The name of element $\mathbf M$ originates from Greek mythology. The element $\mathbf M$ is usually prepared from a certain natural mineral, the main component of which is $\mathbf G$, containing $31.56 \%$ of $\mathbf M$. The first step in preparing the element $\mathbf M$ is to treat $\mathbf G$ with elemental carbon and excess chlorine gas at high temperature. The element $\mathbf M$ can directly combine with nitrogen gas to form a stable golden-yellow solid $\mathbf A$, and $\mathbf A$ is isostructural with $\mathrm{NaCl}$. $\mathbf A$ dissolves in aqua regia to form $\mathbf B$. Adding $\mathrm{Zn}$ powder to a concentrated $\mathrm{HCl}$ solution of $\mathbf B$ causes the solution to turn purple, and the solute is $\mathbf C$. $\mathbf B$ is easily converted to $\mathbf D$ upon exposure to air. $\mathbf D$ is also very stable, insoluble in sulfuric acid or nitric acid, but soluble in hydrofluoric acid to form $\mathbf E$, and $\mathbf E$ can also be used for the preparation of $\mathbf M$. Dissolving $\mathbf B$ in anhydrous ethanol and passing ammonia gas through it can prepare $\mathbf F$, and the mass fraction of $\mathbf M$ in $\mathbf F$ is $20.99 \%$. In the actual existing form of $\mathbf F$, $\mathbf M$ has two chemical environments, and the molecular weight does not exceed 1000.

The following statements are correct:
元素$\mathbf M$的名称来源于希腊神话。$\mathbf M$单质常从某天然矿物制备，该矿物主要成分为$\mathbf G$，含$\mathbf M$ $3 1 . 5 6 \%$ 。制备$\mathbf M$单质的第一步为$\mathbf G$与碳单质一起用过量氯气高温处理。$\mathbf M$单质可直接与氮气化合生成稳定的金黄色固体$\mathbf A$，$\mathbf A$与$\mathrm{NaCl}$类质同晶。$\mathbf A$溶于王水生成$\mathbf B$。向$\mathbf B$的浓$\mathrm{HCl}$溶液中加入$\mathrm {Zn}$粉，溶液变为紫色，其中溶质为$\mathbf C$。$\mathbf B$暴露在空气中易转化为$\mathbf D$，$\mathbf D$同样十分稳定，不溶于硫酸或硝酸，但可以溶于氢氟酸生成$\mathbf E$，$\mathbf E$也可用于$\mathbf M$的制备。将$\mathbf B$溶于无水乙醇并通入氨气可制备$\mathbf F$，$\mathbf F$中$\mathbf M$质量分数为 $2 0 . 9 9 \%$。$\mathbf F$实际存在形态中$\mathbf M$有两种化学环境，分子量不超过1000。

