Given the existence of a heteroatom-containing cage-like polyvanadate, where $\mathrm{V}: \mathrm{O}=3: 7$, the content of non-metallic elements is $46.56 \%$, bearing 5 negative charges, but not all $\mathrm{V}$ are in their highest oxidation state and the heteroatom is located at the center of the cage, and it is known that all $\mathrm{V}$ are in square pyramidal coordination. The descriptions of vertices, edges, and faces in the options refer to polyhedra with vanadium atoms and non-terminal oxygen atoms as vertices.

Which of the following statements is correct:
已知存在一种含有杂原子的笼状多钒酸根, 其中 $\mathrm{V}: \mathrm{O}=3: 7$, 非金属元素含量为 $46.56 \%$, 带 5 个负电荷, 但 $\mathrm{V}$ 并不是均为最高价且杂原子位于笼的中心, 已知$\mathrm{V}$均为四方锥型配位。选项中关于顶点、棱、面的描述，其对象是以钒原子和非端基氧原子为顶点的多面体。

下列说法正确的是：
All other options are incorrect
The heteroatom contained in this acid radical is located in the fourth period of the periodic table.
The heteroatom contained in this acid radical is a short-period element.
The mass fraction of oxygen in this acid radical is $\rm 37.8 \%$
The mass fraction of oxygen in the acid radical is $\rm 44.8 \%$
The polyvanadate has 64 edges.
The polyoxovanadate has 28 edges.
The polyvanadate has 26 faces.
The polyoxovanadate has 32 faces.
The polyoxovanadate has 18 vertices.
其他选项均不正确
该酸根中含有的杂原子位于元素周期表第四周期
该酸根中含有的杂原子为短周期元素
该酸根中氧的质量分数为$\rm 37.8 \%$
该酸根中氧的质量分数为$\rm 44.8 \%$
该多钒酸根有64条棱
该多钒酸根有28条棱
该多钒酸根有26个面
该多钒酸根有32个面
该多钒酸根有18个顶点
Since the heteroatom is located at the center of the cage, and $\mathrm{V}$ has a square pyramidal coordination, <Checkpoint>[due to the geometric configuration, the terminal $\mathrm{O}$ can only face outwards](1)</Checkpoint>, according to Coulombic interactions, it can be inferred that the heteroatom carries a negative charge, that is, a non-metallic ion. (In addition, the mass fraction of the metal element in the vanadium-oxygen framework is $57.71\%>1-46.56\%$, so the heteroatom must be a non-metal)

<Checkpoint>[According to Coulombic interactions, it can be inferred that the heteroatom carries a negative charge, that is, a non-metallic ion](1)</Checkpoint>

Therefore, assume the chemical formula of the radical is $\rm (V_3O_7)_x\bf{M}\rm ^{5-}$, at this time, the atomic weight $m$ of the non-metallic element $\bf M$ satisfies the equation:

$\frac{16\times7x + t}{(50.94\times3+16\times7)x + t}=0.4656$

Substituting $x=1,2,3\cdots$ respectively, it is inferred that when $x=6$, there is the most reasonable solution $t=126.9$, which is the $\rm I$ element. Therefore, the heteroatom is $\rm I$, and options B and C are both incorrect.

<Checkpoint>[Calculations and reasoning show that the heteroatom is $\rm I$](3)</Checkpoint>

Thus, the chemical formula of the radical is $\rm V_{18}O_{42}I^{5-}$, and the calculated mass fraction of oxygen is $\rm 39.16\%$, so options D and E are both incorrect.

<Checkpoint>[The chemical formula of the radical is $\rm V_{18}O_{42}I^{5-}$](1)</Checkpoint>

Since each $\mathrm{V}$ has a square pyramidal coordination, each $\mathrm{V}$ has one terminal oxygen coordination, and the number of vertices is the number of oxygen atoms, 42;

<Checkpoint>[Number of vertices is 42](1)</Checkpoint>

The ratio of non-terminal oxygen atoms to vanadium atoms is $(42-18):18=4:3$, each $\rm V$ connects four non-terminal oxygen atoms, so the non-terminal oxygen atom is $4\times \frac 34=3$ coordinated; therefore, the number of edges is $(42-18)\times3=72$.

<Checkpoint>[Non-terminal oxygen atom is 3-coordinated](1)</Checkpoint>

<Checkpoint>[Number of edges is $72$](1)</Checkpoint>

According to Euler's theorem, the number of faces is $72-42+2=32$, option I is correct, and F, J, H, G are incorrect.

<Checkpoint>[According to Euler's theorem, the number of faces is $72-42+2=32$](1)</Checkpoint>
由于该杂原子位于笼的中心, $\mathrm{V}$为四方锥型配位，由于几何构型限制端基$\mathrm{O}$只能朝向外侧, 根据库伦作用，可以猜测该杂原子带有负电荷, 也即一非金属离子。（另外，钒氧骨架中金属元素的质量分数为$57.71\%>1-46.56\%$，所以该杂原子必为非金属）

<Checkpoint>[由于几何构型限制端基$\mathrm{O}$只能朝向外侧](1)</Checkpoint>

<Checkpoint>[ 根据库伦作用，可以猜测该杂原子带有负电荷, 也即一非金属离子](1)</Checkpoint>

因此设该酸根的化学式为 $\rm (V_3O_7)_x\bf{M}\rm ^{5-}$, 此时非金属元素$\bf M$的原子量$m$满足方程：

$\frac{16\times7x + t}{(50.94\times3+16\times7)x + t}=0.4656$

将 $x=1,2,3\cdots$ 等分别代入, 推知 $x=6$ 时有最合理解$t=126.9$，为$\rm I$元素。因此该杂原子为$\rm I$，选项B,C均错误。

<Checkpoint>[计算推理得该杂原子为$\rm I$](3)</Checkpoint>

由此该酸根的化学式为$\rm V_{18}O_{42}I^{5-}$，计算可得氧的质量分数为$\rm 39.16\%$，选项D,E均错误。

<Checkpoint>[该酸根的化学式为$\rm V_{18}O_{42}I^{5-}$](1)</Checkpoint>

由于每个$\mathrm{V}$均为四方锥配位，故每个$\mathrm{V}$均有一个端基氧配位，顶点数为氧原子个数42；

<Checkpoint>[顶点数为42](1)</Checkpoint>

非端基氧原子与钒原子比例为$(42-18):18=4:3$，每个$\rm V$连接四个非端基氧原子，故非端基氧原子为$4\times \frac 34=3$配位；因此棱数为$(42-18)\times3=72$。

<Checkpoint>[非端基氧原子为3配位](1)</Checkpoint>

<Checkpoint>[棱数为$72$](1)</Checkpoint>

根据欧拉定理计算可知面数为$72-42+2=32$，选项I正确，F,J,H,G错误。

<Checkpoint>[根据欧拉定理计算可知面数为$72-42+2=32$](1)</Checkpoint>