The element $\bf{A}$ exists as a gray simple substance with relatively active chemical properties. Mixing the simple substance of $\bf{A}$ with a binary compound $\bf{B}$ containing $\mathrm{35.77\%}$ of $\bf{A}$ in a sealed tube at a certain stoichiometric ratio and heating to $\mathrm{393 ~K}$ yields a brownish-red oily liquid $\bf{C}$ containing $\mathrm{69.01\%}$ of $\bf{A}$; $\bf{B}$ can be prepared by passing $\mathrm{Cl_2}$ into a solution of $\bf{C}$ in $\mathrm{CCl_4}$. Anhydrous liquid ammonia can react with $\bf{C}$ to produce the simple substance of $\bf{A}$, while an ammonia solution in diethyl ether reacts with $\bf{C}$ at $\mathrm{193~K}$ to produce an explosive substance $\bf{D}$ with an $\bf{A}$ content of $84.93\%$. $\bf{E}$, which is isoelectronic with $\bf{C}$ in a broader sense, can also react with $\mathrm{NH_3}$ to produce a similarly explosive $\bf{F}$ with a nitrogen content of $30.40\%$.

Reacting $\bf{D}$ with $\bf{B}$ in $\mathrm{SOCl_2}$ at $313 \mathrm{~K}$ yields a yellow precipitate. After washing and recrystallization, a yellow compound $\bf{G}$ is obtained. Its cation contains only one $\bf{A}$ and the $\bf{A}$ content in the cation is $36.35\%$, and the nitrogen content is $19.35\%$; the overall $\bf{A}$ content in the entire compound is $32.62\%$.

Which of the following statements is correct:
$\bf{A}$元素的单质为灰色,化学性质较为活泼。将$\bf{A}$单质与$\bf{A}$含量为$\mathrm{35.77\%}$的二元化合物$\bf{B}$按一定的化学计量比混合于封管中,并加热至$\mathrm{393 ~K}$,可以得到$\bf{A}$含量为$\mathrm{69.01\%}$的棕红色油状液体$\bf{C}$; 而将 $\mathrm{Cl_2}$ 通入 $\bf{C}$ 在 $\mathrm{CCl_4}$ 的溶液中可以制得 $\bf{B}$。无水液氨可以与 $\bf{C}$ 反应得到 $\bf{A}$ 单质, 而乙醚的氨溶液与 $\bf{C}$ 在 $\mathrm{193~K}$ 下反应可以得到有爆炸性的$\bf{D}$, $\bf{D}$中$\bf{A}$含量为 $84.93\%$ 。与 $\bf{C}$ 互为广义等电子体的 $\bf{E}$ 也可以与 $\mathrm{NH_3}$ 反应得到同样有爆炸性的含氮量为 $30.40 \%$ 的 $\bf{F}$。

将$\bf{D}$与$\bf{B}$于 $\mathrm{SOCl_2}$ 中在 $313 \mathrm{~K}$ 下反应, 可以得到一种黄色沉淀。经洗涤、重结晶之后得到一种黄色化合物$\bf{G}$。其阳离子中只含有一个$\bf{A}$，且$\bf{A}$含量为$36.35 \%$, 氮元素含量为$19.35 \%$; 整个化合物中$\bf{A}$含量为$32.62 \%$。

