$\bf M$ is a metallic element, $\bf A$ is a 1:1 type salt containing $\bf M$, with a molar mass of $\rm 223.0~g/mol$. When $\rm BF_3$ is passed into an anhydrous hydrogen fluoride solution of $\bf A$, an diamagnetic red solid $\bf B$ is obtained, and a $\rm 1:1$ type salt $\bf C$ precipitates simultaneously. $\rm 0.1227~g$ of $\bf A$ is dissolved in anhydrous hydrogen fluoride, and allowed to react fully under a $\rm 2 ~atm$ atmosphere of $\rm Xe$, yielding $\rm 0.1862~g$ of solid, the components of which are a mixture of white solid $\bf D$, dark brown paramagnetic solid $\bf E$, and $\rm 2KF\cdot 5HF$. $\bf B$ can also react directly with $\rm Xe$ gas to obtain $\bf D$ and $\bf E$. $\bf E$ has extremely strong oxidizing power, for example, when $\bf E$ is added to $\rm NaOH$ solution, it decomposes violently, producing a dark brown precipitate $\bf F$ and releasing gas; $\bf E$ reacts with chloroform to produce $\bf G$ (also known as Freon-11) and yellow solid $\bf H$, etc.; $\bf H$ is easily soluble in water $(\rm 1800~g/L)$. When $\bf H$ is dissolved in anhydrous hydrogen fluoride, and colloidal $\bf M$ element is added, a brown solid $\bf I$ precipitates, $\bf I$ has a cadmium iodide inverse structure; $\bf I$ undergoes disproportionation upon contact with water, yielding $\bf M$ element and solid $\bf H$ again.

It is known that $\bf B$ exhibits a chain structure.

The following statements are correct:
$\bf M$为金属元素, $\bf A$为含$\bf M$的1:1型盐, 摩尔质量$\rm 223.0~g/mol$。向$\bf A$的无水氟化氢溶液中通入$\rm BF_3$, 得到抗磁性红色固体$\bf B$, 同时析出$\rm 1:1$型盐$\bf C$。将$\rm 0.1227~g \bf A$溶于无水氟化氢中, 置于$\rm 2 ~atm$的$\rm Xe$气氛下充分反应, 得到$\rm 0.1862~g$固体, 其成分为白色固体$\bf D$、暗棕色顺磁性固体$\bf E$以及$\rm 2KF\cdot 5HF$的混合物。$\bf B$与$\rm Xe$气直接反应也可得到$\bf D$和$\bf E$。$\bf E$具有极强的氧化性, 例如将$\bf E$投入$\rm NaOH$溶液中时剧烈分解,生成暗棕色沉淀$\bf F$并放出气体; $\bf E$与氯仿反应, 生成$\bf G$(别称氟利昂-11)和黄色固体$\bf H$等产物; $\bf H$易溶于水$(\rm 1800~g/L)$。将$\bf H$溶于无水氟化氢中, 加入胶体状$\bf M$单质, 析出棕色固体$\bf I$, $\bf I$具有反式碘化镉结构; $\bf I$遇水发生歧化, 又得回$\bf M$单质和固体$\bf H$。

已知$\bf B$呈现链状结构。

下列说法正确的是：
All other options are incorrect
$\bf M$ The ground state has two single electrons.
The constituent elements of $\bf A$ include two short-period elements.
The number of binary compounds in $\bf A-I$ is greater than 6.
$\bf A-I$ contains a compound composed of four elements.
$\bf A-I$ all contain the same short-period element.
The chemical equation for the decomposition of $\bf E$ in $\rm NaOH$, balanced with the simplest coefficient ratio, has a greatest common divisor of 3 between the sum of the product coefficients and the sum of the reactant coefficients.
The $\bf d_{xy}$ orbital of the $\rm M$ element in $\bf B$ is in a half-filled state.
The $\rm d_{x^2+y^2}$ orbital of element $\bf M$ in $\bf B$ is fully occupied.
其他选项均不正确
$\bf M$基态有两个单电子
$\bf A$的组成元素中有两种短周期元素
$\bf A-I$中二元化合物的数量大于6个
$\bf A-I$中存在由四种元素组成的化合物
$\bf A-I$中均含有同一种短周期元素
$\bf E$在$\rm NaOH$中分解的化学方程式，以最简系数比配平后，产物的系数和和反应物的系数和的最大公约数为3
$\bf B$中$\bf M$元素的$\rm d_{xy}$轨道为半充满状态
$\bf B$中$\bf M$元素的$\rm d_{x^2+y^2}$轨道为全充满状态
The starting point of this question is that $\bf H$ is easily soluble in water $(\rm 1800~g/L)$. According to element knowledge, it can be immediately concluded that $\bf H$ is $\rm AgF$.

