A certain compound $\mathbf X$ contains only two elements, $\mathbf A$ and $\mathbf B$.

Under high pressure, $\mathbf X$ forms an atomic crystal. This crystal is orthorhombic, in which $\mathbf A$ is four-coordinated and $\mathbf B$ is two-coordinated, $a=622~\rm{pm},b=435~\rm{pm}, c=607~\rm{pm}, \rho=3.56~g\cdot cm^{-3}, 
Z=8$.

$\mathbf X$ can form a cubic molecular crystal under low temperature and high pressure, $\rm{Z = 4}$, and the molecules cannot rotate freely.

Under high pressure, $\mathbf X$ can also form a tetragonal molecular crystal with cell parameters $a = 354~\rm{pm}, c = 413~\rm{pm}, Z = 2$. In a proper unit cell, if the $\mathbf A$ atoms are at the vertices of the unit cell, then the coordinates of two $\mathbf B$ atoms are ($0.266,0.266,0$) and ($0.766,0.234,0.5$), respectively.

Please select all the correct options from the following propositions:

1.  $\mathbf X$ molecule has a center of symmetry
2.  The lattice type of the cubic crystal of $\mathbf X$ is simple cubic
3.  The tetragonal crystal of $\mathbf X$ has a center of symmetry
4.  The tetragonal crystal of $\mathbf X$ has a twofold rotation axis passing through the $\mathbf B$ atoms
5.  The tetragonal crystal of $\mathbf X$ does not have a twofold rotation axis not passing through the $\mathbf A$ atoms
某化合物$\mathbf X$仅含$\mathbf A$和$\mathbf B$两种元素。

在高压下，$\mathbf X$形成了一种原子晶体。此晶体为正交晶系，其中$\mathbf A$均为四配位，$\mathbf B$均为二配位，$a=622~\rm{pm},b=435~\rm{pm}, c=607~\rm{pm}, \rho=3.56~g\cdot cm^{-3}, 
Z=8$。

$\mathbf X$在低温高压下可形成一种立方晶系的分子晶体，$\rm{Z = 4}$，分子无法自由转动。

在高压下，$\mathbf X$还能形成一种四方晶系的分子晶体，晶胞参数$a = 354~\rm{pm}, c = 413~\rm{pm}, Z = 2$。在一个正当晶胞中，若以$\mathbf A$原子为晶胞顶点，则其中两个$\mathbf B$的坐标分别为($0.266,0.266,0$)和($0.766,0.234,0.5$)。

请选出以下命题中所有正确的选项：

1. $\mathbf X$分子有对称中心
2. $\mathbf X$的立方晶体的点阵类型是简单立方
3. $\mathbf X$的四方晶体有对称中心
4.  $\mathbf X$的四方晶体有穿过$\mathbf B$原子的二重旋转轴
5.  $\mathbf X$的四方晶体没有不穿过$\mathbf A$原子的二重旋转轴
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
1,2,3,4
1,2,3,5
1,2,4,5
1,3,4,5
1,2,3,4,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
1,2,3,4
1,2,3,5
1,2,4,5
1,3,4,5
1,2,3,4,5
First, deduce the chemical formula of $\mathbf X$:

Unit cell volume $V = a \times b \times c = 622~\rm{pm} \times 435~\rm{pm} \times 607~\rm{pm} = (1.643 \times 10^8) ~pm^3 \approx (1.643 \times 10^{-22})~\rm{cm}^3$

Molar mass of $\mathbf X$ $M = \frac{\rho \times V \times N_A}{Z} = \frac{3.56 \times 1.643 \times 10^{-22} \times 6.022 \times 10^{23}}{8} \approx 44.0 ~\rm{g }\cdot mol^{-1}$

<Checkpoint>[Molar mass of $\mathbf X$ $M \approx 44.0 ~\rm{g} \cdot mol^{-1}$](2)</Checkpoint>

Since the total number of chemical bonds is conserved, the coordination number of $\mathbf A$ to $\mathbf B$ is equal to the coordination number of $\mathbf B$ to $\mathbf A$, hence $4 \times n_A = 2 \times n_B$.

