A metallic element $\mathbf{M}$ can form a $1:1:1$ compound crystal with fluorine and oxygen. The structure of this crystal is as follows: $\mathrm{O}$ atoms are arranged in an ordered manner, forming parallel stacked oxygen layers; a group of $\mathbf{M}$ atoms and $\mathrm{F}$ atoms together fill the elongated cubic voids formed by the $\mathrm{O}$ atoms, with $\mathbf{M}$ adopting a mono-capped square antiprismatic coordination. If the $\mathrm{O}$ atoms are taken as the vertices of the unit cell, the coordinates of one $\mathrm{F}$ atom are $\left(\frac{1}{2},0,0.3476\right)$, and the $\mathrm{F}$ atoms form a distorted tetrahedron within this unit cell. In this crystal, the distance between adjacent oxygen atoms is $d_\mathrm{O-O}=265.6\ \mathrm{pm}$, and the lattice parameter $a=375.6\ \mathrm{pm}$.  

If the distorted tetrahedron formed by the $\mathrm{F}$ atoms exhibits two types of $\mathrm{\angle F-F-F}$ angles, the larger of which is $70.2\degree$, calculate the value of the lattice parameter $c$.
某金属元素$\mathbf{M}$可与氟、氧两种元素构成$1:1:1$化合物晶体。该晶体的结构为：$\mathrm{O}$原子有序排布，形成平行堆积的氧原子层；一组$\mathbf{M}$原子与$\mathrm{F}$原子共同填入$\mathrm{O}$原子形成的拉长立方体空隙中，$\mathbf{M}$为单帽四方反棱柱配位；若以$\mathrm{O}$原子为晶胞顶点，其中一个$\mathrm{F}$原子的坐标为$\left(\frac{1}{2},0,0.3476\right)$，$\mathrm{F}$在该晶胞中构成变形四面体。该晶体中，相邻氧原子间的距离为$d_\mathrm{O-O}=265.6\ \mathrm{pm}$，晶胞参数$a=375.6\ \mathrm{pm}$。

若$\mathrm{F}$原子构成的变形四面体中存在两种$\mathrm{\angle F-F-F}$角度，较大的一个为$70.2\degree$，求晶胞参数$c$的值。
$316.8\ \mathrm{pm}$
$623.6\ \mathrm{pm}$
$1247.2\ \mathrm{pm}$
$2594\ \mathrm{pm}$
$399.2\ \mathrm{pm}$
$798.3\ \mathrm{pm}$
$1596.6\ \mathrm{pm}$
$3193\ \mathrm{pm}$
All other options are incorrect.
$316.8\ \mathrm{pm}$
$623.6\ \mathrm{pm}$
$1247.2\ \mathrm{pm}$
$2594\ \mathrm{pm}$
$399.2\ \mathrm{pm}$
$798.3\ \mathrm{pm}$
$1596.6\ \mathrm{pm}$
$3193\ \mathrm{pm}$
其他选项均不正确
Since $\frac{d_\mathrm{O-O}}{a}=\frac{1}{\sqrt{2}}$, it can be reasonably inferred that when one $\mathrm{O}$ atom is located at the vertex of the unit cell, there is another $\mathrm{O}$ atom located at the face center of the bottom face, with coordinates $\left(\frac{1}{2},\frac{1}{2},0\right)$.

<Checkpoint>[$\frac{d_\mathrm{O-O}}{a}=\frac{1}{\sqrt{2}}$](1)</Checkpoint>

<Checkpoint>[There is an $\mathrm{O}$ atom located at $\left(\frac{1}{2},\frac{1}{2},0\right)$](1)</Checkpoint>

Therefore, the number of formula units in this unit cell is at least 2, and the crystal has tetragonal symmetry.

