Sodium salt $\mathrm{Na_2\mathbf{X}}$ contains a unique cyclic anion, in which each atom in the anion has the same chemical environment. Sodium salt $\mathrm{Na_2\mathbf{X}}$ is often used to test metal cations such as $\mathrm{Ba}^{2+} 、 \mathrm{Pb}^{2+}$, and even when in contact with the surface of certain insoluble lead salts (such as PbS), it can seize Pb²⁺ from them and form a characteristic purplish-red substance. $\mathrm{Na_2\mathbf{X}}$ is unstable to both light and heat and is very sensitive to oxidizing environments. Under acidic conditions, $\mathrm{I_2}$ reacts with $\mathrm{Na_2\mathbf{X}}$ in equal proportions, yielding only $\mathrm{NaI}$ and gas $\mathbf{M} ; \mathrm{KMnO_4}$ reacts with $\mathrm{Na_2\mathbf{X}}$, yielding gas $\mathbf{N}$. If the gas $\mathbf{M}$ obtained from the reaction of $1.498 \mathrm{~g} ~\mathrm{Na_2} \mathbf{X}$ with $\mathrm{I_2}$ is passed through sufficient $\mathrm{I_2O_5}, \mathrm{I_2O_5}$ decreases in mass by $0.672 \mathrm{~g}$. A weakly acidic $\mathrm{PbAc_2}$ solution is divided into two portions, one with $\mathrm{Na_2\mathbf{X}}$, forming a basic salt precipitate $\mathbf{A}$ containing one water of crystallization (in which the stoichiometric ratio of each ion contained is 2 : 1 : 2), precipitate $\mathbf{A}$ is scarlet; the other
is added with sufficient $\mathrm{Na_2CO_3}$ to obtain precipitate $\mathbf{B}$, heating precipitate $\mathbf{B}$ in aqueous solution yields precipitate $\mathrm{Pb_3(OH)_2(CO_3)_2}$ with the same elemental composition as $\mathbf{A}$.

Please select the option below that best fits the meaning of the question:
钠盐 $\mathrm{Na_2\mathbf{X}}$ 中含有独特的环状阴离子, 其阴离子中每种原子的化学环境相同。钠盐 $\mathrm{Na_2\mathbf{X}}$ 常用于检验 $\mathrm{Ba}^{2+} 、 \mathrm{Pb}^{2+}$ 等金属阳离子, 甚至在接触某些难溶铅盐（如 PbS）表面时，可夺取其中的 Pb²⁺ 并与之生成特征性紫红色物质。$\mathrm{Na_2\mathbf{X}}$ 对光热均不稳定, 对氧化性环境也十分敏感。在酸性条件下 $\mathrm{I_2}$ 与 $\mathrm{Na_2\mathbf{X}}$
等比例反应, 只得到 $\mathrm{NaI}$ 和气体 $\mathbf{M} ; \mathrm{KMnO_4}$ 与 $\mathrm{Na_2\mathbf{X}}$ 反应, 得到气体 $\mathbf{N}$ 。若将 $1.498 \mathrm{~g} ~\mathrm{Na_2} \mathbf{X}$ 与 $\mathrm{I_2}$ 反应得到的气体 $\mathbf{M}$ 通过足量 $\mathrm{I_2O_5}, \mathrm{I_2O_5}$ 的质量减少 $0.672 \mathrm{~g}$ 。将呈弱酸性的 $\mathrm{PbAc_2}$ 溶液分为两份, 一份加入 $\mathrm{Na_2\mathbf{X}}$, 形成含一个结晶水的碱式盐沉淀 $\mathbf{A}$（其中所含各离子的化学计量比为2 : 1 : 2）, 沉淀 $\mathbf{A}$ 呈猩红色; 另一份
加入足量 $\mathrm{Na_2CO_3}$ 得到沉淀 $\mathbf{B}$, 在水溶液中加热沉淀 $\mathbf{B}$ 得到与 $\mathbf{A}$ 组成元素相同的沉淀 $\mathrm{Pb_3(OH)_2(CO_3)_2}$。

