$\bf{N}$'s simple substance is an important semiconductor material and can be used as a dopant for silicon. In nature, one form in which $\bf{N}$ exists is compound $\bf{A}$; $\bf{A}$ is a readily sublimable, sparingly soluble solid containing $60.9 \%$ $\bf{N}$. $\bf{A}$ dissolves in concentrated nitric acid to form $\bf{B}$; passing gas $\bf{C}$ into the acidic solution of $\bf{B}$ yields acid $\bf{D}$ and a yellow precipitate $\bf{E}$; heating $\bf{D}$ to remove water yields oxide $\bf{F}$. Heating $\bf{A}$ in air can also yield oxide $\bf{F}$; heating a mixture of $\bf{F}$ and $\mathrm{S_2 Cl_2}$ in a chlorine atmosphere yields a colorless polar liquid $\bf{G}$ and an irritating gas $\bf{H}$. An aqueous solution of $\bf{H}$ and $\bf{C}$ reacts to form $\bf{E}$ and water. Furthermore, $\bf{N}$ can also form an oxoacid $\bf{Z}$; $1 \mathrm{~mol}$ of $\bf{Z}$ hydrolyzes and consumes $2 \mathrm{~mol}$ of water, yielding only $4 \mathrm{~mol}$ of $\bf{B}$. $\bf{Z}$ is a centrosymmetric molecule with two $C_{2}$ axes passing through the $\bf{N}$ atoms; in the $\bf{Z}$ molecule, $\bf{N}$ has two chemical environments and two coordination numbers.

The sum of the serial numbers of the correct options is:

$1$. Acid $\bf{D}$ is more acidic than $\bf{B}$.

$2$. $\bf{A}$ and $\bf{F}$ have the same number of atoms in their simplest chemical formulas.

$3$. $\bf{C}$ and $\bf{H}$ molecules have different symmetries.

$4$. Molecule $\bf{G}$ contains only one lone pair of electrons.

$5$. $\mathrm{O}$ in $\bf{Z}$ has four chemical environments.

$6$. Ideally, if all hydroxyl groups can be completely ionized, molecule $\bf{Z}$ can ionize a maximum of $6$ hydrogen ions.
$\bf{N}$ 的单质是一种重要的半导体材料,可作为硅的掺杂物。在自然界中, $\bf{N}$ 存在的一种形式是           化合物$\bf{A}$ ; $\bf{A}$  是易升华的难溶性固体,其中含 $\bf{N}$  $60.9 \%$。$\bf{A}$ 溶于浓硝酸生成 $\bf{B}$ ; 向 $\bf{B}$ 的酸性溶液中通入气体 $\bf{C}$,可生成酸 $\bf{D}$ 和黄色沉淀 $\bf{E}$; $\bf{D}$ 加热脱水可得到氧化物 $\bf{F}$。将 $\bf{A}$ 在空气中加热也可得到氧化物 $\bf{F}$; 在氯气气氛下加热 $\bf{F}$ 与 $\mathrm{S_2 Cl_2}$ 的混合物，可以得到无色极性液体 $\bf{G}$ 和刺激性气体 $\bf{H}$, $\bf{H}$ 和 $\bf{C}$ 的水溶液反应生成 $\bf{E}$ 和水。不仅如此，$\bf{N}$ 还可以形成含氧酸 $\bf{Z}$; $1 \mathrm{~mol} \bf{Z}$ 水解消耗 $2 \mathrm{~mol}$ 水,仅得到 $4 \mathrm{~mol} \bf{B}$ 。 $\bf{Z}$ 是中心对称的分子,具有两根穿过 $\bf{N}$ 原子的 $C_{2}$ 轴; $\bf{Z}$ 分子中, $\bf{N}$ 有两种化学环境和两种配位数。

下列选项中，正确选项序号之和为：

$1$.酸 $\bf{D}$ 的酸性比 $\bf{B}$ 强。

$2$. $\bf{A}$ 和 $\bf{F}$ 最简化学式的原子数相同。

$3$. $\bf{C}$ 和 $\bf{H}$ 分子的对称性不同。

$4$. $\bf{G}$ 分子仅含有一对孤对电子。

$5$. $\bf{Z}$ 中$\mathrm{O}$ 有四种化学环境。

$6$. 理想情况下羟基均可完全电离，$\bf{Z}$ 分子最多可电离出 $6$ 个氢离子。
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$\bf{N}$'s simple substance is a semiconductor material and can be used as a dopant for silicon: Common semiconductor doping elements are mainly elements of Group IIIA and Group VA, including boron, phosphorus, arsenic, etc.

