A certain short-period element $\bf{X}$ is often used in nuclear power reactors. A certain halide of it reacts fully in $\rm{N_2O_4}$ to obtain a white crystal $\bf{A}$.

Now, thermogravimetric analysis is performed on $\bf{A}$. It is known that there are two weight loss peaks. First, it loses $\rm{58.04\%}$ of its weight to obtain a white substance $\bf{B}$. Continuing heating, it loses another $\rm{8.52\%}$ of its weight to obtain a substance $\bf{C}$ with a special structure. $\bf{C}$ contains 4 $\rm{C_3}$ axes, all $\bf{X}$ chemical environments are the same, and the coordination number is 4.

Coincidentally, people have also discovered substances with structures similar to $\bf{C}$ in many transition metals. In a similar particle $\bf{M}$ composed of a certain metal from the fourth period, the ligand contains acetate, and the mass fraction of the metal is $\rm{26.88\%}$. All metal atoms chemical environments are the the same, and all are 6-coordinated. It is known that this particle $\bf{M}$ can be prepared in aqueous solution.

The following statements are correct:
某种短周期元素$\bf{X}$常应用于核动力反应堆中。其某种卤化物在$\rm{N_2O_4}$中充分反应,得到一种白色晶体$\bf{A}$。

现在对$\bf{A}$进行热重分析,已知其有两个失重峰,首先失重$\rm{58.04\%}$得到白色物质$\bf{B}$,继续加热,再失重$\rm{8.52\%}$得到一种具有特殊结构的物质$\bf{C}$。$\bf{C}$中含有4个$\rm{C_3}$轴,所有$\bf{X}$化学环境相同,且配位数均为4。

无独有偶,人们在很多过渡金属中也发现了具有类似$\bf{C}$的结构的物质。对于第四周期的某金属组成的类似粒子$\bf{M}$中 ,配体含有乙酸根，该金属的质量分数为$\rm{26.88\%}$, 所有金属原子化学环境相同, 且均为6配位。已知该粒子$\bf{M}$可以在水溶液中制备。

下列说法正确的是：
All other options are incorrect
$\bf{X}$ atom's ground state contains two unpaired electrons.
The number of atoms of $\bf{A}$ is greater than $\bf{C}$.
The mass fraction of oxygen in $\bf{M}$ is $\rm{39.5\%}$
The mass fraction of oxygen in $\bf{M}$ is $\rm{44.1\%}$
The symmetry of $\bf{M}$ is greater than $\bf{C}$
$\bf{M}$ is a neutral particle
其他选项均不正确
$\bf{X}$原子基态含有两个单电子
$\bf{A}$的原子数大于$\bf{C}$
$\bf{M}$中氧的质量分数为$\rm{39.5\%}$
$\bf{M}$中氧的质量分数为$\rm{44.1\%}$
$\bf{M}$的对称性大于$\bf{C}$
$\bf{M}$为中性粒子
Short-period element $\bf{X}$ is often used in nuclear power reactors. In nuclear reactors, $\rm{Be}$ is often used as a neutron moderator and reflector, and $\rm{B}$ is often used as a neutron absorber. Therefore, $\bf{X}$ may be $\rm{B,Be}$.

<Checkpoint>[$\bf{X}$ may be $\rm{B,Be}$](1)</Checkpoint>

The halide of $\bf{X}$ reacts with $\rm{N_2O_4}$, generally yielding anhydrous nitrate and an adduct of $\rm{N_2O_4}$; considering that the molecular weight of the loss in the first step is very large, it is very likely that the added $\rm{N_2O_4}$ is lost.

<Checkpoint>[Halide reacts with $\rm{N_2O_4}$, generally yielding anhydrous nitrate and an adduct of $\rm{N_2O_4}$](1)</Checkpoint>

<Checkpoint>[The added $\rm{N_2O_4}$ is lost in the first step of weight loss](1)</Checkpoint>

If $\bf{X}$ is $\rm{B}$, $\bf{A}$ can be written as $\rm{B(NO_3)_3\cdot xN_2O_4}$. First of all, nitrates of $\rm{B}$ are not common; secondly, no matter what value x takes, it cannot correspond to this weight loss peak. Therefore, $\bf{X}$ is not $\rm{B}$.

<Checkpoint>[Based on the weight loss ratio, it is determined that $\bf{X}$ is not $\rm{B}$](1)</Checkpoint>

If $\bf{X}$ is $\rm{Be}$, $\bf{A}$ can be written as $\rm{Be(NO_3)_2\cdot xN_2O_4}$. At this time, when x is 2, the first step weight loss ratio is $\rm{184/317=58.0\%}$, which meets the requirements of the question.
Therefore, x is 2, and the chemical formula of $\bf{A}$ is $\rm{Be(NO_3)_2\cdot 2N_2O_4}$. $\bf{X}$ is $\rm{Be}$, option B is incorrect.

