$\bf{M}$ is a silver-gray, hard and brittle transition metal with metallic luster. It is the main additive of stainless steel, which can make steel have high corrosion resistance. The reaction of nitric acid and $\bf{M}$ amalgam can obtain a deliquescent black oxide powder $\bf{A}$, which can burn in the air to obtain $\bf{B}$. A 1:1 mixture of carbon monoxide and chlorine gas at high temperature can convert $\bf{B}$ into $\bf{C}$. The fusion reaction of $\bf{B}$ with potassium chlorate can obtain a yellow solid $\mathbf{D}$. After dissolving $\mathbf{D}$ in water and acidifying, the solution will turn orange due to the formation of $\mathbf{E}$. Further addition of concentrated sulfuric acid to the solution will form an orange-red solid $\bf{F}$.

It is known that there is the following process:
1. $\bf{C}$ can react with $\rm{Ph_4AsCl}$ and sodium azide under certain conditions to generate a mononuclear complex, in which the mass fraction of nitrogen is $\rm{17.34\%}$.
2. Powdered $\mathbf{E}$ is heated to boiling with excess concentrated hydrofluoric acid and then cooled to obtain a substance $\mathbf{G}$ containing $\mathbf{M}$ $32.89\%$.
3. $\mathbf{E}$ reacts with $\mathbf{C}$ under molten conditions, and one of the products is similar to $\mathbf{G}$.

The following statement is correct:
$\bf{M}$是一种银灰色,具金属光泽且硬而脆的过渡金属,是不锈钢的主要添加物,可使钢材具有高度耐腐蚀的特性。硝酸和$\bf{M}$的汞齐反应可以得到易潮解的黑色氧化物粉末$\bf{A}$, $\bf{A}$能够在空气中燃烧从而得到$\bf{B}$。高温下1:1的一氧化碳和氯气的混合物可使$\bf{B}$转化为$\bf{C}$。$\bf{B}$与氯酸钾熔融反应, 则可以得到黄色固体$\mathbf{D}$, 将$\mathbf{D}$溶于水之后酸化, 溶液会因为$\mathbf{E}$的形成而变为橙色,进一步向溶液中加入浓硫酸,则会有桔红色固体$\bf{F}$形成。

已知有如下过程：
1. $\bf{C}$可与$\rm{Ph_4AsCl}$和叠氮化钠在一定条件下反应生成一单核配合物，其中氮元素的质量分数为$\rm{17.34\%}$。
2. 粉状的$\mathbf{E}$与过量的浓氢氟酸加热至沸后放冷,可以得到一种含$\mathbf{M}$ $32.89\%$的物质$\mathbf{G}$。
3. $\mathbf{E}$与$\mathbf{C}$在熔融条件下反应，产物之一与$\mathbf{G}$相似。

下列说法正确的是：
All other options are incorrect
The mass fraction of $\bf{M}$ in $\bf{A}$ is $\rm{68.4\%}$
The mass fraction of $\bf{M}$ in $\bf{C}$ is $\rm{33.6\%}$
The mass fraction of $\bf{M}$ in $\bf{F}$ is $\rm{76.5\%}$
The complex in Process 1 is a 1:1 type ionic compound.
In Process 1, the hybridization of $\bf{M}$ in the complex is $\rm{sp^3}$ hybridization.
The sum of the coefficients of all species in the chemical equation for Process 3, balanced to the simplest integer ratio, is 12.
The sum of the coefficients of all species in the chemical equation of Process 2, when balanced to the simplest integer ratio, is 8.
其他选项均不正确
$\bf{A}$中$\bf{M}$的质量分数为$\rm{68.4\%}$
$\bf{C}$中$\bf{M}$的质量分数为$\rm{33.6\%}$
$\bf{F}$中$\bf{M}$的质量分数为$\rm{76.5\%}$
过程1中的配合物为1：1型离子化合物
过程1中配合物中$\bf{M}$杂化方式为$\rm{sp^3}$杂化
过程3的化学方程式，配平为最简整数比时，所有物种的系数和为12
过程2的化学方程式，配平为最简整数比时，所有物种的系数和为8
$\bf{M}$ is the main additive of stainless steel, and is hard and brittle, generally $\rm{Ni}$ or $\rm{Cr}$.

