Thermogravimetric analysis was performed on a certain ternary compound **G** containing $\mathrm{Be}$, and $1.000\ \mathrm{g}$ of the compound was heated under an inert atmosphere:

- The first weight loss occurred at $80\degree \mathrm{C}$, with $0.789\ \mathrm{g}$ of solid remaining;
- The second weight loss occurred at $150\degree \mathrm{C}$, with $0.683\ \mathrm{g}$ of solid remaining;
- The third weight loss occurred at $400\degree \mathrm{C}$, with $0.149\ \mathrm{g}$ of solid remaining;
- Further heating to higher temperatures resulted in continued weight loss, but it was not significant.

It is known that only two small molecules **A** and **B** are considered to be produced during the weight loss process (**A** and **B** may continue to decompose later, but this is not relevant to the question). Both **A** and **B** are composed of two elements, **X** and **Y**. The ratio of the number of **X** to **Y** in **A** is the same as the ratio of the number of **Y** to **X** in **B**, and the mass fraction of element **X** in **A** does not exceed $3 \%$.

Deduce the element symbols or chemical formulas of **X**, **Y**, **A**, **B**, and **G**, and select the correct option from the following options.
对于含 $\mathrm{Be}$ 的某三元化合物 **G** 进行热重分析，将 $1.000\ \mathrm{g}$ 该化合物在惰性气氛下进行加热：

- $80\degree \mathrm{C}$ 时发生第一次失重，残余 $0.789\ \mathrm{g}$ 固体；
- $150\degree \mathrm{C}$ 时发生第二次失重，残余 $0.683\ \mathrm{g}$  固体；
- $400\degree \mathrm{C}$ 时发生第三次失重，残余 $0.149\ \mathrm{g}$ 固体；
- 加热至更高的温度，仍存在失重，但不明显。

已知失重过程中可视为仅产生两种小分子 **A** 和 **B**（**A** 和 **B**后续可能继续分解，但与本题无关），**A** 与 **B** 均由 **X**、**Y** 两种元素组成，**A** 中 **X** 与 **Y** 的个数比与 **B** 中 **Y** 与 **X** 的个数比相同，且 **A** 中 **X** 元素的质量分数不超过 $3 \%$。

试推出 **X**、**Y**、**A**、**B**、**G** 的元素符号或化学式，并从以下选项中选出正确选项。
All other options are incorrect
The mass fraction of **X** in **B** does not exceed 7 times the mass fraction of **X** in **A**
Both **A** and **B** are highly soluble in water, and their aqueous solutions exhibit alkalinity.
The first two instances of weightlessness produced only the same small molecule, whereas the third instance of weight loss yielded two different small molecules.
The ratio of the numbers of **X** and **Y** in **G** is $4 : 5$
The ratio of the numbers of **X** and **Y** in **G** is $5 : 4$
If **G** is an ionic compound, the ratio of anions to cations is $2 : 1$
If **G** is an ionic compound, then the cation has two different ligands.
After the third weightlessness, the residual compound is a hydrolyzable binary compound.
其他选项均不正确
**B** 中 **X** 的质量分数不超过 **A** 中 **X** 的质量分数的7倍
**A** 和 **B** 均易溶于水，且溶于水均显碱性
前两次失重只产生了同一种小分子，而第三次失重产生了两种不同的小分子
**G** 中 **X**  和 **Y** 的个数比是$4 : 5$
**G** 中 **X**  和 **Y** 的个数比是$5 : 4$
如果 **G** 是离子化合物，则阴离子和阳离子的比例为$2 : 1$
如果 **G** 是离子化合物，则阳离子有两种不同的配体
第三次失重后，残余的化合物是一种易水解的二元化合物
The mass fraction of **X** in **A** does not exceed $3\%$. Assuming the chemical formula of **A** is $\mathrm{\mathbf{XY}_a}$ (where a may not be an integer), the following relationship can be calculated: (where $M$ represents atomic/molecular weight)

$\frac{M(\mathrm{\mathbf{Y}})}{M(\mathrm{\mathbf{X}})} > 32.33 \times \frac{1}{a}$

This indicates that either $M(\mathrm{\mathbf{Y}})$ is much larger than $M(\mathrm{\mathbf{X}})$, or $a$ is particularly large. Since **A** is a small molecule, and considering the first three periods of elements, it can be inferred that **X** is most likely $\mathrm{H}$, otherwise the formula weight of **Y** would be particularly large, which is obviously not a common small molecule. Therefore, first consider that **X** is $\mathrm{H}$.