下列说法正确的是：
All other options are incorrect
$\bf{A}$ atomic ground state only has one single electron
$\bf{B}$ and $\bf{E}$ chemical formulas have the same number of atoms.
$\bf{B}$ and $\bf{D}$ have the same number of atoms in their chemical formulas.
$\bf{D}$ and $\bf{F}$ have different numbers of atoms in their chemical formulas.
$\bf{G}$ is a 1:1 type ionic compound
$\bf{G}$ anion possesses a tetrahedral configuration.
The chemical formula of $\bf{G}$ contains 18 atoms after removing atom $\bf{A}$.
其他选项均不正确
$\bf{A}$原子基态只具有一个单电子
$\bf{B}$和$\bf{E}$化学式原子数相同
$\bf{B}$和$\bf{D}$化学式原子数相同
$\bf{D}$和$\bf{F}$化学式原子数不同
$\bf{G}$为1：1型离子化合物
$\bf{G}$阴离子具有四面体构型
$\bf{G}$的化学式去掉$\bf{A}$原子之后含有18个原子
$\mathrm{Cl_2}$ is passed into $\bf{C}$ to obtain a binary compound $\bf{B}$. It can be determined that $\bf{B}$ is a chloride; assume the chemical formula of $\bf{B}$ is $\rm{\bf{A}\rm{Cl_n}}$, calculate based on the mass fraction:

<Checkpoint>[$\bf{B}$ is a chloride](1)</Checkpoint>

If n is 1, the relative atomic mass of $\bf{A}$ is $\rm{35.45/(1-0.3577)-35.45=19.7}$, no reasonable element.

If n is 2, the relative atomic mass of $\bf{A}$ is $\rm{35.45\times2/(1-0.3577)-35.45\times2=39.5}$, no reasonable element.

If n is 3, the relative atomic mass of $\bf{A}$ is $\rm{35.45\times3/(1-0.3577)-35.45\times3=59.2}$, no reasonable element.

If n is 4, the relative atomic mass of $\bf{A}$ is $\rm{35.45\times4/(1-0.3577)-35.45\times4=78.9}$, consider it to be $\rm{Se}$, the valence state is reasonable.

No reasonable element exists when n is greater than or equal to 5.

In summary, it is determined that $\bf{A}$ is $\rm{Se}$, then $\bf{B}$ is $\rm{SeCl_4}$. $\rm{Se}$ has two unpaired electrons, so option B is incorrect.

<Checkpoint>[$\bf{A}$ is $\rm{Se}$](2)</Checkpoint>

<Checkpoint>[$\bf{B}$ is $\rm{SeCl_4}$](1)</Checkpoint>

The content of $\rm{Se}$ in $\bf{C}$ is $\mathrm{69.01\%}$ and should only contain $\rm{Se，Cl}$, it can be deduced that $\bf{C}$ is $\rm{Se_2Cl_2}$.

<Checkpoint>[$\bf{C}$ is $\rm{Se_2Cl_2}$](1)</Checkpoint>

The content of $\rm{Se}$ in $\bf{D}$ is $84.93\%$; since $\bf{D}$ is an explosive, consider it to be a nitride; according to the mass fraction, it is easy to obtain the simplest chemical formula of $\bf{D}$ as $\rm{SeN}$, according to the properties of chalcogen elements, it is known that this compound is $\rm{Se_4N_4}$.

<Checkpoint>[$\bf{D}$ is an explosive, consider it to be a nitride](1)</Checkpoint>

<Checkpoint>[The simplest chemical formula of $\bf{D}$ is $\rm{SeN}$, actually this compound is $\rm{Se_4N_4}$](1)</Checkpoint>

$\bf{E}$ is the generalized isoelectronic species of $\rm{Se_2Cl_2}$, then $\bf{E}$ is likely to be $\rm{S_2Cl_2}$ or $\rm{Te_2Cl_2}$ or the uncommon $\rm{O_2Cl_2}$; $\bf{E}$ can also generate the explosive substance $\bf{F}$, with a nitrogen content of $30.40 \%$, calculation yields the simplest chemical formula as $\rm{SN}$, actually $\rm{S_4N_4}$; then $\bf{E}$ is $\rm{S_2Cl_2}$. From this, it can be seen that options C, D, and E are all incorrect.