Therefore, the atomic ratio of $\mathbf A$ and $\mathbf B$ is 1:2

<Checkpoint>[The atomic ratio of $\mathbf A$ and $\mathbf B$ is 1:2](1)</Checkpoint>

Compounds that meet the above conditions may be $\rm{CO_2}$ or $\rm{N_2O}$, but it is difficult for $N_2O$ to achieve four-coordination of oxygen elements, so the chemical formula of $\mathbf X$ is $\rm{CO_2}$, $\mathbf A$ is $\rm{C}$, and $\mathbf B$ is $\rm{O}$

<Checkpoint>[$\mathbf X$ is $\rm{CO_2}$, $\mathbf A$ is $\rm{C}$, $\mathbf B$ is $\rm{O}$](1)</Checkpoint>

The $\rm{CO_2}$ molecule has a center of symmetry, so proposition 1 is correct.

<Checkpoint>[The $\rm{CO_2}$ molecule has a center of symmetry, proposition 1 is correct](1)</Checkpoint>

The molecular crystal of the cubic crystal system formed by $\rm{CO_2}$ has 4 $\rm{CO_2}$ molecules in its proper unit cell. If it is a face-centered cubic lattice, the spatial orientation of the 4 $\rm{CO_2}$ molecules must be consistent, which will lead to the loss of symmetry of the cubic crystal system. Therefore, under the conditions of the problem, the cubic crystal of $\mathbf X$ can only be a simple cubic lattice. Proposition 2 is correct.

<Checkpoint>[The cubic crystal of $\mathbf X$ is a simple cubic lattice, proposition 2 is correct](1)</Checkpoint>

Next, analyze the tetragonal crystal of $\mathbf X$. The ungiven atomic coordinates can be obtained with the help of the symmetry of the tetragonal crystal system and the molecular symmetry of $\rm{CO_2}$.

Let $\mathbf {C1} (0,0,0)$, $\mathbf {O1} (0.266,0.266,0)$, $\mathbf {O2} (0.766,0.234,0.5)$

$\mathbf {C1}$ and $\mathbf {O1}$ belong to the same molecule, then we can get $\mathbf {O3} (0.734,0.734,0)$

According to the characteristic symmetry elements of the tetragonal crystal system, the crystal must have a four-fold rotation axis, inversion axis, or screw axis in the $z$ direction. The $\rm{CO_2}$ molecule does not satisfy the requirements of the four-fold rotation axis or inversion axis in the $z$ direction, so it can only be a $4_2$ screw axis in the $z$ direction. From this, we can determine that $\mathbf {C2} (0.5,0.5,0.5)$, $\mathbf {O4} (0.234,0.766,0.5)$

<Checkpoint>[$\mathbf {C2} (0.5,0.5,0.5)$, $\mathbf {O3} (0.734,0.734,0)$, $\mathbf {O4} (0.234,0.766,0.5)$](2)</Checkpoint>

Then analyze the symmetry elements. The positions of all $\rm{C}$ atoms in the crystal are centers of symmetry, so proposition 3 is correct.

<Checkpoint>[The tetragonal crystal has a center of symmetry, proposition 3 is correct](1)</Checkpoint>

The straight line where the $\rm{CO_2}$ molecule is located is a two-fold rotation axis, so proposition 4 is correct.

<Checkpoint>[Proposition 4 is correct](1)</Checkpoint>

The straight line through $x = 0,y=0.5$ or $x=0.5,y=0$, and parallel to the $z$ axis is a two-fold rotation axis, which does not pass through $\mathbf A$, so proposition 5 is incorrect.