以下说法正确的是
The metallic valence in $\mathbf A$ is $\mathrm {+4}$.
$\mathbf B$ and $\mathbf C$ are both metal chlorides, wherein the metal valence states are $\mathrm {+4}$ and $\mathrm {+2}$, respectively.
$\mathbf D$ has multiple crystal structures, in one of which the anions form an approximate hexagonal close-packed structure, and the cations fill all the octahedral voids formed by the anions.
$\mathbf E$ substance has more than 1 possible structure, and none of them are harmful to the environment.
The chemical formula for preparing $\mathbf F$ mentioned in the problem, after balancing and simplifying the coefficients, the sum of the coefficients on both sides of the equation are equal.
In the true structure of $\mathbf F$, the $\mathrm O$ element can be connected to a maximum of 2 $\mathbf M$ [elements].
In the real structure of $\mathbf F$, the coordination polyhedron formed by ligands around the central atom is a trigonal bipyramidal structure.
$\mathbf G$ ore exhibits magnetism comparable to magnetite, which is due to their similar crystal structures.
In the equation for the first step of the $\mathbf G$ preparation of $\mathbf M$, the product contains metals with valence states of $\mathrm {+2}$ and $\mathrm {+4}$, respectively.
In the equation for the first step of $\mathbf G$ preparation of $\mathbf M$, after balancing and simplifying coefficients, has the sum of reactant coefficients on the left side less than the sum of product coefficients on the right side.
All of the above options are incorrect.
$\mathbf A$中金属价态为$\mathrm {+4}$价
$\mathbf B$和$\mathbf C$都为金属氯化物，其中金属价态分别为$\mathrm {+4}$价和$\mathrm {+2}$价
$\mathbf D$有多种晶体结构，其中一种晶体结构阴离子作近似六方密堆积，阳离子填充在由阴离子构成的所有八面体空隙中
$\mathbf E$物质有大于1种的可能结构，它们都对环境没有危害
题目中提到的制备$\mathbf F$的化学式，配平和简化系数后，方程式左右两边系数和相等
$\mathbf F$的真实结构中，$\mathrm O$元素最多可以和2个$\mathbf M$相连
$\mathbf F$的真实结构中，配体围绕中心原子形成的配位多面体为三角双锥结构
$\mathbf G$矿石与磁铁矿磁性相当，这是因为他们的晶体结构类似
$\mathbf G$制备$\mathbf M$第一步的方程式中，产物包含的金属价态分别为$\mathrm {+2}$价和$\mathrm {+4}$价
$\mathbf G$制备$\mathbf M$第一步的方程式中，配平和化简系数后，左边反应物系数和小于右边产物系数和
以上选项均不正确
The element $\mathbf M$ from mythology is commonly found in $\mathrm {V、Ti、Nb、Ta}$, etc. According to the direct combination of elemental $\mathbf M$ with nitrogen gas to produce a stable golden yellow solid $\mathbf A$, $\mathbf A$ is isostructural with $\mathrm {NaCl}$, it can be inferred that $\mathbf A$ has a face-centered cubic structure, and the ratio of anions to cations is 1:1, so $\mathbf A$ should be $\mathrm {MN}$, where $\mathrm {TiN}$ is golden yellow. Finally, it is inferred that $\mathbf M$ is $\mathrm {Ti}$ and $\mathbf A$ is $\mathrm {TiN}$. The metal valence in $\mathbf A$ is $+3$, so A is incorrect.

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

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

$\mathbf A$ dissolves in aqua regia to form $\mathbf B$. Zinc powder is added to the concentrated hydrochloric acid solution of $\mathbf B$, and the solution turns purple, and the solute is $\mathbf C$, which is purple.

Aqua regia is a strong oxidant that oxidizes $\mathbf M$ to its highest stable oxidation state and forms chlorides. Therefore, $\mathbf B$ is $\mathrm {TiCl_4}$, and $\mathbf C$ is $\mathrm {TiCl_3}$, where $\mathrm{Ti^{3+}}$ is purple in solution, which meets the requirements of the question, and the metal valences are $+4$ and $+3$ respectively, so B is incorrect.

<Checkpoint>[$\mathbf B$ is $\mathrm {TiCl_4}$](1)</Checkpoint>

<Checkpoint>[$\mathbf C$ is $\mathrm {TiCl_3}$](1)</Checkpoint>

$\mathbf B$ is easily converted to $\mathbf D$ when exposed to air. In addition, $\mathbf D$ is also very stable, insoluble in sulfuric acid or nitric acid, but can be dissolved in hydrofluoric acid to form $\mathbf E$. It is inferred that $\mathbf D$ is $\mathrm {TiO_2}$. $\mathrm {TiO_2}$ has three crystal structures, namely rutile, anatase, and brookite. Among them, in the rutile structure, the anions are approximately cubic packed, and the cations fill half of the octahedral voids formed by the anions, so C is incorrect.

<Checkpoint>[$\mathbf D$ is $\mathrm {TiO_2}$](1)</Checkpoint>

<Checkpoint>[In the rutile structure, the anions are approximately cubic packed, and the cations fill half of the octahedral voids formed by the anions](1)</Checkpoint>

$\mathbf D$ reacts with hydrofluoric acid to form $\mathbf E$. $\mathbf E$ is $\mathrm {H_2[TiF_6]}$ or $\mathrm {TiF_4}$. Fluorotitanic acid is polluting to the environment, so D is incorrect.