<Checkpoint>[$\bf{E}$ is $\rm{S_2Cl_2}$](1)</Checkpoint>

<Checkpoint>[The simplest chemical formula of $\bf{F}$ is $\rm{SN}$, actually this compound is $\rm{S_4N_4}$](1)</Checkpoint>

$\rm{Se_4N_4}$ and $\rm{SeCl_4}$ can synthesize $\bf{G}$. Its cation contains only one $\rm{Se}$; and the $\rm{Se}$ content is $36.35 \%$, the nitrogen content is $19.35 \%$; based on this, the ratio of the number of $\rm{Se}$ and $\rm{N}$ in the cation can be calculated as $\rm{78.96/0.3635 : 14.01/0.1935 = 1:3}$, containing only one $\rm{Se}$, and the chemical formula of the cation can be calculated as $\rm{SeN_3S_3^{n+}}$ according to the synthesis method.

<Checkpoint>[In $\bf{G}$ cation, the ratio of the number of $\rm{Se}$ and $\rm{N}$ is $\rm{1:3}$](1)</Checkpoint>

<Checkpoint>[The chemical formula of $\bf{G}$ cation is $\rm{SeN_3S_3}$](1)</Checkpoint>

The $\rm{Se}$ content in the entire compound is $32.62 \%$, which is smaller than the $\rm{Se}$ content in the cation, meaning that the anion must contain $\rm{Se}$; assuming the anion contains 1 $\rm{Se}$, $\bf{G}$ contains one cation and n anions, then the relative molecular mass of the anion is $\rm{78.96\times(1+n)/0.3262-78.96/0.3635=242n+25}$, subtracting the mass of n $\rm{Se}$ atoms, when n is 0.5, exactly the relative molecular mass of three chlorines remain; therefore, it is obvious that the anion is $\mathrm{SeCl_6^{2-}}$; at this time, the cation must be $\rm{SeN_3S_3^{+}}$ to match the mass fraction. Therefore, the chemical formula of $\bf{G}$ is $\rm{[SeN_3S_3]_2[SeCl_6]}$, option F is incorrect, and H is correct. $\mathrm{SeCl_6^{2-}}$ has an octahedral configuration, so option G is incorrect.

<Checkpoint>[$\bf{G}$ anion is $\mathrm{SeCl_6^{2-}}$, octahedral configuration](2)</Checkpoint>

<Checkpoint>[$\bf{G}$ cation is $\rm{SeN_3S_3^{+}}$](1)</Checkpoint>

<Checkpoint>[The chemical formula of $\bf{G}$ is $\rm{[SeN_3S_3]_2[SeCl_6]}$, which is a 1:2 type ionic compound](2)</Checkpoint>
$\mathrm{Cl_2}$通入$\bf{C}$得到二元化合物$\bf{B}$，可以判断$\bf{B}$是氯化物；设$\bf{B}$的化学式为$\rm{\bf{A}\rm{Cl_n}}$，根据质量分数求算：

<Checkpoint>[$\bf{B}$是氯化物](1)</Checkpoint>

若n取1，$\bf{A}$的相对原子质量为$\rm{35.45/(1-0.3577)-35.45=19.7}$，无合理元素。

若n取2，$\bf{A}$的相对原子质量为$\rm{35.45\times2/(1-0.3577)-35.45\times2=39.5}$，无合理元素。

若n取3，$\bf{A}$的相对原子质量为$\rm{35.45\times3/(1-0.3577)-35.45\times3=59.2}$，无合理元素。

若n取4，$\bf{A}$的相对原子质量为$\rm{35.45\times4/(1-0.3577)-35.45\times4=78.9}$，考虑为$\rm{Se}$，价态合理。

n取5以上无合理元素。

综上判断$\bf{A}$为$\rm{Se}$，则$\bf{B}$为$\rm{SeCl_4}$。$\rm{Se}$有两个单电子，选项B错误。

<Checkpoint>[$\bf{A}$为$\rm{Se}$](2)</Checkpoint>

<Checkpoint>[$\bf{B}$为$\rm{SeCl_4}$](1)</Checkpoint>

$\bf{C}$中$\rm{Se}$含量为$\mathrm{69.01\%}$且应当只含$\rm{Se，Cl}$，可以推出$\bf{C}$为$\rm{Se_2Cl_2}$。

