Salt $\mathbf{A}$ contains the metallic element $\mathbf{Z}$ with a mass fraction of 57.77%. When heated to $400^\circ C$, salt $\mathbf{A}$ decomposes with a weight loss of 11.144%, releasing the elemental gas $\mathbf{X}$. The solid elemental substance $\mathbf{Y}$ can react with gas $\mathbf{X}$ in different proportions to form two binary compounds, $\mathbf{B}$ and $\mathbf{C}$. $\mathbf{B}$ is a liquid with a mass ratio of 1.336 between its two elements, while $\mathbf{C}$ is a gas with a density approximately 9 times that of air under STP conditions.  

$\mathbf{A}$ can dissolve in $\mathbf{B}$ to form a solution (denoted as Solution 1). When gas $\mathbf{D}$ is passed into Solution 1 (Reaction 1), $\mathbf{D}$ is oxidized to the cation $\mathbf{D}^+$, and the elemental form of $\mathbf{Z}$ precipitates. The remaining part, after vacuum distillation, yields salt $\mathbf{E}$, which shares the same cation $\mathbf{D}^+$ as $\mathbf{A}$.  

If $\mathbf{F}$ is added to Solution 1 instead, $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{F}$ react in a 1:1:1 ratio to form the ionic compound $\mathbf{G}$ (Reaction 2). The structure of $\mathbf{G}$ is distinctive, as both its cation and anion belong to the $O_{h}$ point group and are isoelectronic with each other.  

Salt $\mathbf{H}$ has the same anion as $\mathbf{A}$, but its molar mass is 2 g/mol greater than that of $\mathbf{A}$. When $\mathbf{H}$ reacts with $\mathbf{B}$, it releases the colorless gas $\mathbf{W}$ (Reaction 3). At room temperature, $\mathbf{W}$ can combine with $\mathbf{D}$ to form a reddish-brown gas, which can further dimerize.  

The above transformations can be represented by the following reaction equations:  

1. $\mathbf{A} \xrightarrow{400^\circ C} \cdots \mathbf{X}$  

2. $\mathbf{X} \xrightarrow{\mathbf{Y}} \mathbf{B} + \mathbf{C}$  

3. $\mathbf{A} + \mathbf{B} \xrightarrow[\mathbf{D}]{\text{Rxn 1}} \mathbf{E} + \mathbf{Z}$  

4. $\mathbf{A} + \mathbf{B} \xrightarrow[\mathbf{F}]{\text{Rxn 2}} \mathbf{G}$  

5. $\mathbf{B} + \mathbf{H} \xrightarrow{\text{Rxn 3}} \mathbf{W}$  

6. $\mathbf{W} + \mathbf{D} \rightarrow \text{brown gas}$  

Based on the above information, select the most appropriate option from the following choices.
盐$\mathbf{A}$中含有金属元素$\mathbf{Z}$，质量分数为57.77%。盐$\mathbf{A}$在加热至 $400^\circ C$ 时分解失重11.144 % 并释放出单质气体$\mathbf{X}$。固体单质$\mathbf{Y}$与气体$\mathbf{X}$可以按不同的比例反应得到两种二元化合物$\mathbf{B}$和$\mathbf{C}$。$\mathbf{B}$是液体，其中两种元素的质量比为1.336，$\mathbf{C}$是气体，STP密度下约为空气密度的9倍。$\mathbf{A}$可以溶于$\mathbf{B}$形成溶液(记为溶液1)，将气体$\mathbf{D}$通入溶液1中(反应1)，$\mathbf{D}$被氧化成阳离子$\mathbf{D}^+$并析出$\mathbf{Z}$元素的单质。剩余部分经真空蒸馏后得到盐$\mathbf{E}$，$\mathbf{E}$与$\mathbf{A}$有相同的阳离子 $\mathbf{ D } ^{ + }$ 。如果是将$\mathbf{F}$加入溶液1中，则$\mathbf{A}$,$\mathbf{B}$, $\mathbf{F}$会以1:1:1的比例反应得到离子化合物$\mathbf{G}$(反应2)。$\mathbf{G}$的结构很有特点，其阴阳离子均属于 $ O_{ h }$ 点群，且互为等电子体。盐$\mathbf{H}$和盐$\mathbf{A}$有相同的阴离子，$\mathbf{H}$摩尔质量比$\mathbf{A}$大2g/mol，$\mathbf{H}$与$\mathbf{B}$作用可以释放出无色气体$\mathbf{W}$(反应3)，$\mathbf{W}$与$\mathbf{D}$在室温下就可以化合得到红棕色气体，该气体还能二聚。上述转化可用如下的多个反应方程式表示：

