It is known that $\mathbf{M}$ and $\mathbf{N}$ are adjacent elements in the periodic table with increasing atomic numbers, and they can form elemental gases $\mathbf{A}$ and $\mathbf{B}$, respectively. At $25^\circ \mathrm{C}$, heating a mixture of gold, elemental gas $\mathbf{A}$, and elemental gas $\mathbf{B}$ yields a 1:1 type salt $\mathbf{X}$. The cation of $\mathbf{X}$ is a dinuclear cation exhibiting paramagnetism, while the anion is a diamagnetic octahedral ion. When $\mathbf{X}$ is heated under vacuum conditions, it decomposes into $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$, where $\mathbf{C}$ is a dinuclear complex in which the oxidation state of gold is the same as in $\mathbf{X}$.  

Based on the above information, select the correct option.
已知 $\mathbf{M}$， $\mathbf{N}$是元素周期表中相邻的元素且原子序数依次增大，分别可形成单质气体 $\mathbf{A}$， $\mathbf{B}$. 在 $25^\circ \mathrm{C} $,下，加热金、单质气体 $\mathbf{A}$、单质气体 $\mathbf{B}$的混合物，得到一种1:1型盐 $\mathbf{X}$， $\mathbf{X}$的阳离子为双核阳离子，呈顺磁性，阴离子为抗磁性八面体型离子，将$\mathbf{X}$在真空条件下加热， $\mathbf{X}$可分解为 $\mathbf{A}$、 $\mathbf{B}$和 $\mathbf{C}$， $\mathbf{C}$为双核配合物，其中金的氧化态与 $\mathbf{X}$中相同。

请根据以上信息，选择正确的选项。
The bond order of the $\mathbf{X}$ cation is 1.5
The sum of the atomic numbers of $\mathbf{M}$ and $\mathbf{N}$ is 19.
According to the crystal field theory, the electron configuration of the central atom in the $\mathbf{X}$ anion is $t_{2g}^4e_g^2$.
The molecular point group of the $\mathbf{C}$ molecule is $D_{2d}$
The relative molecular mass of $\mathbf{X}$ molecule is $343 g/mol$
In the reaction equation for the high-temperature decomposition of $\mathbf{X}$, assuming the equation coefficients have been reduced to their simplest form, the sum of the coefficients of reactants and products is 5.
None of the above options is correct
 $\mathbf{X}$阳离子的键级为1.5
 $\mathbf{M}$和 $\mathbf{N}$的原子序数之和为19
 根据晶体场理论，$\mathbf{X}$阴离子中心原子的电子组态为$t_{2g}^4e_g^2$
 $\mathbf{C}$分子的分子点群为$D_{2d}$
 $\mathbf{X}$分子的相对分子质量为$343 g/mol$
在 $\mathbf{X}$高温分解的反应方程式中，假设方程式系数约化到了最简，则反应物生成物系数之和为5
以上选项均不正确
According to the problem, there are not many pairs of adjacent elements in the periodic table that can form elemental gases. Considering the chemical properties of gold, it can be inferred that these two gases are highly oxidizing. It can be speculated that $\mathbf{N}$ is the element $F$, $\mathbf{A}$ is $O_2$, and $\mathbf{B}$ is $F_2$. The sum of the atomic numbers of these two elements is 17, making option B incorrect.

<Checkpoint>[$\mathbf{A}$ is $O_2$, $\mathbf{B}$ is $F_2$](1)</Checkpoint>

<Checkpoint>[The sum of the atomic numbers of the two elements is 17](1)</Checkpoint>

The product $\mathbf{X}$ formed by the reaction of gold with gases $\mathbf{A}$ and $\mathbf{B}$ is an ionic compound. The anion is an octahedral complex, with the central element of the anion being $Au$ surrounded by six ligands. The ligands cannot be oxygen, as this would result in an excessively high oxidation state for gold. Therefore, the composition of the anion should be $AuF_6$, with the oxidation state unknown. As for the cation, it is dinuclear and should be $O_2$, exhibiting paramagnetism. It carries a single positive charge, forming $O_2^+$, having lost one antibonding electron, which increases the bond order by 0.5, resulting in a bond order of 2.5. Thus, option A is incorrect.

<Checkpoint>[The cation of $\mathbf{X}$ is $O_2^+$, with a bond order of 2.5](1)</Checkpoint>

The anion of $\mathbf{X}$ is $AuF_6^-$, where $Au$ is in the +5 oxidation state. The chemical formula of $\mathbf{X}$ is $O_2AuF_6$, with a relative molecular mass of $343 \text{ g/mol}$, making option E correct.

<Checkpoint>[The chemical formula of $\mathbf{X}$ is $O_2AuF_6$, with a relative molecular mass of $343 \text{ g/mol}$](1)</Checkpoint>

According to crystal field theory, the d orbitals of $Au$ will split into three $t_{2g}$ orbitals and two $e_g$ orbitals in an octahedral field. Based on the oxidation state of $Au$, the valence electron configuration of the $Au$ atom can be inferred as $(5d^6)$. If it is high-spin, the configuration would be $t_{2g}^4e_g^2$, with four unpaired electrons, exhibiting paramagnetism. If it is low-spin, the configuration would be $t_{2g}^6$, with no unpaired electrons, making it diamagnetic. Given the diamagnetic condition, it can be inferred that the electrons of the $Au$ element are in a low-spin state, with a hybridization of $d^2sp^3$, making option C incorrect.