<Checkpoint>[The chemical formula of $\bf{A}$ is $\rm{Be(NO_3)_2\cdot 2N_2O_4}$](1)</Checkpoint>

$\bf{C}$ contains 4 $\rm{C_3}$ axes, all $\bf{X}$ chemical environments are the same, and the coordination number is 4. First, it is judged that the compound has a tetrahedral structure; therefore, it is highly probable that it contains four $\rm{Be}$. It is easy to think of the classic structure of $\rm{Be}$ compounds $\rm{Be_4O(NO_3)_6}$. According to the weight loss peak, it can be judged that the chemical formula is correct. Therefore, $\bf{C}$ is $\rm{Be_4O(NO_3)_6}$. According to the chemical formula, option C is incorrect.

<Checkpoint>[$\bf{C}$ has a tetrahedral structure; contains four $\rm{Be}$](1)</Checkpoint>

<Checkpoint>[$\bf{C}$ is $\rm{Be_4O(NO_3)_6}$](1)</Checkpoint>

According to the question, the particles formed by $\bf{M}$ should have a similar metal-oxygen cluster structure and have acetate as a ligand. The chemical formula can be tentatively written as $\rm{M_xO(CH_3COO)_y}$;

<Checkpoint>[The particles formed by $\bf{M}$ have a similar metal-oxygen cluster structure](1)</Checkpoint>

Considering the mass fraction of $\bf{M}$, since $\bf{M}$ is hexacoordinate, if the oxygen cluster provides 1 coordination number, the ligand needs to provide 5 coordination numbers; if the structure of this particle is the same as $\rm{Be_4O(NO_3)_6}$ and is a tetrahedral structure, and acetate is a bridging ligand, the coordination number of $\bf{M}$ is at most 4, so there are other ligands that provide coordination numbers.

Considering that the particles are prepared in water, consider that the ligand is water;

<Checkpoint>[There are other ligands that provide coordination numbers](1)</Checkpoint>

<Checkpoint>[There is a ligand that is water](1)</Checkpoint>

The chemical formula can be tentatively written as $\rm{M_xO(CH_3COO)_y(H_2O)_z}$; considering that x is 4, at this time, acetate is a bridging ligand located on the edge of the tetrahedron, that is, y=6; in order to make the coordination number of $\bf{M}$ 6, each $\bf{M}$ needs to be additionally coordinated with two water molecules. Therefore, z is 8;

According to the mass fraction of $\rm{26.88\%}$, the relative atomic mass of $\bf{M}$ is calculated to be 47.2, and there is no reasonable metal; therefore, the particle does not contain four $\bf{M}$.

<Checkpoint>[Through mass fraction inference, the particle does not contain four $\bf{M}$](1)</Checkpoint>

Considering the decrease in symmetry, the particle contains three $\bf{M}$. At this time, acetate acts as a bridging ligand, and two acetate groups can exist on the side of the planar triangle. Therefore, y can be 6 or 3, and the corresponding z that satisfies the coordination number can be 3 or 9.

According to the mass fraction, the relative atomic mass of $\bf{M}$ is calculated: when y=6, z=3, the atomic weight is 52.08, which is $\rm{Cr}$; when y=3, z=9, it is 43.5, and there is no reasonable metal.

Therefore, the chemical formula of the particle is $\rm{Cr_3O(CH_3COO)_6(H_2O)_3}$. Considering the common oxidation state of +3 for hexacoordinate $\rm{Cr}$, the particle should carry a +1 charge. Therefore, options F and G are both incorrect.

<Checkpoint>[Considering the decrease in symmetry, the chemical formula of the particle is $\rm{Cr_3O(CH_3COO)_6(H_2O)_3}$](1)</Checkpoint>

<Checkpoint>[Considering the common oxidation state of +3 for hexacoordinate $\rm{Cr}$, the particle should carry a +1 charge](1)</Checkpoint>

The oxygen mass fraction in the particle is calculated to be $\rm{44.1\%}$, option D is incorrect, and E is correct.
短周期元素$\bf{X}$常应用于核动力反应堆中，在核反应堆中，$\rm{Be}$常用作中子慢化剂和反射剂，$\rm{B}$常用作中子吸收剂。因此$\bf{X}$可能为$\rm{B,Be}$。

<Checkpoint>[$\bf{X}$可能为$\rm{B,Be}$](1)</Checkpoint>

$\bf{X}$的卤化物与$\rm{N_2O_4}$反应，一般得到无水硝酸盐与$\rm{N_2O_4}$的加合物；考虑第一步失重分子量很大，很有可能失去加和的$\rm{N_2O_4}$。