<Checkpoint>[The crystal has tetragonal symmetry](1)</Checkpoint>

The coordinates of one $\mathrm{F}$ atom are $\left(\frac{1}{2},0,0.3476\right)$. Considering that the 9 vertices of the monocapped square antiprism can be divided into 3 layers, including two squares staggered by $45°$ and a single vertex located in the $z$ direction. Therefore, the coordinates of one $\mathbf{M}$ atom can be set as $\left(\frac{1}{2},0,z\right)$, which is located on the midline of one side of the unit cell, aligned with the known coordinates of one $\mathrm{F}$ atom. $\mathrm{F}$ forms the single vertex of the monocapped square antiprism, and 4 $\mathrm{O}$ atoms constitute one square layer of the monocapped square antiprism, with the direction of the remaining square layer differing by $45°$.

<Checkpoint>[One $\mathbf{M}$ is located at $\left(\frac{1}{2},0,z\right)$](1)</Checkpoint>

<Checkpoint>[The single vertex of the monocapped square antiprism is $\mathrm{F}$, and one square layer is composed of $\mathrm{O}$](2)</Checkpoint>

Now consider the other square layer of the monocapped square antiprism. Since the coordinates of two $\mathrm{O}$ atoms in the unit cell are already known, if both the upper and lower layers are composed of $\mathrm{O}$, then an $\mathrm{O-F-F-O}$ layer will inevitably exist. Continuous stacking of four layers of anions is very unstable, and it is difficult to satisfy the $1:1:1$ ratio if both the upper and lower layers are $\mathrm{O}$. Therefore, consider using $\mathrm{F}$ to construct the other square layer.

<Checkpoint>[The other square layer of the monocapped square antiprism is composed of $\mathrm{F}$](1)</Checkpoint>

Since the directions of the two square layers in the monocapped square antiprism differ by $45°$, and the $\mathrm{O}$ square is $\left(0,0,0\right),\left(1,0,0\right),\left(\frac{1}{2},\frac{1}{2},0\right),\left(\frac{1}{2},-\frac{1}{2},0\right)$, with its sides along the face diagonal direction, the sides of the square composed of $\mathrm{F}$ atoms should be along the coordinate axis direction.

In order to maintain a formula unit number of 2, only one additional $\mathrm{F}$ atom can be introduced into a unit cell, with a reasonable position being $x=0,y=\frac{1}{2}$. Since $\mathrm{F}$ forms a distorted tetrahedron in the unit cell with $\mathrm{O}$ as the vertex, its $z$ coordinate cannot be the same as the known $0.3476$. Due to tetragonal symmetry, a reasonable coordinate is $z=1-0.3476=0.6524$, i.e., the coordinates of the other $\mathrm{F}$ atom are $\left(0,\frac{1}{2},0.6524\right)$.

<Checkpoint>[The coordinates of the other $\mathrm{F}$ atom are $\left(0,\frac{1}{2},0.6534\right)$](1)</Checkpoint>

In this way, two monocapped square antiprisms are formed in the unit cell, one with a single vertex facing up and the other facing down, which exactly accommodates 2 $\mathbf{M}$ atoms, satisfying the $1:1:1$ atomic ratio.

<Checkpoint>[There are 2 monocapped square antiprisms in the unit cell](1)</Checkpoint>

The specific unit cell structure is as follows, where the red spheres are $\mathrm{O}$, the green spheres are $\mathrm{F}$, and the gray spheres are $\mathbf{M}$.

<MultiModal>![图中有一个由灰色线描绘的长方体，其高度大于水平方向的长宽，图中还有红色、灰色、绿色三种球以及连接的黄色线段。其中，大量红色球和灰色球按1:1的数量比构成两个相同的、上下起伏的层，分别位于长方体的顶面和底面，同一层的红色球高度一致，占据长方体的顶点以及相应顶面或底面的面心，灰色球位于红色球形成的平面上下两侧，每对相邻的灰色球与红色球之间通过一条黄色线段相连，每个灰色球与4个红色球相连，每个红色球与4个灰色球相连，以长方体的侧面为参照，位于前后侧面的灰色球位置高于所连接的红色球，位于左右侧面的灰色球低于所连接的红色球。图中画出了4个绿色球分别位于前后侧面的中部偏上和左右侧面的中部偏下。](/media/uploads/d058d568-b3b5-47db-95f0-923a536b6215.png)</MultiModal>

In this unit cell, the tetrahedron formed by $\mathrm{F}$ consists of 4 congruent isosceles triangles, belonging to the $D_\mathrm{2d}$ point group. The $\angle\mathrm{F-F-F}$ bond angles are of two types, namely the apex angle and the base angle of the isosceles triangle.