1. $\mathbf{A} \xrightarrow{400^\circ C} \cdots \mathbf{X}$

2. $\mathbf{X} \xrightarrow{\mathbf{Y}} \mathbf{B} + \mathbf{C}$

3. $\mathbf{A} + \mathbf{B} \xrightarrow[\mathbf{D}]{\text{Rxn 1}} \mathbf{E} + \mathbf{Z}$

4. $\mathbf{A} + \mathbf{B} \xrightarrow[\mathbf{F}]{\text{Rxn 2}} \mathbf{G}$

5. $\mathbf{B} + \mathbf{H} \xrightarrow{\text{Rxn 3}} \mathbf{W}$

6. $\mathbf{W} + \mathbf{D} \rightarrow \text{brown gas}$

根据以上信息，选出下列选项中最符合题意的选项
The elemental gas produced by the decomposition of $\mathbf{A}$ is $O_2$, and the solid elemental $\mathbf{Y}$ belongs to a transition metal element.
The monatomic gas produced by the decomposition of $\mathbf{A}$ is $O_2$, and the solid monatomic element $\mathbf{Y}$ belongs to the group IIIA elements.
The elemental gas produced by the decomposition of $\mathbf{A}$ is $N_2$, and the solid elemental $\mathbf{Y}$ belongs to a transition metal element.
The elemental gas produced by the decomposition of $\mathbf{A}$ is $N_2$, and the solid elemental $\mathbf{Y}$ belongs to the group IVA elements.
The elemental gas produced by the decomposition of $\mathbf{A}$ is $Cl_2$, and the solid elemental $\mathbf{Y}$ belongs to the alkali metal or alkaline earth metal elements.
The monoatomic gas produced by the decomposition of $\mathbf{A}$ is $Cl_2$, and the solid monoatomic $\mathbf{Y}$ belongs to a transition metal element.
The monatomic gas produced by the decomposition of $\mathbf{A}$ is $Cl_2$, and the solid monatomic element $\mathbf{Y}$ belongs to the VA group elements.
The monoatomic gas produced by the decomposition of $\mathbf{A}$ is $F_2$, and the solid monoatomic element $\mathbf{Y}$ belongs to the alkali metal or alkaline earth metal elements.
The elemental gas produced by the decomposition of $\mathbf{A}$ is $F_2$, and the solid elemental $\mathbf{Y}$ belongs to the transition metal elements.
The elemental gas produced by the decomposition of $\mathbf{A}$ is $F_2$, and the solid elemental $\mathbf{Y}$ belongs to the VIA group elements.
The monatomic gas produced by the decomposition of $\mathbf{A}$ is $O_3$, and the solid monatomic $\mathbf{Y}$ belongs to an alkali metal or alkaline earth metal element.
The monoatomic gas produced by the decomposition of $\mathbf{A}$ is $O_3$, and the solid monoatomic $\mathbf{Y}$ belongs to the Group VIIA elements.
The monatomic gas produced by the decomposition of $\mathbf{A}$ is a noble gas.
The cations in $\mathbf{G}$ are composed of Group VA and Group VIA elements, while the anions are composed of Group IVA and Group VIIA elements.
The cations in $\mathbf{G}$ are composed of Group VA and Group VIIA elements, while the anions are composed of Group VIA and Group VIIA elements.
The cations in $\mathbf{G}$ are composed of Group IVA and Group VIA elements, while the anions are composed of Group VA and Group VIA elements.
