Sparingly soluble salt $\bf A$ decomposes upon heating to produce four substances $\bf B, C, D, E$ (reaction 1). $\bf B$ is a black solid, insoluble in dilute hydrochloric acid and dilute sulfuric acid, but soluble in sodium sulfide solution; $\bf B$ is also soluble in nitric acid, accompanied by the formation of $\bf F$; $\bf B$ reacts with a certain non-metallic element upon heating to $\rm 1000\degree C$, and cooling yields $\bf C$. $\bf C$ is a liquid at room temperature; $\bf C$ reacts with chlorine gas at high temperature to produce a relatively common organic solvent $\bf G$, with byproduct $\bf H$ also obtained (reaction 2), $\bf H$ is an orange-yellow liquid at room temperature; $\bf H$ can continue to react with $\bf C$ to obtain $\bf G$ (reaction 3). $\bf D$ is a gas at room temperature and can be prepared by reacting a highly toxic potassium salt with copper sulfate solution (reaction 4); upon adding ferric chloride solution to the resulting suspension, gas $\bf D$ is still generated (reaction 5); $\bf D$ burns in an appropriate amount of pure oxygen to produce $\rm CO$ and $\bf E$; $\bf E$ is a gas at room temperature with a density slightly less than that of air.

Reactions 1-3 are chemical equations, reactions 4 and 5 are ionic equations, all balanced with the simplest stoichiometric coefficients. The following statements are made:

1. The chemical formula of $\bf A$ has an even number of atoms.
2. In the chemical equation for reaction 1, the sum of the coefficients of all species is a prime number.
3. The number of binary compounds among $\bf A-H$ is equal to 4.
4. In the ionic equations for reactions 4 and 5, the sum of the reactant coefficients are equal.
5. The sum of the coefficients of all species in the chemical equation for reaction 2 is less than the sum of the product coefficients in the chemical equation for reaction 3.

The option that contains all the correct statements is:
难溶盐$\bf A$受热分解生成$\bf B，C，D，E$四种物质(反应1)。$\bf B$为黑色固体，难溶于稀盐酸、稀硫酸，可溶于硫化钠溶液；$\bf B$也可溶于硝酸，同时伴有$\bf F$生成；$\bf F$与某非金属单质加热至$\rm 1000\degree C$反应，冷却可得到$\bf C$。$\bf C$在常温下为液体；$\bf C$与氯气高温反应，可用于生产一种较为常见的有机溶剂$\bf G$，同时得到副产物$\bf H$(反应2)，$\bf H$在常温下为橙黄色液体；$\bf H$可继续与$\bf C$反应得到$\bf G$(反应3)。$\bf D$在常温下为气体，可由一种剧毒钾盐与硫酸铜溶液反应制得(反应4)；向反应后的悬浊液中滴加氯化铁溶液，仍有气体$\bf D$生成(反应5)；$\bf D$在适量纯氧中燃烧，生成$\rm CO$和$\bf E$；$\bf E$在常温下为气体，密度略小于空气。

反应1~3为化学方程式，反应4，5为离子方程式，均以最简系数比配平的情况下，存在下列说法：

1. $\bf A$的化学式原子个数为偶数
2. 反应1的化学方程式，所有物种系数和为质数
3. $\bf A-H$中二元化合物的数目等于4
4. 反应4，5的离子方程式，反应物系数和相等
5. 反应2的化学方程式所有物种系数和小于反应3化学方程式的产物系数和。

包含所有正确说法的选项是：
All other options are incorrect
1，2
1，3
1，4
1，5
2，3
2，4
3，4
3，5
2，5
4，5
1，2，4
1，3，4
1，3，5
2，3，4
2，3，5
2，4，5
3，4，5
1，2，3，4
1，2，3，5
2，3，4，5
1，2，4，5
1，2，3
1，4，5
1，2，3，4，5
其他选项均不正确
1，2
1，3
1，4
1，5
2，3
2，4
3，4
3，5
2，5
4，5
1，2，4
1，3，4
1，3，5
2，3，4
2，3，5
2，4，5
3，4，5
1，2，3，4
1，2，3，5
2，3，4，5
1，2，4，5
1，2，3
1，4，5
1，2，3，4，5
The entry point of this question lies in the preparation of $\bf D$. Highly toxic potassium salts are generally considered to be $\rm KCN$, then $\bf D$ is the gas $\rm (CN)_2$. Since a suspension is formed, $\rm CuCN$ precipitate is produced. Therefore, the ionic equation for reaction 4 is:

<Checkpoint>[Highly toxic potassium salts are generally considered to be $\rm KCN$](1)</Checkpoint>