<Checkpoint>[The high-spin configuration is $t_{2g}^4e_g^2$, exhibiting paramagnetism](1)</Checkpoint>

<Checkpoint>[The low-spin configuration is $t_{2g}^6$, exhibiting diamagnetism](1)</Checkpoint>

<Checkpoint>[The electron configuration of the $\mathbf{X}$ anion should be $t_{2g}^6$](1)</Checkpoint>

When $\mathbf{X}$ is heated in a vacuum to decompose it, the oxidation state of $Au$ in $\mathbf{C}$ is the same as in $\mathbf{X}$, also +5. $\mathbf{C}$ should be a fluoride of $Au$, and since it is a dinuclear complex, it should be $Au_2F_{10}$. Two $[AuF_6]$ octahedra are connected by sharing an edge, and the molecular point group is $D_{2h}$, making option D incorrect.

<Checkpoint>[$\mathbf{C}$ is $Au_2F_{10}$, with a molecular point group of $D_{2h}$](1)</Checkpoint>

The decomposition equation is written as $2O_2AuF_6 = Au_2F_{10} + 2O_2 + F_2$. The sum of the coefficients of the reactants and products is $2+1+2+1=6$, making option F incorrect.

<Checkpoint>[$2O_2AuF_6 = Au_2F_{10} + 2O_2 + F_2$, option F is incorrect](1)</Checkpoint>
根据题意，元素周期表中相邻且能形成单质气体的两种元素并不多，结合金元素的化学性质可以猜测这两种气体具有极强的氧化性，可以猜测， $\mathbf{N}$为$F$元素， $\mathbf{A}$为$O_2$， $\mathbf{B}$为$F_2$，两元素原子序数之和为17，选项B错误

<Checkpoint>[$\mathbf{A}$为$O_2$，$\mathbf{B}$为$F_2$](1)</Checkpoint>

<Checkpoint>[两元素原子序数之和为17](1)</Checkpoint>

金与$\mathbf{A}$,$\mathbf{B}$气体反应的生成物$\mathbf{X}$  为离子化合物，阴离子为八面体配合物，则阴离子中心元素为$Au$，周围有6个配体，配体不能为氧否则会导致金的氧化态过高，从而阴离子原子组成应为$AuF_6$，价态未知；再看阳离子，阳离子为双核，应为$O_2$，呈顺磁性，因此带一个正电荷，为$O_2^+$,失去了一个反键电子，键级增加0.5，则键级为2.5，选项A错误.

<Checkpoint>[$\mathbf{X}$阳离子为$O_2^+$，键级为2.5](1)</Checkpoint>

$\mathbf{X}$阴离子为$AuF_6^-$，其中$Au$为正五价，$\mathbf{X}$的化学式为$O_2AuF_6$，相对分子质量为$343 g/mol$，E选项正确；

<Checkpoint>[$\mathbf{X}$的化学式为$O_2AuF_6$，相对分子质量为$343 g/mol$](1)</Checkpoint>

根据晶体场理论，$Au$的d轨道会在八面体场中分裂为三个$t_{2g}$轨道和两个$e_g$轨道，根据$Au$的价态可以推测$Au$原子的价电子组态为$(5d^6)$，如果是高自旋，则组态为 $t_{2g}^4e_g^2$ ，有4个未成对电子，属于顺磁性。如果是低自旋，则组态为$t_{2g}^6$，没有未成对电子，为抗磁性。根据抗磁性的条件，可以推测$Au$元素的电子为低自旋，杂化方式为$d^2sp^3$，C选项错误

<Checkpoint>[高自旋组态为 $t_{2g}^4e_g^2$，呈顺磁性](1)</Checkpoint>

<Checkpoint>[低自旋组态为$t_{2g}^6$，为抗磁性](1)</Checkpoint>

<Checkpoint>[$\mathbf{X}$阴离子电子组态应为$t_{2g}^6$](1)</Checkpoint>

在真空中加热$\mathbf{X}$使其分解，$\mathbf{C}$中$Au$价态与$\mathbf{X}$众相同，也为正五价，应为$Au$的氟化物，又因为是双核配合物，应为$Au_2F_{10}$，两个$[AuF_6]$八面体共棱连接，分子点群为$D_{2h}$，D选项错误，

<Checkpoint>[$\mathbf{C}$为$Au_2F_{10}$，分子点群为$D_{2h}$](1)</Checkpoint>

分解方程式写作$2O_2AuF_6 = Au_2F_{10} + 2O_2 + F_2$，反应物与生成物系数之和为$2+1+2+1=6$，选项F错误

<Checkpoint>[$2O_2AuF_6 = Au_2F_{10} + 2O_2 + F_2$，选项F错误](1)</Checkpoint>