Yellow solid $\mathbf{A}$ is composed of two elements, a metallic element $\mathbf{X}$ and a non-metallic element $\mathbf{Y}$, and is one of the important minerals of element $\mathbf{X}$. Calcination of $\mathbf{A}$ in air can produce solid $\mathbf{B}$ and colorless gas $\mathbf{C}$. $\mathbf{B}$ contains $69.94\%$ of element $\mathbf{X}$. $\mathbf{C}$ reacts with $\mathrm{Cl_2}$ to obtain a colorless fuming liquid $\mathbf{D}$. $\mathbf{D}$ is easily hydrolyzed to produce $\mathbf{E}$ and $\mathbf{F}$. $\mathbf{C}$ can react with $\mathrm{O_2}$ to produce $\mathbf{G}$ under the action of a heated catalyst. $\mathbf{G}$ is soluble in water and also yields $\mathbf{E}$.

$\mathbf{C}$ is an excellent Lewis acid. Researchers recently discovered that the reaction of $0.5 \mathrm{~mL}$ of anhydrous $\mathbf{C}$ with $0.02 \mathrm{~g}$ of a certain metal azide $\mathbf{I}$ at low temperature can produce an adduct $\mathbf{H}$. The nitrogen content in $\mathbf{H}$ is $13.87\%$. If the orientation of $\mathbf{C}$ in $\mathbf{H}$ is ignored, its anionic symmetry is the same as that of water.

The following statements are correct:
黄色固体 $\mathbf{A}$ 由金属元素 $\mathbf{X}$ 和非金属元素 $\mathbf{Y}$ 两种元素构成,是$\mathbf{X}$元素的重要矿物之一。 $\mathbf{A}$ 在空气中煅烧可生成固体 $\mathbf{B}$ 与无色气体 $\mathbf{C}$, $\mathbf{B}$ 中含$\mathbf{X}$元素 $69.94\%$。$\mathbf{C}$ 与 $\mathrm{Cl_2}$ 反应得到无色发烟液体 $\mathbf{D}, \mathbf{D}$ 极易水解生成 $\mathbf{E}$ 和 $\mathbf{F}$, $\mathbf{C}$ 可以在催化剂加热作用下和 $\mathrm{O_2}$ 反应生成 $\mathbf{G}$, $\mathbf{G}$ 可溶于水也得到 $\mathbf{E}$ 。

$\mathbf{C}$ 是一种优良的 Lewis 酸,科研人员日前发现将 $0.5 \mathrm{~mL}$ 无水 $\mathbf{C}$ 在低温下与 $0.02 \mathrm{~g}$
某金属叠氮盐 $\mathbf{I}$ 反应可以得到加成物 $\mathbf{H}, \mathbf{H}$ 中氮的含量为 $13.87 \%$, 若忽略 $\mathbf{H}$ 中 $\mathbf{C}$ 的取向,则其阴离子对称性与水相同。

下列说法正确的是：
All other options are incorrect
$\mathbf{A}$ is a 1:2 type ionic compound.
The highest oxidation state that element $\mathbf{X}$ can form is +4.
$\mathbf{D}$ is a regular tetrahedral molecule
G has a certain reducing property.
The mass fraction of nitrogen in $\mathrm{I}$ is approximately $28\%$
The mass fraction of $\mathrm{H}$ cation is $43.9\%$
H is a 2:1 type ionic compound.
其他选项均不正确
$\mathbf{A}$是1：2型离子化合物
$\mathbf{X}$元素可以形成的最高氧化态为+4
$\mathbf{D}$为正四面体构型分子
$\mathrm{G}$具有一定的还原性
$\mathrm{I}$中氮的质量分数约为$28\%$
$\mathrm{H}$中阳离子的质量分数为$43.9\%$
$\mathrm{H}$为2：1型离子化合物。
$\mathbf{B}$ is obtained by calcining $\mathbf{A}$, it can be determined that $\mathbf{B}$ is the stable oxide of element $\mathbf{X}$; let the chemical formula of $\mathbf{B}$ be $\mathrm{\bf{X}\rm{O_n}}$, containing $69.94\%$ of element $\mathbf{X}$.

<Checkpoint>[$\mathbf{B}$ is the stable oxide of element $\mathbf{X}$](1)</Checkpoint>

If it is $\mathrm{\bf{X}\rm{O}}$, the relative atomic mass of $\mathbf{X}$ is $\mathrm{16/(1-0.6994)-16=37.2}$, no reasonable metal.

If it is $\mathrm{\bf{X_2}\rm{O_3}}$, the relative atomic mass of $\mathbf{X}$ is $\mathrm{(16\times3/(1-0.6994)-16\times3)/2=55.9}$, which is $\rm{Fe}$.

If it is $\mathrm{\bf{X}\rm{O_2}}$, the relative atomic mass of $\mathbf{X}$ is $\mathrm{16\times2/(1-0.6994)-16\times2=74.5}$, no reasonable metal.

