Metal $\mathbf{X}$ is soluble in aqua regia to obtain $\mathbf{A}$, which is then treated with $SO_2$ to obtain $\mathbf{B}$. Saturated solution of $\mathbf{Y}$ is added to the above mixture, and orange-red crystals of $\mathbf{C}$ precipitate. $\mathbf{C}$ can be reduced by $H_2$ under heating conditions to obtain elemental $\mathbf{X}$. A dilute solution of $\mathbf{C}$ is colorless. When a concentrated solution of the highly toxic salt $\mathbf{Z}$ is added to it, a light yellow precipitate $\mathbf{D}$ precipitates. $1.00 g$ of $\mathbf{D}$ can absorb $8.9 mL$ ($1 atm, 0\degree \mathrm{C}$) of chlorine gas, resulting in a non-stoichiometric compound $\mathbf{E}$ with electrical conductivity, and the color changes from pale yellow to bronze. $\mathbf{E}$ can also be obtained by "combining" $\mathbf{D}$ and a certain complex $\mathbf{F}$ in solution in an approximate 5:1 ratio and precipitating. (The valence state of the metal in $\mathbf{F}$ is higher than that in $\mathbf{E}$)

The crystallographic information of $\mathbf{C}$ and $\mathbf{Y}$ is as follows:

$\mathbf{C}$ is a tetragonal unit cell with three types of atoms. The atomic coordinates of all three types of atoms are as follows:

The first type of atom is located at $(0,0,0)$

The second type of atom is located at $(0,\frac{1}{2},\frac{1}{2})$ and $(\frac{1}{2},0,\frac{1}{2})$

The third type of atom is located at $(\frac{1}{4},\frac{1}{4},0)$, $(\frac{3}{4},\frac{1}{4},0)$, $(\frac{1}{4},\frac{3}{4},0)$, $(\frac{3}{4},\frac{3}{4},0)$.

The unit cell parameters of $\mathbf{C}$ are $a=b=702.4 \mathrm{pm}, c=414.7 \mathrm{pm}$, and the density is $3.369 \mathrm{~g} / \mathrm{cm}^{3}$

$\mathbf{Y}$ has a rock salt structure with a unit cell parameter of $a=629.2 \mathrm{pm}$ and a density of $1.987 \mathrm{~g} / \mathrm{cm}^{3}$

The crystal of $\mathbf{Z}$ is isostructural with $\mathbf{Y}$ at room temperature, and its hot water solution has a bitter almond taste.

Determine the chemical formulas of $\mathbf{X}$, $\mathbf{Y}$, $\mathbf{Z}$, and $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$, $\mathbf{E}$, and select the correct options from the following.
金属$\mathbf{X}$可溶于王水得到$\mathbf{A}$, $\mathbf{A}$在$SO_2$ 的作用下得到$\mathbf{B}$。将上述混合物中加入饱和的$\mathbf{Y}$溶液, 析出橙红色晶体$\mathbf{C}$, $\mathbf{C}$可以在加热的条件下被$H_2$还原得到$\mathbf{X}$单质。$\mathbf{C}$的稀溶液是无色的, 向其中加入剧毒的盐$\mathbf{Z}$的浓溶液, 析出浅黄色沉淀$\mathbf{D}$。$1.00 g$的$\mathbf{D}$可以吸收 $8.9 mL$($1 atm, 0\degree \mathrm{C}$)的氯气, 得到具有导电能力的非整比化合物$\mathbf{E}$, 且颜色由淡黄色转变为青铜色。$\mathbf{E}$还可以通过$\mathbf{D}$和某种配合物$\mathbf{F}$在溶液中大约以 5:1 的比例 “化合”并析出。（$\mathbf{F}$中金属的价态比$\mathbf{E}$中高）

$\mathbf{C}$和$\mathbf{Y}$的正当晶胞信息如下：

$\mathbf{C}$为四方晶胞，共有三种原子，三种原子的全部原子坐标如下：

第一种原子位于$(0,0,0)$

第二种原子位于$(0,\frac{1}{2},\frac{1}{2})$和$(\frac{1}{2},0,\frac{1}{2})$

第三种原子位于$(\frac{1}{4},\frac{1}{4},0)$，$(\frac{3}{4},\frac{1}{4},0)$，$(\frac{1}{4},\frac{3}{4},0)$，$(\frac{3}{4},\frac{3}{4},0)$。