Group IIIA elements may exist in nature as acids, oxides, and hydroxides, while Group VA elements often exist in nature as oxides or sulfides.

<Checkpoint>[Determined to be Group IIIA or Group VA element.](1)</Checkpoint>

And because they are all main group elements, their oxidation states in compounds are relatively fixed. Group IIIA elements mainly exist in nature in the $+3$ valence state, and Group VA elements mainly exist in nature in the $+3$ and $+5$ valence states.

According to the question, compound $\bf{A}$ gives the mass fraction and has the property of being easily sublimated, that is, it is mostly covalent. Group IIIA oxides, hydroxides, and oxyacid salts are not easily sublimated, but some halides of aluminum, such as $\mathrm{AlCl_3}$, have sublimability.

For Group VA, certain oxides and sulfides of phosphorus, arsenic, and antimony, as well as white phosphorus, have the property of sublimation. The possible compounds listed above are all binary compounds, which are relatively easy to verify by returning the data.

Using the given data for enumeration and calculation, it can be known that:

$\frac{74.9\times2}{74.9\times2+32.06\times3}=60.9\%$

Therefore, it is known that $\bf{N}$ is $\mathrm{As}$, and the simplest formula of $\bf{A}$ is $\mathrm{As_2S_3}$.

<Checkpoint>[$\bf{N}$ is $\mathrm{As}$, and the simplest formula of $\bf{A}$ is $\mathrm{As_2S_3}$.](2)</Checkpoint>

$\bf{A}$ dissolves in concentrated nitric acid to produce $\bf{B}$. Concentrated nitric acid has oxidizing properties and can oxidize all $\mathrm{As}$ and $\mathrm{S}$ to the highest valence state. Therefore, $\bf{B}$ is the highest valence oxyacid salt of $\mathrm{As}$, namely $\mathrm{H_3AsO_4}$.

<Checkpoint>[$\bf{B}$ is $\mathrm{H_3AsO_4}$.](1)</Checkpoint>

Gas $\bf{C}$ is passed into the acidic solution of $\bf{B}$ to generate acid $\bf{D}$ and yellow precipitate $\bf{E}$.

Among the elements given in the question, $\mathrm{S}$ simple substance and some $\mathrm{S}$ compounds of $\mathrm{As}$ are yellow, and $\bf{B}$ is $\mathrm{H_3AsO_4}$, the highest valence oxyacid of $\mathrm{As}$, so the added gas $\bf{C}$ may have reducing properties.

Among the reducing gases, the oxidation product of $\mathrm{H_2S}$ can be $\mathrm{S}$ simple substance, which is just yellow, and the logic is self-consistent. Suppose $\bf{C}$ is $\mathrm{H_2S}$, and the reaction is:

$$
2 \mathrm{H_3AsO_4}+2 \mathrm{H_2S} \rightarrow 2 \mathrm{H_3AsO_3}+2 \mathrm{S}+2 \mathrm{H_2O}
$$

Consistent with the above assumptions.

And $\bf{D}(\mathrm{H_3AsO_3})$ is heated to dehydrate to generate oxide. The following conditions are met.

<Checkpoint>[Therefore, $\bf{C}$ is $\mathrm{H_2S}$, and $\bf{D}$ is $\mathrm{H_3AsO_3}$.](1)</Checkpoint>

$\mathrm{H_3AsO_3}$ is dehydrated to produce arsenic trioxide:

$2 \mathrm{H_3AsO_3} \rightarrow \mathrm{As_2O_3}+3 \mathrm{H_2O}$

Therefore, $\bf{F}$ is $\mathrm{As_2O_3}$.

<Checkpoint>[$\bf{F}$ is $\mathrm{As_2O_3}$.](1)</Checkpoint>

Similarly, it can be verified that when $\mathrm{As_2S_3}$ is heated in the air, because the oxidation capacity of air is weaker than that of oxygen, the generation of $\mathrm{As_2O_3}$ is a thermodynamically more stable product under this condition.

