$\bf{X、Y}$ are elements in the same group and have similar chemical properties. The dioxide of $\bf{X}$, $\bf{A}$, has widespread applications in organic chemistry, and its aqueous solution can be reduced by sulfur dioxide to generate the element $\bf{X}$ (reaction 1); $\bf{A}$ can react with $\bf{H}$ to obtain ion $\bf{B}$ and ion $\bf{C}$ (reaction 2), $\bf{B}$ has one more $\bf{X}$ atom than $\bf{C}$ and the charge remains unchanged. The chloride of $\bf{X}$, $\bf{D}$, reacts with $\mathrm { N H } _ { 3 }$ to produce an explosive substance $\bf{E}$ (reaction 3). $\bf{D}$ reacts with $\bf{A}$ to obtain $\bf{F}$, and $\bf{F}$ reacts with Cu element, and the products include $\bf{A}$ and $\bf{G}$ (reaction 4). $\bf{F}$ can also react with excess sulfur element (reaction 5), and the products include $\bf{A}$ and $\bf{G}$, as well as $\bf{I}$, which has a similar structure to $\bf{G}$. $\pmb { \bf{Z} } ^ { 2 -} $ containing 57.01% $\bf{Y}$ reacts with sulfite to generate the element $\bf{Y}$ and the oxygen-containing anion $\bf{H}$ (reaction 6). The sodium salt of $\bf{H}$ is a common reagent in analytical chemistry and reacts with $\mathrm { I } _ { 2 }$ to convert to $\bf{C}$.
Which of the following statements are correct?
1. $\bf{B}$ contains 11 atoms
2. Molecule $\bf{E}$ contains a 4-fold rotation axis
3. The ratio of reaction coefficients of each reactant and product in reaction 3 is 12:64:3:4:48
4. The ratio of reaction coefficients of each reactant and product in reaction 5 is 6:4:3:2:1
5. Sulfide ions are also produced in reaction 6
6. $\bf{I}$ is a reddish-brown gas
7. Reaction 2 does not require acid participation
8. Reaction 4 also produces a water-insoluble white precipitate.
$\bf{X、Y}$是同族元素，二者具有相似的化学性质。$\bf{X}$的二氧化物$\bf{A}$在有机化学中有广泛应用，其水溶液可被二氧化硫还原生成$\bf{X}$的单质(反应1)；$\bf{A}$可与$\bf{H}$反应得到离子$\bf{B}$和离子$\bf{C}$(反应2)，$\bf{B}$比$\bf{C}$多一个$\bf{X}$原子且电荷不变。$\bf{X}$的氯化物$\bf{D}$与 $\mathrm { N H } _ { 3 }$ 反应会得到一种易爆物质$\bf{E}$(反应3)。$\bf{D}$与$\bf{A}$反应可得到$\bf{F}$，$\bf{F}$与Cu单质反应，产物包括$\bf{A}$和$\bf{G}$(反应4)，$\bf{F}$亦可与过量硫单质反应(反应5)，产物除$\bf{A}$和$\bf{G}$外，还有与$\bf{G}$结构相似的$\bf{I}$。含$\bf{Y}$57.01%的 $\pmb { \bf{Z} } ^ { 2 -} $ 与亚硫酸根反应，生成$\bf{Y}$的单质以及含氧阴离子$\bf{H}$(反应6）。$\bf{H}$的钠盐是分析化学中的一种常用试剂，与 $\mathrm { I } _ { 2 }$ 反应转化为$\bf{C}$。
下面的说法中哪些是正确的？
1. $\bf{B}$中含有11个原子
2. 分子$\bf{E}$含有4次旋转轴
3. 反应3各个反应物与产物的反应系数比为12：64：3：4：48
4. 反应5中各个反应物与产物的反应系数比为6：4：3：2：1
5. 反应6中还生成了硫离子
6. $\bf{I}$是红棕色气体
7. 反应2不需要酸参与
8. 反应4还生成了一种难溶于水的白色沉淀。
All other options are incorrect
2.3.6.
1.2.3.4.6.
1.4.5.
2.3.5.
3.6.8.
2.4.7.8.
1.5.8.
2.4.7.
2.3.
4.5
2.8.
其他选项均不正确
2.3.6.
1.2.3.4.6.
1.4.5.
2.3.5.
3.6.8.
2.4.7.8.
1.5.8.
2.4.7.
2.3.
4.5.
2.8.
According to the knowledge of elemental chemistry, $\bf{X}$ is $Se$, $\bf{Y}$ is $Te$

<Checkpoint>[$\bf{X}$ is $Se$, $\bf{Y}$ is $Te$](1)</Checkpoint>

$\bf{A}$ is the dioxide of $Se$, so $\bf{A}$ is $SeO_2$; from the mass fraction, we can know that $\bf{Z²⁻}$ should be $TeS_3^{2-} $. From the role of the sodium salt of $\bf{H}$ in analytical chemistry, we can know that $\bf{H}$ is $S_2O_3^{2-}$. Thus we can deduce that $\bf{C}$ is $S_4O_6^{2-}$, and from the fact that $\bf{B}$ has one more $Se$ atom than $\bf{C}$, we can deduce that the chemical formula of $\bf{B}$ is $SeS_4O_6^{2-}$.

