Element **X** is mainly derived from associated minerals such as molybdenite. When molybdenite is roasted, **X** is converted to the volatile oxide **A**. Dissolving **A** in ammonia yields salt **B**, which can be reduced with hydrogen to obtain **X**. Reducing **A** with $\rm CO$ produces oxide **C**, whose crystal is cubic and has very low resistivity. Spongy **X** reacts with chlorine at $400\sim500\degree\mathrm{C}$ to obtain **D**, and **D** disproportionates upon contact with water, forming **E** and oxide **F** in a 1:2 ratio. The oxidation state of **X** in **E** is the same as in **A**. The structure of **D** is actually a dimer. In a nitrogen atmosphere, **D** undergoes thermal decomposition to produce **G** (containing **X** $63.65 \% $) and gas **H**.

Deduce the substances represented by the above letters and determine the correctness of the following statements.

1. **X** belongs to Group 8

2. The valence of the metal in **A** is greater than or equal to 6

3. The formula weight of **B** is greater than 250

4. The low resistivity of **C** is due to the presence of metallic bonds in the crystal

5. **D** is stable in air

6. The formula weight of **G** is greater than 300

7. **H** can be used for water purification

Let a be the sum of the squares of the serial numbers of all correct statements, and b be the square of the sum of the serial numbers of all incorrect statements. Find the value of a/b, taking three significant figures.
单质**X**主要来源于辉钼矿等伴生矿物，当焙烧辉钼矿时,**X**转化为挥发性的氧化物**A**。**A**溶于氨水得到盐**B**，通过氢气还原**B**可以得到**X**。用$\rm CO$还原 **A**制得氧化物**C**，**C**的晶体为立方晶系，具有很低的电阻率。海绵状的**X**与氯气在$400\sim500\degree\mathrm{C}$反应可得**D**，而**D**遇水即发生歧化，以1:2生成**E**和氧化物**F**，**E**中**X**的氧化态与**A**相同。**D**的结构实际上为二聚体。在氮气介质中，**D**发生热分解，产生**G**（含**X** $63.65 \% $)和气体**H**。

推断以上字母所代表的物质，判断以下说法的正确性。

1.**X**属于第八族

2.**A**中金属的价态大于等于6

3.**B**的式量大于250

4.**C**的低电阻率源于晶体中存在金属键

5.**D**在空气中稳定

6.**G**式量大于300

7.**H**可用于净水

设a为所有正确说法序号的平方和，b为所有错误说法序号的和的平方，求a/b的值，取三位有效数字
All other options are incorrect
0.242
0.542
0.280
2.10
0.147
其他选项均不正确
0.242
0.542
0.280
2.10
0.147
The entry point of this question is **G**. **D** is obtained from the reaction of **X** element and chlorine gas, and is a chloride of **X**. **D** decomposes upon heating to yield **G** and gas **H**. It is evident that **G** is also a chloride of **X**, while **H** is a chloride of **X** or chlorine gas. Let **G** be $XCl_n$. Given that **G** contains **X** at $63.65\%$, calculate the atomic weight of **X**. Only when n=3 does the atomic weight of **X** equal 186.2, corresponding to $Re$. All other calculation results do not correspond to any element. **G** is $ReCl_3$.

<Checkpoint>[**X** is $Re$](2)</Checkpoint>

<Checkpoint>[**G** is $ReCl_3$](1)</Checkpoint>

$Re$ is an element of the seventh subgroup, so statement 1 is incorrect.

The formula weight of **G** is 293, which is less than 300, so statement 6 is incorrect.

Calcination of a mineral containing **X** yields a volatile **A**. From common elemental knowledge, **A** is $Re_2O_7$. When **A** is dissolved in ammonia water, **B** obtained is $NH_4ReO_4$. The common preparation method for rhenium is the reduction of ammonium perrhenate with hydrogen gas, which is consistent with the description in the question.

The valence state of the metal element in **A** is 7, which is greater than 6, so statement 2 is correct.

The formula weight of **B** is 268, which is greater than 250, so statement 3 is correct.

<Checkpoint>[**A** is $Re_2O_7$](1)</Checkpoint>

<Checkpoint>[**B** is $NH_4ReO_4$](1)</Checkpoint>

**C** is obtained by reducing **A** with $CO$ and is a low-resistivity oxide with a cubic crystal system. Among the oxides of rhenium, only the classic $ReO_3$ possesses such properties. Its crystal structure can be viewed as rhenium atoms at the vertices of the unit cell, with oxygen atoms occupying all the edge centers. Its low resistivity stems from the spatially delocalized electron network formed by the conjugation of rhenium atoms through oxygen atoms. Therefore, statement 4 is incorrect.