Next, infer **Y**. Since the first three periods of elements are preferred, it can be known that $a > 1$. Since common small molecules are preferred, and considering that this molecule is volatilized during thermogravimetric analysis, exclude **Y** being a metal. Among the hydrides of nonmetals in the first three periods, the common hydrides that satisfy $a > 1$ and also have both $\mathrm{\mathbf{XY}_a}$ and $\mathrm{\mathbf{X}_a\mathbf{Y}}$ are only $\mathrm{HN_3}$ and $\mathrm{H_3N}$, and $\mathrm{HO_2}$ and $\mathrm{H_2O}$. It is also found that the mass fraction of $\mathrm{H}$ in $\mathrm{HO_2}$ is $3.05\% > 3\%$, excluding the latter group. Finally, it is deduced that **X** is $\mathrm{H}$ and **Y** is $\mathrm{N}$.

<Checkpoint>[**X** is $\mathrm{H}$, **Y** is $\mathrm{N}$](1)</Checkpoint>

<Checkpoint>[**A** is $\mathrm{HN_3}$, **B** is $\mathrm{NH_3}$](2)</Checkpoint>

Next, analyze options B and C. Calculate that the mass fraction of $\mathrm{H}$ in $\mathrm{NH_3}$ is approximately 7.6 times the mass fraction of $\mathrm{H}$ in $\mathrm{HN_3}$, which is greater than 7 times, so B is incorrect. Both $\mathrm{HN_3}$ and $\mathrm{NH_3}$ are easily soluble in water, but the former is weakly acidic when dissolved in water, and the latter is alkaline, so C is incorrect.

Next, perform thermogravimetric analysis. The total weight loss is $0.851\ \mathrm{g}$. Let the amount of substance of $\mathrm{HN_3}$ be $m\ \mathrm{mol}$, the amount of substance of $\mathrm{NH_3}$ be $n\ \mathrm{mol}$, and the initial amount of substance of the ternary compound be $b\ \mathrm{mol}$. Then the following molecular weight relationship can be obtained:

$\frac{0.851}{b} = m \times 43.03 + n \times 17.01$

The mass of the three weight losses are $0.211\ \mathrm{g}$, $0.106\ \mathrm{g}$, and $0.534\ \mathrm{g}$, respectively. It can be found that the ratio of the first and second weight loss mass is approximately $2 : 1$, which suggests that the two weight losses are the same substance, and the molar ratio is $2 : 1$. Try whether this substance is $\mathrm{HN_3}$ or $\mathrm{NH_3}$ respectively, and $b$ can be calculated, and then substituted into the above formula. After enumeration, it can be found that when the substance is $\mathrm{NH_3}$, $n = 3, m = 2$ is a reasonable solution. Therefore, it is preliminarily determined that this is the inference result.

<Checkpoint>[The first two weight losses are both the loss of $\mathrm{NH_3}$](2)</Checkpoint>

Specifically, $b = \frac{0.106}{17.01} = 6.223 \times 10^{-3}\ \mathrm{mol}$, so the molecular weight of **G** is approximately $1.000 / b = 160.7\ \mathrm{g/mol}$

<Checkpoint>[The molecular weight of **G** is approximately $160.7\ \mathrm{g/mol}$](1)</Checkpoint>

After calculation, it can be concluded that the third weight loss only loses two molecules of $\mathrm{HN_3}$, so option D is incorrect.

<Checkpoint>[The third weight loss only loses two molecules of $\mathrm{HN_3}$](1)</Checkpoint>

Subtracting three molecules of $\mathrm{NH_3}$ and two molecules of $\mathrm{HN_3}$ from the total molecular weight, the remaining molecular weight is $23.52\ \mathrm{g/mol}$, which looks like a $\mathrm{Be}$ and $\mathrm{N}$, but the valence is not correct, so it can be deduced that there is a slight error in the data, and there should be another $\mathrm{H}$. Therefore, the molecular formula of **G** is $\mathrm{BeH_{12}N_{10}}$, or $\mathrm{Be(NH_3)_4(N_3)_2}$.

<Checkpoint>[The molecular formula of **G** is $\mathrm{Be(NH_3)_4(N_3)_2}$](2)</Checkpoint>

From this, it can be known that the ratio of $\mathrm{H}$ to $\mathrm{N}$ is $6 : 5$, so options E and F are both incorrect. It is also known that the product after the third weight loss is $\mathrm{Be(NH)}$, so option I is incorrect.

From the composition of the substance, it is known that if **G** is an ionic compound, the cation should be $\mathrm{[Be(NH_3)_4]^{2+}}$, with only one ligand. The anion is $\mathrm{N_3^-}$. The ratio of cation to anion is $2 : 1$. Therefore, option G is correct and option H is incorrect.