<Checkpoint>[$\bf{A}$ is $SeO_2$](1)</Checkpoint>

<Checkpoint>[$\bf{Z²⁻}$ is $TeS_3^{2-} $](1)</Checkpoint>

<Checkpoint>[$\bf{H}$ is $S_2O_3^{2-}$](1)</Checkpoint>

<Checkpoint>[$\bf{C}$ is $S_4O_6^{2-}$](1)</Checkpoint>

<Checkpoint>[$\bf{B}$ is $SeS_4O_6^{2-}$](1)</Checkpoint>

The chloride of $Se$, $\bf{D}$, should be $SeCl_4$, which corresponds to $\bf{E}$ when reacted with ammonia to yield $Se_4N_4$.

<Checkpoint>[$\bf{D}$ should be $SeCl_4$](1)</Checkpoint>

<Checkpoint>[$\bf{E}$ is $Se_4N_4$](1)</Checkpoint>

$\bf{D}$ reacts with $\bf{A}$ to yield $\bf{F}$, which should be $SeOCl_2$, and can react with Cu to yield $\bf{G}$ as $Se_2Cl_2$, and reacts with sulfur to yield $\bf{I}$ with a similar structure: $S_2Cl_2$.

<Checkpoint>[$\bf{F}$ is $SeOCl_2$](1)</Checkpoint>

<Checkpoint>[$\bf{G}$ is $Se_2Cl_2$](1)</Checkpoint>

<Checkpoint>[$\bf{I}$ is $S_2Cl_2$](1)</Checkpoint>

$\bf{B}$ contains 4+6+1=11 atoms, 1 is correct.
The molecule $\bf{E}$ contains a 4-fold rotoinversion axis but not a rotation axis, 2 is incorrect.
$\bf{I}$ is a red-brown liquid, 6 is incorrect.

Reaction 1. ${SeO_2 + 2SO_2 + 2H_2O = 2H_2SO_4 + Se}$

<Checkpoint>[Reaction 1. ${SeO_2 + 2SO_2 + 2H_2O = 2H_2SO_4 + Se}$](1)</Checkpoint>

Reaction 2. ${4H^+ + SeO_2 + 4S_2O_3^{2-} = S_4O_6^{2-} + SeS_4O_6^{2-} + 2H_2O}$

Reaction 2 requires acid participation, 7 is incorrect.

<Checkpoint>[Reaction 2. ${4H^+ + SeO_2 + 4S_2O_3^{2-} = S_4O_6^{2-} + SeS_4O_6^{2-} + 2H_2O}$](1)</Checkpoint>

Reaction 3. ${12SeCl_4 + 64NH_3 = 3Se_4N_4 + 2N_2 + 48NH_4Cl}$

The ratio of the reaction coefficients of each reactant and product in reaction 3 is 12:64:3:2:48, 3 is incorrect.

<Checkpoint>[Reaction 3. ${12SeCl_4 + 64NH_3 = 3Se_4N_4 + 2N_2 + 48NH_4Cl}$](1)</Checkpoint>

Reaction 4. ${3Cu + 4SeOCl_2 = 3CuCl_2 + 2SeO_2 + Se_2Cl_2}$ 

$CuCl_2$ is soluble in water, 8 is incorrect.

<Checkpoint>[ Reaction 4. ${3Cu + 4SeOCl_2 = 3CuCl_2 + 2SeO_2 + Se_2Cl_2}$ ](1)</Checkpoint>

Reaction 5. ${6S + 4SeOCl_2 = 3S_2Cl_2 + 2SeO_2 + Se_2Cl_2}$

The ratio of the reaction coefficients of each reactant and product in reaction 5 is 6:4:3:2:1, 4 is correct.

<Checkpoint>[Reaction 5. ${6S + 4SeOCl_2 = 3S_2Cl_2 + 2SeO_2 + Se_2Cl_2}$](1)</Checkpoint>

Reaction 6. ${TeS_3^{2-} + 2SO_3^{2-} = 2S_2O_3^{2-} + S^{2-} + Te}$  

<Checkpoint>[Reaction 6. ${TeS_3^{2-} + 2SO_3^{2-} = 2S_2O_3^{2-} + S^{2-} + Te}$ ](1)</Checkpoint>

Sulfide ions are also generated in reaction 6, 5 is correct.