In $\mathbf{G}$, the cations are exclusively composed of Group IVA elements, while the anions are composed of Group VA and Group VIA elements.
In $\mathbf{G}$, the cations are solely composed of Group VA elements, and the anions are composed of Group IVA and Group VIA elements.
The cations in $\mathbf{G}$ are exclusively composed of Group VIA elements, while the anions consist of Group VIA and Group VIIA elements.
The cations in $\mathbf{G}$ are composed solely of Group VIIA elements, while the anions are composed of Group VA and Group VIIA elements.
The cations in $\mathbf{G}$ are composed of Group VA and Group IVA elements, while the anions consist solely of Group IVA elements.
The cations in $\mathbf{G}$ are composed of Group VIA and Group VIIA elements, while the anions consist solely of Group VA elements.
The cations in $\mathbf{G}$ are composed of Group VIIA and Group VA elements, while the anions consist solely of Group VIA elements.
The cations in $\mathbf{G}$ are composed of Group VIIA and Group IIIA elements, while the anions consist solely of Group VIIA elements.
None of the above answers are correct
The above answers have two or more correct options
$\mathbf{A}$分解产生的单质气体是$O_2$，固体单质$\mathbf{Y}$属于过渡金属元素
$\mathbf{A}$分解产生的单质气体是$O_2$，固体单质$\mathbf{Y}$属于IIIA族元素
$\mathbf{A}$分解产生的单质气体是$N_2$，固体单质$\mathbf{Y}$属于过渡金属元素
$\mathbf{A}$分解产生的单质气体是$N_2$，固体单质$\mathbf{Y}$属于IVA族元素
$\mathbf{A}$分解产生的单质气体是$Cl_2$，固体单质$\mathbf{Y}$属于碱金属或碱土金属元素
$\mathbf{A}$分解产生的单质气体是$Cl_2$，固体单质$\mathbf{Y}$属于过渡金属元素
$\mathbf{A}$分解产生的单质气体是$Cl_2$，固体单质$\mathbf{Y}$属于VA族元素
$\mathbf{A}$分解产生的单质气体是$F_2$，固体单质$\mathbf{Y}$属于碱金属或碱土金属元素
$\mathbf{A}$分解产生的单质气体是$F_2$，固体单质$\mathbf{Y}$属于过渡金属元素
$\mathbf{A}$分解产生的单质气体是$F_2$，固体单质$\mathbf{Y}$属于VIA族元素
$\mathbf{A}$分解产生的单质气体是$O_3$，固体单质$\mathbf{Y}$属于碱金属或碱土金属元素
$\mathbf{A}$分解产生的单质气体是$O_3$，固体单质$\mathbf{Y}$属于VIIA族元素
$\mathbf{A}$分解产生的单质气体是一种稀有气体
$\mathbf{G}$中阳离子由VA族元素和VIA族元素构成，阴离子由IVA族元素和VIIA族元素构成
$\mathbf{G}$中阳离子由VA族元素和VIIA族元素构成，阴离子由VIA族元素和VIIA族元素构成
$\mathbf{G}$中阳离子由IVA族元素和VIA族元素构成，阴离子由VA族元素和VIA族元素构成
$\mathbf{G}$中阳离子只由IVA族元素构成，阴离子由VA族元素和VIA族元素构成
$\mathbf{G}$中阳离子只由VA族元素构成，阴离子由IVA族元素和VIA族元素构成
$\mathbf{G}$中阳离子只由VIA族元素构成，阴离子由VIA族元素和VIIA族元素构成
$\mathbf{G}$中阳离子只由VIIA族元素构成，阴离子由VA族元素和VIIA族元素构成
$\mathbf{G}$中阳离子由VA族元素和IVA族元素构成，阴离子只由IVA族元素构成
$\mathbf{G}$中阳离子由VIA族元素和VIIA族元素构成，阴离子只由VA族元素构成
$\mathbf{G}$中阳离子由VIIA族元素和VA族元素构成，阴离子只由VIA族元素构成
$\mathbf{G}$中阳离子由VIIA族元素和IIIA族元素构成，阴离子只由VIIA族元素构成
以上答案均不正确
以上答案有2个及2个以上正确
From the room-temperature combination of $\mathbf{W}$ and $\mathbf{D}$ yielding a red-brown dimeric gas, it can be inferred that dimerizable $NO_2$ gas should have been produced. From this and the fact that gas $\mathbf{D}$ can be oxidized to $\mathbf{D}^+$, it can be deduced that gas $\mathbf{D}$ should be $NO$, and thus $\mathbf{W}$ should be an oxidizing substance. It is easy to assume that $\mathbf{W}$ is $O_2$;

