$\mathbf{A}$、$\mathbf{B}$、$\mathbf{C}$、$\mathbf{D}$ are four non-metallic simple substances, where the corresponding elements of three of them are located in the same period, and the corresponding elements of two of them are located in the same group. $\mathbf{A}$ reacts with oxygen under high temperature conditions to obtain oxide $\mathbf{E}$. $\mathbf{E}$ reacts with urea under high temperature conditions, or $\mathbf{E}$ reacts with $\mathrm{CaB_6}$ under high temperature conditions and a nitrogen atmosphere to obtain a binary compound $\mathbf{F}$, and $\mathbf{F}$ can be used as a component material of rocket engines. At $- 45 ^ { \circ } C$, $\mathbf{C}$ and $\mathbf{D}$ react in $\mathrm{CFCl_3}$ to obtain a yellow solid $ \mathbf{G }$. At $-7 8 ^ { \circ } C$, $\mathbf{D}$ and $\mathbf{G}$ react in $\mathrm{CFCl_3}$ to obtain $\mathbf{H} $. Heating $ \mathbf{H }$ to above $0 ^ { \circ } C$ causes decomposition, yielding $\mathbf{D}$ and $\mathbf{I}$. $\mathbf{I}$ can also be prepared by directly heating $\mathbf{C}$ and $\mathbf{D}$. Mixing $\mathbf{C}$ and $\mathbf{I}$ at $9 0 ^ { \circ } C$, and then heating sufficiently, yields $\mathbf{J}$. $\mathbf{J}$ decomposes at temperatures above $4 50 ^ { \circ } C$, yielding equimolar amounts of $\mathbf{C}$ and $\mathbf{I}$. At low temperatures, $\mathbf{F}$ reacts with $\mathbf{H}$ to obtain $\mathbf{K}$ and a common Lewis acid $\mathbf{L}$. $\mathbf{K}$ decomposes to form $\mathbf{B}$ and $\mathbf{D}$.

Please provide the chemical formulas of each substance and choose the correct option.
$\mathbf{A}$、$\mathbf{B}$、$\mathbf{C}$、$\mathbf{D}$是四种非金属单质，其中某三个对应的元素位于同一周期，某两个对应的元素位于同一族。$\mathbf{A}$在高温条件下与氧气反应，得到氧化物$\mathbf{E}$。$\mathbf{E}$与尿素在高温条件下反应，或在高温条件下、氮气气氛中$\mathbf{E}$与 $\mathrm{CaB_6}$反应，均可得到二元化合物$\mathbf{F}$，$\mathbf{F}$可作为火箭发动机的组成材料。在 $- 4 5 ^ { \circ } C$ 时，$\mathbf{C}$和 $\mathbf{D}$ 在$\mathrm{CFCl_3}$中反应得到黄色固体 $ \mathbf{G }$ 。在 $-7 8 ^ { \circ } C$ 时，$\mathbf{D}$和$\mathbf{G}$在$\mathrm{CFCl_3}$中反应得到 $\mathbf{H} $ 。将 $ \mathbf{H }$ 加热到 $0 ^ { \circ } C$ 以上，发生分解，得到$\mathbf{D}$和$\mathbf{I}$。$\mathbf{I}$也可由$\mathbf{C}$和$\mathbf{D}$直接加热反应制备。在 $9 0 ^ { \circ } C$ 下将$\mathbf{C}$与$\mathbf{I}$混合，然后充分加热，得到$\mathbf{J}$。$\mathbf{J}$在高于 $4 5 0 ^ { \circ } C$ 时发生分解，得到等摩尔数的$\mathbf{C}$和$\mathbf{I}$。低温下，$\mathbf{F}$与$\mathbf{H}$发生反应，得到 $\mathbf{K}$ 和一种常见的Lewis酸$\mathbf{L}$。 $\mathbf{K}$分解会生成$\mathbf{B}$与$\mathbf{D}$。

