Metal $\mathbf{A}$ is a component in many catalysts. Element $\mathbf{A}$ reacts with $\mathrm{F_2}$ and $\mathrm{Cl _ 2}$ to produce $\mathbf{B}$ and $\mathbf{C}$, respectively, with $\mathbf{B}$ containing 65.12% of $\mathbf{A}$. Reacting $\mathbf{B}$ with $\mathrm{S e F _ { 4  }}$ yields $\mathbf{D}$, whose crystal has a rutile structure. $\mathbf{C}$ has two crystalline forms, $\mathbf{C } _ { 1 }$ and $\mathbf{C } _ { 2 }$, containing chain-like polymers and hexamer units, respectively. Adding glacial acetic acid to a sulfate solution of $\mathbf{A}$ produces $\mathbf{E}$, which commonly exists as a trimer $\mathbf{E } _ { 1 }$, but its dimer $\mathbf{E } _ { 2  }$ has also been found. $\mathbf{E}$ reacts with sodium acetate to produce a mononuclear complex $\mathbf{F}$ and a dinuclear complex $\mathbf{G}$.

Now there are the following statements:

1, The valence of $\mathbf{A}$ in $\mathbf{B}$ is +3.

2, $\mathbf{C}$ and $\mathbf{D}$ have similar chemical formulas.

3, The coordination numbers of $\mathbf{A}$ in $\mathbf{C } _ { 1 }$ and $\mathbf{C } _ { 2 }$ are different.

4, $\mathbf{C } _ { 2 }$ belongs to the $T_d$ point group.

5, If hydrogen atoms are ignored, one possible structure of $\mathbf{E } _ { 1 }$ belongs to the $D_{3h}$ point group.

6, If hydrogen atoms are ignored, $\mathbf{E } _ { 2  }$ has the same highest-order rotation axis as $\mathrm{Re_2Cl_8^{2-}}$.

Let the atomic number of $\mathbf{A}$ be a, the smallest serial number of the correct statements be b, and the sum of the serial numbers of all correct statements be c, then the results obtained by performing modulo operations on a with b+1 and c, respectively, are:
金属$\mathbf { A }$ 是许多催化剂中的成分。$\mathbf { A }$单质与 $\mathrm{F_2}$ 和 $\mathrm{Cl _ 2}$ 反应，分别生成 $\mathbf { B }$ 和$\mathbf { C}$，$\mathbf { B }$中含 $\mathbf { A}$ 65.12% 。将 $\mathbf { B }$ 与 $\mathrm{S e F _ { 4  }}$ 反应生成 $\mathbf { D }$ ，其晶体为金红石结构。 $\mathbf { C }$ 有两种晶型 $\mathbf { C } _ { 1 }$ 和 $\mathbf { C } _ { 2 }$ ，其中分别含有链状多聚体和六聚体单元。将冰醋酸加入 $\mathbf { A }$ 的硫酸盐溶液，产生$\mathbf{E}$， $\mathbf { E }$ 常以三聚体 $\mathbf { E } _ { 1 }$ 的形式存在，但已经发现其二聚体 $\mathbf { E } _ { 2  }$。$\mathbf {  E }$ 与醋酸钠反应，产生单核配合物 $\mathbf { F }$ 和双核配合物$\mathbf{G}$。

现在有下列几个说法：

1，$\mathbf { B }$中$\mathbf { A }$的化合价为+3。

2，$\mathbf { C }$和$\mathbf { D }$具有相似的化学式。

3，$\mathbf { C } _ { 1 }$ 和 $\mathbf { C } _ { 2 }$中$\mathbf { A }$的配位数不同。

4，$\mathbf { C } _ { 2 }$属于$T_d$点群。

5，若忽略氢原子，则$\mathbf { E } _ { 1 }$的一种可能结构属于$D_{3h}$点群。

6，若忽略氢原子，则 $\mathbf { E } _ { 2  }$与$\mathrm{Re_2Cl_8^{2-}}$最高旋转轴次相同。

记$\mathbf { A }$的原子序数为a，正确的说法中最小的序号为b，所有正确说法的序号之和为c，则a分别对b+1和c进行取模运算所得到的结果分别是：
All other options are incorrect
0，5
0，6
0，7
0，8
1，5
1，6
1，7
1，8
2，5
2，6
2，7
2，8
3，5
3，6
3，7
3，8
其他选项均不正确
0，5
0，6
0，7
0，8
1，5
1，6
1，7
1，8
2，5
2，6
2，7
2，8
3，5
3，6
3，7
3，8
Metal $\mathbf { A }$ is a component in many catalysts, presumed to be a transition metal. $\mathbf { B }$ and $\mathbf { C }$ are fluoride and chloride, respectively. Assume the chemical formula of $\mathbf{B}$ is $\mathbf{A}\mathrm{F_x}$. According to the mass fraction, the following equation can be listed:

$\frac{M}{M+19x}=65.12\%$

Try to solve the equation. When $x=3$, $M =106.4g/mol$, which is element $\mathrm{Pd}$. Accordingly, $\mathbf { B }$ seems to be $\mathrm{PdF_3}$, but $\mathrm{Pd}$ generally exists in +2 and +4 valence states rather than +3 valence state.