<Checkpoint>[**C** is $ReO_3$](1)</Checkpoint>

<Checkpoint>[The low resistivity of $ReO_3$ is due to the formation of a delocalized electron network through the conjugation of metal atoms through oxygen atoms in the crystal](1)</Checkpoint>

Chloride **D** disproportionates upon contact with water, forming **E** and oxide **F** in a 1:2 ratio. The oxidation state of **X** in **E** is the same as in **A**, so **E** is $HReO_4$. There is water vapor in the air, so **D** cannot exist stably in the air, and statement 5 is incorrect.

Based on the metal valence state and the disproportionation ratio, there are only two possibilities: either **D** is $ReCl_5$ and **F** is $ReO_2$, or **D** is $ReCl_3$ and **F** is $Re_2O$. However, **G** is already $ReCl_3$. Therefore, **D** is $ReCl_5$ and **F** is $ReO_2$.

<Checkpoint>[**D** is $ReCl_5$](1)</Checkpoint>

<Checkpoint>[**F** is $ReO_2$](1)</Checkpoint>

The gas **H** produced during the decomposition of $ReCl_5$ is $Cl_2$, because rhenium chlorides are not gaseous under normal heating conditions. Chlorine gas can be used for water purification, so statement 7 is correct.

<Checkpoint>[**H** is $Cl_2$](1)</Checkpoint>

In summary, statements 2, 3, and 7 are correct, and statements 1, 4, 5, and 6 are incorrect. a is 62, b is 256, a/b = 0.242, B is correct.

<Checkpoint>[Statements 2, 3, and 7 are correct](1)</Checkpoint>
本题的入手点是**G**。**D**由**X**单质和氯气反应得到，是**X**的氯化物。**D**加热分解得到**G**和气体**H**。可见**G**也是**X**的氯化物，而**H**是**X**的氯化物或氯气。设**G**为$XCl_n$，由**G**含**X** $63.65 \% $，计算**X**的原子量，仅有n=3时，**X**原子量为186.2对应$Re$，其余计算结果均无对应元素。**G**是$ReCl_3$。

<Checkpoint>[**X**是$Re$](2)</Checkpoint>

<Checkpoint>[**G**是$ReCl_3$](1)</Checkpoint>

$Re$是第七副族的元素，说法1错误。

**G**的式量为293，小于300，说法6错误。

含**X**的矿物煅烧得到挥发性的**A**，由元素常识可知**A**是$Re_2O_7$，**A**溶于氨水则得到**B**为$NH_4ReO_4$。铼的常见制备方法即为氢气还原高铼酸铵，与题目中的描述相符。

**A**中金属元素价态为7大于6，说法2正确。

**B**的式量268大于250，说法3正确。

<Checkpoint>[**A**为$Re_2O_7$](1)</Checkpoint>

<Checkpoint>[**B**为$NH_4ReO_4$](1)</Checkpoint>

**C**由$CO$还原**A**得到，是立方晶系的低电阻率氧化物，铼的氧化物中仅有经典的$ReO_3$具有这样的性质。其晶体结构可视作铼原子在晶胞顶点，氧原子占据所有楞心，其低电阻率来源于铼原子间通过与氧原子的共轭形成了空间离域电子网络。故说法4错误。

<Checkpoint>[**C**是$ReO_3$](1)</Checkpoint>

<Checkpoint>[$ReO_3$的低电阻率是由于晶体中金属原子间通过氧原子共轭形成离域电子网络](1)</Checkpoint>

氯化物**D**遇水即歧化，以1:2生成**E**和氧化物**F**，**E**中**X**的氧化态与**A**相同，可知**E**是$HReO_4$。空气中有水蒸气，**D**无法在空气中稳定存在，说法5错误。

由金属价态和歧化比例，只有两种可能，即**D**为$ReCl_5$，**F**为$ReO_2$，或**D**为$ReCl_3$，**F**为$Re_2O$，而**G**已经是$ReCl_3$了。则**D**为$ReCl_5$，**F**为$ReO_2$。

<Checkpoint>[**D**为$ReCl_5$](1)</Checkpoint>

<Checkpoint>[**F**为$ReO_2$](1)</Checkpoint>

$ReCl_5$分解时产生的气体**H**为$Cl_2$，因为铼的氯化物在正常加热条件下不为气态。氯气可用于净水，说法7正确。

<Checkpoint>[**H**为$Cl_2$](1)</Checkpoint>

综上所述，说法2,3,7正确，说法1,4,5,6错误。a为62，b为256，a/b=0.242, B正确。

<Checkpoint>[说法2,3,7正确](1)</Checkpoint>