$\mathbf { A }$ , $\mathbf { B }$ and $\mathbf { C }$ represent an unknown element, respectively. The reaction of $\mathbf { A } _ { 3 } \mathbf { C } _ { 2 }$ with $\mathbf { BC }$ yields a crystal $\bf{X}$ containing $\mathbf { A }$ , $\mathbf { B }$ and $\mathbf { C }$. $\bf{X}$ belongs to the orthorhombic crystal system with unit cell parameters: $a = 8 5 0 . 3$ pm, $b =$ 1028.4pm, $c = 5 0 3 . 4 ~ \mathrm { p m }$, where the arrangement of $\mathbf { A }$ and $\mathbf { C }$ is similar to that of $\mathrm {NaCl}$. It can be regarded as two deformed hexahedrons connected coplanarly along the $a$ axis to form the unit shown in the figure below. The two hexahedrons in each unit are then infinitely extended along the $c$ axis in a coplanar manner to form an infinite chain structure; the chains are not directly connected, but form an infinite three-dimensional structure by sharing $\bf C$ atoms with the $[ \mathbf { BC } _ { 3 } ]$ unit. The $[ \mathbf { BC } _ { 3 } ]$ units in $\bf{X}$ are arranged perpendicular to the $c$ axis, the $\mathbf{B}$ center has a planar triangular configuration with two orientations, and each $[ \mathbf { BC } _ { 3 } ]$ unit has a $\mathbf { B - C}$ bond parallel to the $b$ axis. It is known that all $\bf C$ atoms in the crystal participate in the coordination of $\bf B$.

<MultiModal>![“A schematic diagram. On the left is a structure consisting of two cubes connected coplanarly in the horizontal direction. On the right is a three-dimensional coordinate system with the a-axis pointing to the right, the b-axis pointing upwards, and the c-axis pointing to the front left.”](/media/uploads/a6a4257d-f0a4-4789-9ed1-d5089dcd9468.png)</MultiModal>

According to the crystal diffraction results, the coordinates of some $\mathbf {C}$ atoms in the crystal are: (0.500,0.624,0.750), (0.808,0.371,0.750); the bond length of the $\mathbf {B-C}$ bond parallel to the $b$ axis is $1 8 6 . 3 ~ \mathrm { p m }$. Select the correct $\mathbf {C-B-C}$ bond angle from the following options.
$\mathbf { A }$ 、$\mathbf { B }$ 和 $\mathbf { C }$ 分别表示一种未知元素，$\mathbf { A } _ { 3 } \mathbf { C } _ { 2 }$ 与 $\mathbf { BC }$ 反应可得到含 $\mathbf { A }$ 、$\mathbf { B }$ 和 $\mathbf { C }$的晶体$\bf{X}$，$\bf{X}$属于正交晶系，晶胞参数为： $a = 8 5 0 . 3$ pm， $b =$ 1028.4pm， $c = 5 0 3 . 4 ~ \mathrm { p m }$ ，其中$\mathbf { A }$、$\mathbf {C}$的排布情况与$\mathrm {NaCl}$类似，可视作两个变形六面体沿$a$ 轴方向共面连接形成下图所示单元，每个单元中的两个六面体再分别沿 $c$ 轴通过共面的方式无限延伸形成无限链状结构；链间不直接相连，而是通过与 $[ \mathbf { BC } _ { 3 } ]$ 单元共用$\bf C$原子形成无限三维结构。$\bf{X}$中 $[ \mathbf { BC } _ { 3 } ]$ 单元垂直于 $c$ 轴排布，$\mathbf{B}$中心为平面三角形构型，有2种取向，每个 $[ \mathbf { BC } _ { 3 } ]$ 单元中均有一根 $\mathbf { B - C}$ 键与 $b$ 轴平行。已知，该晶体中所有$\bf C$原子均参与对$\bf B$的配位。

<MultiModal>![“一张示意图。左侧是一个由两个沿水平方向共面连接的立方体组成的结构。右侧是一个三维坐标系，a轴指向右方，b轴指向上方，c轴指向左前方。”](/media/uploads/a6a4257d-f0a4-4789-9ed1-d5089dcd9468.png)</MultiModal>

根据晶体衍射结果，该晶体中部分$\mathbf {C}$原子的坐标为：（0.500,0.624,0.750)，(0.808,0.371,0.750)；与 $b$ 轴平行的 $\mathbf {B-C}$ 键键长为 $1 8 6 . 3 ~ \mathrm { p m }$。从下列选项中选出正确的$\mathbf {C-B-C}$ 键角大小。
All other options are incorrect.
113° and 124°
113° and 134°
118° and 124°
118° and 121°
113° and 118°
124° and 134°
121° and 134°
121° and 124°
其他选项均不正确。
113°和124°
113°和134°
118°和124°
118°和121°
113° 和118°
124° 和134°
121° 和134°
121° 和124°
According to the question, the coordination number of $\bf C$ on the shared face of the two parallel hexahedral chains is 6, and the coordinating atoms are 1 $\bf B$ and 5 $\bf A$, corresponding to the formation of $\mathbf {B-C}$ bonds parallel to the $b$ axis; the coordination number of $\bf C$ on the side faces is 5, and the coordinating atoms are 1 $\bf B$ and 4 $\bf A$.
Therefore, each $\bf C$ atom in the crystal is connected to 1 $\bf B$ atom, and there are 3 $\bf C$ atoms around 1 $\bf B$ atom, so the ratio of the number of $\bf B$ atoms to $\bf C$ atoms is 1 : 3. Since $\bf A$ and $\bf C$ are alternately arranged in the parallelepiped, the ratio of the two is 1 : 1. In summary, the crystal chemical formula is $\mathbf{A_3BC_3}$.

