In general atomic fullerene inclusion compounds, atoms can either form bonds with the fullerene or have no bonding interaction with the fullerene. Reacting $\mathrm{K_3As_7, Ni(COD)_2, Bu_4PBr}$ in ethylenediamine solution yields a precipitate of tetrabutylphosphonium salt. Single-crystal diffraction confirms that the structure of the anion in this salt can be represented as $\mathrm{{[As@Ni_m@As_n]}^{3-}}$. (The structure before the @ symbol is "included" within the structure after the symbol.) This anion has the same symmetry as a regular icosahedron and consists solely of $\mathrm{As}$ and $\mathrm{Ni}$ elements, with a nickel mass fraction of $30.92\%$. Treating the outermost $\mathrm{As_n}$ layer as an independent molecule,

Given: the covalent radius of arsenic is 119 pm, and its van der Waals radius is 185 pm; the nickel atoms in the $\mathrm{Ni_m}$ cluster are in contact with each other, and the distance between the farthest-spaced nickel atoms is 516 pm, while the van der Waals radius of Ni is 163 pm; the As-As distance in $\mathrm{As_n}$ is 275.2 pm, and the distance between the farthest-spaced arsenic atoms is 771.2 pm.

Calculate the distance between each vertex $\mathrm{Ni}$ atom in the $\mathrm{Ni_m}$ cluster and the nearest $\mathrm{As}$ atom on the $\mathrm{As_n}$ layer, and select the closest answer.
在一般的原子富勒烯包合物中，原子可以与富勒烯成键，也可以与富勒烯没有成键作用。在乙二胺溶液中使$\mathrm{K_3As_7, Ni(COD)_2,Bu_4PBr}$反应可以得到一种四丁基鏻盐沉淀。沉淀。单晶衍射证实该盐中阴离子的结构可以表示为$\mathrm{{[As@Ni_m@As_n]}^{3-}}$。（@符号前的结构被"包合"在符号后的结构中)。该阴离子有与正二十面体相同的对称性，且仅由$\mathrm{As}$和$\mathrm{Ni}$两种元素组成，其中镍的质量分数为$30.92\%$ 。将最外层的$\mathrm{As_n }$层看作一个独立的分子，

已知：砷的共价半径为119 pm，范德华半径为185 pm；$\mathrm{Ni_m}$  簇中的镍原子互相接触，间隔最远的镍原子之间的距离为516 pm,Ni的范德华半径为163 pm；$\mathrm{As_n }$ 中 As-As 距离为275.2 pm，间隔最远的砷原子之间的距离为 771.2 pm。

计算 $\mathrm{Ni_m}$ 簇每个顶点$\mathrm{Ni}$原子与$\mathrm{As_n }$层上最近的$\mathrm{As}$原子的距离，选取最接近的答案。
239 pm
234 pm
306 pm
307 pm
613 pm
163 pm
185 pm
239 pm
246 pm
237 pm
234 pm
306 pm
307 pm
613 pm
163 pm
185 pm
239 pm
246 pm
First, it is necessary to deduce the structure of this fullerene inclusion compound, i.e., to determine the values of m and n.  
Since the mass fraction of nickel in the anion is known, the ratio of the number of arsenic atoms to nickel atoms in the anion can be derived as $(0.3092/58.69):((1-0.3092)/74.92)=1:1.75=4:7$.  

<Checkpoint>[The ratio of arsenic atoms to nickel atoms in the anion is 4:7](1)</Checkpoint>  

The anion has the same symmetry as a regular icosahedron, so the m vertices and n vertices should form a polyhedron consistent with icosahedral symmetry. In practice, only the regular dodecahedron and icosahedron satisfy this condition.  

<Checkpoint>[The m vertices and n vertices should form a polyhedron consistent with icosahedral symmetry](1)</Checkpoint>  

Thus, there are at least 12 nickel atoms, with 21 arsenic atoms, where 20 arsenic atoms form a regular icosahedron. Therefore, m is 12 and n is 20.  

<Checkpoint>[m is 12, n is 20](1)</Checkpoint>  

To satisfy the symmetry of a regular icosahedron, the vertices of the $\mathrm{Ni}$ polyhedron must align with the face centers of the $\mathrm{As}$ polyhedron.  