<Checkpoint>[$\mathbf E$ is $\mathrm {H_2[TiF_6]}$ or $\mathrm {TiF_4}$](1)</Checkpoint>

$\mathbf B$ is dissolved in anhydrous ethanol and ammonia gas is passed in to prepare $\mathbf F$. The mass fraction of $\mathbf M$ in $\mathbf F$ is $20.99\%$. $\mathbf F$ is $\mathrm {Ti(OC_2H_5)_4}$. The mass fraction of $\mathrm{Ti}$ is $\omega = 47.87 / (8×12.01 + 20×1.01 + 4×16.00 + 47.87) = 20.99\%$.

<Checkpoint>[$\mathbf F$ is $\mathrm {Ti(OC_2H_5)_4}$](1)</Checkpoint>

The preparation process of $\mathbf F$ can be written as $
\mathrm { TiCl _ { 4 } } + 4 \mathrm { N H _ { 3 } } + 4 \mathrm { C _ { 2 } H _ { 5 } O H } = \mathrm { T i } ( \mathrm { O C _ { 2 } H _ { 5 } } ) _ { 4 } + 4 \mathrm { N H _ { 4 } C l }$. The sum of the coefficients on both sides is not equal, so E is incorrect.

<Checkpoint>[The preparation process of $\mathbf F$ can be written as$$ \mathrm { TiC l _ { 4 } } + 4 \mathrm { N H _ { 3 } } + 4 \mathrm { C _ { 2 } H _ { 5 } O H } = \mathrm { T i } ( \mathrm { O C _ { 2 } H _ { 5 } } ) _ { 4 } + 4 \mathrm { N H _ { 4 } C l }$$](1)</Checkpoint>

The real structure of $\mathbf F$ is shown in the figure, where $\mathrm O$ represents the $\mathrm {OEt}$ group, and the chemical formula is $\mathrm {Ti_4(OC_2H_5)_{16}}$, the mass is $228.11×4=912.44$, which is less than 1000, and there are two different environments of $\mathrm {Ti}$.

<MultiModal>![Real structure of substance F, where $\mathrm O$ represents the $\mathrm {OEt}$ group, Ti are connected by oxygen bridges, ethoxy groups form an octahedron surrounding $\mathrm {Ti}$, two octahedrons on top, two staggered octahedrons on the bottom, chemical formula is $\mathrm {Ti_4(OC_2H_5)_{16}}$](/media/uploads/84627396-413c-443e-b89e-98168f889ea7.png)</MultiModal>

In the real structure of $\mathbf F$, the $\mathrm O$ element acts as an oxygen bridge and connects at most 3 $\mathrm {Ti}$. The coordination polyhedron formed by the ligands around the central atom is an octahedral structure. The 4 octahedrons are divided into two layers, so F and G are incorrect.

<Checkpoint>[In the real structure of $\mathbf F$, the $\mathrm O$ element acts as an oxygen bridge and connects at most 3 $\mathrm {Ti}$. The coordination polyhedron formed by the ligands around the central atom is an octahedral structure. The 4 octahedrons are divided into two layers](1)</Checkpoint>

$\mathbf M$ is prepared from the mineral $\mathbf G$, and the mass fraction of $\mathbf M$ is $\mathrm {31.56\%}$. Common $\mathrm {Ti}$ minerals mainly include rutile $\mathrm {TiO_2}$ and ilmenite $\mathrm {FeTiO_3}$.

If $\mathbf G$ is $\mathrm {TiO_2}$, $\mathrm {Ti\%= 47.87/(47.87 + 2×16.00) = 59.9\%}$, which is inconsistent.

If $\mathbf G$ is ilmenite $\mathrm {FeTiO_3}$ , $\mathrm {Ti\% = 47.87 / (55.85 + 47.87 + 3×16.00) = 47.87 / 151.72 = 31.55\%}$. This ore has a perovskite structure, while magnetite has a spinel structure, so the considerable magnetism does not come from structural similarity, so H is incorrect.

