**A** is a monoprotic weak acid that is liquid at room temperature, which reacts with sodium hydroxide to obtain salt **B**. **A** reacts with sodium borohydride to obtain salt **D**, while **B** reacts with a chloride **C** of a short-period main group element **M** to obtain salt **E**. It is known that the mass fractions of nitrogen in **D** and **E** are both greater than $\mathrm{80 \%}$.

A fluoride **F** of another short-period main group element reacts with antimony pentafluoride in liquid hydrogen fluoride to obtain **G**. The mass fraction of $\mathrm{Sb}$ in **G** is $\mathrm{43.06 \%}$. **G** reacts with **A** in liquid hydrogen fluoride to obtain an unstable high-energy material **H**. **H** reacts with **D** and **E** in liquid sulfur dioxide to obtain **I** and **J** respectively. The cations of these two substances are the same, and the mass fractions of nitrogen in the anions are $\mathrm{95.66 \%}$ and $\mathrm{91.23 \%}$ respectively, and the stability of **I** and **J** is lower than that of **H**.

From the above, the following options are correct:
**A**是一种常温下为液态的一元弱酸，其与氢氧化钠反应得到盐**B**。**A**与硼氢化钠反应得到盐**D**，而**B**与一种短周期主族元素**M**的氯化物**C**反应得到盐**E**。已知**D，E**中氮元素质量分数均大于$\mathrm{80 \%}$。

另一种短周期主族元素的氟化物**F**与五氟化锑在液态氟化氢中反应得到**G**，**G**中$\mathrm{Sb}$的质量分数为$\mathrm{43.06 \%}$，**G**与**A**在液态氟化氢中反应得到一种不稳定的高能材料**H**。**H**在液态二氧化硫中分别与**D，E**反应得到**I，J**，这两种物质的阳离子相同，阴离子中氮元素质量分数分别为$\mathrm{95.66 \%}$和$\mathrm{91.23 \%}$，且**I，J**的稳定性比**H**更低。

由上，下列选项正确的是：
The chemical formula of **D** contains only two elements.
The structure of **C** is a trigonal pyramidal structure.
The mass fraction of nitrogen in **E** is approximately $\mathrm{85 \%}$
**F** has four conformational isomers.
There are no $\pi$ bonds in the cation of **G**.
The mass fraction of nitrogen in **H** is approximately $\mathrm{27 \%}$.
The cations of **I,J** have a linear structure.
**J** and **E** have different anions.
All of the above options are incorrect.
**D**的化学式只含有两种元素
**C**的结构为三角锥形结构
**E**中氮元素的质量分数约为$\mathrm{85 \%}$
**F**有四种构象异构体
**G**的阳离子中不存在$\pi$键
**H**中氮元素的质量分数约为$\mathrm{27 \%}$
**I,J**的阳离子为直线型结构
**J**和**E**具有不同的阴离子
以上选项均不正确
First, the question states that nitrogen element exists in **D**, which can only come from **A**; the only monobasic liquid weak acids containing nitrogen are $\mathrm{HCN, HN_3, HNO2}$. Since the mass fraction of nitrogen in **D** is greater than $\mathrm{80 \%}$, **A** can only be $\mathrm{HN_3}$ to satisfy this mass fraction.

<Checkpoint>[The only monobasic liquid weak acids containing nitrogen are $\mathrm{HCN, HN_3, HNO2}$](1)</Checkpoint>

<Checkpoint>[**A** can only be $\mathrm{HN_3}$ to satisfy the mass fraction.](1)</Checkpoint>

$\mathrm{HN_3}$ reacts with sodium hydroxide to produce **B**, sodium azide, and reacts with sodium borohydride to obtain **D**. Since the nitrogen content of **D** is extremely high, the azide radical will certainly not be reduced. Thus, hydrogen ions are reduced by negative hydrogen, and the azide radical coordinates with the boron atom. The apparent result of the reaction is that negative hydrogen is replaced by the azide radical:

$\mathrm{4HN_3 + NaBH_4 = NaB(N_3)_4 + 4H_2}$

<Checkpoint>[ $\mathrm{4HN_3 + NaBH_4 = NaB(N_3)_4 + 4H_2}$](1)</Checkpoint>

**D** is $\mathrm{NaB(N_3)_4}$, containing three elements, so option A is incorrect.

<Checkpoint>[**D** is $\mathrm{NaB(N_3)_4}$, containing three elements, so option A is incorrect.](1)</Checkpoint>