It can be found that only 1, 4, and 5 are correct, choose D.
根据元素化学知识可以知道，$\bf{X}$是$Se$,$\bf{Y}$是$Te$

<Checkpoint>[$\bf{X}$是$Se$,$\bf{Y}$是$Te$](1)</Checkpoint>

$\bf{A}$是$Se$的二氧化物，故$\bf{A}$为$SeO_2$；由质量分数可以知道，$\bf{Z²⁻}$应为$TeS_3^{2-} $。由$\bf{H}$的钠盐在分析化学中的作用可以知道，$\bf{H}$为$S_2O_3^{2-}$.从而我们可以推出$\bf{C}$为$S_4O_6^{2-}$,由$\bf{B}$比$\bf{C}$多一个$Se$原子可以推出$\bf{B}$的化学式为$SeS_4O_6^{2-}$。

<Checkpoint>[$\bf{A}$为$SeO_2$](1)</Checkpoint>

<Checkpoint>[$\bf{Z²⁻}$为$TeS_3^{2-} $](1)</Checkpoint>

<Checkpoint>[$\bf{H}$为$S_2O_3^{2-}$](1)</Checkpoint>

<Checkpoint>[$\bf{C}$为$S_4O_6^{2-}$](1)</Checkpoint>

<Checkpoint>[$\bf{B}$为$SeS_4O_6^{2-}$](1)</Checkpoint>

$Se$的氯化物$\bf{D}$应为$SeCl_4$，与氨反应得到$Se_4N_4$对应$\bf{E}$.

<Checkpoint>[$\bf{D}$应为$SeCl_4$](1)</Checkpoint>

<Checkpoint>[$\bf{E}$为$Se_4N_4$](1)</Checkpoint>

$\bf{D}$与$\bf{A}$反应可得到$\bf{F}$,应为$SeOCl_2$,可以与Cu反应得到$\bf{G}$为$Se_2Cl_2$,与硫反应得到与之结构类似的$\bf{I}$: $S_2Cl_2$。

<Checkpoint>[$\bf{F}$为$SeOCl_2$](1)</Checkpoint>

<Checkpoint>[$\bf{G}$为$Se_2Cl_2$](1)</Checkpoint>

<Checkpoint>[$\bf{I}$为$S_2Cl_2$](1)</Checkpoint>

$\bf{B}$中含有4+6+1=11个原子,1正确。
分子$\bf{E}$含有4次反轴而非旋转轴，2错误。
$\bf{I}$是红棕色液体，6错误。

反应1. ${SeO_2 + 2SO_2 + 2H_2O = 2H_2SO_4 + Se}$

<Checkpoint>[反应1. ${SeO_2 + 2SO_2 + 2H_2O = 2H_2SO_4 + Se}$](1)</Checkpoint>

反应2. ${4H^+ + SeO_2 + 4S_2O_3^{2-} = S_4O_6^{2-} + SeS_4O_6^{2-} + 2H_2O}$

 反应2需要酸参与，7错误。

<Checkpoint>[反应2. ${4H^+ + SeO_2 + 4S_2O_3^{2-} = S_4O_6^{2-} + SeS_4O_6^{2-} + 2H_2O}$](1)</Checkpoint>

反应3. ${12SeCl_4 + 64NH_3 = 3Se_4N_4 + 2N_2 + 48NH_4Cl}$

反应3各个反应物与产物的反应系数比为12：64：3：2：48，3错误。

<Checkpoint>[反应3. ${12SeCl_4 + 64NH_3 = 3Se_4N_4 + 2N_2 + 48NH_4Cl}$](1)</Checkpoint>

反应4. ${3Cu + 4SeOCl_2 = 3CuCl_2 + 2SeO_2 + Se_2Cl_2}$ 

$CuCl_2$溶于水，8错误。

<Checkpoint>[ 反应4. ${3Cu + 4SeOCl_2 = 3CuCl_2 + 2SeO_2 + Se_2Cl_2}$ ](1)</Checkpoint>

反应5. ${6S + 4SeOCl_2 = 3S_2Cl_2 + 2SeO_2 + Se_2Cl_2}$

反应5中各个反应物与产物的反应系数比为6：4：3：2：1，4正确。

<Checkpoint>[反应5. ${6S + 4SeOCl_2 = 3S_2Cl_2 + 2SeO_2 + Se_2Cl_2}$](1)</Checkpoint>

反应6. ${TeS_3^{2-} + 2SO_3^{2-} = 2S_2O_3^{2-} + S^{2-} + Te}$  

<Checkpoint>[反应6. ${TeS_3^{2-} + 2SO_3^{2-} = 2S_2O_3^{2-} + S^{2-} + Te}$ ](1)</Checkpoint>

反应6中还生成了硫离子，5正确。

可以发现，只有1，4，5是正确的，选D。