<Checkpoint>[$\bf D$ is the gas $\rm (CN)_2$](1)</Checkpoint>

$\rm 2Cu^{2+} + 4CN^- = 2CuCN + (CN)_2$

<Checkpoint>[Reaction 4: $\rm 2Cu^{2+} + 4CN^- = 2CuCN + (CN)_2$](1)</Checkpoint>

Upon adding ferric chloride solution, the gas $\rm (CN)_2$ is still generated, which is obviously an oxidation reaction of $\rm Fe(III)$; therefore, the equation for reaction 5 is

$\rm 2Fe^{3+} + 2CuCN + 2Cl^- = (CN)_2 + 2CuCl + 2Fe^{2+} $

<Checkpoint>[Reaction 5: $\rm 2Fe^{3+} + 2CuCN + 2Cl^- = (CN)_2 + 2CuCl + 2Fe^{2+} $](1)</Checkpoint>

For the ionic equations of reactions 4 and 5, the sum of the reactant coefficients is 6, so statement 4 is correct.

$\rm (CN)_2$ combustion produces $\rm CO$ and a gas with a density slightly less than air, so its relative molecular mass is less than 29, which is obviously $\rm N_2$. Therefore, $\bf E$ is $\rm N_2$.

<Checkpoint>[$\bf E$ is $\rm N_2$](1)</Checkpoint>

$\bf A$ decomposes to produce $\rm N_2$ and $\rm (CN)_2$, so it is obviously a thiocyanate; $\bf B$ is insoluble in acid but soluble in sodium sulfide solution, which is obviously a metal sulfide. Common metal thiocyanates with black sulfides are considered to be $\rm Cu, Hg$; however, $\rm CuS$ is insoluble in nitric acid, so $\bf A$ is $\rm Hg(SCN)_2$ and $\bf B$ is $\rm HgS$.

<Checkpoint>[$\bf A$ decomposes to produce $\rm N_2$ and $\rm (CN)_2$, which is a thiocyanate](1)</Checkpoint>

<Checkpoint>[$\rm CuS$ is insoluble in nitric acid, therefore $\bf A$ is $\rm Hg(SCN)_2$](1)</Checkpoint>

<Checkpoint>[$\bf B$ is $\rm HgS$](1)</Checkpoint>

$\rm HgS$ dissolves in nitric acid to produce $\bf F$, and $\bf F$ reacts with a non-metallic element to obtain liquid $\bf C$, which can react with chlorine to produce an organic solvent. Therefore, the non-metallic element must be carbon; thus, the organic solvent $\bf G$ is likely to be $\rm CCl_4$. Considering that $\bf F$ contains sulfur, it can be assumed that the liquid $\bf C$ is $\bf CS_2$, and the orange-yellow liquid $\bf H$ is $\rm S_2Cl_2$. $\bf F$ should be elemental $\rm S$.

<Checkpoint>[Organic solvent $\bf G$ is $\rm CCl_4$](1)</Checkpoint>

<Checkpoint>[Orange-yellow liquid $\bf H$ is $\rm S_2Cl_2$](1)</Checkpoint>

<Checkpoint>[Liquid $\bf C$ is $\bf CS_2$](1)</Checkpoint>

<Checkpoint>[$\bf F$ should be $\rm S$](1)</Checkpoint>

According to the chemical formulas, there are 5 binary compounds among $\bf A-H$, so statement 3 is incorrect. $\rm Hg(SCN)_2$ contains 7 atoms, so statement 1 is incorrect.

$\rm Hg(SCN)_2$ decomposes to produce $\rm CS_2$, $\rm HgS$, $\rm N_2$ and $\rm (CN)_2$; writing reaction 1:

$\rm 4Hg(SCN)_2 = 4HgS + 2CS_2 + 3(CN)_2 + N_2$

<Checkpoint>[Reaction 1: $\rm 4Hg(SCN)_2 = 4HgS + 2CS_2 + 3(CN)_2 + N_2$](1)</Checkpoint>

The sum of the equation coefficients is 14, which is not a prime number, so statement 2 is incorrect.

Reactions 2 and 3 can be written according to the chemical formulas:

<Checkpoint>[Reaction 2: $\rm CS_2 + 3Cl_2 = CCl_4 + S_2Cl_2$](1)</Checkpoint>

$\rm CS_2 + 2S_2Cl_2 = CCl_4 + 6S$

<Checkpoint>[Reaction 3: $\rm CS_2 + 2S_2Cl_2 = CCl_4 + 6S$](1)</Checkpoint>

According to the coefficients, statement 5 is correct.