Heating a mixture of $\bf{F}$ and $\mathrm{S_2 Cl_2}$ in a chlorine atmosphere can obtain a colorless polar liquid $\bf{G}$ and an irritating gas $\bf{H}$. $\mathrm{S_2 Cl_2}$ is mostly used for chlorination to provide chlorine atoms, while a chlorine atmosphere is an oxidation condition, so this step is an oxidative chlorination reaction.

Sulfur is oxidized to sulfur dioxide, which conforms to the description of an irritating gas.

$4 \mathrm{As_2O_3}+9 \mathrm{Cl_2}+3 \mathrm{S_2Cl_2}\to8 \mathrm{AsCl_3}+6 \mathrm{SO_2}$

<Checkpoint>[Therefore, $\bf{G}$ is $\mathrm{AsCl_3}$, and $\bf{H}$ is $\mathrm{SO_2}$](2)</Checkpoint>

$\bf{Z}$ is an oxyacid, 1 mol of $\bf{Z}$ hydrolyzes and consumes 2 mol of water, only yielding 4 mol of $\bf{B}$:

$\bf{B}$ is $\mathrm{H_3AsO_4}$ .

Hydrolysis reaction: $\mathrm{\bf{Z}}+2 \mathrm{H_2O} \rightarrow 4 \mathrm{H_3AsO_4}$ .

Right side: $4 \mathrm{~mol} \mathrm{H_3AsO_4}$ contains $4 \mathrm{~mol} \mathrm{As} 、 12 \mathrm{~mol} \mathrm{H} 、 16 \mathrm{~mol} \mathrm{O}$ .

Left side: $\mathrm{\bf{Z}}+2 \mathrm{~mol} \mathrm{H_2O}$ (containing $4 \mathrm{~mol} \mathrm{H} 、 2 \mathrm{~mol} \mathrm{O}$ ) .

Therefore, Z provides $4 \mathrm{~mol}\mathrm{As}$、$8 \mathrm{~mol}\mathrm{H}$、$14 \mathrm{~mol}\mathrm{O}$ , and the molecular formula is $\mathrm{H_8As_4O_{14}}$ .

The molecular formula is $\mathrm{H_8As_4O_{14}}$ .

<Checkpoint>[The molecular formula of $\bf{Z}$ is $\mathrm{H_8As_4O_{14}}$ .](1)</Checkpoint>

Simple valence calculation shows that the total valence of $\mathrm{As}$ is $+20$, each $\mathrm{As}$ is $+5$ valence, there are no lone pair electrons, As has two chemical environments and two coordination numbers, indicating that one coordination number is 4 and the other coordination number is 6.

Assuming that there is no double-bonded oxygen and all oxygen is dicoordinate, $\mathrm{H_8As_4O_{14}}$ is composed of two tetrahedrally coordinated $\mathrm{As}$ and two octahedrally coordinated $\mathrm{As}$, connected by oxygen bridges. There are a total of six bridging oxygens and eight hydroxyl oxygens.

Centrosymmetric molecule, with two $C_{2}$ axes passing through the $\mathrm{As}$ atoms.

According to the above conditions and the difficulty of ring formation, its structure can be deduced:

<MultiModal>![O[As]1(O)O[As](O2)(O3)(O)(O)O[As](O)(O)O[As]23(O)(O)O1](/media/uploads/c73ed93e-8cf7-4622-9a42-d42fb168aedf.png)</MultiModal>

<Checkpoint>[$\mathrm{H_8As_4O_{14}}$ structure is O[As]1(O)O[As](O2)(O3)(O)(O)O[As](O)(O)O[As]23(O)(O)O1](1)</Checkpoint>

<Checkpoint>[$\mathrm{H_8As_4O_{14}}$ is composed of two tetrahedrally coordinated $\mathrm{As}$ and two octahedrally coordinated $\mathrm{As}$, connected by oxygen bridges. There are a total of six bridging oxygens and eight hydroxyl oxygens.](1)</Checkpoint>

$1$. The acidity of acid $\bf{D}$ is weaker than that of $\bf{B}$. Incorrect.

$2$. The simplest formulas of $\bf{A}$ and $\bf{F}$ both contain 5 atoms, correct.

$3$. The symmetry of molecules $\bf{C}$ and $\bf{H}$ are the same. Both are bent molecules, $C_{2v}$ point group, incorrect.

$4$. The central atom of molecule $\bf{G}$ contains a lone pair of electrons, and each chlorine atom contains three lone pairs of electrons, for a total of 10 pairs. Incorrect.