Currently, there is too little information about **E** to deduce, so let's continue to deduce further:

The fluoride **F** of a short-period main group element reacts with antimony pentafluoride in liquid hydrogen fluoride to obtain **G**. The mass fraction of $\mathrm{Sb}$ in **G** is $\mathrm{43.06 \%}$. Antimony pentafluoride, as a strong Lewis acid, readily forms $\mathrm{SbF_6^-}$ anions in liquid hydrogen fluoride. Considering that **G** contains one $\mathrm{SbF_6^-}$, the relative molecular mass of the remaining cation is

<Checkpoint>[Antimony pentafluoride readily forms $\mathrm{SbF_6^-}$ in liquid hydrogen fluoride](1)</Checkpoint>

$\mathrm{121.8/0.4306-121.8-19\times6=47}$

Since **F** is a fluoride, consider that the cation contains fluorine atoms:
If it contains one fluorine atom, the remaining mass is 28, which is exactly two nitrogen elements. If it contains two fluorine atoms, the remaining mass is 9, which corresponds to no element. Therefore, the chemical formula of the cation of **G** is $\mathrm{N_2F^+}$, and the chemical formula of **G** is $\mathrm{[N_2F^+][SbF_6^-]}$.

<Checkpoint>[The chemical formula of the cation of **G** is $\mathrm{N_2F^+}$, and the chemical formula of **G** is $\mathrm{[N_2F^+][SbF_6^-]}$.](2)</Checkpoint>

In $\mathrm{N_2F^+}$, in order to satisfy the charge, a double bond needs to be formed between the nitrogen elements, thus there is a $\pi$ bond, and option E is incorrect.

<Checkpoint>[$\mathrm{N_2F^+}$ has a $\pi$ bond, and option E is incorrect](1)</Checkpoint>

**F** is a fluoride and $\mathrm{N_2F^+}$ is generated, so its structure is obviously $\mathrm{N_2F_2}$. To satisfy the Lewis structure, its bonding mode can be expressed as $\mathrm{F-N=N-F}$, which has cis and trans isomers, so option D is incorrect.

<Checkpoint>[**F** is $\mathrm{N_2F_2}$, which has cis and trans isomers, so option D is incorrect](1)</Checkpoint>

$\mathrm{[N_2F^+][SbF_6^-]}$ reacts with hydrazoic acid in liquid hydrogen fluoride to generate **H**. Since the anion is relatively stable, it should not change; therefore, consider the reaction of the cation. High-energy materials easily lead to the idea that the cation contains an azide radical structure, and then infer that the cation is $\mathrm{N_3^-}$ replacing the fluorine of $\mathrm{N_2F^+}$, generating $\mathrm{N_5^+}$ ions, so the preliminary inference is that **H** is $\mathrm{[N_5^+][SbF_6^-]}$, and its correctness remains to be verified below.

<Checkpoint>[The anion is relatively stable, so it should not change](1)</Checkpoint>

<Checkpoint>[The cation may be $\mathrm{N_5^+}$ ions](1)</Checkpoint>

Further infer that **H** reacts with **D**, i.e., $\mathrm{NaB(N_3)_4}$, to produce **I** with a nitrogen content of $\mathrm{95.66 \%}$ and lower stability. Thus, **I** has a large number of nitrogen atoms, and the structure of $\mathrm{B(N_3)_4^-}$ should be retained;

<Checkpoint>[The structure of $\mathrm{B(N_3)_4^-}$ should be retained](1)</Checkpoint>

If the $\mathrm{N_5^+}$ ion inferred above replaces the sodium ion, the nitrogen content of the generated $\mathrm{[N_5^+][B(N_3)_4^-]}$ is exactly $\mathrm{95.66 \%}$, and the substance is more unstable than **H** due to its excessive nitrogen content, which is in line with the meaning of the question. Thus, substance **I** is $\mathrm{[N_5^+][B(N_3)_4^-]}$. Therefore, the chemical formula of **H** is also verified to be correct.

<Checkpoint>[Substance **I** is $\mathrm{[N_5^+][B(N_3)_4^-]}$](2)</Checkpoint>

<Checkpoint>[The chemical formula of **H** is $\mathrm{[N_5^+][SbF_6^-]}$, and is verified to be correct](1)</Checkpoint>

The mass fraction of nitrogen in $\mathrm{[N_5^+][SbF_6^-]}$ is $\mathrm{22.9 \%}$, so option F is incorrect.