<Checkpoint>[Gas $\mathbf{D}$ should be $NO$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{W}$ is $O_2$](0.5)</Checkpoint>

$\mathbf{B, C}$ are two binary compounds, with the former having a mass ratio of 1.336; the latter has a density under standard atmospheric pressure that is 9 times that of air, which essentially means its molecular weight is approximately 9 times that of air's molecular weight (~29), i.e., the molecular weight of $\mathbf{C}$ is $261 g/mol$.

<Checkpoint>[The molecular weight of $\mathbf{C}$ is $261 g/mol$](0.5)</Checkpoint>

Gas $\mathbf{X}$ and solid elemental $\mathbf{Y}$ can combine to form $\mathbf{B, C}$. Since $\mathbf{X}$ is not oxygen, the commonly considered elemental gases are $F_2$ or $Cl_2$, as nitrogen gas rarely forms variable-valency binary compounds. A simple table calculation suggests that the corresponding elemental gas should be $F_2$.

| Number of F atoms x | Total relative atomic mass of F 19 x | Remaining mass 261 – 19 | Possible combinations (other atom count y, other atom molecular mass M) | Matches elemental? | Mass ratio verification (xM : 19x′ = 1.336)    |
| ------ | ------------- | -------------- | ---------------------------- | ------- | ---------------------------- |
| 1      | 19            | 242            | (1, 242)                     | ✗       | —                            |
| 2      | 38            | 223            | (1, 223)                     | ✗       | —                            |
| 3      | 57            | 204            | (1, 204)                     | ✗       | —                            |
| 4      | 76            | 185            | (1, 185)                     | ✗       | —                            |
| 5 | 95        | 166       | (2, 83)                  | ✗       | —                            |
| 5  | 95        | 166        | (1, 127)                 | ✗ |—                            |
| 6      | 114           | 147            | (1, 147)                     | ✗       | —                            |
| 6      | 114           | 147            | (3, 49)                      | ✗       | —                            |
| 7  | 133      | 128        | (1, 128)                 | ✓ (I) | 1\*127 : 5\*19 $\approx$ 1.336 (IF₅) |
| 8      | 152           | 109            | (1, 109)                     | ✗       | —                            |
| 9      | 171           | 90             | (1, 90)                      | ✗       | —                            |
| 10     | 190           | 71             | (1, 71)                      | ✗       | —                            |



<Checkpoint>[The elemental gas $\mathbf{X}$ produced by the decomposition of $\mathbf{A}$ is $F_2$](1)</Checkpoint>

From the mass ratio of $\mathbf{B}$, it can be deduced that $\mathbf{B, C}$ should be $IF_5$ and $IF_7$.

<Checkpoint>[$\mathbf{B}$ is $IF_5$.](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{C}$ is $IF_7$.](0.5)</Checkpoint>

Based on the above information, the mass fraction of 57.77% can be used to calculate that the metal element $\mathbf{Z}$ is $Au$, and the salt $\mathbf{A}$ is $NO^+AuF_6^-$.

<Checkpoint>[$\mathbf{A}$ is $NO^+AuF_6^-$](1.0)</Checkpoint>

Since salt $\mathbf{H}$ and salt $\mathbf{A}$ share the same anion, and $\mathbf{H}$ has a molar mass 2 g/mol greater than $\mathbf{A}$, and $\mathbf{H}$ reacts with $\mathbf{B}$ to release colorless oxygen gas, it can be deduced that $O_2^+$ is the cation.

<Checkpoint>[$O_2^+$ is the cation of salt $\mathbf{H}$](1)</Checkpoint>

Salt $\mathbf{E}$ has the same cation as $\mathbf{A}$, which is $NO^+$, and its anion should originate from $\mathbf{B}$. Thus, $\mathbf{E}$ is $NO^+IF_6^-$.

<Checkpoint>[Salt $\mathbf{E}$ is $NO^+IF_6^-$](1.0)</Checkpoint>

Both the cation and anion of $\mathbf{G}$ belong to the $O_{ h }$ point group and are isoelectronic. Since both the cation and anion belong to the $O_h$ point group, they should both be of the form $XF_6^{\pm}$ (where X represents some element).

Based on the reaction equation for the formation of $\mathbf{G}$, $\mathbf{G}$ must contain either $IF_6^-$ or $IF_6^+$. If it were $IF_6^-$, there is no reasonable isoelectronic cation, so it can be inferred that the cation of $\mathbf{G}$ is $IF_6^+$.

<Checkpoint>[The cation of $\mathbf{G}$ is $IF_6^+$](1)</Checkpoint>

The anion is an isoelectronic counterpart of $IF_6^+$. Since the reaction equation is 1:1:1, $\mathbf{G}$ must be a 1:1 salt, meaning the anion can only be $SbF_6^-$. Thus, $\mathbf{G}$ is $IF_6^+SbF_6^-$.

<Checkpoint>[$\mathbf{G}$ must be a 1:1 salt](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{G}$ is $IF_6^+SbF_6^-$.](1.0)</Checkpoint>

Based on the material inference results, only option T is correct.
从$\mathbf{W}$和$\mathbf{D}$室温化合得到红棕色二聚气体可以推断应该产生了可以二聚的$NO_2$气体。由此和气体$\mathbf{D}$可以被氧化成$\mathbf{D}^+$可以推测出，气体$\mathbf{D}$应当是$NO$，那么$\mathbf{W}$应该是一个有氧化性的物质，很容易想到$\mathbf{W}$是$O_2$；

<Checkpoint>[气体$\mathbf{D}$应当是$NO$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{W}$是$O_2$](0.5)</Checkpoint>