<Checkpoint>[There are two types of $\angle\mathrm{F-F-F}$ bond angles, namely the apex angle and the base angle of the isosceles triangle](1)</Checkpoint>

In this tetrahedron, there is a pair of mutually perpendicular edges with a length of $a$. The other 4 oblique edge lengths are $\sqrt{\left(\frac{a}{\sqrt{2}}\right)^2 + \left((1-2 \times 0.3476) c\right)^2} = \sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}$.

<Checkpoint>[The two types of edge lengths of the tetrahedron are $a$ and $\sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}$](1)</Checkpoint>

From geometric relationships, if the apex angle of the isosceles triangle is $70.2°$, then $\sin{\frac{70.2°}{2}}=\frac{\frac{a}{2}}{\sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}}$. Substituting and solving gives $c=623.6\ \mathrm{pm}$.

<Checkpoint>[If the apex angle is $70.2°$, then $c=623.6\ \mathrm{pm}$](1)</Checkpoint>

If the base angle of the isosceles triangle is $70.2°$, then $\cos{70.2°}=\frac{\frac{a}{2}}{\sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}}$. Substituting and solving gives $c=1596.6\ \mathrm{pm}$.

<Checkpoint>[If the base angle is $70.2°$, then $c=1596.6\ \mathrm{pm}$](1)</Checkpoint>

Observing the options, it can be seen that the correct option should be one of $\mathtt{B,G}$. To determine which of the above classifications is physically realistic, consider the distance between each vertex in the coordination polyhedron and $\mathbf{M}$. It has been previously assumed that the coordinates of one $\mathbf{M}$ atom are $\left(\frac{1}{2},0,z\right)$. From the previous discussion, it is known that in order to satisfy the structure of the 9-coordinate monocapped square antiprism, it is necessary that $0.3476 \le z \le 1$, but the specific coordinates cannot be deduced.

<Checkpoint>[$0.3476 \le z \le 1$](1)</Checkpoint>

From the atomic coordinates in the crystal, there are two types of $\mathbf{M}-\mathrm{F}$ bond lengths (axial and non-axial) and one type of $\mathbf{M}-\mathrm{O}$ bond length, which are $\left(z-0.3476\right)c$ (axial $\mathbf{M}-\mathrm{F}$), $\sqrt{\frac{a^2}{2}+\left(\left(z-0.6524\right)c\right)^2}$ (non-axial $\mathbf{M}-\mathrm{F}$) and $\sqrt{\left(\frac{a}{2}\right)^2+\left(\left(1-z\right)c\right)^2}$ ($\mathbf{M}-\mathrm{O}$).

<Checkpoint>[There are two types of $\mathbf{M}-\mathrm{F}$ bond lengths (axial and non-axial) and one type of $\mathbf{M}-\mathrm{O}$ bond length](1)</Checkpoint>

<Checkpoint>[The axial $\mathbf{M}-\mathrm{F}$ bond length is $\left(z-0.3476\right)c$](1)</Checkpoint>

<Checkpoint>[The non-axial $\mathbf{M}-\mathrm{F}$ bond length is $\sqrt{\frac{a^2}{2}+\left(\left(z-0.6524\right)c\right)^2}$](1)</Checkpoint>

<Checkpoint>[The $\mathbf{M}-\mathrm{O}$ bond length is $\sqrt{\left(\frac{a}{2}\right)^2+\left(\left(1-z\right)c\right)^2}$](3)</Checkpoint>

Let $c=623.6\ \mathrm{pm}$ and $c=1596.6\ \mathrm{pm}$ respectively, and plot the three bond lengths against z.