<Checkpoint>[Proposition 5 is incorrect](1)</Checkpoint>

Based on the above analysis, the correct options are 1, 2, 3, 4
首先推断$\mathbf X$的化学式：

晶胞体积$V = a \times b \times c = 622~\rm{pm} \times 435~\rm{pm} \times 607~\rm{pm} = (1.643 \times 10^8) ~pm^3 \approx (1.643 \times 10^{-22})~\rm{cm}^3$

$\mathbf X$的摩尔质量$M = \frac{\rho \times V \times N_A}{Z} = \frac{3.56 \times 1.643 \times 10^{-22} \times 6.022 \times 10^{23}}{8} \approx 44.0 ~\rm{g }\cdot mol^{-1}$

<Checkpoint>[$\mathbf X$的摩尔质量$M \approx 44.0 ~\rm{g} \cdot mol^{-1}$](2)</Checkpoint>

由于化学键的总数是守恒的，$\mathbf A$对$\mathbf B$的配位数等于$\mathbf B$对$\mathbf A$的配位数，故有$4 \times n_A = 2 \times n_B$。

所以$\mathbf A$和$\mathbf B$的原子个数比为1:2

<Checkpoint>[$\mathbf A$和$\mathbf B$的原子个数比为1:2](1)</Checkpoint>

符合以上条件的化合物可能是$\rm{CO_2}$或$\rm{N_2O}$，但$N_2O$难以实现氧元素的四配位，所以$\mathbf X$的化学式是$\rm{CO_2}$，$\mathbf A$是$\rm{C}$，$\mathbf 
B$是$\rm{O}$

<Checkpoint>[$\mathbf X$是$\rm{CO_2}$，$\mathbf A$是$\rm{C}$，$\mathbf B$是$\rm{O}$](1)</Checkpoint>

$\rm{CO_2}$分子具有对称中心，所以命题1正确。

<Checkpoint>[$\rm{CO_2}$分子具有对称中心，命题1正确](1)</Checkpoint>

$\rm{CO_2}$形成的立方晶系的分子晶体，其正当晶胞包含4个$\rm{CO_2}$分子，如果为面心立方点阵，则4个$\rm{CO_2}$分子的空间取向必须一致，这将导致立方晶系对称性的丧失。所以，在题设条件下，$\mathbf X$的立方晶体只能是简单立方点阵。命题2正确。

<Checkpoint>[$\mathbf X$的立方晶体是简单立方点阵，命题2正确](1)</Checkpoint>

接下来分析$\mathbf X$的四方晶体。可以借助四方晶系的对称性和$\rm{CO_2}$的分子对称性，得到未给出的原子坐标。

设$\mathbf {C1} (0,0,0)$, $\mathbf {O1} (0.266,0.266,0)$, $\mathbf {O2} (0.766,0.234,0.5)$

$\mathbf {C1}$和$\mathbf {O1}$属于同一个分子，则可得$\mathbf {O3} (0.734,0.734,0)$

根据四方晶系的特征对称元素，该晶体必须具有$z$方向的四重旋转轴、反轴或螺旋轴。$\rm{CO_2}$分子在$z$方向不满足四重旋转轴或反轴的要求，则只能是$z$方向的$4_2$螺旋轴。由此可以确定，$\mathbf {C2} (0.5,0.5,0.5)$, $\mathbf {O4} (0.234,0.766,0.5)$

<Checkpoint>[$\mathbf {C2} (0.5,0.5,0.5)$, $\mathbf {O3} (0.734,0.734,0)$, $\mathbf {O4} (0.234,0.766,0.5)$](2)</Checkpoint>

接着分析对称元素，该晶体中所有$\rm{C}$原子的位置都是对称中心，所以命题3正确。

<Checkpoint>[四方晶体存在对称中心，命题3正确](1)</Checkpoint>

$\rm{CO_2}$分子所在的直线是二重旋转轴，所以命题4正确。

<Checkpoint>[命题4正确](1)</Checkpoint>

通过$x = 0,y=0.5$或$x=0.5,y=0$，且平行于$z$轴的直线是二重旋转轴，其并不通过$\mathbf A$，故命题5错误。

<Checkpoint>[命题5错误](1)</Checkpoint>

综合以上分析，正确的选项是1,2,3,4