<Checkpoint>[$\mathbf G$ is ilmenite $\mathrm {FeTiO_3}$](1)</Checkpoint>

<Checkpoint>[Ilmenite has a perovskite structure, while magnetite has a spinel structure](1)</Checkpoint>

The first step in preparing $\mathbf M$ from $\mathbf G$ is the high-temperature reaction of $\mathbf G$ with carbon in chlorine gas (carbon chlorination method).
The equation is $\mathrm{2FeTiO_3+6C+7Cl_2=2FeCl_3 + 2TiCl_4 + 6CO}$, the sum of the coefficients on the left side is greater than the sum of the coefficients on the right side, and the metal valences of the products are $+3$ and $+4$ respectively, so I and J are incorrect.

<Checkpoint>[Preparing $\mathbf M$ from $\mathbf G$ The equation is $\mathrm{2FeTiO_3+6C+7Cl_2=2FeCl_3 + 2TiCl_4 + 6CO}$](1)</Checkpoint>

The final answer is K.
来自神话的元素$\mathbf M$较为常见的有$\mathrm {V、Ti、Nb、Ta}$等，根据$\mathbf M$单质与氮气直接化合生成稳定的金黄色固体 $\mathbf A$, $\mathbf A$与 $\mathrm {NaCl}$ 类质同晶可以推断 $\mathbf A$ 为面心立方结构，阴离子阳离子比例为1：1，因此 $\mathbf A$应该为 $\mathrm {MN}$,其中$\mathrm {TiN}$为金黄色，最终推断出$\mathbf M$为$\mathrm {Ti}$，$\mathbf A$为$\mathrm {TiN}$，$\mathbf A$中金属价态为$+3$价，A错误。

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

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

$\mathbf A$ 溶于王水生成$\mathbf B$。$\mathbf B$的浓盐酸溶液中加入锌粉，溶液变为紫色，溶质为 $\mathbf C$，颜色为紫色。

王水是强氧化剂，会将 $\mathbf M$ 氧化到其最高稳定氧化态，并形成氯化物。所以 ，$\mathbf B$为$\mathrm {TiCl_4}$，$\mathbf C$为$\mathrm {TiCl_3}$，其中$\mathrm{Ti^{3+}}$在溶液中为紫色，符合题目要求，金属价态分别为$+4、+3$价，B错误。

<Checkpoint>[$\mathbf B$为$\mathrm {TiCl_4}$](1)</Checkpoint>

<Checkpoint>[$\mathbf C$为$\mathrm {TiCl_3}$](1)</Checkpoint>

$\mathbf B$ 暴露在空气中易转化为$\mathbf D$，结合$\mathbf D$同样十分稳定，不溶于硫酸或硝酸，但可以溶于氢氟酸生成$\mathbf E$，推断$\mathbf D$为$\mathrm {TiO_2}$，$\mathrm {TiO_2}$有三种晶体结构，分别是金红石型、锐钛矿型、板钛矿型，其中金红石结构阴离子作近似立方堆积，阳离子填充在阴离子构成的八面体空隙的一半，C错误。

<Checkpoint>[$\mathbf D$为$\mathrm {TiO_2}$](1)</Checkpoint>

<Checkpoint>[金红石结构阴离子作近似立方堆积，阳离子填充在阴离子构成的八面体空隙的一半](1)</Checkpoint>

$\mathbf D$与氢氟酸反应生成$\mathbf E$，$\mathbf E$为$\mathrm {H_2[TiF_6]}$或者$\mathrm {TiF_4}$，氟钛酸对环境有污染，D错误。

<Checkpoint>[$\mathbf E$为$\mathrm {H_2[TiF_6]}$或者$\mathrm {TiF_4}$](1)</Checkpoint>

将$\mathbf B$溶于无水乙醇并通入氨气可制备$\mathbf F$，$\mathbf F$中$\mathbf M$质量分数为 $20.99\%$，$\mathbf F$为$\mathrm {Ti(OC_2H_5)_4}$,$\mathrm{Ti}$ 的质量分数 $\omega = 47.87 / (8×12.01 + 20×1.01 + 4×16.00 + 47.87) = 20.99\%$。