$\mathbf{C}$ 的晶胞参数 $a=b=702.4 \mathrm{pm}, c=414.7 \mathrm{pm}$, 密度为 $3.369 \mathrm{~g} / \mathrm{cm}^{3}$

$\mathbf{Y}$ 具有岩盐型结构, 晶胞参数 $a=629.2 \mathrm{pm}$, 密度为 $1.987 \mathrm{~g} / \mathrm{cm}^{3}$

$\mathbf{Z}$的晶体在室温下与$\mathbf{Y}$同构, 其热水溶液有苦杏仁味。

确定$\mathbf{X}$、$\mathbf{Y}$、$\mathbf{Z}$，和$\mathbf{A}$、$\mathbf{B}$、$\mathbf{C}$、$\mathbf{D}$、$\mathbf{E}$的化学式，选出以下正确的选项。
The sum of the reactant coefficients in the chemical equation for the dissolution of $\mathbf{X}$ in aqua regia (note: not the ionic equation) is 23.
The sum of the reactant coefficients in the chemical equation for the dissolution of $\mathbf{X}$ in aqua regia (note: not the ionic equation) is 24.
The sum of the reactant coefficients in the chemical equation for the dissolution of $\mathbf{X}$ in aqua regia (note: not the ionic equation) is 27.
In the crystal structure of $\mathbf{C}$, the sum of the coordination numbers of the elements with positive valence is 10, and the difference in coordination numbers is 4.
In the crystal structure of $\mathbf{C}$, the sum of the coordination numbers of the elements with positive valence is 16, and the difference in coordination numbers is 4.
In the chemical reaction equation where $\mathbf{E}$ is precipitated through $\mathbf{D}$ and a certain complex $\mathbf{F}$ in solution, the actual ratio of $\mathbf{D}$ to $\mathbf{F}$ is $\frac{17}{3}$.
In the chemical reaction equation where $\mathbf{E}$ is precipitated through $\mathbf{D}$ and a certain complex $\mathbf{F}$ in solution, the actual ratio of $\mathbf{D}$ to $\mathbf{F}$ is $\frac{16}{3}$.
None of the above options is correct
$\mathbf{X}$溶于王水的化学方程式中（注意不是离子方程式），反应物系数之和为23。
$\mathbf{X}$溶于王水的化学方程式中（注意不是离子方程式），反应物系数之和为24。
$\mathbf{X}$溶于王水的化学方程式中（注意不是离子方程式），反应物系数之和为27。
$\mathbf{C}$的晶体结构中, 化合价为正的元素的配位数之和为10，配位数之差为4。
$\mathbf{C}$的晶体结构中, 化合价为正的元素的配位数之和为16，配位数之差为4。
$\mathbf{E}$通过$\mathbf{D}$和某种配合物$\mathbf{F}$在溶液中析出的化学反应方程式中实际$\mathbf{D}$和$\mathbf{F}$的比例为$\frac{17}{3}$。
$\mathbf{E}$通过$\mathbf{D}$和某种配合物$\mathbf{F}$在溶液中析出的化学反应方程式中实际$\mathbf{D}$和$\mathbf{F}$的比例为$\frac{16}{3}$。
以上选项均不正确
From $\rho=Z M / V N_{\mathrm{A}}$

$M(\mathbf{Y})=\rho(\mathbf{Y}) V N_{\mathrm{A}} / Z=1.987 \times 629.2^{3} \times 10^{-30} \times 6.022 \times 10^{23} / 4=74.5 \mathrm{~g} / \mathrm{mol}$

Since $\mathbf{Y}$ is a 1:1 form of salt, it can be obtained through molar mass that: $\mathbf{Y}$ is $\mathrm{KCl}$

<Checkpoint>[$\mathbf{Y}$ is $\mathrm{KCl}$](1)</Checkpoint> .


$M(\mathbf{C})=\rho(\mathbf{C}) V N_{\mathrm{A}} / Z=3.369 \times 702.4^{2} \times 414.7 \times 10^{-30} \times 6.022 \times 10^{23} / 1=415.1 \mathrm{~g} / \mathrm{mol} $

According to the given proper unit cell and atomic coordinates of $\mathbf{C}$, the ratio of the three atoms can be obtained. Since $\mathbf{X}$ is a metallic element and usually does not have a negative valence, it can be inferred through valence states that $\mathbf{C}$ contains 1 $X$, 2 $K$, and 4 $Cl$.

