$元素是$\mathrm{Te}$](1)</Checkpoint>

方法1：

<Checkpoint>[$\mathrm{H}_{2} \mathrm{SeO}_{3}+4 \mathrm{HI} \rightarrow \mathrm{Se}+2 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{H}_{2} \mathrm{TeO}_{3}+6 \mathrm{HI} \rightarrow \mathrm{H}_{2} \mathrm{TeI}_{6}+3 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

方法2：

<Checkpoint>[$\mathrm{Se}+\mathrm{Na}_{2} \mathrm{SO}_{3}\rightarrow \mathrm{Na}_{2}\mathrm{Se} \mathrm{SO}_{3}$](1)</Checkpoint>

方法$1$中$\mathrm{Se}$元素价态发生变化，以沉淀形式析出，$\mathrm{Te}$元素价态不变，仍保留在溶液中，过滤可直接分离出$\mathrm{Se}$单质。

方法$2$中$\mathrm{Se}$元素价态发生变化，保留在溶液中，$\mathrm{Te}$不会溶解。

<Checkpoint>[方法1和方法2$\mathrm{Se}$价态均发生变化，方法1$\mathrm{Te}$价态不变](1)</Checkpoint>

由反应方程式，说法1错误，说法2错误。

$\mathrm{EDTA}$ 二钠盐可以与试样中的过渡金属离子络合而掩蔽之,防止其氧化 $3$,$3'$-二氨基联苯胺或与 $3$,$3'$-二氨基联苯胺发生络合形成有色物质,干扰测定。说法3正确。

<Checkpoint>[$\mathrm{EDTA}$ 二钠盐可以掩蔽过渡金属离子防止其干扰测定。](1)</Checkpoint>

化合物$\mathbf{M}$的结构：

<MultiModal>![C12=N[Se]N=C1C=CC(C3=CC4=N[Se]N=C4C=C3)=C2](/media/uploads/bfa0b5c2-2c15-4c50-b067-7e37b7a90592.png)</MultiModal>

<Checkpoint>[化合物M的结构为C12=N[Se]N=C1C=CC(C3=CC4=N[Se]N=C4C=C3)=C2](2)</Checkpoint>

由$\mathbf{M}$的结构可以判断，随着中间碳碳单键的旋转，不可以同时拥有二重旋转轴、相互垂直的$2$个镜面、对称中心；说法4错误。

<Checkpoint>[二重旋转轴，对称中心和相互垂直的$2$个镜面无法同时存在](1)</Checkpoint>

样品分析计算过程：

首先计算利用$\mathrm{1 ~mL}$甲苯萃取$\mathrm{25 ~mL}$ $\mathbf{M}$的溶液的萃取率$\mathrm{E}$：

$\mathrm{E=k_D/(k_D+V_{water}/V_{sample})=25.4/(25.4+25)=0.504}$

<Checkpoint>[利用$\mathrm{1 ~mL}$甲苯萃取$\mathrm{25 ~mL}$ $\mathbf{M}$的溶液的萃取率$\mathrm{E=0.504}$](2)</Checkpoint>

甲苯溶液中硒元素的浓度: $\mathrm{c}_{S e}=\frac{\mathrm{n}_{S e, o}}{V}=\frac{E \times \mathrm{n}_{S e, w}}{V}=\frac{\mathrm{E} m_{1}}{M_{S e} V}$ 

由于 $1$ 分子 $\mathbf{M}$ 中含有 2 个 $\mathrm{Se}$ 原子, $\mathbf{M}$的浓度: $c_{\mathbf{M}}=\frac{1}{2} c_{S e}=\frac{E m_{1}}{2 M_{S e} V}$

吸光度 $\mathrm{A}=\varepsilon l c_{\mathrm{M}}+\mathrm{A}_{0}=\frac{\varepsilon l E m_{1}}{2 M_{s e} V}+\mathrm{A}_{0}$, 推出 $A$ 与 $m_{1}$ 存在线性关系且斜率为 $\frac{\varepsilon l E}{2 M_{s e} V}$ 

<Checkpoint>[$\mathrm{A}=\frac{\varepsilon l E m_{1}}{2 M_{s e} V}+\mathrm{A}_{0}$ ](1)</Checkpoint>

用表 1 中数据拟合:(得 $y=a_{1} x+b_{1}, a_{1}=0.060035\left(\mu \mathrm{g}^{-1}\right), b_{1}=0.031554, r=0.99995$ 

<Checkpoint>[表1拟合结果：$y=0.060035 x+0.031554$](1)</Checkpoint>

$\varepsilon=\frac{2 M_{S e} V}{l E} \times a=1.88 \times 10^{4} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{cm}^{-1}$ 

样品中硒的含量:

对样品有 $\mathrm{A}=\varepsilon l c_{\mathrm{M}}+\mathrm{A}_{0}=\frac{\varepsilon l E\left(m+m_{2}\right)}{2 M_{s e} V}+\mathrm{A}_{0}=\frac{\varepsilon l E}{2 M_{s e} V} \times m_{2}+\frac{\varepsilon l E m}{2 M_{s e} V}+\mathrm{A}_{0}$,
$A$ 与 $m2$ 存在线性关系;

<Checkpoint>[$\mathrm{A}=\frac{\varepsilon l E}{2 M_{s e} V} \times m_{2}+\frac{\varepsilon l E m}{2 M_{s e} V}+\mathrm{A}_{0}$](1)</Checkpoint>

用表中数据拟合: 得 $y=a_{2} x+b_{2}, a_{2}=0.062024\left(\mu \mathrm{g}^{-1}\right), b_{2}=0.037754, \quad r=0.99920$ 

<Checkpoint>[表2拟合结果：$y=0.062024 x+0.037754$](1)</Checkpoint>

由 $b_{1}=\mathrm{A}_{0}, b_{2}=\frac{\varepsilon l E m}{2 M_{s e} V}+\mathrm{A}_{0}$, 可知 $a_{2} = \frac{k^{\prime} \varepsilon l E}{2 M_{s e} V}=\frac{b_{2}-b_{1}}{\mathrm{n}}$,

$\quad m=\frac{b_{2}-b_{1}}{a_{2}}=0.09996 \mu \mathrm{g}$

$\mathrm{w}=\frac{m}{m_{0}} \times 1 \times 10^{6} \mathrm{ppm}=0.100 \mathrm{ppm}$ 说法5正确。

如果利用$a_{1}$进行计算，得到结果为0.103 ppm在误差范围内，同样正确。

<Checkpoint>[ $\mathrm{w}=0.100或0.103 \mathrm{~ppm}$](2)</Checkpoint>

从而说法3，5正确，选择J。