<Checkpoint>[There are two coordination forms of $\bf C$ atoms, one with a coordination number of 6, and the coordinating atoms are 1 $\bf B$ and 5 $\bf A$](1)</Checkpoint>

<Checkpoint>[The other coordination number of $\bf C$ atoms is 5, and the coordinating atoms are 1 $\bf B$ and 4 $\bf A$](1)</Checkpoint>

<Checkpoint>[The crystal chemical formula is $\mathbf{A_3BC_3}$](1)</Checkpoint>

According to the question, the hexahedral chain in the direction perpendicular to the $c$ axis satisfies translational symmetry, so the lattice type of the crystal belongs to base-centered orthorhombic. Combining the translational symmetry of the space lattice, the coordinates of the $\bf C$ atom with the same spatial environment as the $\bf C$ atom at the position (0.808,0.371,0.750) are (0.308,0.871,0.750).

<Checkpoint>[The lattice type of the crystal is base-centered orthorhombic](2)</Checkpoint>

Assuming the coordinate of a $\bf B$ atom in the crystal is $x$, the bond length of the $\mathbf{B-C}$ bond parallel to the $b$ axis can be obtained as:

$186.3 \mathrm{pm}=(x-0.624)\times1028.4 \mathrm{pm}$

Substituting to solve for $x=0.805$, so the coordinate of the $\bf B$ atom is (0.500,0.805,0.750).

<Checkpoint>[The coordinate of the $\bf B$ atom connected to the $\bf C$ atom with coordinate (0.500,0.624,0.750) is (0.500,0.805,0.750)](2)</Checkpoint>

Now calculate the bond angle based on the coordinates:

$\theta=\arctan[(0.871-0.805) b /(0.500-0.308)a]+90^{\circ}=113^{\circ}$

<Checkpoint>[One $\bf C-B-C$ bond angle is $113^{\circ}$](2)</Checkpoint>

Since $\bf B$ is planar triangular coordination, the other $\bf C-B-C$ bond angle is:

$360^{\circ}-113^{\circ} \times 2=134^{\circ}$

<Checkpoint>[The other $\bf C-B-C$ bond angle is $134^{\circ}$](2)</Checkpoint>

Actually, $\mathbf { A }$ , $\mathbf { B }$ and $\mathbf { C }$ represent elements $\mathrm { Ca }$ , $\mathrm { Cr }$ and $\mathrm { N }$ respectively.
根据题意，两条平行六面体链所共用面上的$\bf C$配位数为6，配位原子为1个$\bf B$、5个$\bf A$，对应形成与$b$轴平行的$\mathbf {B-C}$ 键；两侧面上的$\bf C$配位数为5，配位原子为1个$\bf B$、4个$\bf A$。
因此该晶体中每个$\bf C$原子均与1个$\bf B$原子相连，而1个$\bf B$原子周围有3个$\bf C$原子，因此$\bf B$原子和$\bf C$原子个数比为1 : 3。而平行六面体中$\bf A$、$\bf C$交替排布，因此二者个数比为1 : 1，综上晶体化学式为$\mathbf{A_3BC_3}$。

<Checkpoint>[$\bf C$原子有两种配位形式，一种配位数为6，配位原子为1个$\bf B$、5个$\bf A$](1)</Checkpoint>

<Checkpoint>[$\bf C$原子的另一种配位数为5，配位原子为1个$\bf B$、4个$\bf A$](1)</Checkpoint>

<Checkpoint>[晶体化学式为$\mathbf{A_3BC_3}$](1)</Checkpoint>

再根据题意，在垂直于 $c$ 轴方向六面体链满足平移对称性，因此该晶体所属的空间点阵型式为底心正交。结合空间点阵的平移对称性可知与 (0.808,0.371,0.750)位置处$\bf C$原子空间环境相同的$\bf C$原子坐标为 (0.308,0.871,0.750)。 

<Checkpoint>[晶体的点阵型式为底心正交](2)</Checkpoint>

假设晶体中一个$\bf B$原子的坐标为 $x$, 则根据与 $b$ 轴平行的$\mathbf{B-C}$键键长可得: 

$186.3 \mathrm{pm}=(x-0.624)\times1028.4 \mathrm{pm}$

代入解得 $x=0.805$, 因此该$\bf B$原子的坐标为(0.500,0.805,0.750)。 

<Checkpoint>[与坐标为（0.500,0.624,0.750)的$\bf C$原子相连的$\bf B$原子的坐标为(0.500,0.805,0.750)](2)</Checkpoint>

下面再结合坐标计算键角: 

$\theta=\arctan[(0.871-0.805) b /(0.500-0.308)a]+90^{\circ}=113^{\circ}$

<Checkpoint>[一种$\bf C-B-C$键角为$113^{\circ}$](2)</Checkpoint>

由于$\bf B$为平面三角形配位, 另一种$\bf C-B-C$键角为: 

$360^{\circ}-113^{\circ} \times 2=134^{\circ}$

<Checkpoint>[另一种$\bf C-B-C$键角为$134^{\circ}$](2)</Checkpoint>


实际上，$\mathbf { A }$ 、$\mathbf { B }$ 和 $\mathbf { C }$ 分别表示元素$\mathrm { Ca }$ 、$\mathrm { Cr }$ 和 $\mathrm { N }$ 。