请给出各个物质的化学式，并选择正确的选项。
All other options are incorrect.
Element $\mathbf{A}$ is a gas at room temperature.
In the process of generating $\mathbf{F}$, the roles of urea and $\mathrm{CaB_6}$ are completely consistent.
$\mathbf{E}$ is an amphoteric oxide
The central atom of $\mathbf{J}$ is $sp^3d$ hybridized.
$\mathbf{F}$ and $\mathbf{H}$ react in a 1:1 ratio.
The boiling point of $\mathbf{I}$ is higher than $\mathbf{J}$.
$\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$, the elements corresponding to the four kinds of simple substances all have only one stable hydride.
其他选项均不正确
元素$\mathbf{A}$在室温下为气体
生成$\mathbf{F}$的过程中,尿素和$\mathrm{CaB_6}$的作用是完全一致的
$\mathbf{E}$为两性氧化物
$\mathbf{J}$的中心原子为$sp^3d$杂化
$\mathbf{F}$和$\mathbf{H}$反应为1：1反应
$\mathbf{I}$的沸点高于$\mathbf{J}$
$\mathbf{A}$，$\mathbf{B}$，$\mathbf{C}$，$\mathbf{D}$ ，四种单质对应的元素均只有一种稳定氢化物
Since $\mathbf{L}$ is a common Lewis acid and the elements involved are all non-metallic elements, it can be known that it should contain boron and halogen elements. Moreover, it is generated by the reaction of $\mathbf{H}$ and $\mathbf{F}$, and $\mathbf{G}$, $\mathbf{J}$, and $\mathbf{I}$ have the same constituent elements as the compound of $\mathbf{H}$. It does not conform to the properties of boron, so $\mathbf{F}$ should contain boron. Next, analyze the preparation process of the binary compound $\mathbf{F}$: One way is the reaction of oxide $\mathbf{E}$ with urea. Since there is no other source of boron other than $\mathbf{E}$, it indicates that it is boron oxide $B_2O_3$, and the role of urea in the reaction is usually a carbon source or a nitrogen source, and it is more reasonable to act as a nitrogen source here; another way is the reaction of $B_2O_3$ with $\mathrm{CaB_6}$ in a nitrogen atmosphere at high temperature. Combining the reaction conditions, it can be known that the existence form of the $\mathrm{Ca}$ element in the product is $\mathrm{CaO}$, then the corresponding binary compound is a boron-nitrogen compound. In summary, starting from multiple conditions, it can be concluded that $\mathbf{F}$ should be $BN$. $BN$ has a high melting point and high hardness, which is also consistent with the statement in the question "used in rocket engines".

In the process of generating $BN$, urea acts as a nitrogen source, and $\mathrm{CaB_6}$ acts as a reducing agent (and also as a partial boron source), so the functions are different, and option C is incorrect.

$B_2O_3$ is a typical acidic oxide, and option D is incorrect.

<Checkpoint>[The chemical formula of compound $\mathbf{E}$ is $B_2O_3$](1)</Checkpoint>

<Checkpoint>[The chemical formula of compound $\mathbf{F}$ is $BN$](1)</Checkpoint>

<Checkpoint>[In the process of generating $BN$, the role of urea is a nitrogen source](1)</Checkpoint>

<Checkpoint>[In the process of generating $BN$, the role of $\mathrm{CaB_6}$ is a reducing agent and partial boron source](1)</Checkpoint>

<Checkpoint>[$B_2O_3$ is a typical acidic oxide](1)</Checkpoint>

The compound $\mathbf{E}$ is generated by the reaction of the element $\mathbf{A}$ with oxygen, indicating that $\mathbf{A}$ is elemental boron, and elemental boron is solid at room temperature, so option B is incorrect.

<Checkpoint>[$\mathbf{A}$ is elemental boron, which is solid at room temperature](1)</Checkpoint>

Since $\mathbf{F}$ is $BN$, $\mathbf{L}$ generated by its reaction with $\mathbf{H}$ does not contain nitrogen, so $\mathbf{K}$ should contain nitrogen, and one of $\mathbf{B}$ and $\mathbf{D}$ generated by its decomposition should be nitrogen gas. $\mathbf{D}$ and $\mathbf{C}$ can generate four compounds with different ratios ($\mathbf{G}$, $\mathbf{H}$, $\mathbf{I}$, $\mathbf{J}$) under different conditions, which does not conform to the properties of nitrogen, so the element $\mathbf{B}$ is nitrogen gas.

<Checkpoint>[The chemical formula of $\mathbf{B}$ is $N_2$](1)</Checkpoint>

Combined with the fact that three elements are in the same period, it is known that there are nitrogen and boron elements, so one of $\mathbf{C}$ and $\mathbf{D}$ must be a second-period element, and the element corresponding to $\mathbf{C}$ and boron form a common Lewis acid, which is a halogen. According to the question, $\mathbf{C}$ and $\mathbf{D}$ can form 4 binary compounds with different central element oxidation states. Only interhalogen compounds meet the conditions, which also conforms to the information "same group" mentioned in the question. Bromine can only generate two stable and confirmed binary compounds ($BrF$ can only be observed in the spectrum, and the existence of $BrF_7$ is controversial), while iodine meets the conditions and can generate four fluorides, so the element corresponding to $\mathbf{D}$ is iodine, and correspondingly $\mathbf{C}$ is fluorine, and $\mathbf{L}$ is $BF_3$.