<Checkpoint>[$\bf{M}$ is the main additive of stainless steel, generally $\rm{Ni}$ or $\rm{Cr}$](1)</Checkpoint>

The yellow solid $\mathbf{D}$ dissolves in acidic solution to produce an orange color. Based on element knowledge, it is easy to think of chromate $\rm{CrO_4^{2-}}$ and dichromate $\rm{Cr_2O_7^{2-}}$. Therefore, $\bf{M}$ is $\rm{Cr}$, $\mathbf{D}$ is potassium chromate $\rm{K_2CrO_4}$, and $\mathbf{E}$ is potassium dichromate $\rm{K_2Cr_2O_7}$.

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

<Checkpoint>[$\mathbf{D}$ is $\rm{K_2CrO_4}$, $\mathbf{E}$ is $\rm{K_2Cr_2O_7}$](1)</Checkpoint>

$\bf{A}$ can burn to produce $\bf{B}$, and $\bf{B}$ can be fused with potassium chlorate to produce $\rm{K_2CrO_4}$, then $\bf{B}$ should be the stable oxide $\rm{Cr_2O_3}$. At this time, $\bf{A}$ can only be the low-valent oxide $\rm{CrO}$.

<Checkpoint>[$\bf{B}$ should be $\rm{Cr_2O_3}$](1)</Checkpoint>

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

$\rm{Cr_2O_3}$ is converted to $\bf{C}$ by 1:1 carbon monoxide and chlorine gas, so $\bf{C}$ should be the chloride $\rm{CrCl_3}$, and $\rm{CrO_2Cl_2}$ will not be generated under reducing conditions.

<Checkpoint>[$\bf{C}$ is the chloride $\rm{CrCl_3}$](1)</Checkpoint>

The orange-red solid $\bf{F}$ produced by adding concentrated sulfuric acid to the $\rm{K_2Cr_2O_7}$ solution is $\rm{CrO_3}$.

<Checkpoint>[$\bf{F}$ is $\rm{CrO_3}$](1)</Checkpoint>

According to the chemical formula, options B, C, and D are all incorrect.

The following studies three processes:

Process 1, $\rm{Ph_4AsCl}$ and sodium azide form a complex with $\rm{CrCl_3}$. According to the Lewis acid-base reaction, the cation is likely to be $\rm{Ph_4As^+}$. The species that can be used for coordination in the system are likely to be $\rm{N_3^-}$ or $\rm{Cl^-}$ and can only be monodentate ligands. Considering that $\rm{Cr(III)}$ is commonly hexacoordinate, the chemical formula of this mononuclear complex may be $\rm{(Ph_4As)_3[Cr(N_3)_xCl_{6-x}]}$. Considering the mass fraction of nitrogen, try the specific value of x. When x is 6, it satisfies the mass fraction of nitrogen $\rm{17.34\%}$ given in the question. Therefore, the chemical formula of the complex is $\rm{(Ph_4As)_3[Cr(N_3)_6]}$, which is a 3:1 type ionic compound, and option E is incorrect. $\mathrm{Cr(III)}$ is $\rm{d^2sp^3}$ hybridized, and option F is incorrect.

