The metal $\mathrm{M}$ is chemically reactive and dissolves in dilute sulfuric acid to form compound **A**. Reacting **A** with a $\mathrm{KHCO_3}$ solution under a $\mathrm{CO_2}$ atmosphere at $10\,\degree \mathrm{C}$ for 7 days yields **B**, which contains only one type of anion. Pyrolysis of **B** produces oxide **C**. When water vapor passes over the surface of **C** at $1300\,\degree \mathrm{C}$, compound **D** is formed. Compound **D** dissolves in excess $\mathrm{NaOH}$ solution to form sodium salt **E**. At $150\,\degree \mathrm{C}$, the reaction of ammonia gas with $\mathrm{Et_2M}$ yields compound **F** and gas **G**. **F** dissolves in a liquid ammonia solution of potassium amide to form **H**, in which the mass fraction of $\mathrm{M}$ is $w_\mathrm{M} = 31.48\%$. The iodide **I** of $\mathrm{M}$ reacts with $\mathrm{Et_2M}$ in $\mathrm{THF}$ to form organometallic compound **J**.

The correct statements include:

1. Metal $\mathrm{M}$ is an element from the fifth period.  
2. Water is generated during the pyrolysis of **B**.  
3. **C** is a thermochromic material.  
4. In **E**, $\mathrm{M}$ is hexacoordinated.  
5. The molecular weight of **J** is approximately $\mathrm{180~g/mol}$.
金属 $\mathrm{M}$ 的化学性质较为活泼，溶于稀硫酸得到化合物 **A**。将 **A** 与 $\mathrm{KHCO_3}$ 溶液在 $\mathrm{CO_2}$ 气氛中于 $10\,\degree \mathrm{C}$ 作用 7 天制得 **B**，**B** 只有一种阴离子。热解 **B** 得氧化物 **C**。当水蒸气于 $1300\,\degree \mathrm{C}$ 下通过 **C** 表面可生成化合物 **D**。化合物 **D** 溶于过量 $\mathrm{NaOH}$ 溶液生成钠盐 **E**。在 $150\,\degree \mathrm{C}$ 时氨气通过 $\mathrm{Et_2M}$ 反应得到化合物 **F** 和气体 **G**。**F** 溶解于氨基钾的液氨溶液生成 **H**，**H** 中 $w_\mathrm{M} = 31.48\%$。$\mathrm{M}$ 的碘化物 **I** 与 $\mathrm{Et_2M}$ 在 $\mathrm{THF}$ 中反应得到有机金属化合物 **J**。

下列说法正确的有：

1.金属 $\mathrm{M}$ 为第五周期元素

2.热解 **B** 时会生成水

3.**C** 是一种热致变色材料

4.**E** 中 $\mathrm{M}$ 为六配位

5.**J** 的分子量约为 $\mathrm{180~g/mol}$
1,4
2.5
3
1,3,4
2,3,4
1, 2, 3
1, 2, 3, 4
2,4
3,4
3,5
2,3
None of the above options are correct
1,4
2,5
3
1,3,4
2,3,4
1,2,3
1,2,3,4
2,4
3,4
3,5
2,3
以上选项均不正确
The fact that $\mathrm{M}$ dissolves in dilute sulfuric acid to form a soluble sulfate and releases hydrogen gas indicates it is an active metal. Its salt **A** reacts with $\mathrm{KHCO_3}$ under a $\mathrm{CO_2}$ atmosphere at low temperature to form carbonate **B**, which can be thermally decomposed to yield oxide **C**. **C** reacts with high-temperature water vapor to form hydroxide **D**, which dissolves in excess $\mathrm{NaOH}$ to form complex **E**, demonstrating amphoteric hydroxide characteristics. Finally, based on the formation of compound **H** in a solution of potassium amide in liquid ammonia, where $w_\mathrm{M} = 31.48\%$, it is speculated that **H** is an amido complex with potassium as the cation. Calculation of the mass fraction confirms that $\mathrm{M}$ is $\mathrm{Zn}$.  

<Checkpoint>[It is speculated that **H** is an amido complex with potassium as the cation](1)</Checkpoint>  

<Checkpoint>[$\mathrm{M}$ is $\mathrm{Zn}$, statement 1 is incorrect](2)</Checkpoint>  

**H** is $\mathrm{K_2[Zn(NH_2)_4]}$.  

<Checkpoint>[**H** is $\mathrm{K_2[Zn(NH_2)_4]}$](1)</Checkpoint>  

Since $\mathrm{M}$ reacts with dilute sulfuric acid to form a soluble sulfate and releases hydrogen gas, **A** is $\mathrm{ZnSO_4}$.  

<Checkpoint>[**A** is $\mathrm{ZnSO_4}$](1)</Checkpoint>  

**A** reacts with $\mathrm{KHCO_3}$ under a $\mathrm{CO_2}$ atmosphere at low temperature to precipitate insoluble **B**, and **B** contains only one type of anion, so **B** is $\mathrm{ZnCO_3}$.  

<Checkpoint>[**B** is $\mathrm{ZnCO_3}$, and thermal decomposition of **B** produces only carbon dioxide, statement 2 is incorrect](1)</Checkpoint>  

**B** decomposes upon heating, releasing $\mathrm{CO_2}$ to form an oxide, so **C** is $\mathrm{ZnO}$. Under high-temperature conditions, zinc oxide loses a small amount of oxygen to form a non-stoichiometric compound. Due to lattice defects, color centers are generated, leading to absorption in the visible light region and resulting in color changes, making it a thermochromic material.  

