The element $\bf{X}$'s simple substance reacts with oxygen at high temperature to produce a yellow solid $\bf{A}$. Dissolving $\bf{A}$ in $\rm{HClO_4}$ yields a perchlorate. Adding $\rm{NaOH}$ solution until neutral results in a hexanuclear cation $\bf{B}$ with a +6 charge in the solution. This ion possesses high symmetry and has three fourfold axes. Continuing to add $\rm{NaOH}$ produces a white solid. Adding $\rm{NaOH}$ further until alkaline and then introducing $\rm{Cl_2}$ yields a yellow solid $\bf{C}$.

The element $\bf{Y}$'s simple substance can react with hot water to produce a white solid $\bf{D}$, which immediately transforms into $\bf{E}$ containing a mass fraction of $\bf{Y}$ of approximately $52.35\%$ upon contact with air.

The following statements are made:

1. The atomic numbers of $\bf{X}$ and $\bf{Y}$ are both odd numbers.

2. In the cation $\bf{B}$, there exists $\rm{OH^-}$ coordinated to only one $\bf{X}$ atom.

3. $\bf{D}$ can react with solid $\bf{C}$ after dissolving in dilute sulfuric acid, and the sum of the coefficients on the reactant side and the sum of the coefficients on the product side of the ionic equation (with the simplest coefficients and one neutral component in both reactants and products) are both odd numbers.

4. $\bf{E}$ is a white solid.

What is the sum of the numbers of the correct statements?
元素$\bf{X}$的单质高温下和氧气反应生成黄色固体$\bf{A}$，将$\bf{A}$溶于 $\rm{HClO_4}$ 中得到高氯酸盐，滴加 $\rm{NaOH}$ 溶液至中性，溶液中得到一个带6个正电荷的6核阳离子$\bf{B}$，该离子具有高度对称性，存在三根四重轴。继续滴加 $\rm{NaOH}$，得到一白色固体，再滴加 $\rm{NaOH}$ 至碱性，通入 $\rm{Cl_2}$，得到黄色固体$\bf{C}$。

元素$\bf{Y}$的单质能与热水反应生成白色的固体$\bf{D}$，和空气接触会立刻转化为含$\bf{Y}$质量分数约为 $52.35\%$ 的$\bf{E}$。

有如下说法：

1.$\bf{X}$和$\bf{Y}$的原子序数均为奇数。

2.阳离子$\bf{B}$中存在只对1个$\bf{X}$原子配位的 $\rm{OH^-}$。

3.$\bf{D}$溶于稀硫酸后可以与固体$\bf{C}$反应，反应的离子方程式（系数化至最简，反应物与产物中都有一个电中性组分）反应物一侧系数之和以及产物一侧系数之和均为奇数。

4.$\bf{E}$为白色固体。

正确说法编号之和为多少？
1
3
4
5
6
8
10
None of the above options are correct.
1
3
4
5
6
8
10
以上选项均不对
First, deduce element $\bf{X}$: Combining the information (1) the simple substance reacts with oxygen at high temperature to produce a yellow solid; (2) a highly symmetric 6-core cation; (3) a yellow solid is obtained under alkaline oxidation conditions, it can be comprehensively determined that $\bf{X}$ is bismuth.

<Checkpoint>[$\bf{X}$ is bismuth](1)</Checkpoint>

Cation $\bf{B}$ contains 6 $\rm{Bi}$ and carries six positive charges, indicating that it contains $6\times3-6=12$ $\rm{OH^-}$. From the three four-fold axes, it can be known that it is a regular octahedral shape, and 12 $\rm{OH^-}$ are located on the 12 edges of the $\rm{Bi_6}$ octahedron, and there is no $\rm{OH^-}$ coordinated to only one $\rm{Bi}$ atom. Statement 2 is incorrect.

<Checkpoint>[There is no $\rm{OH^-}$ coordinated to only one $\bf{X}$ atom in cation $\bf{B}$](1)</Checkpoint>

$\bf{C}$ is produced under alkaline and oxidizing conditions, and is yellow, indicating that it is $\rm{NaBiO_3}$.

<Checkpoint>[$\bf{C}$ is $\rm{NaBiO_3}$](0.5)</Checkpoint>

Then, deduce element $\bf{Y}$: It is easy to know that $\bf{D}$ is a hydroxide. $\bf{E}$ is the product formed after $\bf{D}$ comes into contact with air. The phrase "immediately generated" rules out basic carbonates, and $\bf{E}$ is likely to be a hydroxide or a hydroxy oxide. Assume the chemical formula of $\bf{E}$ is $\bf{Y}\rm{O_a(OH)_b}$, try different non-negative integer combinations of $\rm{a,b}$, it can be known that $\rm{a=1,b=2},\bf{Y}=\rm{Mn}$ is the only reasonable solution. Manganese can react with hot water, which also fits the description. From this, it is known that the atomic numbers of $\bf{X,Y}$ are both odd numbers, and statement 1 is correct. $\bf{E}$ is $\rm{MnO(OH)_2}$, which is a brown or black solid, and statement 4 is incorrect.