<Checkpoint>[The mass fraction of nitrogen in $\mathrm{[N_5^+][SbF_6^-]}$ is $\mathrm{22.9 \%}$, so option F is incorrect.](1)</Checkpoint>

At this point, let's go back and infer the chemical formulas of **E** and **J**. First of all, the cation of **J** is also $\mathrm{[N_5^+]}$. At the same time, to satisfy the nitrogen content of $\mathrm{91.23 \%}$, the anion must contain nitrogen.

It is known that **E** is generated by the reaction of sodium azide with a chloride of a short-period main group element $\mathrm{M}$, and the nitrogen content in **E** is also greater than $\mathrm{80 \%}$. Therefore, there should also be multiple azide radicals similar to **D** in **E**, and the cation is a sodium ion. The reaction of **D** and **H** should be similar to that of **E**, which is a double decomposition reaction, that is, the anion of **J** is the same as **E**, so option H is incorrect.

<Checkpoint>[There should also be multiple azide radicals similar to **D** in **E**, and the cation is a sodium ion.](1)</Checkpoint>

<Checkpoint>[The anion of **J** is the same as **E**, so option H is incorrect](1)</Checkpoint>

Therefore, the chemical formula of **I** can be set as $\mathrm{[N_5^+][M(N_3)_m^-]}$, where according to the question, $\mathrm{M}$ is a short-period main group element. Iterate through the values of m:

When m is 1, the remaining relative molecular mass is $\mathrm{(5+3\times1)\times14/0.9123 - (5+3\times1)\times14=10.8}$, which corresponds to no element.

When m is 2, with the same method, the remaining relative molecular mass is 14.8; when m is 3, the remaining is 18.8; when m is 4, the remaining is 22.9; when m is 5, the remaining is 26.9, all of which correspond to no element.

When m is 6, the remaining is 31, which is exactly the phosphorus element, and according to the valence, the phosphorus element is +5, which also conforms to the element properties. Therefore, $\mathrm{M}$ is the phosphorus element.

<Checkpoint>[$\mathrm{M}$ is the phosphorus element, m=6](2)</Checkpoint>

Therefore, the common anion of **J** and **E** is $\mathrm{[P(N_3)_6^-]}$, the chemical formula of **E** is $\mathrm{[Na^+][P(N_3)_6^-]}$, and the chemical formula of **J** is $\mathrm{[N_5^+][P(N_3)_6^-]}$.

<Checkpoint>[The chemical formula of **E** is $\mathrm{[Na^+][P(N_3)_6^-]}$, and the chemical formula of **J** is $\mathrm{[N_5^+][P(N_3)_6^-]}$](2)</Checkpoint>

The mass fraction of nitrogen in $\mathrm{[Na^+][P(N_3)_6^-]}$ is $\mathrm{82.4 \%}$, so option C is incorrect.

<Checkpoint>[The mass fraction of nitrogen in $\mathrm{[Na^+][P(N_3)_6^-]}$ is $\mathrm{82.4 \%}$, so option C is incorrect.](1)</Checkpoint>

Since $\mathrm{M}$ is the phosphorus element and its valence is +5, its chloride $\mathrm{C}$ is obviously phosphorus pentachloride, which has a trigonal bipyramidal structure, so option B is incorrect.

<Checkpoint>[$\mathrm{C}$ is phosphorus pentachloride, which has a trigonal bipyramidal structure, so option B is incorrect.](1)</Checkpoint>

The cations of **I** and **J** are both $\mathrm{[N_5^+]}$, and its structure is, according to the above inference, the azide radical replacing $\mathrm[N_2F^+]$, so it can be understood as two nitrogen-nitrogen triple bonds connected by a nitrogen atom as a bridging group. The bridging nitrogen atom is $sp^3$ hybridized, so the entire ion structure is bent, and option G is incorrect.

<Checkpoint>[**I** and **J** cations are both $\mathrm{[N_5^+]}$, which is bent, and option G is incorrect](1)</Checkpoint>

In summary, all options A-H are incorrect, so option I of this question is correct.