<Checkpoint>[The chemical formula of $\mathbf{C}$ is $F_2$](1)</Checkpoint>

<Checkpoint>[The chemical formula of $\mathbf{D}$ is $I_2$](1)</Checkpoint>

<Checkpoint>[The chemical formula of $\mathbf{L}$ is $BF_3$](1)</Checkpoint>

Finally, the four compounds of iodine and fluorine ($\mathbf{G}$, $\mathbf{H}$, $\mathbf{I}$, $\mathbf{J}$) should be $IF$, $IF_3$, $IF_5$, and $IF_7$, respectively. According to the change in the ratio between the two elements that transform into each other (fluorine content $\mathbf{H}$ is less than $\mathbf{G}$, $\mathbf{I}$ is greater than $\mathbf{H}$, $\mathbf{J}$ is greater than $\mathbf{I}$ and only has two more fluorine atoms), it can be known that $\mathbf{G}$, $\mathbf{H}$, $\mathbf{I}$, and $\mathbf{J}$ are $IF_3$, $IF$, $IF_5$, and $IF_7$, respectively. The central atom of $IF_7$ is $sp^3d^3$ hybridized, and option E is incorrect.

<Checkpoint>[The chemical formula of compound $\mathbf{G}$ is $IF_3$](1)</Checkpoint>

<Checkpoint>[The chemical formula of compound $\mathbf{H}$ is $IF$](1)</Checkpoint>

<Checkpoint>[The chemical formula of compound $\mathbf{I}$ is $IF_5$](1)</Checkpoint>

<Checkpoint>[The chemical formula of compound $\mathbf{J}$ is $IF_7$](1)</Checkpoint>

<Checkpoint>[$IF_7$ has $sp^3d^3$ hybridization at the central atom](1)</Checkpoint>

Finally, combining the fact that the decomposition of $\mathbf{K}$ will generate $\mathbf{B}$ and $\mathbf{D}$, it can be known that the compound $\mathbf{K}$ is $NI_3$. $BN$ and $IF$ react in a stoichiometric ratio of 1 : 3 to obtain $BF_3$ and $NI_3$, so option F is incorrect.

<Checkpoint>[The chemical formula of compound $\mathbf{K}$ is $NI_3$](1)</Checkpoint>

<Checkpoint>[$BN$ and $IF$ react in a stoichiometric ratio of 1 : 3 to obtain $BF_3$ and $NI_3$](1)</Checkpoint>

For option G, $IF_5$ is more polar than $IF_7$, and the orientation force is greater, and its boiling point is higher.

<Checkpoint>[$IF_5$ is more polar than $IF_7$, and the orientation force is greater. Its boiling point is higher.](1)</Checkpoint>

For option H, if hydrazoic acid is not considered, the element $N$ has two stable hydrides, $NH_3$ and $N_2H_4$, and the option is incorrect.

<Checkpoint>[If hydrazoic acid is not considered, the element $N$ has two stable hydrides, $NH_3$ and $N_2H_4$](1)</Checkpoint>

In summary, G is correct.
由$\mathbf{L}$为常见Lewis酸，且涉及元素均为非金属元素，可知其应含有硼元素和卤素。并且，其为$\mathbf{H}$与$\mathbf{F}$反应生成，且与$\mathbf{H}$化合物组成元素相同的还有$\mathbf{G}$、$\mathbf{J}$、$\mathbf{I}$，其不符合硼元素的性质，故而$\mathbf{F}$中应含有硼元素。再来分析二元化合物$\mathbf{F}$的制备过程：一种方式是氧化物$\mathbf{E}$与尿素反应，而除去$\mathbf{E}$以外没有其他硼的来源，说明其为硼的氧化物$B_2O_3$，而尿素在反应中的作用通常是碳源或氮源，此处做氮源更合理；另一种方式是在高温条件下、氮气气氛中$B_2O_3$与 $\mathrm{CaB_6}$反应，结合反应条件可知产物中$\mathrm{Ca}$元素的存在形体是$\mathrm{CaO}$，那么对应的二元化合物便是硼氮化合物。综上，从多个条件出发可得出$\mathbf{F}$应为$BN$，$BN$具有高熔点、高硬度，这也和题目所述“用于火箭发动机中”相符。