<Checkpoint>[The cation is likely to be $\rm{Ph_4As^+}$](1)</Checkpoint>

<Checkpoint>[$\rm{Cr(III)}$ is commonly hexacoordinate, the chemical formula of this mononuclear complex may be $\rm{(Ph_4As)_3[Cr(N_3)_xCl_{6-x}]}$](1)</Checkpoint>

<Checkpoint>[The chemical formula of the complex is $\rm{(Ph_4As)_3[Cr(N_3)_6]}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{Cr(III)}$ is $\rm{d^2sp^3}$ hybridized](1)</Checkpoint>

Process 2, $\rm{K_2Cr_2O_7}$ reacts with $\rm{HF}$ to obtain a substance $\mathbf{G}$ containing $\mathrm{Cr}$ $32.89\%$. Since $\rm{HF}$ does not exhibit reducing properties, this reaction is not a redox reaction, so $\mathbf{G}$ is likely to be a fluoro complex. Consider the chemical formula of the complex as $\rm{CrO_xF_y}$, and containing $\mathrm{Cr(VI)}$. Common compounds may be $\rm{KCrO_3F}$, $\rm{CrO_2F_2}$ and $\rm{KCrOF_5}$. After calculation, it can be seen that $\rm{KCrO_3F}$ satisfies the mass fraction given in the question. Therefore, $\mathbf{G}$ is $\rm{KCrO_3F}$, balance the chemical equation:

<Checkpoint>[$\rm{HF}$ does not exhibit reducing properties, therefore this reaction is not a redox reaction](1)</Checkpoint>

<Checkpoint>[$\mathbf{G}$ is $\rm{KCrO_3F}$](1)</Checkpoint>

$\rm{K_2Cr_2O_7 + 2HF = 2KCrO_3F + H_2O}$

<Checkpoint>[$\rm{K_2Cr_2O_7 + 2HF = 2KCrO_3F + H_2O}$](1)</Checkpoint>

The sum of the coefficients is 6, and option H is incorrect.

Process 3: $\rm{K_2Cr_2O_7}$ reacts with $\rm{CrCl_3}$ in a molten state. The product is similar to $\rm{KCrO_3F}$, so it is obvious that the product is $\rm{KCrO_3Cl}$. Under molten conditions, a stable oxide $\rm{Cr_2O_3}$ should be generated, so balance the equation:

<Checkpoint>[The products are $\rm{KCrO_3Cl}$ and $\rm{Cr_2O_3}$](1)</Checkpoint>

$\rm{3K_2Cr_2O_7 + 2CrCl_3 = 6KCrO_3Cl + Cr_2O_3}$

<Checkpoint>[$\rm{3K_2Cr_2O_7 + 2CrCl_3 = 6KCrO_3Cl + Cr_2O_3}$](1)</Checkpoint>

The sum of the reaction coefficients is 12, and option G is correct.
$\bf{M}$是不锈钢的主要添加物，且硬而脆，一般为$\rm{Ni}$或者$\rm{Cr}$。

<Checkpoint>[$\bf{M}$是不锈钢的主要添加物，一般为$\rm{Ni}$或者$\rm{Cr}$](1)</Checkpoint>

黄色固体$\mathbf{D}$溶于酸性溶液产生橙色，根据元素知识很容易想到是铬酸根$\rm{CrO_4^{2-}}$和重铬酸根$\rm{Cr_2O_7^{2-}}$。因此$\bf{M}$为$\rm{Cr}$，$\mathbf{D}$为铬酸钾$\rm{K_2CrO_4}$，$\mathbf{E}$为重铬酸钾$\rm{K_2Cr_2O_7}$。

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

<Checkpoint>[$\mathbf{D}$为$\rm{K_2CrO_4}$，$\mathbf{E}$为$\rm{K_2Cr_2O_7}$](1)</Checkpoint>

$\bf{A}$能燃烧生成$\bf{B}$且$\bf{B}$可与氯酸钾熔融生成$\rm{K_2CrO_4}$，则$\bf{B}$应当为稳定的氧化物$\rm{Cr_2O_3}$。此时$\bf{A}$只能为低价态氧化物$\rm{CrO}$。

<Checkpoint>[$\bf{B}$应当为$\rm{Cr_2O_3}$](1)</Checkpoint>

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

$\rm{Cr_2O_3}$被1:1的一氧化碳和氯气转化为$\bf{C}$，因此$\bf{C}$应当为氯化物$\rm{CrCl_3}$，还原性环境下不会生成$\rm{CrO_2Cl_2}$。