Corresponding to the element mass, $\mathbf{X}$ is $Pt$, and $\mathbf{C}$ is $K_2PtCl_4$.

<Checkpoint>[$\mathbf{X}$ is $Pt$](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$ is $K_2PtCl_4$](1)</Checkpoint>

Based on the description of bitter almond smell, it can be determined that $\mathbf{Z}$ is $KCN$, so $\mathbf{D}$ should be $K_{2}Pt(CN)_{4}$

<Checkpoint>[$\mathbf{D}$ is $K_{2}Pt(CN)_{4}$](1)</Checkpoint>

According to the question, $1.00 \mathrm{~mol} \mathbf{D}$ can absorb $377.4 \times 8.9 \times 10^{-3} / 22.4=0.15\mathrm{~mol}$ of chlorine gas, that is, $1.00 \mathrm{~mol} \mathbf{D}$ contains $0.3\mathrm{~mol}$ of $Cl$ atoms. $\mathbf{E}$ is $\mathrm{K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$

<Checkpoint>[$\mathbf{E}$ is $\mathrm{K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$](1)</Checkpoint>

Combining the information in other reactions, we get:

$\mathbf{A}$ is $K_2Pt(CN)_4$, $\mathbf{B}$ is $H_2Pt(CN)_4$, $\mathbf{C}$ is $K_2PtCl_4$

<Checkpoint>[$\mathbf{A}$ is $K_2Pt(CN)_4$, $\mathbf{B}$ is $H_2Pt(CN)_4$](1)</Checkpoint>

The chemical reaction equation for $\mathbf{X}$ dissolving in aqua regia is:

$3 \mathrm{Pt}+4 \mathrm{HNO}_{3}+18 \mathrm{HCl}=3 \mathrm{H}_{2} \mathrm{PtCl}_{6}+4 \mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O}$

<Checkpoint>[$3 \mathrm{Pt}+4 \mathrm{HNO}_{3}+18 \mathrm{HCl}=3 \mathrm{H}_{2} \mathrm{PtCl}_{6}+4 \mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

The sum of the reactant coefficients is 25. Options A, B, and C are all incorrect.

<Checkpoint>[The sum of the reactant coefficients is 25. Options A, B, and C are all incorrect](1)</Checkpoint>

From the given proper unit cell and atomic coordinates of $\mathbf{C}$, it can be obtained that the coordination number of potassium is 8 and the coordination number of platinum is 4. That is, the sum of the coordination numbers is 12, and the difference in the coordination numbers is 4. Options D and E are both incorrect.

<Checkpoint>[The sum of the coordination numbers of the two elements is 12, and the difference in the coordination numbers is 4. Options D and E are both incorrect](1)</Checkpoint>

$\mathbf{E}$ can also be "combined" from $\mathbf{D}$ and a certain complex $\mathbf{F}$ in solution in an approximate ratio of 5:1 and precipitated. The valence state of the metal in $\mathbf{F}$ is higher than that in $E$. The valence state of the metal in $\mathbf{E}$ is +2, and the valence state of the metal in $\mathbf{F}$ should be another common valence state of Pt, +4. According to "combination", it can be determined that $\mathbf{F}$ contains chloride ions, and then try to find a complex with a suitable ratio. It is obtained that $\mathbf{F}$ is $K_2Pt(CN)_4Cl_2$.

<Checkpoint>[$\mathbf{F}$ is $K_2Pt(CN)_4Cl_2$](1)</Checkpoint>

The chemical reaction equation is:

$17 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4}+3 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{2}=20 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$

<Checkpoint>[$17 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4}+3 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{2}=20 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$](1)</Checkpoint>

The actual ratio is $\frac{17}{3}$, option F is correct, and option G is incorrect.

<Checkpoint>[The actual ratio is $\frac{17}{3}$, option F is correct, and option G is incorrect.](1)</Checkpoint>
由 $\rho=Z M / V N_{\mathrm{A}}$

$M(\mathbf{Y})=\rho(\mathbf{Y}) V N_{\mathrm{A}} / Z=1.987 \times 629.2^{3} \times 10^{-30} \times 6.022 \times 10^{23} / 4=74.5 \mathrm{~g} / \mathrm{mol}$