<Checkpoint>[$\bf{C}$为$\rm{Se_2Cl_2}$](1)</Checkpoint>

$\bf{D}$中$\rm{Se}$含量为 $84.93\%$；由于$\bf{D}$为易爆物，考虑为氮化物；根据质量分数易得$\bf{D}$的最简化学式为$\rm{SeN}$，根据硫族元素性质可知该化合物为$\rm{Se_4N_4}$。

<Checkpoint>[$\bf{D}$为易爆物，考虑为氮化物](1)</Checkpoint>

<Checkpoint>[$\bf{D}$的最简化学式为$\rm{SeN}$，实际该化合物为$\rm{Se_4N_4}$](1)</Checkpoint>

$\bf{E}$为$\rm{Se_2Cl_2}$的广义等电子体，则$\bf{E}$大概率为$\rm{S_2Cl_2}$或者$\rm{Te_2Cl_2}$或者不常见的$\rm{O_2Cl_2}$；$\bf{E}$也可生成爆炸性物质 $\bf{F}$，含氮量为 $30.40 \%$，计算可得最简化学式为$\rm{SN}$，实际为$\rm{S_4N_4}$；那么$\bf{E}$为$\rm{S_2Cl_2}$。由此可知选项C,D,E均错误。

<Checkpoint>[$\bf{E}$为$\rm{S_2Cl_2}$](1)</Checkpoint>

<Checkpoint>[$\bf{F}$的最简化学式为$\rm{SN}$，实际该化合物为$\rm{S_4N_4}$](1)</Checkpoint>

$\rm{Se_4N_4}$和$\rm{SeCl_4}$可合成$\bf{G}$。其阳离子中只含有一个$\rm{Se}$; 且$\rm{Se}$含量为$36.35 \%$, 氮元素含量为$19.35 \%$; 根据此可算出阳离子中
$\rm{Se}$和$\rm{N}$的个数比为$\rm{78.96/0.3635 : 14.01/0.1935 = 1:3}$，只含有一个$\rm{Se}$，根据合成方式可计算得阳离子化学式为$\rm{SeN_3S_3^{n+}}$。

<Checkpoint>[$\bf{G}$阳离子中 $\rm{Se}$和$\rm{N}$的个数比为$\rm{1:3}$](1)</Checkpoint>

<Checkpoint>[$\bf{G}$阳离子化学式为$\rm{SeN_3S_3}$](1)</Checkpoint>

整个化合物中$\rm{Se}$含量为$32.62 \%$，比阳离子中$\rm{Se}$含量更小，意味着阴离子一定含$\rm{Se}$；假设阴离子含1个$\rm{Se}$，$\bf{G}$含有一个阳离子，n个阴离子，则阴离子的相对分子质量为$\rm{78.96\times(1+n)/0.3262-78.96/0.3635=242n+25}$，减去n个$\rm{Se}$的质量后，当n取0.5时刚好剩余三个氯的相对分子质量；因此很明显阴离子为$\mathrm{SeCl_6^{2-}}$；此时阳离子必须为$\rm{SeN_3S_3^{+}}$才符合质量分数。故$\bf{G}$化学式为$\rm{[SeN_3S_3]_2[SeCl_6]}$，选项F错误，H正确。$\mathrm{SeCl_6^{2-}}$为八面体构型，选项G错误。

<Checkpoint>[$\bf{G}$阴离子为$\mathrm{SeCl_6^{2-}}$，八面体构型](2)</Checkpoint>

<Checkpoint>[$\bf{G}$阳离子为$\rm{SeN_3S_3^{+}}$](1)</Checkpoint>

<Checkpoint>[$\bf{G}$化学式为$\rm{[SeN_3S_3]_2[SeCl_6]}$，为1：2型离子化合物](2)</Checkpoint>