<Checkpoint>[If **G** is an ionic compound, the cation is $\mathrm{[Be(NH_3)_4]^{2+}}$, the anion is $\mathrm{N_3^-}$](1)</Checkpoint>

In summary, option G should be selected.
由 **A** 中 **X** 的质量分数不超过 $3 \%$，假设 **A** 的化学式为$\mathrm{\mathbf{XY}_a}$（其中a可以不为整数），可以计算出下列关系式：（其中$M$表示原子/分子量）

$\frac{M(\mathrm{\mathbf{Y}})}{M(\mathrm{\mathbf{X}})} > 32.33 \times \frac{1}{a}$

这说明要么$M(\mathrm{\mathbf{Y}})$远大于$M(\mathrm{\mathbf{X}})$，要么$a$特别大。又因为 **A** 是小分子，优先考虑前三周期元素，可推知 **X** 极大概率是$\mathrm{H}$，否则 **Y** 式量会特别大，显然不是常见小分子。故首先考虑 **X** 是$\mathrm{H}$。

接下来推断 **Y** ，由于优先考虑前三周期元素，可知$a > 1$。由于优先考虑常见小分子，考虑到这个分子是热重分析时挥发出来的，排除 **Y** 是金属。在前三周期非金属的氢化物中，满足$a > 1$，且同时存在$\mathrm{\mathbf{XY}_a}$和$\mathrm{\mathbf{X}_a\mathbf{Y}}$的常见氢化物，只有$\mathrm{HN_3}$和$\mathrm{H_3N}$，以及$\mathrm{HO_2}$和$\mathrm{H_2O}$两组。又发现$\mathrm{HO_2}$中$\mathrm{H}$的质量分数为$3.05\% > 3\%$，排除后一组。最终推出 **X** 是$\mathrm{H}$，**Y** 是$\mathrm{N}$。

<Checkpoint>[**X** 是$\mathrm{H}$，**Y** 是$\mathrm{N}$](1)</Checkpoint>

<Checkpoint>[**A** 是$\mathrm{HN_3}$， **B** 是$\mathrm{NH_3}$](2)</Checkpoint>

接下来分析选项B和C，计算出$\mathrm{NH_3}$中$\mathrm{H}$的质量分数约为$\mathrm{HN_3}$中$\mathrm{H}$的质量分数的7.6倍，大于7倍，故B错误。而$\mathrm{HN_3}$和$\mathrm{NH_3}$均易溶于水，但前者溶于水呈弱酸性，后者呈碱性，故C错误。

接下来进行热重分析，总的失重质量为$0.851\ \mathrm{g}$，设其中$\mathrm{HN_3}$物质的量为$m\ \mathrm{mol}$，$\mathrm{NH_3}$物质的量为$n\ \mathrm{mol}$，起始三元化合物物质的量为$b\ \mathrm{mol}$。则可得出以下分子量的关系式：

$\frac{0.851}{b} = m \times 43.03 + n \times 17.01$

三次失重的质量分别为$0.211\ \mathrm{g}$，$0.106\ \mathrm{g}$，$0.534\ \mathrm{g}$。可发现第一次与第二次失重质量之比约为$2 : 1$，这暗示这两次失重是同一种物质，且摩尔比为$2 : 1$。分别尝试这种物质是$\mathrm{HN_3}$和$\mathrm{NH_3}$，即可算出$b$，然后代入上式。经枚举，可发现当该物质是$\mathrm{NH_3}$时，$n = 3, m = 2$是合理的解。故初步断定这就是推断结果。

<Checkpoint>[前两次失重失去的均是$\mathrm{NH_3}$](2)</Checkpoint>

具体而言，$b = \frac{0.106}{17.01} = 6.223 \times 10^{-3}\ \mathrm{mol}$，故 **G** 的分子量约为$1.000 / b = 160.7\ \mathrm{g/mol}$

<Checkpoint>[**G** 的分子量约为$160.7\ \mathrm{g/mol}$](1)</Checkpoint>

经计算可得出，第三次失重只失去了两分子$\mathrm{HN_3}$，故选项D错误。

<Checkpoint>[第三次失重只失去了两分子$\mathrm{HN_3}$](1)</Checkpoint>

将总分子量扣除三分子$\mathrm{NH_3}$和两分子$\mathrm{HN_3}$，剩余的分子量为$23.52\ \mathrm{g/mol}$，看上去是一个$\mathrm{Be}$和$\mathrm{N}$，但这样价态不对，故可推出数据有微小的误差，应还有一个$\mathrm{H}$。故 **G** 的分子式为$\mathrm{BeH_{12}N_{10}}$，或$\mathrm{Be(NH_3)_4(N_3)_2}$。

<Checkpoint>[**G** 的分子式为$\mathrm{Be(NH_3)_4(N_3)_2}$](2)</Checkpoint>

由此可知，$\mathrm{H}$与$\mathrm{N}$的比例为$6 : 5$，故选项E和F均错误。亦知第三次失重后的产物是$\mathrm{Be(NH)}$，故选项I错误。

由物质组成，知道如果 **G** 是离子化合物，阳离子应为$\mathrm{[Be(NH_3)_4]^{2+}}$，只有一种配体。而阴离子为$\mathrm{N_3^-}$。阴阳离子比例为$2 : 1$。故选项G正确，选项H错误。

<Checkpoint>[如果 **G** 是离子化合物，阳离子为$\mathrm{[Be(NH_3)_4]^{2+}}$，阴离子为$\mathrm{N_3^-}$](1)</Checkpoint>

综上，应选择选项G。