For the case of $c=623.6\ \mathrm{pm}$, we have:

<MultiModal>![该图为二维坐标图。横坐标标签为z，范围0~1.5，每0.2有一深灰线并附带坐标，每0.05有一浅灰线；纵坐标标签为d/nm，范围为0~0.5，在0.2和0.4处有一深灰线并附带坐标，每0.05有一浅灰线。图中有蓝、绿、紫三条彩色线，三条线的左端均位于横坐标0.375附近且略向左超出，右端均位于横坐标1处。从左往右看，蓝色线为上升的直线段，左端纵坐标为0，右端纵坐标略高于0.4。绿色线先降后升，两个端点均位于纵坐标0.3~0.35之间，右端点略高，绿色线在横坐标0.65附近达到最低点，其纵坐标略高于0.25。紫色线先直线下降后向右弯曲至水平，左端点纵坐标约为0.45，右端点纵坐标略低于0.2。三条彩色线在横坐标约0.7~0.8，纵坐标约0.25至0.3范围内形成3个交点，蓝-紫两线的交点位于绿色线下方。](/media/uploads/ba1e0714-d7cc-4305-bc54-080c681ad785.png)</MultiModal>

For the case of $c=1596.6\ \mathrm{pm}$, we have:

<MultiModal>![该图为二维坐标图。横坐标标签为z，范围0~1.5，每0.2有一深灰线并附带坐标，每0.05有一浅灰线；纵坐标标签为d/nm，范围为0~1.1，每0.2有一深灰线并附带坐标，每0.05有一浅灰线。图中有蓝、绿、紫三条彩色线，三条线的左端均位于横坐标0.375附近且略向左超出，右端均位于横坐标1处。从左往右看，蓝色线为上升的直线段，左端纵坐标为0，右端纵坐标略低于1.05。绿色线先降后升，左端点纵坐标略高于0.55，右端点纵坐标略高于0.6，绿色线在横坐标0.65附近达到最低点，其纵坐标略高于0.25。紫色线先直线下降后向右弯曲至水平，直线下降段斜率与蓝色线斜率的负值相近，左端点纵坐标略高于1.05，右端点纵坐标略低于0.2。三条彩色线在横坐标略低于0.45至略高于0.8，纵坐标约0.3至0.55范围内形成3个交点，蓝-紫两线的交点位于绿色线相应横坐标处上方。](/media/uploads/b5d8ff57-7c7e-4f69-a0ef-701591240454.png)</MultiModal>

In the two figures, the blue line is the axial $\mathbf{M}-\mathrm{F}$ bond length, the green line is the non-axial $\mathbf{M}-\mathrm{F}$ bond length, and the purple line is the $\mathbf{M}-\mathrm{O}$ bond length.

From the figures, it can be seen that when $c=623.6\ \mathrm{pm}$, a suitable $z$ value can be found such that the bond lengths are in the range of $0.25\ \mathrm{nm}$ to $0.3\ \mathrm{nm}$ and the bond length difference is small, which is consistent with the general bond length range of actual ionic crystals formed by $\mathrm{F,O}$. However, when $c=1596.6\ \mathrm{pm}$, regardless of the value of $z$ within the range, the difference in bond lengths reaches $1.7$ to $1.8$ times, which is inconsistent with objective facts. Therefore, the only correct option is $\mathtt{B}$.