请选择下列选项中最符合题意的一项：
All other options are incorrect
$\mathbf{A}$ The chemical formula after removing water of crystallization contains 16 atoms.
$\mathbf{X^{2-}}$ ion does not possess aromaticity.
The chemical equation for the reaction of $\mathrm{Na_2\mathbf{X}}$ with $\mathrm{I_2}$, when balanced with the simplest integer ratio, has a sum of product coefficients equal to 6.
The ion equation for the reaction of $\mathrm{Na_2\mathbf{X}}$ with acidic $\mathrm{KMnO_4}$, when balanced to the simplest integer ratio, has a sum of product coefficients equal to 60.
Ions $\mathbf{Y^{2-}}$ that are structurally similar to $\mathbf{X^{2-}}$ and have a relative molecular mass 28.01 smaller exhibit extremely low stability.
Multiple options are correct among options B-F.
其他选项均不正确
$\mathbf{A}$去掉结晶水后的化学式含有16个原子
$\mathbf{X^{2-}}$离子不具有芳香性
$\mathrm{Na_2\mathbf{X}}$ 与$\mathrm{I_2}$反应的化学方程式，配平为最简整数比时，产物系数和为6。
$\mathrm{Na_2\mathbf{X}}$ 与酸性$\mathrm{KMnO_4}$反应的离子方程式，配平为最简整数比时，产物系数和为60。
与$\mathbf{X^{2-}}$结构相似且相对分子质量小28.01的$\mathbf{Y^{2-}}$离子稳定性极低。
选项B-F中有多项正确
The starting point of this question is to use $\mathrm{I_2O_5}$ to test the gas $\mathbf{M}$. According to chemical knowledge, the gas $\mathbf{M}$ is $\mathrm{CO}$, and the reaction is quantitative:

<Checkpoint>[$\mathbf{M}$ is $\mathrm{CO}$](1)</Checkpoint>

$\mathrm{I_2O_5 + 5CO = I_2 + 5CO_2}$

<Checkpoint>[$\mathrm{I_2O_5 + 5CO = I_2 + 5CO_2}$](1)</Checkpoint>

According to the mass given in the question, the amount of substance of $\mathrm{CO}$ produced is

$\mathrm{0.672/16\times5=0.042 ~mol}$

$\mathrm{I_2}$ reacts with $\mathrm{Na_2\mathbf{X}}$ in equal proportions to obtain $\mathrm{NaI}$ and $\mathrm{CO}$, indicating that the cyclic anion $\mathbf{X^{2-}}$ is composed only of carbon and oxygen elements;

<Checkpoint>[Cyclic anion $\mathbf{X^{2-}}$ is composed only of carbon and oxygen elements;](1)</Checkpoint>

$1.498 \mathrm{~g} ~\mathrm{Na_2} \mathbf{X}$ can produce $\mathrm{0.042 ~mol}$ of $\mathrm{CO}$, that is, the content of sodium element in $\mathrm{Na_2} \mathbf{X}$ is $\mathrm{1.498 - 0.042\times28.01=0.322 ~g = 0.014 ~mol}$, thus the relative molecular mass of $\mathbf{X^{2-}}$ is calculated as $\mathrm{(1.498-0.322)/(0.014/2)=168}$, which is six $\mathrm{CO}$, so the chemical formula of $\mathbf{X^{2-}}$ is
$\mathrm{C_6O_6^{2-}}$.

<Checkpoint>[The relative molecular mass of $\mathbf{X^{2-}}$ is 168](1)</Checkpoint>

<Checkpoint>[The chemical formula of $\mathbf{X^{2-}}$ is $\mathrm{C_6O_6^{2-}}$](1)</Checkpoint>

Since $\mathbf{X^{2-}}$ is a cyclic anion, it can be deduced that it contains a six-membered carbon ring, composed of six carbonyl groups; the structure is O=C(C([O-])=C([O-])C(C1=O)=O)C1=O, the six carbons/oxygens have the same chemical environment due to tautomerism, which is consistent with the description of the question.

<Checkpoint>[The structure of $\mathbf{X^{2-}}$ is O=C(C([O-])=C([O-])C(C1=O)=O)C1=O](1)</Checkpoint>

<MultiModal>![The structure of $\mathbf{X^{2-}}$: O=C(C([O-])=C([O-])C(C1=O)=O)C1=O](/media/uploads/eda40c24-ff14-43a1-b96d-289338845ba4.png)</MultiModal>

In the structure of O=C(C([O-])=C([O-])C(C1=O)=O)C1=O, there is a pair of electrons delocalized on the six-membered ring, which conforms to the 4n + 2 rule (n = 0) and has certain aromaticity, so option C is incorrect.