<Checkpoint>[Solubility information indicates that $\bf H$ is $\rm AgF$](1)</Checkpoint>

Therefore, $\bf M$ is $\rm Ag$.

<Checkpoint>[$\bf M$ is $\rm Ag$](1)</Checkpoint>

$\bf G$ is also known as Freon-11, then $\bf G$ is $\rm CFCl_3$, and $\bf E$ contains $\rm F$;

<Checkpoint>[$\bf G$ is $\rm CFCl_3$](1)</Checkpoint>

$\bf A$ is soluble in anhydrous hydrogen fluoride and reacts with $\rm Xe$ to produce $\rm 2KF\cdot 5HF$, then $\bf A$ contains $\rm K$ and $\rm Ag$, and most likely contains $\rm F$. Since it is a $\rm 1：1$ type salt, the remaining relative molecular mass is $\rm 223.0-107.9-39.1=76.0=19.0\times4$, which is four fluorine atoms. Thus, the chemical formula of $\bf A$ is $\rm KAgF_4$.

<Checkpoint>[The chemical formula of $\bf A$ is $\rm KAgF_4$](1)</Checkpoint>

$\bf B$ is diamagnetic, then $\bf B$ contains $\rm d^8$ configuration of $\rm Ag^{3+}$; therefore, $\bf B$ is $\rm AgF_3$. The 1:1 type salt $\bf C$ must contain $\rm K$. Since $\rm BF_3$ is added, $\bf C$ is $\rm KBF_4$.

<Checkpoint>[$\bf B$ is $\rm AgF_3$](1)</Checkpoint>

<Checkpoint>[$\bf C$ is $\rm KBF_4$](1)</Checkpoint>

$\bf A$ is soluble in anhydrous hydrogen fluoride and reacts with $\rm Xe$. Since $\rm Ag(III)$ has strong oxidizing properties, $\rm KAgF_4$ is a strong fluorinating agent and can oxidize $\rm Xe$. The white solid $\bf D$ should be $\rm XeF_2$. $\rm Ag(III)$ is reduced to a paramagnetic species, which can only be $\rm d^9$ configuration of $\rm Ag^{2+}$, so $\bf E$ is $\rm AgF_2$.

<Checkpoint>[$\bf D$ is $\rm XeF_2$](1)</Checkpoint>

<Checkpoint>[Paramagnetic species, can only be $\rm d^9$ configuration of $\rm Ag^{2+}$](1)</Checkpoint>

<Checkpoint>[$\bf E$ is $\rm AgF_2$](1)</Checkpoint>

$\rm AgF_2$ decomposes in $\rm NaOH$ solution. $\rm Ag(II)$ is also a highly unstable valence state, so it can oxidize water to produce oxygen, and itself is reduced. The dark brown precipitate $\bf F$ is obviously $\rm Ag_2O$.

<Checkpoint>[$\bf F$ is $\rm Ag_2O$](1)</Checkpoint>

$\rm AgF$ reacts with $\rm Ag$, which can only be a comproportionation reaction. Associating with the structure of trans-$\rm CdI_2$, the generated $\bf I$ should be $\rm Ag_2F$.

<Checkpoint>[$\bf I$ is $\rm Ag_2F$](1)</Checkpoint>

Thus, determine the options:

The ground state configuration of $\rm Ag$ is $\rm 4d^{10}5s^1$, with one single electron, so B is incorrect.

$\rm KAgF_4$ only has one short-period element, $\rm F$, so C is incorrect.

According to the chemical formula, options D, E, and F are all incorrect.

$\rm AgF_2$ decomposes to produce $\rm Ag_2O$ and oxygen, and the equation can be written as:

$\rm 4AgF_2 + 8NaOH = 8NaF + 2Ag_2O + O_2 + 4H_2O$

<Checkpoint>[$\rm 4AgF_2 + 8NaOH = 8NaF + 2Ag_2O + O_2 + 4H_2O$](1)</Checkpoint>

The sum of the product coefficients is 15, and the sum of the reactant coefficients is 12. The greatest common divisor is 3, so option G is correct.

$\rm AgF_3$ has a chain structure. It can be seen that $\rm Ag(III)$ in $\rm AgF_3$ is coordinated by four $\rm F$ in a square planar configuration, and the $\rm AgF_4$ units are connected by sharing $\rm F$. $\rm Ag(III)$ is now in a square planar field. Under the $\rm d^8$ configuration, $\rm d_{x^2+y^2}$ has no electrons, and $\rm d_{xy}$ is fully filled, so options H and I are incorrect.