<Checkpoint>[**C** is $\mathrm{ZnO}$, which is a thermochromic material, statement 3 is correct](1)</Checkpoint>  

**C** reacts with high-temperature water vapor to form a hydroxide, so **D** is $\mathrm{Zn(OH)_2}$.  

<Checkpoint>[**D** is $\mathrm{Zn(OH)_2}$](1)</Checkpoint>  

**D** dissolves in excess $\mathrm{NaOH}$ to form a coordination salt, so **E** is $\mathrm{Na_2[Zn(OH)_4]}$.  

<Checkpoint>[**E** is $\mathrm{Na_2[Zn(OH)_4]}$, which is 4-coordinate, statement 4 is incorrect](1)</Checkpoint>  

$\mathrm{Et_2Zn}$ undergoes ligand exchange with $\mathrm{NH_3}$ to form an alkane, so **F** is $\mathrm{Zn(NH_2)_2}$ and **G** is $\mathrm{C_2H_6}$.  

<Checkpoint>[**F** is $\mathrm{Zn(NH_2)_2}$](1)</Checkpoint>  

<Checkpoint>[**G** is $\mathrm{C_2H_6}$](1)</Checkpoint>  

The iodide **I** of $\mathrm{M}$ is $\mathrm{ZnI_2}$.  

<Checkpoint>[**I** is $\mathrm{ZnI_2}$](1)</Checkpoint>  

$\mathrm{ZnI_2}$ undergoes ligand exchange with $\mathrm{Et_2Zn}$ to form an organometallic compound, so **J** is $\mathrm{EtZnI}$.  

<Checkpoint>[**J** is $\mathrm{EtZnI}$, with a molecular weight of approximately $\mathrm{221~g/mol}$, statement 5 is incorrect](1)</Checkpoint>  

In conclusion, the answer is option C.  

<Checkpoint>[The answer is option C](1)</Checkpoint>
根据 $\mathrm{M}$ 溶于稀硫酸生成可溶性硫酸盐并放出氢气，表明其为活泼金属；其盐 **A** 与 $\mathrm{KHCO_3}$ 在 $\mathrm{CO_2}$ 氛围低温下生成碳酸盐 **B** 并可热分解得氧化物 **C**，**C** 与高温水蒸气生成氢氧化物 **D** 并可溶于过量 $\mathrm{NaOH}$ 得到配合物 **E**，体现两性氢氧化物特征；最后根据溶解于氨基钾的液氨溶液生成的 **H** 中 $w_\mathrm{M} = 31.48\%$，猜测**H**为氨基配合物，阳离子为钾离子，通过质量分数计算可确认 $\mathrm{M}$ 为 $\mathrm{Zn}$。  

<Checkpoint>[猜测**H**为氨基配合物，阳离子为钾离子](1)</Checkpoint>
 
<Checkpoint>[$\mathrm{M}$ 是 $\mathrm{Zn}$，说法1错误](2)</Checkpoint>

 **H** 是 $\mathrm{K_2[Zn(NH_2)_4]}$。

<Checkpoint>[**H** 是 $\mathrm{K_2[Zn(NH_2)_4]}$](1)</Checkpoint>

$\mathrm{M}$ 与稀硫酸反应生成可溶性硫酸盐并放出氢气，所以 **A** 是 $\mathrm{ZnSO_4}$

<Checkpoint>[**A** 是 $\mathrm{ZnSO_4}$](1)</Checkpoint>

**A** 与 $\mathrm{KHCO_3}$ 在 $\mathrm{CO_2}$ 氛围、低温下沉淀出难溶的 **B**，并且 **B** 只有一种阴离子，所以 **B** 是 $\mathrm{ZnCO_3}$

<Checkpoint>[**B** 是 $\mathrm{ZnCO_3}$，热解 **B** 时只产生二氧化碳，说法2错误](1)</Checkpoint>

**B** 加热分解脱 $\mathrm{CO_2}$ 生成氧化物，所以 **C** 是 $\mathrm{ZnO}$，在高温环境下氧化锌失去一小部分氧形成非整比化合物，由于晶格的缺陷产生了色心，在可见光区有吸收，因而颜色出现变化，是一种热致变色材料

<Checkpoint>[**C** 是 $\mathrm{ZnO}$，是一种热致变色材料，说法3正确](1)</Checkpoint>

**C** 与高温水蒸气作用生成氢氧化物，所以 **D** 是 $\mathrm{Zn(OH)_2}$

<Checkpoint>[**D** 是 $\mathrm{Zn(OH)_2}$](1)</Checkpoint>

**D** 溶于过量 $\mathrm{NaOH}$ 给出配位盐，所以 **E** 是 $\mathrm{Na_2[Zn(OH)_4]}$

<Checkpoint>[**E** 是 $\mathrm{Na_2[Zn(OH)_4]}$，为4配位，说法4错误](1)</Checkpoint>

$\mathrm{Et_2Zn}$ 与 $\mathrm{NH_3}$ 发生配体交换并生成烷烃，所以 **F** 是 $\mathrm{Zn(NH_2)_2}$ 且 **G** 为 $\mathrm{C_2H_6}$

<Checkpoint>[**F** 是 $\mathrm{Zn(NH_2)_2}$](1)</Checkpoint>

<Checkpoint>[**G** 为 $\mathrm{C_2H_6}$](1)</Checkpoint>

$\mathrm{M}$ 的碘化物 **I** 是$\mathrm{ZnI_2}$

<Checkpoint>[**I** 是$\mathrm{ZnI_2}$](1)</Checkpoint>

$\mathrm{ZnI_2}$ 与 $\mathrm{Et_2Zn}$ 发生配体交换，得到有机金属化合物，所以 **J** 是 $\mathrm{EtZnI}$。

<Checkpoint>[**J** 是 $\mathrm{EtZnI}$，分子量约为 $\mathrm{221~g/mol}$，说法5错误](1)</Checkpoint>

综上，答案为选项C

<Checkpoint>[答案为选项C](1)</Checkpoint>