<Checkpoint>[The vertices of the $\mathrm{Ni}$ polyhedron must align with the face centers of the $\mathrm{As}$ polyhedron.](1)</Checkpoint>  

Let the distance between the farthest face centers of the $\mathrm{As}_{20}$ polyhedron be $d$, and the distance from a vertex to the face center of each $\mathrm{As}_{5}$ face be $r$.  
Then  

$$r=(275.2 / 2) / \sin 36^{\circ} \mathrm{pm}=234.1 \mathrm{pm}$$  

<Checkpoint>[The distance from a vertex to the face center of each $\mathrm{As}_{5}$ face is 234.1 pm](1)</Checkpoint>  

$d / 2=\left[\left(771.2 / 2\right)^{2}-234.1^{2}\right]^{1 / 2} \mathrm{pm}=306.4 \mathrm{pm}$  

$d=612.8 \mathrm{pm}$  

The distance from a nickel atom to the nearest face center of the $\mathrm{As}_{20}$ polyhedron is  

$(612.8-516) / 2 \mathrm{pm}=48.4 \mathrm{pm}$  

<Checkpoint>[The distance from a nickel atom to the nearest face center of the $\mathrm{As}_{20}$ polyhedron is 48.4 pm](1)</Checkpoint>  

Based on the right triangle relationship, $d(\mathrm{Ni}-\mathrm{As})=\left(48.4^{2}+234.1^{2}\right) \mathrm{pm}=239 \mathrm{pm}$  

<Checkpoint>[$d(\mathrm{Ni}-\mathrm{As})=\left(48.4^{2}+234.1^{2}\right) \mathrm{pm}=239 \mathrm{pm}$](1)</Checkpoint>  

Thus, option H is correct.
首先需要推出该富勒烯包合物的结构，也就是解出m和n的值。
由于知道阴离子中镍的质量分数，从而可以得到该阴离子中，砷元素和镍元素原子个数之比为$(0.3092/58.69):((1-0.3092)/74.92)=1:1.75=4:7$。

<Checkpoint>[该阴离子中，砷元素和镍元素原子个数之比为4：7](1)</Checkpoint>

该阴离子有与正二十面体相同的对称性，从而m个顶点和n个顶点应当能构成符合正二十面体对称性的多面体，实际上也只有正十二面体和正二十面体。

<Checkpoint>[m个顶点和n个顶点应当能构成符合正二十面体对称性的多面体](1)</Checkpoint>

所以镍元素至少有12个，此时砷元素有21个，20个砷元素刚好构成正二十面体；因此m为12，n为20。

<Checkpoint>[m为12，n为20](1)</Checkpoint>

若要满足正二十面体的对称性, 则 $\mathrm{Ni}$ 多面体的顶点需要正对 $\mathrm{As}$ 多面体的面心.

<Checkpoint>[$\mathrm{Ni}$ 多面体的顶点需要正对 $\mathrm{As}$ 多面体的面心.](1)</Checkpoint>

设 $\mathrm{As}_{20}$ 多面体相距最远的面心间距为 $d$, 每个 $\mathrm{As}_{5}$ 面上顶点到面心的距离为 $r$
则 

$$r=(275.2 / 2) / \sin 36^{\circ} \mathrm{pm}=234.1 \mathrm{pm}$$ 

<Checkpoint>[每个 $\mathrm{As}_{5}$ 面上顶点到面心的距离为 234.1 pm](1)</Checkpoint>

$d / 2=\left[\left(771.2 / 2\right)^{2}-234.1^{2}\right]^{1 / 2} \mathrm{pm}=306.4 \mathrm{pm}$

$d=612.8 \mathrm{pm}$ 

则镍原子到距离最近的 $\mathrm{As}_{20}$ 多面体面心之间的距离为 

$(612.8-516) / 2 \mathrm{pm}=48.4 \mathrm{pm}$ 

<Checkpoint>[镍原子到距离最近的 $\mathrm{As}_{20}$ 多面体面心之间的距离为 48.4 pm ](1)</Checkpoint>

根据直角三角形关系，$d(\mathrm{Ni}-\mathrm{As})=\left(48.4^{2}+234.1^{2}\right) \mathrm{pm}=239 \mathrm{pm}$

<Checkpoint>[$d(\mathrm{Ni}-\mathrm{As})=\left(48.4^{2}+234.1^{2}\right) \mathrm{pm}=239 \mathrm{pm}$](1)</Checkpoint>

从而选项H正确。