$\bf N$ is a rare element whose mineral resources are sparsely distributed, often occurring as associated minerals or as impurities in other minerals. Elemental $\bf N$ is insoluble in water or hydrochloric acid, but can be oxidized to $\bf A$ by fuming nitric acid; $\bf A$ is a white amorphous powder, slightly soluble in water, and easily dehydrated; compound $\bf B$ is obtained after complete dehydration of $\bf A$. $\bf B$ reacts with carbon tetrachloride at $\rm 723~K$ to give a white hygroscopic solid $\bf C$. $\bf C$ combines with elemental $\bf N$ to form $\bf D$; $\bf D$ can also be obtained by reacting elemental $\bf N$ with sulfuryl chloride at $\rm 310K$ in a $1:1$ molar ratio. $\bf D$ reacts with liquid non-metallic element $\bf E$ to obtain a yellow solid $\bf F$, in which the mass fraction of $\bf N$ is $\rm 35.61\%$. When isolated from air, elemental $\bf N$ reacts with molten $\rm KCN$ to produce $\bf G$ and release gas $\bf H$; $\bf G$ reacts with acid to produce a highly toxic, foul-smelling gas $\bf I$; $\bf I$ spontaneously decomposes to elemental $\bf N$ at $\rm 273~K$.

The following options are correct:
$\bf N$ 是一种稀有元素,其矿产资源分布稀散,多伴生在其它矿物中或以杂质形式存在于其它矿物中。$\bf N$ 单质不溶于水或盐酸,但可被发烟硝酸氧化为 $\bf A$；$\bf A$ 为白色无定形粉末，在水中溶解度小，易脱水；$\bf A$ 完全脱水后得到化合物 $\bf B$。$\bf B$ 与四氯化碳在 $\rm 723~K$ 下反应，得到白色吸湿固体 $\bf C$。$\bf C$ 与 $\bf N$ 单质化合生成$\bf D$；$\bf D$ 也可由 $\bf N$ 单质与磺酰氯在 $\rm 310K$ 下按摩尔比 $1:1$ 反应得到。$\bf D$ 与液态非金属单质 $\bf E$ 反应可得到黄色固体 $\bf F$，$\bf F$ 中 $\bf N$ 的质量分数为 $\rm 35.61\%$。在隔绝空气时，$\bf N$ 单质与 $\rm KCN$ 熔融反应，得到 $\bf G$ 并放出气体 $\bf H$；$\bf G$ 与酸生成剧毒有恶臭的气体 $\bf I$；$\bf I$ 在 $\rm 273~K$ 下即可自发分解得到 $\bf N$ 单质。

下列选项正确的是：
All other options are incorrect
Element $\bf N$ is a subgroup metal element.
Element $\bf N$ is a main group metal element.
The number of binary compounds in $\bf A-I$ is 5
The number of binary compounds in $\bf A-I$ is 7.
$\bf B-I$ contains 4 substances whose molecular formula contains greater than or equal to 4 atoms.
$\bf B-I$ contains 4 substances with molecular formulas containing 3 atoms.
其他选项均不正确
元素 $\bf N$ 为副族金属元素
元素 $\bf N$ 为主族金属元素
$\bf A-I$ 中二元化合物的数目为5
$\bf A-I$ 中二元化合物的数目为7
$\bf B-I$ 中分子式含有大于等于4个原子的有4种物质
$\bf B-I$ 中分子式含有等于3个原子的有4种物质
$\bf N$ is a rare element, insoluble in hydrochloric acid and water, exhibiting properties of an inert metal or a high-period main group nonmetal element;

When $\bf N$ is fused with $\rm KCN$, a gas is produced, then the remaining $\rm G$ is basically certain to be a potassium salt containing $\bf N$; therefore, $\bf N$ is likely negatively charged, excluding metallic elements.

<Checkpoint>[$\bf N$ is likely negatively charged, excluding metallic elements](1)</Checkpoint>

$\bf G$ reacts with acid to produce the highly toxic gas $\bf I$, then $\bf I$ is a hydride of $\bf N$, containing a highly toxic substance, likely a high-period chalcogen and pnictogen element.

<Checkpoint>[$\bf I$ is a hydride of $\bf N$](1)</Checkpoint>

$\bf I$ decomposes at room temperature, indicating instability, therefore it can be judged that the nonmetallic property of $\bf N$ is very weak, belonging to a high period; therefore, it can be basically judged that $\bf N$ is $\rm Te$ or $\rm Se$, but $\rm H_2Se$ does not decompose at room temperature, so $\bf N$ can only be $\rm Te$. Therefore, options B and C are incorrect.

<Checkpoint>[$\bf N$'s nonmetallic property is very weak, belonging to a high-period element](1)</Checkpoint>

<Checkpoint>[Judging $\bf N$ to be $\rm Te$ based on the properties of the hydride](2)</Checkpoint>

$\rm Te$ dissolves in fuming nitric acid to produce $\bf A$, then $\bf A$ is $\rm TeO_2\cdot xH_2O$. The dehydrated product $\bf B$ is $\rm TeO_2$; $\rm TeO_2$ reacts with carbon tetrachloride to obtain a hygroscopic solid $\bf C$, which can be judged to be a common hydrolyzable chloride $\rm TeCl_4$. $\rm TeCl_4$ combines with $\rm Te$ in a comproportionation reaction, and the resulting $\bf D$ is $\rm TeCl_2$.