<Checkpoint>[$\mathrm{Pd}$ in $\mathbf { B }$ has +2 and +4 valences](1)</Checkpoint>

Therefore, $\mathbf { B }$ is $\mathrm{Pd_2F_6}$, which can be represented as $\mathrm{Pd^{\mathrm{II}}[Pd^{\mathrm{IV}}F_6]}$.

After obtaining that $\mathbf { A }$ is element $\mathrm{Pd}$, it is natural to conclude that $\mathbf{C}$ is $\mathrm{PdCl_2}$. $\mathbf { D }$ crystal has a rutile structure, indicating that its atomic ratio is 1:2. Therefore, $\mathbf { D }$ is $\mathrm{PdF_2}$.

Thus, 1 is incorrect, and 2 is correct.

$\mathrm{Pd}$ in $\mathbf{C}$ has a +2 valence. When tetracoordinated, it tends to have a square planar configuration. It can be deduced that its chain polymer structure is:

<MultiModal>![The figure shows the chain polymer structure of $\mathrm{PdCl_2}$. Each $\mathrm{Pd}$ is coordinated by four $\mathrm{Cl}$, and the coordination configuration is square planar. Two adjacent $\mathrm{Pd}$ share 2 $\mathrm{Cl}$, and the $\mathrm{Pd-Cl-Pd}$ bond angle is about $90^{\circ}$.](/media/uploads/cbeb94b2-13d7-4957-893b-8c1f4e9ab7af.png)</MultiModal>

Its hexamer structure is:

<MultiModal>![The figure shows the hexamer structure of $\mathrm{PdCl_2}$. The black ball is $\mathrm{Pd}$, and the white ball is $\mathrm{Cl}$. Six $\mathrm{Pd}$ atoms form a regular octahedron, and each edge of the regular octahedron has a $\mathrm{Cl}$ atom bridging.](/media/uploads/545d9736-295c-493a-811f-30dbcb3bd9cd.png)</MultiModal>

<Checkpoint>[In both polymers, $\mathrm{Pd}$ is tetracoordinated, and $\mathbf { C } _ { 2 }$ has $O_h$ symmetry](2)</Checkpoint>

Therefore, 3 and 4 are both incorrect.

Adding glacial acetic acid to the sulfate solution of $\mathbf { A }$ produces $\mathbf{E}$. Then $\mathbf{E}$ is $\mathrm{Pd(OAc)_2}$. Its trimer $\mathbf { E } _ { 1 }$ has a $D_{3h}$ structure, as shown in the figure below:

<MultiModal>![The figure shows the trimer structure of $\mathrm{Pd(OAc)_2}$. The black ball is $\mathrm{Pd}$, and the white ball is $\mathrm{AcO}$. Three $\mathrm{Pd}$ atoms form a triangle, and $\mathrm{AcO}$ acts as a bidentate ligand to connect two adjacent $\mathrm{Pd}$. Every two $\mathrm{Pd}$ are connected by two $\mathrm{AcO}$, and the coordination configuration is square planar.](/media/uploads/334c358c-5533-4df7-bf81-17aec1612d14.png)</MultiModal>

<Checkpoint>[This structure belongs to the $D_{3h}$ point group](1)</Checkpoint>

The structure of its dimer $\mathbf { E } _ { 2  }$ is shown in the figure below:

<MultiModal>![The figure shows the dimer structure of $\mathrm{Pd(OAc)_2}$. Two $\mathrm{Pd}$ are bonded to each other, and a total of 4 $\mathrm{AcO}$ are used as bridging ligands to connect $\mathrm{Pd}$. The orientations of two adjacent $\mathrm{AcO}$ are perpendicular to each other.](/media/uploads/2c0f7b7e-ebe5-4688-af8f-0064c03827f3.png)</MultiModal>

The structure of $\mathrm{Re_2Cl_8^{2-}}$ is similar to it, and the highest order of the axis is 4.

<Checkpoint>[Ignoring $\mathrm{H}$, $\mathbf { E } _ { 2  }$ has a four-fold axis.](1)</Checkpoint>

Ignoring hydrogen atoms, the highest rotational axis order of $\mathbf { E } _ { 2  }$ is the same as that of $\mathrm{Re_2Cl_8^{2-}}$.