<Checkpoint>[$\mathbf F$为$\mathrm {Ti(OC_2H_5)_4}$](1)</Checkpoint>

$\mathbf F$的制备过程可以写为$
\mathrm { TiCl _ { 4 } } + 4 \mathrm { N H _ { 3 } } + 4 \mathrm { C _ { 2 } H _ { 5 } O H } = \mathrm { T i } ( \mathrm { O C _ { 2 } H _ { 5 } } ) _ { 4 } + 4 \mathrm { N H _ { 4 } C l }$，两边系数和不相等，E错误。

<Checkpoint>[$\mathbf F$的制备过程可以写为$$ \mathrm { TiC l _ { 4 } } + 4 \mathrm { N H _ { 3 } } + 4 \mathrm { C _ { 2 } H _ { 5 } O H } = \mathrm { T i } ( \mathrm { O C _ { 2 } H _ { 5 } } ) _ { 4 } + 4 \mathrm { N H _ { 4 } C l }$$](1)</Checkpoint>

$\mathbf F$的真实结构如图所示，其中$\mathrm O$代表$\mathrm {OEt}$基团，化学式为$\mathrm {Ti_4(OC_2H_5)_{16}}$，质量为$228.11×4=912.44$，小于1000，有两种不同环境的$\mathrm {Ti}$。

<MultiModal>![物质F的真实结构，其中$\mathrm O$代表$\mathrm {OEt}$基团,Ti之间通过氧桥连接，乙氧基组成八面体将$\mathrm {Ti}$围在中间，上面两个八面体，下面两个交错的八面体，化学式为$\mathrm {Ti_4(OC_2H_5)_{16}}$](/media/uploads/84627396-413c-443e-b89e-98168f889ea7.png)</MultiModal>

$\mathbf F$的真实结构中，$\mathrm O$元素作为氧桥，最多连接3个$\mathrm {Ti}$，配体围绕中心原子形成的配位多面体是八面体结构，4个八面体分为两层，F、G错误。

<Checkpoint>[$\mathbf F$的真实结构中，$\mathrm O$元素作为氧桥，最多连接3个$\mathrm {Ti}$，配体围绕中心原子形成的配位多面体是八面体结构，4个八面体分为两层](1)</Checkpoint>

$\mathbf M$从矿物$\mathbf G$制备，$\mathbf M$的质量分数为 $\mathrm {31.56\%}$。常见的$\mathrm {Ti}$的矿物主要有金红石$\mathrm {TiO_2}$、钛铁矿$\mathrm {FeTiO_3}$。

如果 $\mathbf G$是 $\mathrm {TiO_2}$，$\mathrm {Ti\%= 47.87/(47.87 + 2×16.00) = 59.9\%}$不符。

如果$\mathbf G$是钛铁矿$\mathrm {FeTiO_3}$ ,$\mathrm {Ti\% = 47.87 / (55.85 + 47.87 + 3×16.00) = 47.87 / 151.72 = 31.55\%}$。该矿石为钙钛矿型结构，而磁铁矿为尖晶石型结构，因此磁性相当不来自于结构相似，H错误。

<Checkpoint>[$\mathbf G$是钛铁矿$\mathrm {FeTiO_3}$](1)</Checkpoint>

<Checkpoint>[钛铁矿为钙钛矿型结构，而磁铁矿为尖晶石型结构](1)</Checkpoint>

通过$\mathbf G$制备 $\mathbf M$ 的第一步是 $\mathbf G$与碳在氯气中高温反应（碳氯化法）。
方程式为$\mathrm{2FeTiO_3+6C+7Cl_2=2FeCl_3 + 2TiCl_4 + 6CO}$，左边系数和大于右边系数和，产物金属价态分别为$+3、+4$价，I、J错误。

<Checkpoint>[通过$\mathbf G$制备 $\mathbf M$ 方程式为$\mathrm{2FeTiO_3+6C+7Cl_2=2FeCl_3 + 2TiCl_4 + 6CO}$](1)</Checkpoint>

最终答案为K。