<Checkpoint>[$\bf{E}$ is a hydroxide or hydroxy oxide](0.5)</Checkpoint>

<Checkpoint>[$\bf{Y}$ is manganese](1)</Checkpoint>

<Checkpoint>[$\bf{E}$ is $\rm{MnO(OH)_2}$](1)</Checkpoint>

<Checkpoint>[$\bf{E}$ is a brown or black solid](0.5)</Checkpoint>

The reaction in statement 3 is the reaction of $\rm{NaBiO_3}$ and $\rm{Mn^{2+}}$ under acidic conditions. The reaction equation is

$\rm{2Mn^{2+}+5NaBiO_3+14H^+\rightarrow 2MnO_4^-+5Bi^{3+}+7H_2O+5Na^+}$

The sum of the coefficients on the reactant side and the sum of the coefficients on the product side are both odd numbers. Statement 3 is correct.

<Checkpoint>[$\rm{2Mn^{2+}+5NaBiO_3+14H^+\rightarrow 2MnO_4^-+5Bi^{3+}+7H_2O+5Na^+}$](1)</Checkpoint>

In summary, the correct statements are 1 and 3, and the answer is option C.
先推断元素$\bf{X}$：结合 (1)单质高温下和氧气反应生成黄色固体；(2)高对称性6核阳离子；(3)碱中氧化条件得到黄色固体 等信息，综合确定$\bf{X}$为铋。

<Checkpoint>[$\bf{X}$为铋](1)</Checkpoint>

阳离子$\bf{B}$含有6个 $\rm{Bi}$，带六个正电荷，可知含有 $6\times3-6=12$ 个 $\rm{OH^-}$。由三根四重轴可知其为正八面体形，12个 $\rm{OH^-}$处于 $\rm{Bi_6}$ 八面体的12条棱上，没有只对1个 $\rm{Bi}$ 原子配位的 $\rm{OH^-}$。说法2错误。

<Checkpoint>[阳离子$\bf{B}$中不存在只对1个$\bf{X}$原子配位的 $\rm{OH^-}$](1)</Checkpoint>

$\bf{C}$在碱性、氧化性的条件中生成，且为黄色，可知为 $\rm{NaBiO_3}$。

<Checkpoint>[$\bf{C}$为 $\rm{NaBiO_3}$](0.5)</Checkpoint>

再推断元素$\bf{Y}$：易知$\bf{D}$为氢氧化物。$\bf{E}$为$\bf{D}$与空气接触后生成的产物，由“立即生成”排除碱式碳酸盐，$\bf{E}$很可能为氢氧化物或羟基氧化物。设$\bf{E}$化学式为 $\bf{Y}\rm{O_a(OH)_b}$，尝试 $\rm{a,b}$ 的不同非负整数组合，可知 $\rm{a=1,b=2},\bf{Y}=\rm{Mn}$ 为唯一合理解。锰可以与热水反应，也符合题述。由此知$\bf{X,Y}$的原子序数均为奇数，说法1正确。$\bf{E}$为 $\rm{MnO(OH)_2}$，是棕色或黑色固体，说法4错误。

<Checkpoint>[$\bf{E}$为氢氧化物或羟基氧化物](0.5)</Checkpoint>

<Checkpoint>[$\bf{Y}$为锰](1)</Checkpoint>

<Checkpoint>[$\bf{E}$为 $\rm{MnO(OH)_2}$](1)</Checkpoint>

<Checkpoint>[$\bf{E}$为棕色或黑色固体](0.5)</Checkpoint>

说法3的反应是 $\rm{NaBiO_3}$ 和 $\rm{Mn^{2+}}$在酸性条件的反应，反应方程式为

$\rm{2Mn^{2+}+5NaBiO_3+14H^+\rightarrow 2MnO_4^-+5Bi^{3+}+7H_2O+5Na^+}$

反应物一侧系数之和以及产物一侧系数之和均为奇数。说法3正确。

<Checkpoint>[$\rm{2Mn^{2+}+5NaBiO_3+14H^+\rightarrow 2MnO_4^-+5Bi^{3+}+7H_2O+5Na^+}$](1)</Checkpoint>

综上所述，正确的说法有1和3，答案为选项C。