$\mathbf{B, C}$是两种二元化合物，前者质量比是1.336； 后者在标准大气压强下密度是空气9倍，实际上相当于说分子量是空气分子量(约29)的约9倍，也就是$\mathbf{C}$的分子量是$261 g/mol$。

<Checkpoint>[$\mathbf{C}$的分子量是$261 g/mol$](0.5)</Checkpoint>

气体$\mathbf{X}$和固体单质$\mathbf{Y}$可化合为$\mathbf{B, C}$，由于$\mathbf{X}$不是氧气，常见的能够考虑的单质气体就是$F_2$或$Cl_2$，氮气很难形成变价二元化合物。简单如下打表计算可以推测出对应的单质气体应该是$F_2$。

| F的个数 x | F的总相对原子质量 19 x| 剩余质量 261 – 19 | (另一原子个数y, 另一原子分子质量M) 可能组合| 是否匹配单质  | 质量比验证 (xM : 19x′ = 1.336)    |
| ------ | ------------- | -------------- | ---------------------------- | ------- | ---------------------------- |
| 1      | 19            | 242            | (1, 242)                     | ✗       | —                            |
| 2      | 38            | 223            | (1, 223)                     | ✗       | —                            |
| 3      | 57            | 204            | (1, 204)                     | ✗       | —                            |
| 4      | 76            | 185            | (1, 185)                     | ✗       | —                            |
| 5 | 95        | 166       | (2, 83)                  | ✗       | —                            |
| 5  | 95        | 166        | (1, 127)                 | ✗ |—                            |
| 6      | 114           | 147            | (1, 147)                     | ✗       | —                            |
| 6      | 114           | 147            | (3, 49)                      | ✗       | —                            |
| 7  | 133      | 128        | (1, 128)                 | ✓ (I) | 1\*127 : 5\*19 $\approx$ 1.336 (IF₅) |
| 8      | 152           | 109            | (1, 109)                     | ✗       | —                            |
| 9      | 171           | 90             | (1, 90)                      | ✗       | —                            |
| 10     | 190           | 71             | (1, 71)                      | ✗       | —                            |



<Checkpoint>[$\mathbf{A}$分解产生的单质气体$\mathbf{X}$是$F_2$](1)</Checkpoint>

根据$\mathbf{B}$的质量比可以推出$\mathbf{B, C}$ 应当是$IF_5$和$IF_7$。

<Checkpoint>[$\mathbf{B}$ 是$IF_5$。](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{C}$ 是$IF_7$。](0.5)</Checkpoint>

根据上述信息，可以利用质量分数57.77%计算出金属元素$\mathbf{Z}$是$Au$，盐$\mathbf{A}$是$NO^+AuF_6^-$。

<Checkpoint>[$\mathbf{A}$是$NO^+AuF_6^-$](1.0)</Checkpoint>

从盐$\mathbf{H}$和盐$\mathbf{A}$有相同的阴离子，$\mathbf{H}$摩尔质量比$\mathbf{A}$大2g/mol，$\mathbf{H}$与$\mathbf{B}$作用可以释放出无色气体氧气可推出$O_2^+$为阳离子。

<Checkpoint>[$O_2^+$为盐$\mathbf{H}$的阳离子](1)</Checkpoint>

盐$\mathbf{E}$与$\mathbf{A}$阳离子相同，为$NO^+$，阴离子应当来源于$\mathbf{B}$，故$\mathbf{E}$为$NO^+IF_6^-$。

<Checkpoint>[盐$\mathbf{E}$是$NO^+IF_6^-$](1.0)</Checkpoint>

$\mathbf{G}$的阴阳离子均属于 $ O_{ h }$ 点群，且互为等电子体。阴阳离子都属于$O_h$点群，应该均为$XF_6^{\pm}$（X代表某种元素），

根据生成$\mathbf{G}$的方程式，$\mathbf{G}$一定存在$IF_6^-$或者$IF_6^+$，如果为$IF_6^-$，不存在合理的阳离子等电子体，那么可以推测出$\mathbf{G}$阳离子是$IF_6^+$，

<Checkpoint>[$\mathbf{G}$阳离子是$IF_6^+$](1)</Checkpoint>

阴离子为$IF_6^+$的等电子体，由于方程式为1：1：1反应，$\mathbf{G}$一定是1：1型盐，那么阴离子只能是$SbF_6^-$，那么$\mathbf{G}$是$IF_6^+SbF_6^-$。

<Checkpoint>[$\mathbf{G}$一定是1：1型盐](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{G}$是$IF_6^+SbF_6^-$。](1.0)</Checkpoint>

根据物质推断结果，只有选项T正确。