在生成$BN$的过程中，尿素的作用是氮源，$\mathrm{CaB_6}$的作用是还原剂（同时也做部分硼源），因此作用不同，选项C错误。

$B_2O_3$是典型的酸性氧化物，选项D错误。

<Checkpoint>[化合物$\mathbf{E}$的化学式为$B_2O_3$](1)</Checkpoint>

<Checkpoint>[化合物$\mathbf{F}$的化学式为$BN$](1)</Checkpoint>

<Checkpoint>[在生成$BN$的过程中，尿素的作用是氮源](1)</Checkpoint>

<Checkpoint>[在生成$BN$的过程中，$\mathrm{CaB_6}$的作用是还原剂和部分硼源](1)</Checkpoint>

<Checkpoint>[$B_2O_3$是典型的酸性氧化物](1)</Checkpoint>

而化合物$\mathbf{E}$为单质$\mathbf{A}$与氧气反应生成，说明$\mathbf{A}$为硼单质，硼单质在室温下为固体，选项B错误。

<Checkpoint>[$\mathbf{A}$为硼单质，在室温下为固体](1)</Checkpoint>

由于$\mathbf{F}$为$BN$，其与$\mathbf{H}$反应生成的$\mathbf{L}$中不含有氮元素，故而$\mathbf{K}$中应含有氮元素，其分解生成的$\mathbf{B}$与$\mathbf{D}$中应有一个为氮气。而$\mathbf{D}$与$\mathbf{C}$在不同条件下可以生成四种比例的化合物（$\mathbf{G}$、$\mathbf{H}$、$\mathbf{I}$、$\mathbf{J}$）其不符合氮元素的性质，故而单质$\mathbf{B}$为氮气。

<Checkpoint>[$\mathbf{B}$的化学式为$N_2$](1)</Checkpoint>

结合有三种元素同周期，此时已知有氮元素和硼元素，所以$\mathbf{C}$与$\mathbf{D}$之一必定为二周期元素，且$\mathbf{C}$对应的元素与硼元素组成了常见路易斯酸、为卤素。根据题意，$\mathbf{C}$与$\mathbf{D}$之间可形成4种中心元素氧化态不同的二元化合物，符合条件的只有卤素互化物，这也符合题目所述“同族”的信息。溴仅能生成两种稳定存在且被证实的二元化合物（$BrF$仅能在光谱中观察，$BrF_7$存在有争议），而碘符合条件，能生成四种氟化物，故而$\mathbf{D}$对应的元素为碘元素，相应地$\mathbf{C}$为氟气，$\mathbf{L}$为$BF_3$。

<Checkpoint>[$\mathbf{C}$的化学式为$F_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{D}$的化学式为$I_2$](1)</Checkpoint>

<Checkpoint>[$\mathbf{L}$的化学式为$BF_3$](1)</Checkpoint>

最后碘与氟生成的四种比例的化合物（$\mathbf{G}$、$\mathbf{H}$、$\mathbf{I}$、$\mathbf{J}$）由于四种化合物分别应该为$IF$、$IF_3$、$IF_5$、$IF_7$，根据其中互相转化的两种元素之间的比例关系变化（氟含量$\mathbf{H}$小于$\mathbf{G}$，$\mathbf{I}$大于$\mathbf{H}$，$\mathbf{J}$大于$\mathbf{I}$且仅多两个氟原子）可知，$\mathbf{G}$、$\mathbf{H}$、$\mathbf{I}$、$\mathbf{J}$分别为$IF_3$、$IF$、$IF_5$、$IF_7$。$IF_7$的中心原子为$sp^3d^3$杂化，选项E错误。

<Checkpoint>[化合物$\mathbf{G}$的化学式为$IF_3$](1)</Checkpoint>

<Checkpoint>[化合物$\mathbf{H}$的化学式为$IF$](1)</Checkpoint>

<Checkpoint>[化合物$\mathbf{I}$的化学式为$IF_5$](1)</Checkpoint>

<Checkpoint>[化合物$\mathbf{J}$的化学式为$IF_7$](1)</Checkpoint>

<Checkpoint>[$IF_7$的中心原子为$sp^3d^3$杂化](1)</Checkpoint>

最后，结合$\mathbf{K}$分解会生成$\mathbf{B}$与$\mathbf{D}$，可知化合物$\mathbf{K}$为$NI_3$。$BN$和$IF$按照1 : 3的计量比反应得到$BF_3$和$NI_3$，因此选项F错误。

<Checkpoint>[化合物$\mathbf{K}$的化学式为$NI_3$](1)</Checkpoint>

<Checkpoint>[$BN$和$IF$按照1 : 3的计量比反应得到$BF_3$和$NI_3$](1)</Checkpoint>

对于G选项，$IF_5$相较于$IF_7$有极性，取向力大，其沸点更高。

<Checkpoint>[$IF_5$相较于$IF_7$有极性，取向力大。其沸点更高。](1)</Checkpoint>

对于H选项，若不考虑叠氮酸，$N$元素有$NH_3$和$N_2H_4$两种稳定氢化物，选项错误。

<Checkpoint>[若不考虑叠氮酸，$N$元素有$NH_3$和$N_2H_4$两种稳定氢化物](1)</Checkpoint>

综上，G正确。