由于 $\mathbf{Y}$ 是 1:1 形式的盐，通过摩尔质量可以得到：$\mathbf{Y}$ 是 $\mathrm{KCl}$ 

<Checkpoint>[$\mathbf{Y}$ 是 $\mathrm{KCl}$](1)</Checkpoint> 。


$M(\mathbf{C})=\rho(\mathbf{C}) V N_{\mathrm{A}} / Z=3.369 \times 702.4^{2} \times 414.7 \times 10^{-30} \times 6.022 \times 10^{23} / 1=415.1 \mathrm{~g} / \mathrm{mol} $

根据给出的$\mathbf{C}$的正当晶胞和原子坐标可以得到三种原子的比例。由于$\mathbf{X}$是金属元素，通常不会为负价，通过价态推断，$\mathbf{C}$中含有1个$X$，2个$K$，4个$Cl$。

对应元素质量得到$\mathbf{X}$是$Pt$，$\mathbf{C}$是$K_2PtCl_4$。

<Checkpoint>[$\mathbf{X}$是$Pt$](1)</Checkpoint>

<Checkpoint>[$\mathbf{C}$是$K_2PtCl_4$](1)</Checkpoint>

根据描述苦杏仁味，可以判断$\mathbf{Z}$为$KCN$，故$\mathbf{D}$应为$K_{2}Pt(CN)_{4}$

<Checkpoint>[$\mathbf{D}$是$K_{2}Pt(CN)_{4}$](1)</Checkpoint>

根据题意, $1.00 \mathrm{~mol} \mathbf{D}$ 可以吸收 $377.4 \times 8.9 \times 10^{-3} / 22.4=0.15\mathrm{~mol}$ 的氯气，即$1.00 \mathrm{~mol} \mathbf{D}$中含有$0.3\mathrm{~mol}$的$Cl$原子。 $\mathbf{E}$ 为 $\mathrm{K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$ 

<Checkpoint>[$\mathbf{E}$ 为 $\mathrm{K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$](1)</Checkpoint>

结合其他反应中的信息得到：

$\mathbf{A}$是$K_2Pt(CN)_4$，$\mathbf{B}$是$H_2Pt(CN)_4$，$\mathbf{C}$是$K_2PtCl_4$

<Checkpoint>[$\mathbf{A}$是$K_2Pt(CN)_4$，$\mathbf{B}$是$H_2Pt(CN)_4$](1)</Checkpoint>

$\mathbf{X}$溶于王水的化学反应方程式为：

$3 \mathrm{Pt}+4 \mathrm{HNO}_{3}+18 \mathrm{HCl}=3 \mathrm{H}_{2} \mathrm{PtCl}_{6}+4 \mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O}$

<Checkpoint>[$3 \mathrm{Pt}+4 \mathrm{HNO}_{3}+18 \mathrm{HCl}=3 \mathrm{H}_{2} \mathrm{PtCl}_{6}+4 \mathrm{NO}+8 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

反应物系数之和为25。选项A、B、C均错误

<Checkpoint>[反应物系数之和为25。选项A、B、C均错误](1)</Checkpoint>

通过给出的$\mathbf{C}$的正当晶胞和原子坐标可以得到钾的配位数为8，铂的配位数为4。即配位数之和为12，配位数之差为4，选项D、E均错误。

<Checkpoint>[两种元素的配位数之和为12，配位数之差为4。选项D、E均错误](1)</Checkpoint>

$\mathbf{E}$还可以通过$\mathbf{D}$和某种配合物$\mathbf{F}$在溶液中大约以5:1的比例 “化合”并析出。$\mathbf{F}$中金属的价态比$E$中高。$\mathbf{E}$中金属的价态为+2，$\mathbf{F}$中金属的价态应当为Pt的另一种常见价态+4。根据“化合”，可以判断$\mathbf{F}$中含有氯离子，再尝试找到比例适宜的配合物。得到$\mathbf{F}$是$K_2Pt(CN)_4Cl_2$。

<Checkpoint>[$\mathbf{F}$是$K_2Pt(CN)_4Cl_2$](1)</Checkpoint>

化学反应方程式为：

$17 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4}+3 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{2}=20 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$

<Checkpoint>[$17 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4}+3 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{2}=20 \mathrm{~K}_{2} \mathrm{Pt}(\mathrm{CN})_{4} \mathrm{Cl}_{0.3}$](1)</Checkpoint>

实际比例为$\frac{17}{3}$，选项F正确，选项G错误。

<Checkpoint>[实际比例为$\frac{17}{3}$，选项F正确，选项G错误。](1)</Checkpoint>