Other oxide types also have no reasonable results; therefore $\mathbf{X}$ is $\rm{Fe}$, $\mathbf{B}$ is $\rm{Fe_2O_3}$. $\rm{Fe}$ exists in the +6 valence state, option C is incorrect.

<Checkpoint>[$\mathbf{X}$ is $\rm{Fe}$, exists in the +6 valence state](2)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ is $\rm{Fe_2O_3}$](1)</Checkpoint>

Minerals generally contain sulfur or carbon elements; gas $\bf{C}$ may be $\rm{SO_2}$ or $\rm{CO_2}$; but $\bf{C}$ can react with $\mathrm{Cl_2}$, so $\rm{CO_2}$ is excluded; $\bf{C}$ is likely to be $\rm{SO_2}$.

<Checkpoint>[Minerals generally contain sulfur or carbon elements](1)</Checkpoint>

<Checkpoint>[$\bf{C}$ is likely to be $\rm{SO_2}$](1)</Checkpoint>

$\mathbf{C}$ reacts with $\mathrm{O_2}$ to produce $\mathbf{G}$, then $\mathbf{G}$ is $\rm{SO_3}$; dissolving in water to obtain $\mathbf{E}$, then $\mathbf{E}$ is $\mathrm{H_2SO_4}$. $\rm{SO_3}$ is the highest valence oxide and has no reducing property, option E is incorrect.

<Checkpoint>[$\mathbf{G}$ is $\rm{SO_3}$, has no reducing property](1)</Checkpoint>

At this time, $\mathbf{Y}$ should be the sulfur element, so $\mathbf{A}$ should be pyrite $\rm{FeS_2}$. The $\rm{FeS_2}$ anion is $\rm{S_2^{2-}}$, which is a 1:1 type ionic compound, option B is incorrect.

<Checkpoint>[$\mathbf{Y}$ should be the sulfur element, so $\mathbf{A}$ should be pyrite $\rm{FeS_2}$](2)</Checkpoint>

<Checkpoint>[$\rm{FeS_2}$ anion is $\rm{S_2^{2-}}$, which is a 1:1 type ionic compound](1)</Checkpoint>

$\rm{SO_2}$ reacts with $\mathrm{Cl_2}$ to obtain a colorless fuming liquid, according to the elemental properties, it is sulfuryl chloride, that is, $\mathbf{D}$ is $\mathrm{SO_2Cl_2}$, which hydrolyzes to produce sulfuric acid $\mathbf{E}$ and hydrochloric acid $\mathbf{F}$. $\mathrm{SO_2Cl_2}$ is not a regular tetrahedron, but a distorted tetrahedral molecule, option D is incorrect.

<Checkpoint>[$\mathbf{D}$ is $\mathrm{SO_2Cl_2}$, which is a distorted tetrahedral molecule](1)</Checkpoint>

$\mathbf{I}$ is a metal azide, which reacts with sulfur dioxide to obtain an adduct, so there should be no redox reaction; the chemical formula of the adduct can be written as $\mathrm{M(SO_2)_x(N_3)_y}$, considering that the anion symmetry is the same as water, x and y should be 1 and 2 respectively;

<Checkpoint>[Reacts with sulfur dioxide to obtain an adduct, so there should be no redox reaction](1)</Checkpoint>

<Checkpoint>[Anion symmetry is the same as water, the ratio of $\rm{SO_2, N_3}$ is 1:2 or 2:1](1)</Checkpoint>

When x is 1 and y is 2, the atomic mass of metal $\rm{M}$ is $14\times6/0.1387-14\times6-64=457.6$, the atomic mass is too large.

x is 2; at this time, the atomic mass of metal $\rm{M}$ is $14\times3/0.1387-14\times3-64\times2=132.8$, which is $\rm{Cs}$, just in line with the valence.

Therefore, the chemical formula of $\bf{H}$ is $\rm{Cs[(SO_2)_2N_3]}$, and $\bf{I}$ is $\rm{CsN_3}$. $\rm{Cs[(SO_2)_2N_3]}$ is a 1:1 type ionic compound, option H is incorrect.