Therefore, statements 4 and 5 are correct, and option K is correct.
本题入手点在$\bf D$的制备，剧毒钾盐一般考虑为$\rm KCN$，则$\bf D$为气体$\rm (CN)_2$。由于生成悬浊液，因此产生$\rm CuCN$沉淀。因此反应4的离子方程式为：

<Checkpoint>[剧毒钾盐一般考虑为$\rm KCN$](1)</Checkpoint>

<Checkpoint>[$\bf D$为气体$\rm (CN)_2$](1)</Checkpoint>

$\rm 2Cu^{2+} + 4CN^- = 2CuCN + (CN)_2$

<Checkpoint>[反应4：$\rm 2Cu^{2+} + 4CN^- = 2CuCN + (CN)_2$](1)</Checkpoint>

滴加氯化铁溶液,仍有气体$\rm (CN)_2$生成，明显为$\rm Fe(III)$的氧化反应；因此反应5方程式为

$\rm 2Fe^{3+} + 2CuCN + 2Cl^- = (CN)_2 + 2CuCl + 2Fe^{2+} $

<Checkpoint>[反应5：$\rm 2Fe^{3+} + 2CuCN + 2Cl^- = (CN)_2 + 2CuCl + 2Fe^{2+} $](1)</Checkpoint>

反应4，5的离子方程式，反应物系数和均为6，说法4正确。

$\rm (CN)_2$燃烧生成$\rm CO$和密度略小于空气的气体，则其相对分子质量小于29，很明显为$\rm N_2$。因此$\bf E$为$\rm N_2$。

<Checkpoint>[$\bf E$为$\rm N_2$](1)</Checkpoint>

$\bf A$分解产生$\rm N_2$和$\rm (CN)_2$，则很明显为硫氰酸盐；$\bf B$难溶于酸可溶于硫化钠溶液，很明显为金属硫化物。常见的硫氰酸金属盐且硫化物为黑色，考虑$\rm Cu,Hg$；但$\rm CuS$不溶于硝酸，因此$\bf A$为$\rm Hg(SCN)_2$，$\bf B$为$\rm HgS$。

<Checkpoint>[$\bf A$分解产生$\rm N_2$和$\rm (CN)_2$，为硫氰酸盐](1)</Checkpoint>

<Checkpoint>[$\rm CuS$不溶于硝酸，因此$\bf A$为$\rm Hg(SCN)_2$](1)</Checkpoint>

<Checkpoint>[$\bf B$为$\rm HgS$](1)</Checkpoint>

$\rm HgS$溶于硝酸生成$\bf F$，$\bf F$与非金属单质反应得到液体$\bf C$，该液体与氯气反应可产生有机溶剂。因此该非金属单质一定为碳；从而有机溶剂$\bf G$大概率为$\rm CCl_4$。考虑$\bf F$含有硫元素，则可以想到液体$\bf C$为$\bf CS_2$，橙黄色液体$\bf H$为$\rm S_2Cl_2$。$\bf F$应当为$\rm S$单质。

<Checkpoint>[有机溶剂$\bf G$为$\rm CCl_4$](1)</Checkpoint>

<Checkpoint>[橙黄色液体$\bf H$为$\rm S_2Cl_2$](1)</Checkpoint>

<Checkpoint>[液体$\bf C$为$\bf CS_2$](1)</Checkpoint>

<Checkpoint>[$\bf F$应当为$\rm S$](1)</Checkpoint>

根据化学式，$\bf A-H$中有5个二元化合物，说法3错误。$\rm Hg(SCN)_2$含7个原子，说法1错误。

$\rm Hg(SCN)_2$分解产生$\rm CS_2$，$\rm HgS$，$\rm N_2$和$\rm (CN)_2$；书写反应1：

$\rm 4Hg(SCN)_2 = 4HgS + 2CS_2 + 3(CN)_2 + N_2$

<Checkpoint>[反应1：$\rm 4Hg(SCN)_2 = 4HgS + 2CS_2 + 3(CN)_2 + N_2$](1)</Checkpoint>

方程式系数和为14，不为质数，说法2错误。

根据化学式即可写出反应2与反应3：

<Checkpoint>[反应2：$\rm CS_2 + 3Cl_2 = CCl_4 + S_2Cl_2$](1)</Checkpoint>

$\rm CS_2 + 2S_2Cl_2 = CCl_4 + 6S$

<Checkpoint>[反应3：$\rm CS_2 + 2S_2Cl_2 = CCl_4 + 6S$](1)</Checkpoint>

根据系数可知说法5正确。

因此最终说法4，5正确，选项K正确。