The correct statements are 2, 5, and 6. The serial number and c=2+5+6=13.
Element $\bf{A}$ is $\mathrm{Pd}$, and the atomic number a=46.
The smallest correct serial number b=2. Calculate a%
(b+1) and a%c:

46 % (2+1) = 46 % 3 = 1

46 % 13 = 7

Therefore, choose H
金属$\mathbf { A }$ 是许多催化剂中的成分，推测其为某副族金属，$\mathbf { B }$ 和$\mathbf { C}$分别为氟化物和氯化物，设$\mathbf { B }$的化学式为$\mathbf{A}\mathrm{F_x}$,根据质量分数可以列出方程：

$\frac{M}{M+19x}=65.12\%$

试解方程，当$x=3$时$M =106.4g/mol$，为元素$\mathrm{Pd}$。据此$\mathbf { B }$似乎为$\mathrm{PdF_3}$，但$\mathrm{Pd}$一般以+2和+4价态存在而非+3价态，

<Checkpoint>[$\mathbf { B }$中的$\mathrm{Pd}$为+2和+4价](1)</Checkpoint>

因此$\mathbf { B }$为$\mathrm{Pd_2F_6}$,可表示为$\mathrm{Pd^{\mathrm{II}}[Pd^{\mathrm{IV}}F_6]}$。

得到$\mathbf { A }$为元素$\mathrm{Pd}$后，自然得出$\mathbf{C}$为$\mathrm{PdCl_2}$。$\mathbf { D }$ 晶体为金红石结构，表示其原子比例为1：2,因此$\mathbf { D }$为$\mathrm{PdF_2}$，

于是1错误，2正确。

$\mathbf{C}$中$\mathrm{Pd}$为+2价，四配位时倾向于平面四边形构型，可推出其链状多聚体结构为：

<MultiModal>![图中为$\mathrm{PdCl_2}$链状多聚体结构，每个$\mathrm{Pd}$周围有四个$\mathrm{Cl}$进行配位，配位构型为平面四边形，相邻的两个$\mathrm{Pd}$之间共用2个$\mathrm{Cl}$，$\mathrm{Pd-Cl-Pd}$键角约为$90^{\circ}$。](/media/uploads/cbeb94b2-13d7-4957-893b-8c1f4e9ab7af.png)</MultiModal>

其六聚体结构为：

<MultiModal>![图中为$\mathrm{PdCl_2}$的六聚体结构，黑球为$\mathrm{Pd}$，白球为$\mathrm{Cl}$，六个$\mathrm{Pd}$原子组成正八面体，正八面体的每一条棱上有一个$\mathrm{Cl}$原子进行桥连。](/media/uploads/545d9736-295c-493a-811f-30dbcb3bd9cd.png)</MultiModal>

<Checkpoint>[两种多聚体中$\mathrm{Pd}$均为四配位，且$\mathbf { C } _ { 2 }$为$O_h$对称性](2)</Checkpoint>

因此3，4均错误。

将冰醋酸加入 $\mathbf { A }$ 的硫酸盐溶液，产生$\mathbf{E}$，则$\mathbf{E}$为$\mathrm{Pd(OAc)_2}$， 其三聚体 $\mathbf { E } _ { 1 }$ 存在一种$D_{3h}$
结构，如下图：

<MultiModal>![图中为$\mathrm{Pd(OAc)_2}$的三聚体结构，黑球为$\mathrm{Pd}$，白球为$\mathrm{AcO}$，三个$\mathrm{Pd}$原子组成三角形，$\mathrm{AcO}$作为双齿配体连接相邻的两个$\mathrm{Pd}$，每两个$\mathrm{Pd}$之间由两个$\mathrm{AcO}$进行连接，配位构型为平面四边形。](/media/uploads/334c358c-5533-4df7-bf81-17aec1612d14.png)</MultiModal>

<Checkpoint>[该结构属于$D_{3h}$点群](1)</Checkpoint>

其二聚体 $\mathbf { E } _ { 2  }$结构如下图：

<MultiModal>![图中为$\mathrm{Pd(OAc)_2}$的二聚体结构，两个$\mathrm{Pd}$之间相互成键，共有4个$\mathrm{AcO}$作为桥连配体连接$\mathrm{Pd}$，相邻的两个$\mathrm{AcO}$取向互相垂直。](/media/uploads/2c0f7b7e-ebe5-4688-af8f-0064c03827f3.png)</MultiModal>

$\mathrm{Re_2Cl_8^{2-}}$的结构与其类似，最高轴次为4，

<Checkpoint>[不考虑$\mathrm{H}$的情况下$\mathbf { E } _ { 2  }$有四次轴。](1)</Checkpoint>

忽略氢原子，则 $\mathbf { E } _ { 2  }$与$\mathrm{Re_2Cl_8^{2-}}$最高旋转轴次相同

正确的说法是2、5、6,序号和c=2+5+6=13。
元素$\bf{A}$是$\mathrm{Pd}$,原子序数a=46。
最小正确序号b=2。计算a%
(b+1)和a%c:

46 % (2+1) = 46 % 3 = 1

46 % 13 = 7

因此选择H