<Checkpoint>[The chemical formula of $\bf{H}$ is $\rm{Cs[(SO_2)_2N_3]}$, which is a 1:1 type ionic compound](2)</Checkpoint>

<Checkpoint>[$\bf{I}$ is $\rm{CsN_3}$](1)</Checkpoint>

Calculation shows that option F is incorrect and G is correct.
$\mathbf{B}$由$\mathbf{A}$煅烧得到，可以判断$\mathbf{B}$为$\mathbf{X}$元素的稳定氧化物；设$\mathbf{B}$的化学式为$\mathrm{\bf{X}\rm{O_n}}$，含$\mathbf{X}$元素 $69.94\%$。

<Checkpoint>[$\mathbf{B}$为$\mathbf{X}$元素的稳定氧化物](1)</Checkpoint>

若为$\mathrm{\bf{X}\rm{O}}$，$\mathbf{X}$的相对原子质量为$\mathrm{16/(1-0.6994)-16=37.2}$，无合理金属。

若为$\mathrm{\bf{X_2}\rm{O_3}}$，$\mathbf{X}$的相对原子质量为$\mathrm{(16\times3/(1-0.6994)-16\times3)/2=55.9}$，为$\rm{Fe}$

若为$\mathrm{\bf{X}\rm{O_2}}$，$\mathbf{X}$的相对原子质量为$\mathrm{16\times2/(1-0.6994)-16\times2=74.5}$，无合理金属。

其他氧化物类型也均无合理结果；因此$\mathbf{X}$为$\rm{Fe}$，$\mathbf{B}$为$\rm{Fe_2O_3}$。$\rm{Fe}$存在+6价，选项C错误。

<Checkpoint>[$\mathbf{X}$为$\rm{Fe}$，存在+6价](2)</Checkpoint>

<Checkpoint>[$\mathbf{B}$为$\rm{Fe_2O_3}$](1)</Checkpoint>

矿物中一般含硫元素或碳元素；气体$\bf{C}$可能是$\rm{SO_2}$或$\rm{CO_2}$；但$\bf{C}$可以与$\mathrm{Cl_2}$反应，因此排除$\rm{CO_2}$；$\bf{C}$大概率是$\rm{SO_2}$。

<Checkpoint>[矿物中一般含硫元素或碳元素](1)</Checkpoint>

<Checkpoint>[$\bf{C}$大概率是$\rm{SO_2}$](1)</Checkpoint>

$\mathbf{C}$ 和 $\mathrm{O_2}$ 反应生成 $\mathbf{G}$，则$\mathbf{G}$为$\rm{SO_3}$；溶于水得到$\mathbf{E}$，则$\mathbf{E}$为$\mathrm{H_2SO_4}$。$\rm{SO_3}$为最高价氧化物，无还原性，选项E错误。

<Checkpoint>[$\mathbf{G}$为$\rm{SO_3}$，无还原性](1)</Checkpoint>

此时$\mathbf{Y}$应为硫元素，故$\mathbf{A}$应当为黄铁矿$\rm{FeS_2}$。$\rm{FeS_2}$阴离子为$\rm{S_2^{2-}}$，为1：1型离子化合物，选项B错误。

<Checkpoint>[$\mathbf{Y}$应为硫元素，故$\mathbf{A}$应当为黄铁矿$\rm{FeS_2}$](2)</Checkpoint>

<Checkpoint>[$\rm{FeS_2}$阴离子为$\rm{S_2^{2-}}$，为1：1型离子化合物](1)</Checkpoint>

$\rm{SO_2}$与$\mathrm{Cl_2}$反应得到无色发烟液体，根据元素性质为硫酰氯即$\mathbf{D}$为$\mathrm{SO_2Cl_2}$，水解生成硫酸$\mathbf{E}$和盐酸$\mathbf{F}$。$\mathrm{SO_2Cl_2}$不为正四面体，为畸变四面体分子，选项D错误。

<Checkpoint>[$\mathbf{D}$为$\mathrm{SO_2Cl_2}$，为畸变四面体分子](1)</Checkpoint>

$\mathbf{I}$为金属叠氮盐，与二氧化硫反应得到加合物，故应当无氧化还原反应发生；加合物化学式可写为$\mathrm{M(SO_2)_x(N_3)_y}$，考虑阴离子对称性与水相同，x与y应当分别取1和2；

<Checkpoint>[与二氧化硫反应得到加合物，故应当无氧化还原反应发生](1)</Checkpoint>

<Checkpoint>[阴离子对称性与水相同，$\rm{SO_2, N_3}$比例为1：2或2：1](1)</Checkpoint>

当x为1，y为2时，金属$\rm{M}$的原子量为$14\times6/0.1387-14\times6-64=457.6$，原子量太大。

x取2；此时金属$\rm{M}$的原子量为$14\times3/0.1387-14\times3-64\times2=132.8$，为$\rm{Cs}$，刚好符合化合价。

因此$\bf{H}$的化学式为$\rm{Cs[(SO_2)_2N_3]}$，$\bf{I}$为$\rm{CsN_3}$。$\rm{Cs[(SO_2)_2N_3]}$为1：1型离子化合物，选项H错误。

<Checkpoint>[$\bf{H}$的化学式为$\rm{Cs[(SO_2)_2N_3]}$，为1：1型离子化合物](2)</Checkpoint>

<Checkpoint>[$\bf{I}$为$\rm{CsN_3}$](1)</Checkpoint>

计算可得选项F错误，G正确。