<Checkpoint>[$\bf{C}$为氯化物$\rm{CrCl_3}$](1)</Checkpoint>

向$\rm{K_2Cr_2O_7}$溶液中加入浓硫酸产生的橙红色固体$\bf{F}$为$\rm{CrO_3}$。

<Checkpoint>[$\bf{F}$为$\rm{CrO_3}$](1)</Checkpoint>

根据化学式可知选项B,C,D均错误。

下面研究三个过程：

过程1，$\rm{Ph_4AsCl}$和叠氮化钠与$\rm{CrCl_3}$形成配合物，根据lewis酸碱反应可知阳离子大概率为$\rm{Ph_4As^+}$。体系中可用于配位的物种大概率为$\rm{N_3^-}$或$\rm{Cl^-}$且只能为单齿配体，考虑到$\rm{Cr(III)}$常见六配位，该单核配合物化学式可能为$\rm{(Ph_4As)_3[Cr(N_3)_xCl_{6-x}]}$。考虑氮的质量分数，尝试x的具体取值，当x为6时满足氮的题设质量分数$\rm{17.34\%}$。因此该配合物化学式为$\rm{(Ph_4As)_3[Cr(N_3)_6]}$，为3：1型离子化合物，选项E错误。$\mathrm{Cr(III)}$为$\rm{d^2sp^3}$杂化，选项F错误。

<Checkpoint>[阳离子大概率为$\rm{Ph_4As^+}$](1)</Checkpoint>

<Checkpoint>[$\rm{Cr(III)}$常见六配位，该单核配合物化学式可能为$\rm{(Ph_4As)_3[Cr(N_3)_xCl_{6-x}]}$](1)</Checkpoint>

<Checkpoint>[配合物化学式为$\rm{(Ph_4As)_3[Cr(N_3)_6]}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{Cr(III)}$为$\rm{d^2sp^3}$杂化](1)</Checkpoint>

过程2，$\rm{K_2Cr_2O_7}$与$\rm{HF}$反应得到含$\mathrm{Cr}$ $32.89\%$的物质$\mathbf{G}$，由于$\rm{HF}$不体现还原性，因此该反应不为氧化还原反应，则$\mathbf{G}$很可能为氟配合物。考虑该配合物化学式为$\rm{CrO_xF_y}$，且含$\mathrm{Cr(VI)}$。常见化合物可能为$\rm{KCrO_3F}$，$\rm{CrO_2F_2}$和$\rm{KCrOF_5}$。经过计算可知，$\rm{KCrO_3F}$满足题设质量分数。因此$\mathbf{G}$为$\rm{KCrO_3F}$，配平化学方程式：

<Checkpoint>[$\rm{HF}$不体现还原性，因此该反应不为氧化还原反应](1)</Checkpoint>

<Checkpoint>[$\mathbf{G}$为$\rm{KCrO_3F}$](1)</Checkpoint>

$\rm{K_2Cr_2O_7 + 2HF = 2KCrO_3F + H_2O}$

<Checkpoint>[$\rm{K_2Cr_2O_7 + 2HF = 2KCrO_3F + H_2O}$](1)</Checkpoint>

系数和为6，选项H错误。

过程3：$\rm{K_2Cr_2O_7}$与$\rm{CrCl_3}$熔融反应，产物与$\rm{KCrO_3F}$类似，则很明显产物为$\rm{KCrO_3Cl}$。熔融条件下应当有稳定氧化物$\rm{Cr_2O_3}$生成，因此配平方程式：

<Checkpoint>[产物为$\rm{KCrO_3Cl}$和$\rm{Cr_2O_3}$](1)</Checkpoint>

$\rm{3K_2Cr_2O_7 + 2CrCl_3 = 6KCrO_3Cl + Cr_2O_3}$

<Checkpoint>[$\rm{3K_2Cr_2O_7 + 2CrCl_3 = 6KCrO_3Cl + Cr_2O_3}$](1)</Checkpoint>

反应系数和为12，选项G正确。