<Checkpoint>[$\rm Ag(III)$ in $\rm AgF_3$ is coordinated by four $\rm F$ in a square planar configuration](1)</Checkpoint>

<Checkpoint>[In a square planar field, under the $\rm d^8$ configuration, $\rm d_{x^2+y^2}$ has no electrons, and $\rm d_{xy}$ is fully filled](1)</Checkpoint>
本题入手点在$\bf H$易溶于水$(\rm 1800~g/L)$，根据元素知识可立即得出$\bf H$为$\rm AgF$。

<Checkpoint>[溶解度信息可知$\bf H$为$\rm AgF$](1)</Checkpoint>

因此，$\bf M$为$\rm Ag$。

<Checkpoint>[$\bf M$为$\rm Ag$](1)</Checkpoint>

$\bf G$别称氟利昂-11，则$\bf G$为$\rm CFCl_3$，且$\bf E$含有$\rm F$；

<Checkpoint>[$\bf G$为$\rm CFCl_3$](1)</Checkpoint>

$\bf A$溶于无水氟化氢与$\rm Xe$反应可产生$\rm 2KF\cdot 5HF$，那么$\bf A$含有$\rm K$和$\rm Ag$，大概率含有$\rm F$。由于为$\rm 1：1$型盐，剩余的相对分子质量为$\rm 223.0-107.9-39.1=76.0=19.0\times4$，为四个氟原子。从而$\bf A$化学式为$\rm KAgF_4$。

<Checkpoint>[$\bf A$化学式为$\rm KAgF_4$](1)</Checkpoint>

$\bf B$为抗磁性，则$\bf B$含有$\rm d^8$构型的$\rm Ag^{3+}$；因此$\bf B$为$\rm AgF_3$。1:1型盐$\bf C$一定含$\rm K$，由加入了$\rm BF_3$可知$\bf C$为$\rm KBF_4$。

<Checkpoint>[$\bf B$为$\rm AgF_3$](1)</Checkpoint>

<Checkpoint>[$\bf C$为$\rm KBF_4$](1)</Checkpoint>

$\bf A$溶于无水氟化氢与$\rm Xe$反应，由于$\rm Ag(III)$具有强氧化性，$\rm KAgF_4$为强氟化剂，可以氧化$\rm Xe$，产生的白色固体$\bf D$应当为$\rm XeF_2$。$\rm Ag(III)$被还原为顺磁性物种，只能为$\rm d^9$构型的$\rm Ag^{2+}$，因此$\bf E$为$\rm AgF_2$。

<Checkpoint>[$\bf D$为$\rm XeF_2$](1)</Checkpoint>

<Checkpoint>[顺磁性物种，只能为$\rm d^9$构型的$\rm Ag^{2+}$](1)</Checkpoint>

<Checkpoint>[$\bf E$为$\rm AgF_2$](1)</Checkpoint>

$\rm AgF_2$在$\rm NaOH$溶液中分解，$\rm Ag(II)$同样为高不稳定价态，因此可氧化水生成氧气，自身被还原，暗棕色沉淀$\bf F$很明显为$\rm Ag_2O$。

<Checkpoint>[$\bf F$为$\rm Ag_2O$](1)</Checkpoint>

$\rm AgF$与$\rm Ag$反应，只能为归中反应，联想到反式$\rm CdI_2$的结构，生成的$\bf I$应当为$\rm Ag_2F$。

<Checkpoint>[$\bf I$为$\rm Ag_2F$](1)</Checkpoint>

从而判断选项：

$\rm Ag$基态构型为$\rm 4d^{10}5s^1$，具有一个单电子，B错误。

$\rm KAgF_4$只有$\rm F$一种短周期元素，C错误。

根据化学式可知选项D,E,F均错误。

$\rm AgF_2$分解生成$\rm Ag_2O$与氧气，可写出方程式：

$\rm 4AgF_2 + 8NaOH = 8NaF + 2Ag_2O + O_2 + 4H_2O$

<Checkpoint>[$\rm 4AgF_2 + 8NaOH = 8NaF + 2Ag_2O + O_2 + 4H_2O$](1)</Checkpoint>

产物系数和为15，反应物系数和为12，最大公约数为3，选项G正确。

$\rm AgF_3$为链状结构，可知$\rm AgF_3$中$\rm Ag(III)$被四个$\rm F$以平面四方型配位，$\rm AgF_4$单元共用$\rm F$进行连接。$\rm Ag(III)$此时为平面四方场，$\rm d^8$组态下$\rm d_{x^2+y^2}$无电子，$\rm d_{xy}$全充满，选项H,I错误。

<Checkpoint>[$\rm AgF_3$中$\rm Ag(III)$被四个$\rm F$以平面四方型配位](1)</Checkpoint>

<Checkpoint>[平面四方场$\rm d^8$组态下$\rm d_{x^2+y^2}$无电子，$\rm d_{xy}$全充满](1)</Checkpoint>