<Checkpoint>[All options A-H are incorrect, so option I of this question is correct](1)</Checkpoint>
首先，题目中给出**D**中存在氮元素，其只能来源于**A**；含氮元素的一元液态弱酸只有$\mathrm{HCN,HN_3,HNO2}$。由于**D**中氮元素的质量分数大于$\mathrm{80 \%}$，则**A**只能为$\mathrm{HN_3}$，才能满足该质量分数。

<Checkpoint>[含氮元素的一元液态弱酸只有$\mathrm{HCN,HN_3,HNO2}$](1)</Checkpoint>

<Checkpoint>[**A**只能为$\mathrm{HN_3}$，才能满足质量分数。](1)</Checkpoint>

$\mathrm{HN_3}$与氢氧化钠反应生成**B**叠氮酸钠，与硼氢化钠反应得到**D**，由于**D**的氮含量极大，叠氮酸根肯定不会被还原，从而氢离子被负氢还原，叠氮酸根与硼原子配位，反应表面结果为负氢被叠氮酸根取代：

$\mathrm{4HN_3 + NaBH_4 = NaB(N_3)_4 + 4H_2}$

<Checkpoint>[ $\mathrm{4HN_3 + NaBH_4 = NaB(N_3)_4 + 4H_2}$](1)</Checkpoint>

**D**为$\mathrm{NaB(N_3)_4}$，含三种元素，A选项错误。

<Checkpoint>[**D**为$\mathrm{NaB(N_3)_4}$，含三种元素，A选项错误。](1)</Checkpoint>

目前**E**的信息过少难以推断，先继续向后推：

短周期主族元素的氟化物**F**与五氟化锑在液态氟化氢中反应得到**G**，**G**中$\mathrm{Sb}$的质量分数为$\mathrm{43.06 \%}$。五氟化锑作为强lewis酸，在液态氟化氢中极易生成$\mathrm{SbF_6^-}$阴离子。考虑**G**含有一个$\mathrm{SbF_6^-}$，剩余的阳离子相对分子质量为

<Checkpoint>[五氟化锑在液态氟化氢中极易生成$\mathrm{SbF_6^-}$](1)</Checkpoint>

$\mathrm{121.8/0.4306-121.8-19\times6=47}$

因**F**为氟化物，考虑阳离子含氟原子：
若含一个氟原子，剩余28的质量，刚好为两个氮元素。若含两个氟原子，剩余质量为9，无元素。故**G**的阳离子化学式为$\mathrm{N_2F^+}$，**G**化学式为$\mathrm{[N_2F^+][SbF_6^-]}$。

<Checkpoint>[**G**的阳离子化学式为$\mathrm{N_2F^+}$，**G**化学式为$\mathrm{[N_2F^+][SbF_6^-]}$。](2)</Checkpoint>

$\mathrm{N_2F^+}$中为满足电荷，氮元素间需形成双键，从而存在$\pi$键，选项E错误。

<Checkpoint>[$\mathrm{N_2F^+}$存在$\pi$键，选项E错误](1)</Checkpoint>

**F**为氟化物且生成了$\mathrm{N_2F^+}$，则其结构很明显为$\mathrm{N_2F_2}$，为满足lewis结构式，其键联方式可表达为$\mathrm{F-N=N-F}$，有顺式和反式两种异构体，选项D错误。

<Checkpoint>[**F**为$\mathrm{N_2F_2}$，有顺式和反式两种异构体，选项D错误](1)</Checkpoint>

$\mathrm{[N_2F^+][SbF_6^-]}$与叠氮酸在液态氟化氢中反应生成**H**，由于阴离子比较稳定，应当不会发生变化；从而考虑阳离子发生反应。高能材料比较容易想到是阳离子含叠氮酸根的结构，进而推断阳离子为$\mathrm{N_3^-}$取代$\mathrm{N_2F^+}$的氟，生成$\mathrm{N_5^+}$离子，从而初步推断**H**为$\mathrm{[N_5^+][SbF_6^-]}$，其正确性有待下面验证。

<Checkpoint>[阴离子比较稳定，应当不会发生变化](1)</Checkpoint>

<Checkpoint>[阳离子可能为$\mathrm{N_5^+}$离子](1)</Checkpoint>

进一步推断，**H**与**D**，即$\mathrm{NaB(N_3)_4}$反应生成含氮元素$\mathrm{95.66 \%}$的**I**且稳定性更低，从而**I**具有大量氮原子，$\mathrm{B(N_3)_4^-}$的结构理应保留；