<Checkpoint>[卤化物与$\rm{N_2O_4}$反应，一般得到无水硝酸盐与$\rm{N_2O_4}$的加合物](1)</Checkpoint>

<Checkpoint>[第一步失重失去加和的$\rm{N_2O_4}$](1)</Checkpoint>

若$\bf{X}$为$\rm{B}$，$\bf{A}$可写为$\rm{B(NO_3)_3\cdot xN_2O_4}$。首先$\rm{B}$的硝酸盐并不常见；其次x无论取任何值都无法对应该失重峰。因此$\bf{X}$不为$\rm{B}$。

<Checkpoint>[根据失重比例判断$\bf{X}$不为$\rm{B}$](1)</Checkpoint>

若$\bf{X}$为$\rm{Be}$，$\bf{A}$可写为$\rm{Be(NO_3)_2\cdot xN_2O_4}$。此时当x取2时，第一步失重比例为$\rm{184/317=58.0\%}$，符合题设。
因此x为2，$\bf{A}$化学式为$\rm{Be(NO_3)_2\cdot 2N_2O_4}$。$\bf{X}$为$\rm{Be}$，选项B错误。

<Checkpoint>[$\bf{A}$化学式为$\rm{Be(NO_3)_2\cdot 2N_2O_4}$](1)</Checkpoint>

$\bf{C}$中含有4个$\rm{C_3}$轴,所有$\bf{X}$化学环境相同,且配位数均为4。首先判断该化合物为四面体结构；因此大概率含有四个$\rm{Be}$。很容易想到$\rm{Be}$化合物的经典结构$\rm{Be_4O(NO_3)_6}$，根据失重峰可判断该化学式正确。因此$\bf{C}$为$\rm{Be_4O(NO_3)_6}$。根据化学式可知选项C错误。

<Checkpoint>[$\bf{C}$为四面体结构；含有四个$\rm{Be}$](1)</Checkpoint>

<Checkpoint>[$\bf{C}$为$\rm{Be_4O(NO_3)_6}$](1)</Checkpoint>

$\bf{M}$形成的粒子根据题设，应当具有类似的金属-氧簇结构，且有乙酸根作为配体。化学式可尝试写为$\rm{M_xO(CH_3COO)_y}$；

<Checkpoint>[$\bf{M}$形成的粒子具有类似的金属-氧簇结构](1)</Checkpoint>

考虑$\bf{M}$的质量分数，由于$\bf{M}$为六配位，在氧簇提供1个配位数的情况下，配体需要提供5配位数；若该粒子与$\rm{Be_4O(NO_3)_6}$结构一致为四面体结构，乙酸根作为桥连配体的情况下，$\bf{M}$的配位数最多为4，所以存在其他配体提供配位数。

考虑该粒子于水中制备，考虑该配体为水；

<Checkpoint>[存在其他配体提供配位数](1)</Checkpoint>

<Checkpoint>[存在配体为水](1)</Checkpoint>

化学式可尝试写为$\rm{M_xO(CH_3COO)_y(H_2O)_z}$；考虑x取4，此时乙酸根作为桥连配体位于四面体的边上，即y=6；为使得$\bf{M}$的配位数为6，每个$\bf{M}$需要额外与两个水配位。因此z取8；

根据质量分数$\rm{26.88\%}$计算$\bf{M}$的相对原子质量为47.2，无合理金属；因此该粒子不含四个$\bf{M}$。

<Checkpoint>[通过质量分数推断，该粒子不含四个$\bf{M}$](1)</Checkpoint>

考虑对称性下降，粒子含有三个$\bf{M}$，此时乙酸根作为桥连配体，平面三角形边可以存在两个乙酸根，因此y可以取6或3，对应满足配位数的z可以取3或9。

根据质量分数计算$\bf{M}$的相对原子质量：当y=6,z=3时，原子量为52.08，为$\rm{Cr}$；当y=3,z=9时为43.5，无合理金属。

因此该粒子的化学式为$\rm{Cr_3O(CH_3COO)_6(H_2O)_3}$，考虑$\rm{Cr}$六配位的常见氧化态+3，该粒子应当带+1电荷。因此选项F,G均错误。

<Checkpoint>[考虑对称性下降，该粒子的化学式为$\rm{Cr_3O(CH_3COO)_6(H_2O)_3}$](1)</Checkpoint>

<Checkpoint>[考虑$\rm{Cr}$六配位的常见氧化态+3，该粒子应当带+1电荷](1)</Checkpoint>

计算该粒子中氧质量分数为$\rm{44.1\%}$，选项D错误，E正确。