$5$. $\mathrm{O}$ in $\bf{Z}$ has four chemical environments. Correct.

$6$. Ideally, all hydroxyl groups can be completely ionized, and $\bf{Z}$ molecule can ionize up to $8$ hydrogen ions. Incorrect.

The sum of the correct option numbers is $7$, choose E.
$\bf{N}$的单质是半导体材料，可作为硅的掺杂物：常见的半导体掺杂元素主要为第三主族和第五主族的元素，包括硼、磷、砷等。
而第三主族元素在自然界中可能以酸、氧化物、氢氧化物的形式存在，第五主族元素在自然界中常以氧化物或硫化物的形式存在。

<Checkpoint>[确定为第三或第五主族元素。](1)</Checkpoint>

且由于均为主族元素，他们在化合物中的氧化态相对固定。第三主族的元素主要在自然界中以$+3$价态存在，第五主族的元素在自然界中主要以$+3$以及$+5$价态的形式存在。

由题意，化合物$\bf{A}$给出了质量分数，且具有易升华的性质，即多以共价性为主。第三主族氧化物、氢氧化物、含氧酸盐均不易升华，但如$\mathrm{AlCl_3}$即铝的部分卤化物具有升华性。

而对于第五主族而言，磷、砷、锑的某些氧化物及硫化物、以及白磷单质均具有升华的性质，以上列举的可能化合物均为二元化合物，带回数据验证较为容易。

利用所给数据进行列举计算，可知：

$\frac{74.9\times2}{74.9\times2+32.06\times3}=60.9\%$

故得知，$\bf{N}$为$\mathrm{As}$，$\bf{A}$最简式为$\mathrm{As_2S_3}$。

<Checkpoint>[$\bf{N}$为$\mathrm{As}$，$\bf{A}$最简式为$\mathrm{As_2S_3}$。](2)</Checkpoint>

$\bf{A}$ 溶于浓硝酸生成 $\bf{B}$，浓硝酸具有氧化性，可将$\mathrm{As}$和$\mathrm{S}$全部氧化至最高价，故$\bf{B}$为$\mathrm{As}$的最高价含氧酸盐，即$\mathrm{H_3AsO_4}$。

<Checkpoint>[$\bf{B}$为$\mathrm{H_3AsO_4}$。](1)</Checkpoint>

向 $\bf{B}$ 的酸性溶液中通入气体 $\bf{C}$，生成酸 $\bf{D}$ 和黄色沉淀 $\bf{E}$。

在题目所给元素中，$\mathrm{S}$单质和部分$\mathrm{As}$的$\mathrm{S}$化物为黄色，且$\bf{B}$为$\mathrm{H_3AsO_4}$，$\mathrm{As}$的最高价含氧酸，故加入的气体$\bf{C}$可能具有还原性。

而还原性气体中，$\mathrm{H_2S}$的氧化产物可以为$\mathrm{S}$单质，刚好为黄色，逻辑自洽。不妨假设$\bf{C}$为$\mathrm{H_2S}$，有反应：

$$
2 \mathrm{H_3AsO_4}+2 \mathrm{H_2S} \rightarrow 2 \mathrm{H_3AsO_3}+2 \mathrm{S}+2 \mathrm{H_2O}
$$

符合上述假设。

且 $\bf{D}(\mathrm{H_3AsO_3})$ 加热脱水生成氧化物。下述条件符合。

<Checkpoint>[故$\bf{C}$为$\mathrm{H_2S}$，$\bf{D}$为$\mathrm{H_3AsO_3}$。](1)</Checkpoint>

$\mathrm{H_3AsO_3}$ 脱水生成三氧化二砷:

$2 \mathrm{H_3AsO_3} \rightarrow \mathrm{As_2O_3}+3 \mathrm{H_2O}$ 

故$\bf{F}$为$\mathrm{As_2O_3}$。

<Checkpoint>[$\bf{F}$为$\mathrm{As_2O_3}$。](1)</Checkpoint>

同样可验证，当$\mathrm{As_2S_3}$在空气中加热，由于空气氧化能力弱于氧气，生成$\mathrm{As_2O_3}$在该条件下是热力学更稳定的产物。