<Checkpoint>[$\mathrm{B(N_3)_4^-}$的结构理应保留](1)</Checkpoint>

上面推断的$\mathrm{N_5^+}$离子如果取代钠离子，生成的$\mathrm{[N_5^+][B(N_3)_4^-]}$的氮元素含量刚好为$\mathrm{95.66 \%}$，且该物质由于含氮量过大，相对于**H**更不稳定，均符合题意，从而物质**I**为$\mathrm{[N_5^+][B(N_3)_4^-]}$。由此**H**的化学式也被验证为正确。

<Checkpoint>[物质**I**为$\mathrm{[N_5^+][B(N_3)_4^-]}$](2)</Checkpoint>

<Checkpoint>[**H**的化学式为$\mathrm{[N_5^+][SbF_6^-]}$，被验证为正确](1)</Checkpoint>

$\mathrm{[N_5^+][SbF_6^-]}$中氮元素质量分数为$\mathrm{22.9 \%}$，选项F错误。

<Checkpoint>[$\mathrm{[N_5^+][SbF_6^-]}$中氮元素质量分数为$\mathrm{22.9 \%}$，选项F错误。](1)</Checkpoint>

此时回过头来推断**E**和**J**的化学式。首先**J**的阳离子也为$\mathrm{[N_5^+]}$，同时为满足氮元素含量为$\mathrm{91.23 \%}$，阴离子一定含氮元素。

已知**E**通过某种短周期主族元素$\mathrm{M}$的氯化物与叠氮化钠生成，且**E**中氮元素含量也大于$\mathrm{80 \%}$，因此**E**中应当也存在与**D**类似的多个叠氮酸根，且阳离子为钠离子。**D**与**H**的反应应当与**E**的类似，为复分解反应，也即**J**的阴离子与**E**相同，选项H错误。

<Checkpoint>[**E**中应当也存在与**D**类似的多个叠氮酸根，且阳离子为钠离子。](1)</Checkpoint>

<Checkpoint>[**J**的阴离子与**E**相同，选项H错误](1)</Checkpoint>

从而可设**I**的化学式为$\mathrm{[N_5^+][M(N_3)_m^-]}$，其中根据题意$\mathrm{M}$为短周期主族元素。遍历m的取值：

m为1时，剩余的相对分子质量为$\mathrm{(5+3\times1)\times14/0.9123 - (5+3\times1)\times14=10.8}$，无对应元素。

m为2时，同样方法，剩余相对分子质量为14.8；m为3时，剩余18.8；m为4时剩余22.9；m为5时剩余26.9，均无对应元素。

m为6时，剩余31，刚好为磷元素，且根据价态磷元素为+5价，也符合元素性质。因此，$\mathrm{M}$为磷元素。

<Checkpoint>[$\mathrm{M}$为磷元素，m=6](2)</Checkpoint>

因而**J**和**E**共同的阴离子为$\mathrm{[P(N_3)_6^-]}$，**E**的化学式为$\mathrm{[Na^+][P(N_3)_6^-]}$，**J**的化学式为$\mathrm{[N_5^+][P(N_3)_6^-]}$。

<Checkpoint>[**E**的化学式为$\mathrm{[Na^+][P(N_3)_6^-]}$，**J**的化学式为$\mathrm{[N_5^+][P(N_3)_6^-]}$](2)</Checkpoint>

$\mathrm{[Na^+][P(N_3)_6^-]}$中氮元素质量分数为$\mathrm{82.4 \%}$，选项C错误。

<Checkpoint>[$\mathrm{[Na^+][P(N_3)_6^-]}$中氮元素质量分数为$\mathrm{82.4 \%}$，选项C错误。](1)</Checkpoint>

由于$\mathrm{M}$为磷元素且其价态为+5，则其氯化物$\mathrm{C}$很明显为五氯化磷，为三角双锥结构，选项B错误。

<Checkpoint>[$\mathrm{C}$为五氯化磷，为三角双锥结构，选项B错误。](1)</Checkpoint>

**I**和**J**阳离子均为$\mathrm{[N_5^+]}$，其结构根据上述推断为叠氮酸根取代$\mathrm[N_2F^+]$，故可理解为两个氮-氮三键通过一个氮原子为桥基连接，桥基氮原子为$sp^3$杂化，故整个离子结构为折线形，选项G错误。

<Checkpoint>[**I**和**J**阳离子均为$\mathrm{[N_5^+]}$，为折线形，选项G错误](1)</Checkpoint>

综上，A-H所有选项均为错误，故该题选项I正确。

<Checkpoint>[A-H所有选项均为错误，故该题选项I正确](1)</Checkpoint>