The metal **M** can react with gas **A** under heating to form a binary compound **B**, which is soluble in both acids and bases. When **B** is dissolved in dilute hydrochloric acid, no gas is released. Dissolving **B** in sodium hydroxide solution and heating yields a solution of **C** and gas **A**, which turns moist red litmus paper blue. Passing hydrogen sulfide gas into the colorless solution **C** produces a white precipitate **D**. **D** is insoluble in water but soluble in hydrochloric acid. Gradually adding dilute hydrochloric acid to solution **C** initially forms a white precipitate **E**, which is soluble in ammonia water. If excess hydrochloric acid is added, the precipitate dissolves to form a colorless solution **F**. **M** can also react with iodoethane to yield a non-iodine-containing compound **G**. When exposed to air, **G** is oxidized by oxygen to form compound **H**, which contains no peroxide bonds. Dissolving 1 mole of **H** in excess sodium hydroxide solution still yields 1 mole of **C** and releases 1 mole of ethane.  

Based on the above information, analyze the chemical formulas of each substance and select the correct option.
金属**M**在加热的条件下，能与气体**A**反应，得到一种既能溶于酸又能溶于碱的二元化合物**B**。**B**溶于稀盐酸时，不放出任何气体。将**B**溶于氢氧化钠溶液并加热，得到C的溶液和气体**A**，**A**能使湿润的红色石蕊试纸变蓝。向无色溶液**C**中通入硫化氢气体，有白色的沉淀**D**生成。**D**不溶于水，但可溶于盐酸。向溶液**C**中逐渐加入稀盐酸，先生成能溶于氨水的白色沉淀**E**，若盐酸过量，则沉淀溶解，得到无色溶液**F**。**M**也能与碘乙烷反应，得到不含碘的化合物**G**，**G**放置在空气中，会被氧气氧化，得到不含过氧键的化合物**H**。将1摩尔**H**溶于过量氢氧化钠溶液仍能得到1摩尔**C**，并放出1摩尔乙烷。

请根据以上信息，分析各个物质的化学式，并选择正确的选项。
All other options are incorrect
Gas **A** is an inert gas.
Metal **M** is commonly used in welding
The molecular weight of compound **B** is approximately 84
Substance **C** contains three elements.
Compound **D** can be used as phosphors and pigments.
Precipitate **E** has a molecular weight of approximately 146.
Substance **F** is not prone to deliquescence.
After heating, compound **G** decomposes to form a compound with a molecular weight of 76.
Compound **H** contains an anion.
其他选项均不正确
气体**A**为惰性气体
金属**M**常用于焊接
化合物**B**分子量约84
物质**C**中有三种元素
化合物**D**可用作荧光粉与颜料
沉淀**E**分子量约146
物质**F**不易潮解
化合物**G**加热后分解生成一种分子量76的化合物
化合物**H**中含有一种负离子
From the fact that **A** can turn litmus blue, it is known to be ammonia ($NH_3$).  

<Checkpoint>[Gas **A** is ammonia](1)</Checkpoint>  

Next, since metal **M** can react with ammonia, it is known to be reactive; since its sulfide precipitate can be transformed by acid, its sulfide solubility is not very poor; since it can form solutions in both acidic and alkaline environments, it is amphoteric; and since it can react with iodoethane to form a compound without iodine, it can form organometallic compounds.  

Therefore, from the aforementioned properties, the one that matches these characteristics is selected, and by further considering the color of its sulfide, Sn is excluded, confirming that metal **M** is zinc (Zn).  

<Checkpoint>[Metal **M** is zinc](5)</Checkpoint>  

Thus, the binary compound **B** is formed by the reaction of ammonia and zinc metal, which should be $Zn_3N_2$.  

<Checkpoint>[Compound **B** is $Zn_3N_2$](1)</Checkpoint>  

Compound **C** is the product dissolved in alkali, so it should be $Na_2Zn(OH)_4$.  

<Checkpoint>[Compound **C** is $Na_2Zn(OH)_4$](1)</Checkpoint>  

Compound **D** is the sulfide precipitate formed, which is $ZnS$, and can be used in pigments and fluorescence.  

<Checkpoint>[Compound **D** is $ZnS$](0.5)</Checkpoint>  

Compound **E** is $Zn(OH)_2$.  

<Checkpoint>[Compound **E** is $Zn(OH)_2$](0.5)</Checkpoint>  

Compound **F** is the chloride, which should be $ZnCl_2$, a substance that is highly hygroscopic.  

<Checkpoint>[Compound **F** is $ZnCl_2$](0.5)</Checkpoint>  

Compound **G** is the organometallic compound formed by the reaction with iodoethane. Since it does not contain iodine, it is identified as $Zn(C_2H_5)_2$. Heating it generates ethylene with a molecular weight of 28.  

<Checkpoint>[Compound **G** is $Zn(C_2H_5)_2$](1)</Checkpoint>  

Compound **H** is the oxidation product of **G**. It dissolves in alkali to produce only one equivalent of ethane and contains no peroxide bond, so it is identified as $Zn(C_2H_5)(OC_2H_5)$.  

<Checkpoint>[Compound **H** is $Zn(C_2H_5)(OC_2H_5)$](1.5)</Checkpoint>  

After deducing all the chemical formulas, it can be concluded based on the properties of the compounds that only option F is correct.
由**A**可以使得石蕊变蓝可知其为氨气（$NH_3$）

<Checkpoint>[气体**A**为氨气](1)</Checkpoint>

接下来，由金属**M**可以与氨气反应，可知其性质活泼；由其硫化物沉淀加酸可转化，可知其硫化物溶解度并非很差；由其可以生成酸碱两种环境的溶液，可知其具有两性；由其可以与碘乙烷反应生成不含碘的化合物，可知其能够生成金属有机化合物。

故而，从前述性质中，选出符合前述性质的，并通过其中硫化物颜色最终排除性质相近的Sn，可知金属**M**为锌（Zn）

<Checkpoint>[金属**M**为锌](5)</Checkpoint>

故而，二元化合物**B**为氨气与锌金属反应生成，应为$Zn_3N_2$。

<Checkpoint>[化合物**B**为$Zn_3N_2$](1)</Checkpoint>

化合物**C**为溶解于碱的产物，故而应为$Na_2Zn(OH)_4$

<Checkpoint>[化合物**C**为$Na_2Zn(OH)_4$](1)</Checkpoint>

化合物**D**为生成的硫化物沉淀，故而为$ZnS$，可用于颜料和荧光。

<Checkpoint>[化合物**D**为$ZnS$](0.5)</Checkpoint>

化合物**E**为$Zn(OH)_2$

<Checkpoint>[化合物**E**为$Zn(OH)_2$](0.5)</Checkpoint>

化合物**F**为氯化物，应为$ZnCl_2$，该物质极易潮解。

<Checkpoint>[化合物**F**为$ZnCl_2$](0.5)</Checkpoint>

化合物**G**为和碘乙烷反应生成的金属有机化合物，因为不含有碘，可知其为$Zn(C_2H_5)_2$。加热生成乙烯，分子量为28。

<Checkpoint>[化合物**G**为$Zn(C_2H_5)_2$](1)</Checkpoint>

化合物**H**为**G**的氧化产物，其溶于碱仅生成一当量的乙烷，不含过氧键，故而可知其为$Zn(C_2H_5)(OC_2H_5)$

<Checkpoint>[化合物**H**为$Zn(C_2H_5)(OC_2H_5)$](1.5)</Checkpoint>

推出所有化学式后，根据化合物性质可以判断只有选项F正确。