<Checkpoint>[When $c=623.6\ \mathrm{pm}$, the bond length range is suitable](1)</Checkpoint>

<Checkpoint>[When $c=1596.6\ \mathrm{pm}$, the bond length difference is too large](1)</Checkpoint>
由于$\frac{d_\mathrm{O-O}}{a}=\frac{1}{\sqrt{2}}$，可合理推断，当一个$\mathrm{O}$原子位于晶胞顶点时，存在另一个$\mathrm{O}$原子位于底面面心，坐标为$\left(\frac{1}{2},\frac{1}{2},0\right)$。

<Checkpoint>[$\frac{d_\mathrm{O-O}}{a}=\frac{1}{\sqrt{2}}$](1)</Checkpoint>

<Checkpoint>[存在$\mathrm{O}$原子位于$\left(\frac{1}{2},\frac{1}{2},0\right)$](1)</Checkpoint>

故该晶胞的式量数至少为2，晶体具有四方对称性。

<Checkpoint>[晶体具有四方对称性](1)</Checkpoint>

一个$\mathrm{F}$原子的坐标为$\left(\frac{1}{2},0,0.3476\right)$，考虑到单帽四方反棱柱的9个顶点可分为3层，包含两个相互错开$45°$的正方形和位于$z$方向的单个顶点。因此，可以设一个$\mathbf{M}$原子的坐标为$\left(\frac{1}{2},0,z\right)$，其位于晶胞的一个侧面的中线上，与已知的1个$\mathrm{F}$原子坐标对齐，$\mathrm{F}$形成单帽四方反棱柱的单个顶点，4个$\mathrm{O}$原子构成单帽四方反棱柱的一个正方形层，剩下一个正方形层的方向与之相差$45°$。

<Checkpoint>[一个$\mathbf{M}$位于$\left(\frac{1}{2},0,z\right)$](1)</Checkpoint>

<Checkpoint>[单帽四方反棱柱的单个顶点为$\mathrm{F}$，一个正方形层由$\mathrm{O}$构成](2)</Checkpoint>

现在考虑单帽四方反棱柱的另一个正方形层，由于晶胞中已有2个$\mathrm{O}$原子的坐标已知，如果上下层都由$\mathrm{O}$构成，那么不可避免的会存在$\mathrm{O-F-F-O}$层，四层阴离子连续堆积很不稳定，并且如果上下层都是$\mathrm{O}$很难满足比例为$1：1：1$，因此考虑用$\mathrm{F}$来构建另一个正方形层。

<Checkpoint>[单帽四方反棱柱的另一个正方形层由$\mathrm{F}$构成](1)</Checkpoint>

由于单帽四方反棱柱中的两个正方形层方向相差$45°$，而$\mathrm{O}$正方形为$\left(0,0,0\right),\left(1,0,0\right),\left(\frac{1}{2},\frac{1}{2},0\right),\left(\frac{1}{2},-\frac{1}{2},0\right)$，其边为面对角线方向，这个由$\mathrm{F}$原子构成的正方形的边应当沿坐标轴方向。

为了保持式量数为2，一个晶胞中只能额外引入1个$\mathrm{F}$原子，合理的位置为$x=0,y=\frac{1}{2}$。由于$\mathrm{F}$在以$\mathrm{O}$为顶点的晶胞中构成变形四面体，其$z$坐标不能与已知的$0.3476$相同，由四方对称性，合理的坐标是$z=1-0.3476=0.6524$，即另一个$\mathrm{F}$原子的坐标为$\left(0,\frac{1}{2},0.6524\right)$。

<Checkpoint>[另一个$\mathrm{F}$原子的坐标为$\left(0,\frac{1}{2},0.6534\right)$](1)</Checkpoint>