<Checkpoint>[$\mathbf{X^{2-}}$ has a pair of electrons delocalized on the six-membered ring, and has certain aromaticity](1)</Checkpoint>

Therefore, it is easy to write the reaction equations of $\mathrm{C_6O_6^{2-}}$ with iodine and acidic potassium permanganate:

$\mathrm{Na_2C_6O_6 + I_2 = 2NaI + 6CO}$

<Checkpoint>[$\mathrm{Na_2C_6O_6 + I_2 = 2NaI + 6CO}$](1)</Checkpoint>

$\mathrm{5C_6O_6^{2-} + 14MnO_4^- + 52H^+ = 14Mn^{2+} + 26H_2O + 30CO_2}$

<Checkpoint>[$\mathrm{5C_6O_6^{2-} + 14MnO_4^- + 52H^+ = 14Mn^{2+} + 26H_2O + 30CO_2}$](1)</Checkpoint>

According to the equation, options D and E are both incorrect.

The ion $\mathbf{Y^{2-}}$ with a relative molecular mass of 28.01 less than that of $\mathrm{C_6O_6^{2-}}$ is obviously $\mathrm{C_5O_5^{2-}}$, which becomes a five-membered ring [O-]C(C(C1=O)=O)=C([O-])C1=O, but it also has a pair of electrons delocalized on the five-membered ring and has certain aromaticity, so the stability is not extremely low, and option F is incorrect.

<Checkpoint>[$\mathbf{Y^{2-}}$ is $\mathrm{C_5O_5^{2-}}$, and the structure is [O-]C(C(C1=O)=O)=C([O-])C1=O](2)</Checkpoint>

<Checkpoint>[$\mathbf{Y^{2-}}$ has a pair of electrons delocalized on the five-membered ring, and has certain aromaticity](1)</Checkpoint>

<MultiModal>![The structure of $\mathbf{Y^{2-}}$: [O-]C(C(C1=O)=O)=C([O-])C1=O](/media/uploads/5f4992ec-91d9-43d7-90e7-faff50787784.png)</MultiModal>

$\mathbf{A}$ and $\mathrm{Pb_3(OH)_2(CO_3)_2}$ have the same constituent elements, so they both contain carbon, hydrogen, and oxygen elements as negative ions; $\mathbf{A}$ contains $\mathrm{C_6O_6^{2-}}$, since it is a basic salt, considering the +2 valence of lead, and combining the stoichiometric ratio of each ion, the chemical formula of $\mathbf{A}$ can only be $\mathrm{Pb_2[C_6O_6](OH)_2\cdot H_2O}$, which contains 18 atoms after removing the crystal water, so option B is incorrect.

<Checkpoint>[The chemical formula of $\mathbf{A}$ is $\mathrm{Pb_2[C_6O_6](OH)_2\cdot H_2O}$](1)</Checkpoint>

In summary, options B-F are all incorrect, and option A is correct.
本题目入手点在利用$\mathrm{I_2O_5}$检验气体$\mathbf{M}$，根据化学常识可知气体$\mathbf{M}$为$\mathrm{CO}$，且反应是定量的：

<Checkpoint>[$\mathbf{M}$为$\mathrm{CO}$](1)</Checkpoint>

$\mathrm{I_2O_5 + 5CO = I_2 + 5CO_2}$

<Checkpoint>[$\mathrm{I_2O_5 + 5CO = I_2 + 5CO_2}$](1)</Checkpoint>

根据题目给出的质量可得，产生的$\mathrm{CO}$物质的量为

$\mathrm{0.672/16\times5=0.042 ~mol}$

$\mathrm{I_2}$ 与 $\mathrm{Na_2\mathbf{X}}$
等比例反应, 得到 $\mathrm{NaI}$ 和 $\mathrm{CO}$，说明环状阴离子$\mathbf{X^{2-}}$只由碳氧元素构成；