<Checkpoint>[$\bf A$ is $\rm TeO_2\cdot xH_2O$](1)</Checkpoint>

<Checkpoint>[$\bf B$ is $\rm TeO_2$](1)</Checkpoint>

<Checkpoint>[$\bf C$ is a common hydrolyzable chloride $\rm TeCl_4$](1)</Checkpoint>

<Checkpoint>[$\bf D$ is $\rm TeCl_2$](1)</Checkpoint>

$\rm TeCl_2$ reacts with the nonmetallic element $\bf E$ to obtain $\bf F$, then $\bf F$ should contain three elements $\rm Te,Cl,\bf E$. Based on the mass fraction, assuming $\bf F$ contains one $\rm Te$, the remaining molecular weight is $127.6/0.3561-127.6=230.7=35.45\times2+79.9\times2$; therefore, $\bf E$ is $\rm Br_2$, and $\bf F$ is $\rm TeCl_2Br_2$.

<Checkpoint>[Inferring that $\bf E$ is $\rm Br_2$ based on the mass fraction](1)</Checkpoint>

<Checkpoint>[$\bf F$ is $\rm TeCl_2Br_2$](1)</Checkpoint>

$\rm Te$ reacts with $\rm KCN$ in a fusion reaction to produce gas and potassium salt, then $\bf G$ is $\rm K_2Te$, $\bf H$ is $\rm (CN)_2$, and the highly toxic substance $\bf I$ is $\rm H_2Te$.

<Checkpoint>[$\bf G$ is $\rm K_2Te$](1)</Checkpoint>

<Checkpoint>[$\bf H$ is $\rm (CN)_2$](1)</Checkpoint>

<Checkpoint>[The highly toxic substance $\bf I$ is $\rm H_2Te$](1)</Checkpoint>

According to the chemical formulas, options D, E, and F are incorrect, and G is correct.
$\bf N$为稀有元素，不溶于盐酸和水，为惰性金属或高周期主族非金属元素的性质；

$\bf N$与$\rm KCN$熔融反应产生气体，则留下的$\rm G$基本肯定为含$\bf N$的钾盐；故$\bf N$大概率为负价，排除金属元素。

<Checkpoint>[$\bf N$大概率为负价，排除金属元素](1)</Checkpoint>

$\bf G$与酸生成剧毒气体$\bf I$，则$\bf I$为$\bf N$的氢化物，含有剧毒，大概率为高周期硫族元素和氮族元素。

<Checkpoint>[$\bf I$为$\bf N$的氢化物](1)</Checkpoint>

$\bf I$常温即分解，不稳定，因此可判断$\bf N$的非金属性很弱，为高周期；因此基本可判断$\bf N$为$\rm Te$或$\rm Se$，但$\rm H_2Se$常温下不分解，因此$\bf N$只能为$\rm Te$。因此选项B,C错误。

<Checkpoint>[$\bf N$的非金属性很弱，为高周期元素](1)</Checkpoint>

<Checkpoint>[根据氢化物性质判断$\bf N$为$\rm Te$](2)</Checkpoint>

$\rm Te$溶于发烟硝酸生成$\bf A$，则$\bf A$为$\rm TeO_2\cdot xH_2O$。脱水后形成的$\bf B$为$\rm TeO_2$；$\rm TeO_2$与四氯化碳反应得到吸湿性固体$\bf C$，可判断$\bf C$为易水解的常见氯化物$\rm TeCl_4$。$\rm TeCl_4$与$\rm Te$化合发生归中反应，生成的$\bf D$为$\rm TeCl_2$。

<Checkpoint>[$\bf A$为$\rm TeO_2\cdot xH_2O$](1)</Checkpoint>

<Checkpoint>[$\bf B$为$\rm TeO_2$](1)</Checkpoint>

<Checkpoint>[$\bf C$为易水解的常见氯化物$\rm TeCl_4$](1)</Checkpoint>

<Checkpoint>[$\bf D$为$\rm TeCl_2$](1)</Checkpoint>

$\rm TeCl_2$与非金属单质$\bf E$反应得到$\bf F$，则$\bf F$应当含有三种元素$\rm Te,Cl,\bf E$。根据质量分数，假设$\bf F$含有一个$\rm Te$，剩余的分子量为$127.6/0.3561-127.6=230.7=35.45\times2+79.9\times2$；故$\bf E$为$\rm Br_2$，$\bf F$为$\rm TeCl_2Br_2$。

<Checkpoint>[根据质量分数推断得到$\bf E$为$\rm Br_2$](1)</Checkpoint>

<Checkpoint>[$\bf F$为$\rm TeCl_2Br_2$](1)</Checkpoint>

$\rm Te$与$\rm KCN$熔融反应产生气体和钾盐，则$\bf G$为$\rm K_2Te$，$\bf H$为$\rm (CN)_2$，剧毒物质$\bf I$为$\rm H_2Te$。

<Checkpoint>[$\bf G$为$\rm K_2Te$](1)</Checkpoint>

<Checkpoint>[$\bf H$为$\rm (CN)_2$](1)</Checkpoint>

<Checkpoint>[剧毒物质$\bf I$为$\rm H_2Te$](1)</Checkpoint>

根据化学式可知，选项D,E,F错误，G正确。