这样，晶胞中就形成了两个单帽四方反棱柱，一个单顶点朝上，另一个朝下，其恰好容纳2个$\mathbf{M}$原子，满足$1:1:1$原子比。

<Checkpoint>[晶胞中有2个单帽四方反棱柱](1)</Checkpoint>

具体的晶胞结构如下，其中红色球为$\mathrm{O}$，绿色球为$\mathrm{F}$，灰色球为$\mathbf{M}$。

<MultiModal>![图中有一个由灰色线描绘的长方体，其高度大于水平方向的长宽，图中还有红色、灰色、绿色三种球以及连接的黄色线段。其中，大量红色球和灰色球按1:1的数量比构成两个相同的、上下起伏的层，分别位于长方体的顶面和底面，同一层的红色球高度一致，占据长方体的顶点以及相应顶面或底面的面心，灰色球位于红色球形成的平面上下两侧，每对相邻的灰色球与红色球之间通过一条黄色线段相连，每个灰色球与4个红色球相连，每个红色球与4个灰色球相连，以长方体的侧面为参照，位于前后侧面的灰色球位置高于所连接的红色球，位于左右侧面的灰色球低于所连接的红色球。图中画出了4个绿色球分别位于前后侧面的中部偏上和左右侧面的中部偏下。](/media/uploads/d058d568-b3b5-47db-95f0-923a536b6215.png)</MultiModal>

在该晶胞中，$\mathrm{F}$形成的四面体由4个全等的等腰三角形构成，属$D_\mathrm{2d}$点群，其$\angle\mathrm{F-F-F}$夹角有两种，分别为等腰三角形的顶角和底角。

<Checkpoint>[$\angle\mathrm{F-F-F}$夹角有两种，分别为等腰三角形的顶角和底角](1)</Checkpoint>

在这个四面体中，存在一对相互垂直的棱，其长度为$a$，另外4条斜向棱长为$\sqrt{\left(\frac{a}{\sqrt{2}}\right)^2 + \left((1-2 \times 0.3476) c\right)^2} = \sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}$。

<Checkpoint>[四面体的两种棱长分别为$a$和$\sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}$](1)</Checkpoint>

由几何关系，若等腰三角形的顶角为$70.2°$，则$\sin{\frac{70.2°}{2}}=\frac{\frac{a}{2}}{\sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}}$，代入解得$c=623.6\ \mathrm{pm}$。

<Checkpoint>[若顶角为$70.2°$，则$c=623.6\ \mathrm{pm}$](1)</Checkpoint>

若等腰三角形的底角为$70.2°$，则$\cos{70.2°}=\frac{\frac{a}{2}}{\sqrt{\frac{a^2}{2} + \left(0.3048c\right)^2}}$，代入解得$c=1596.6\ \mathrm{pm}$。

<Checkpoint>[若底角为$70.2°$，则$c=1596.6\ \mathrm{pm}$](1)</Checkpoint>

观察选项，可知正确选项应为$\mathtt{B,G}$之一。为了确定上述分类讨论中，哪一种是符合物理实际的。考虑配位多面体中各顶点与$\mathbf{M}$的距离。前面已设一个$\mathbf{M}$原子的坐标为$\left(\frac{1}{2},0,z\right)$，由前述讨论可知为了满足9配位单帽四方反棱柱的结构，需要$0.3476 \le z \le 1$，但具体坐标无法推算。

<Checkpoint>[$0.3476 \le z \le 1$](1)</Checkpoint>

由晶体中的原子坐标，有两种$\mathbf{M}-\mathrm{F}$键长（轴向和非轴向）和一种$\mathbf{M}-\mathrm{O}$键长，分别为$\left(z-0.3476\right)c$（轴向$\mathbf{M}-\mathrm{F}$），$\sqrt{\frac{a^2}{2}+\left(\left(z-0.6524\right)c\right)^2}$（非轴向$\mathbf{M}-\mathrm{F}$）和$\sqrt{\left(\frac{a}{2}\right)^2+\left(\left(1-z\right)c\right)^2}$（$\mathbf{M}-\mathrm{O}$）。

<Checkpoint>[有两种$\mathbf{M}-\mathrm{F}$键长（轴向和非轴向）和一种$\mathbf{M}-\mathrm{O}$键长](1)</Checkpoint>

<Checkpoint>[轴向$\mathbf{M}-\mathrm{F}$键长为$\left(z-0.3476\right)c$](1)</Checkpoint>