在氯气气氛下加热 $\bf{F}$ 与 $\mathrm{S_2 Cl_2}$ 的混合物，可以得到无色极性液体 $\bf{G}$ 和刺激性气体 $\bf{H}$。$\mathrm{S_2 Cl_2}$多用于氯代提供氯原子，而在氯气气氛下则是一个氧化的条件，故此步反应为氧化氯代。

硫被氧化成二氧化硫，符合刺激性气体的描述。

$4 \mathrm{As_2O_3}+9 \mathrm{Cl_2}+3 \mathrm{S_2Cl_2}\to8 \mathrm{AsCl_3}+6 \mathrm{SO_2}$

<Checkpoint>[故$\bf{G}$为$\mathrm{AsCl_3}$，$\bf{H}$为$\mathrm{SO_2}$](2)</Checkpoint>

$\bf{Z}$ 是含氧酸, 1 mol $\bf{Z}$ 水解消耗 2 mol 水, 仅得到 4 mol $\bf{B}$:

$\bf{B}$ 是 $\mathrm{H_3AsO_4}$ 。

水解反应: $\mathrm{\bf{Z}}+2 \mathrm{H_2O} \rightarrow 4 \mathrm{H_3AsO_4}$ 。

右侧: $4 \mathrm{~mol} \mathrm{H_3AsO_4}$ 含 $4 \mathrm{~mol} \mathrm{As} 、 12 \mathrm{~mol} \mathrm{H} 、 16 \mathrm{~mol} \mathrm{O}$ 。

左侧: $\mathrm{\bf{Z}}+2 \mathrm{~mol} \mathrm{H_2O}$ (含 $4 \mathrm{~mol} \mathrm{H} 、 2 \mathrm{~mol} \mathrm{O}$ ) 。

因此,，Z提供$4 \mathrm{~mol}\mathrm{As}$、$8 \mathrm{~mol}\mathrm{H}$、$14 \mathrm{~mol}\mathrm{O}$ ，分子式为 $\mathrm{H_8As_4O_{14}}$ 。

分子式为 $\mathrm{H_8As_4O_{14}}$ 。

<Checkpoint>[$\bf{Z}$的分子式为 $\mathrm{H_8As_4O_{14}}$ 。](1)</Checkpoint>

简单化合价计算得到，$\mathrm{As}$的总化合价为$+20$，每个$\mathrm{As}$为$+5$价，无孤对电子，As 有两种化学环境和两种配位数，说明一种配位数 4,一种配位数 6。

假设无双键氧，所有氧均为二配位，$\mathrm{H_8As_4O_{14}}$ 由两个四面体配位的 $\mathrm{As}$ 和两个八面体配位的 $\mathrm{As}$ 组成,通过氧桥连接。共六个桥连氧，八个羟基氧。

中心对称分子,有两根穿过 $\mathrm{As}$ 原子的 $C_{2}$ 轴。

根据上述条件以及成环难度，可推其结构：

<MultiModal>![O[As]1(O)O[As](O2)(O3)(O)(O)O[As](O)(O)O[As]23(O)(O)O1](/media/uploads/c73ed93e-8cf7-4622-9a42-d42fb168aedf.png)</MultiModal>

<Checkpoint>[$\mathrm{H_8As_4O_{14}}$ 结构为O[As]1(O)O[As](O2)(O3)(O)(O)O[As](O)(O)O[As]23(O)(O)O1](1)</Checkpoint>

<Checkpoint>[$\mathrm{H_8As_4O_{14}}$ 由两个四面体配位的 $\mathrm{As}$ 和两个八面体配位的 $\mathrm{As}$ 组成,通过氧桥连接。共六个桥连氧，八个羟基氧。](1)</Checkpoint>

$1$.酸 $\bf{D}$ 的酸性比 $\bf{B}$ 弱。错误。

$2$. $\bf{A}$ 和 $\bf{F}$ 分子最简式均含有5个原子，正确。

$3$. $\bf{C}$ 和 $\bf{H}$ 分子的对称性相同，二者均为折线形分子，$C_{2v}$点群，错误。

$4$. $\bf{G}$ 分子中心原子含有一对孤对电子，每个氯原子含有三对孤对电子，共10对。错误。

$5$. $\bf{Z}$ 中$\mathrm{O}$ 有四种化学环境。正确。

$6$. 理想情况下羟基均可完全电离，$\bf{Z}$ 分子最多可电离出 $8$ 个氢离子。错误。

正确选项序号和为$7$，选择E。