<Checkpoint>[环状阴离子$\mathbf{X^{2-}}$只由碳氧元素构成；](1)</Checkpoint>

$1.498 \mathrm{~g} ~\mathrm{Na_2} \mathbf{X}$可产生$\mathrm{0.042 ~mol}$的$\mathrm{CO}$, 也即$\mathrm{Na_2} \mathbf{X}$中钠元素的含量为$\mathrm{1.498 - 0.042\times28.01=0.322 ~g = 0.014 ~mol}$，从而算出$\mathbf{X^{2-}}$的相对分子质量为$\mathrm{(1.498-0.322)/(0.014/2)=168}$，为六个$\mathrm{CO}$，从而$\mathbf{X^{2-}}$的化学式为
$\mathrm{C_6O_6^{2-}}$。

<Checkpoint>[$\mathbf{X^{2-}}$的相对分子质量为168](1)</Checkpoint>

<Checkpoint>[$\mathbf{X^{2-}}$的化学式为 $\mathrm{C_6O_6^{2-}}$](1)</Checkpoint>

由于$\mathbf{X^{2-}}$为环状阴离子，可以推出其含有碳六元环，由六个羰基组成；结构为O=C(C([O-])=C([O-])C(C1=O)=O)C1=O，六个碳/氧因为互变异构化学环境完全一致，符合题目描述。

<Checkpoint>[$\mathbf{X^{2-}}$结构为O=C(C([O-])=C([O-])C(C1=O)=O)C1=O](1)</Checkpoint>

<MultiModal>![$\mathbf{X^{2-}}$的结构：O=C(C([O-])=C([O-])C(C1=O)=O)C1=O](/media/uploads/eda40c24-ff14-43a1-b96d-289338845ba4.png)</MultiModal>

O=C(C([O-])=C([O-])C(C1=O)=O)C1=O的结构中有一对电子在六元环上离域，符合4n + 2规则(n = 0)，具有一定的芳香性，选项C错误。

<Checkpoint>[$\mathbf{X^{2-}}$有一对电子在六元环上离域，具有一定的芳香性](1)</Checkpoint>

从而很容易写出$\mathrm{C_6O_6^{2-}}$与碘和酸性高锰酸钾的反应方程式：

$\mathrm{Na_2C_6O_6 + I_2 = 2NaI + 6CO}$

<Checkpoint>[$\mathrm{Na_2C_6O_6 + I_2 = 2NaI + 6CO}$](1)</Checkpoint>

$\mathrm{5C_6O_6^{2-} + 14MnO_4^- + 52H^+ = 14Mn^{2+} + 26H_2O + 30CO_2}$

<Checkpoint>[$\mathrm{5C_6O_6^{2-} + 14MnO_4^- + 52H^+ = 14Mn^{2+} + 26H_2O + 30CO_2}$](1)</Checkpoint>

根据方程式，选项D,E均错误。

比$\mathrm{C_6O_6^{2-}}$少28.01相对分子量的离子$\mathbf{Y^{2-}}$很明显为$\mathrm{C_5O_5^{2-}}$，变为五元环[O-]C(C(C1=O)=O)=C([O-])C1=O，但同样有一对电子在五元环上离域，具有一定芳香性，稳定性并不极低，选项F错误。

<Checkpoint>[$\mathbf{Y^{2-}}$为$\mathrm{C_5O_5^{2-}}$，结构为[O-]C(C(C1=O)=O)=C([O-])C1=O](2)</Checkpoint>

<Checkpoint>[$\mathbf{Y^{2-}}$有一对电子在五元环上离域，具有一定芳香性](1)</Checkpoint>

<MultiModal>![$\mathbf{Y^{2-}}$的结构：[O-]C(C(C1=O)=O)=C([O-])C1=O](/media/uploads/5f4992ec-91d9-43d7-90e7-faff50787784.png)</MultiModal>

$\mathbf{A}$与$\mathrm{Pb_3(OH)_2(CO_3)_2}$的组成元素相同，则均含有碳氢氧元素作为负离子；$\mathbf{A}$中含有$\mathrm{C_6O_6^{2-}}$，由于为碱式盐，考虑铅的+2化合价，再结合各离子的计量比可得$\mathbf{A}$的化学式只能为$\mathrm{Pb_2[C_6O_6](OH)_2\cdot H_2O}$，去掉结晶水后含18个原子，选项B错误。

<Checkpoint>[$\mathbf{A}$的化学式为$\mathrm{Pb_2[C_6O_6](OH)_2\cdot H_2O}$](1)</Checkpoint>

综上，选项B-F均不正确，选项A正确。