<Checkpoint>[非轴向$\mathbf{M}-\mathrm{F}$键长为$\sqrt{\frac{a^2}{2}+\left(\left(z-0.6524\right)c\right)^2}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{M}-\mathrm{O}$键长为$\sqrt{\left(\frac{a}{2}\right)^2+\left(\left(1-z\right)c\right)^2}$](3)</Checkpoint>

分别取$c=623.6\ \mathrm{pm}$和$c=1596.6\ \mathrm{pm}$，将三种键长对z作图。

对于$c=623.6\ \mathrm{pm}$的情形，有：

<MultiModal>![该图为二维坐标图。横坐标标签为z，范围0~1.5，每0.2有一深灰线并附带坐标，每0.05有一浅灰线；纵坐标标签为d/nm，范围为0~0.5，在0.2和0.4处有一深灰线并附带坐标，每0.05有一浅灰线。图中有蓝、绿、紫三条彩色线，三条线的左端均位于横坐标0.375附近且略向左超出，右端均位于横坐标1处。从左往右看，蓝色线为上升的直线段，左端纵坐标为0，右端纵坐标略高于0.4。绿色线先降后升，两个端点均位于纵坐标0.3~0.35之间，右端点略高，绿色线在横坐标0.65附近达到最低点，其纵坐标略高于0.25。紫色线先直线下降后向右弯曲至水平，左端点纵坐标约为0.45，右端点纵坐标略低于0.2。三条彩色线在横坐标约0.7~0.8，纵坐标约0.25至0.3范围内形成3个交点，蓝-紫两线的交点位于绿色线下方。](/media/uploads/ba1e0714-d7cc-4305-bc54-080c681ad785.png)</MultiModal>

对于$c=1596.6\ \mathrm{pm}$的情形，有：

<MultiModal>![该图为二维坐标图。横坐标标签为z，范围0~1.5，每0.2有一深灰线并附带坐标，每0.05有一浅灰线；纵坐标标签为d/nm，范围为0~1.1，每0.2有一深灰线并附带坐标，每0.05有一浅灰线。图中有蓝、绿、紫三条彩色线，三条线的左端均位于横坐标0.375附近且略向左超出，右端均位于横坐标1处。从左往右看，蓝色线为上升的直线段，左端纵坐标为0，右端纵坐标略低于1.05。绿色线先降后升，左端点纵坐标略高于0.55，右端点纵坐标略高于0.6，绿色线在横坐标0.65附近达到最低点，其纵坐标略高于0.25。紫色线先直线下降后向右弯曲至水平，直线下降段斜率与蓝色线斜率的负值相近，左端点纵坐标略高于1.05，右端点纵坐标略低于0.2。三条彩色线在横坐标略低于0.45至略高于0.8，纵坐标约0.3至0.55范围内形成3个交点，蓝-紫两线的交点位于绿色线相应横坐标处上方。](/media/uploads/b5d8ff57-7c7e-4f69-a0ef-701591240454.png)</MultiModal>

两图中蓝色线为轴向$\mathbf{M}-\mathrm{F}$键长，绿色线为非轴向$\mathbf{M}-\mathrm{F}$键长，紫色线为$\mathbf{M}-\mathrm{O}$键长。

由图中可以看出，当$c=623.6\ \mathrm{pm}$时，可以找到合适的$z$值，使得键长在$0.25\ \mathrm{nm}$至$0.3\ \mathrm{nm}$范围且键长差距不大，符合$\mathrm{F,O}$参与形成的实际离子晶体的一般键长范围。而当$c=1596.6\ \mathrm{pm}$时，无论$z$在范围内如何取值，键长的差距都达到了$1.7$至$1.8$倍，不符合客观事实。故唯一正确的选项是$\mathtt{B}$。

<Checkpoint>[$c=623.6\ \mathrm{pm}$时，键长范围合适](1)</Checkpoint>

<Checkpoint>[$c=1596.6\ \mathrm{pm}$时，键长差距过大](1)</Checkpoint>