Metal $\bf{X}$ is one of the monetary metals and has relatively strong inertness, but is soluble in aqua regia, forming the anion $\bf{A^-}$. $\mathrm{NBu_4}$ $\bf{A}$ can be reduced by phenylhydrazine hydrochloride to obtain salt $\bf{B}$ and salt $\bf{C}$, which have the same cation and are both 1:1 type salts (reaction 1). The valence state of $\bf{X}$ in $\bf{C}$ is the same as that in $\bf{A^-}$, containing a mass fraction of 31.63% of $\bf{X}$. $\mathrm{NBu_4}$ $\bf{A}$ is reduced by excess acetone, and $\bf{B}$ can also be obtained, and at the same time, the small organic molecule $\bf{D}$ is obtained. One molecule of $\bf{D}$ is produced at the same time as one molecule of $\bf{B}$ is produced.
The following statements are correct:
1. $\bf{C}$ contains oxygen element
2. $\bf{X}$ is the $\mathrm{Pt}$ element
3. The stoichiometric ratio of $\bf{B}$ and $\bf{C}$ in reaction 1 is 1:1
4. Elemental gas is produced in reaction 1
5. $\bf{D}$ can be hydrolyzed
金属$\bf{X}$是货币金属之一，具有比较强的惰性，但是可溶于王水，形成阴离子$\bf{A^-}$。$\mathrm{NBu_4}$ $\bf{A}$可以被苯肼的盐酸盐还原得到盐$\bf{B}$和盐$\bf{C}$，他们具有相同的阳离子且均为1:1型盐（反应1）。其中$\bf{C}$中$\bf{X}$的价态与$\bf{A^-}$相同，含$\bf{X}$质量分数31.63%。$\mathrm{NBu_4}$ $\bf{A}$被过量丙酮还原，同样可以得到$\bf{B}$，同时得到了有机小分子$\bf{D}$。一分子$\bf{B}$产生的同时产生了一分子$\bf{D}$。
下面的说法正确的是：
1. $\bf{C}$中含有氧元素
2. $\bf{X}$为$\mathrm{Pt}$元素
3. 反应1中$\bf{B}$和$\bf{C}$的化学计量比为1：1
4. 反应1中产生了单质气体
5. $\bf{D}$可以被水解
All other options are incorrect
1.2.3.4.5.
2.4.5
2.4.
3.5.
3.4.5.
2.3.4.5.
其他选项均不正确
1.2.3.4.5.
2.4.5
2.4.
3.5.
3.4.5.
2.3.4.5.
Metal $\bf{X}$ is one of the monetary metals and has relatively strong inertness, but it is soluble in aqua regia. Common monetary metals include gold, silver, and copper. Gold fits the description in the question. $\bf{A^-}$ should be $\mathrm{AuCl_4^-}$

<Checkpoint>[$\bf{A^-}$ should be $\mathrm{AuCl_4^-}$](1)</Checkpoint>

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

Therefore, 2 is incorrect.

The valence state of $\bf{X}$ in $\bf{C}$ is the same as that in $\bf{A^-}$, and the mass fraction of $\bf{X}$ is 31.63%. According to the question, it can be inferred that its cation should be $\mathrm{NBu_4^+}$, and the remaining mass corresponds to the anion $\mathrm{AuCl_3Ph^-}$.

1 is incorrect.

<Checkpoint>[$\bf{C}$ is $\mathrm{NBu_4^+AuCl_3Ph^-}$](1)</Checkpoint>

Then $\bf{B}$ should be the product of the reduction of $Au$, which is $\mathrm{NBu_4AuCl_2}$

The stoichiometric ratio of the two is 1:1, and nitrogen gas is produced at the same time, 3.4 are correct.

<Checkpoint>[$\bf{B}$ should be $\mathrm{NBu_4AuCl_2}$](1)</Checkpoint>

The organic small molecule $\bf{D}$ is the product of the 1:1 oxidation of acetone, which should be chloroacetone.

<Checkpoint>[$\bf{D}$ is chloroacetone $\mathrm{ClCH_2COCH_3}$](1)</Checkpoint>

Chlorine can be hydrolyzed and substituted, 5 is correct.

In summary, 3, 4, and 5 are correct, choose F.
金属$\bf{X}$是货币金属之一，具有比较强的惰性，但是可溶于王水。常见的货币金属有金、银和铜，符合题目描述的应为金。$\bf{A^-}$应为$\mathrm{AuCl_4^-}$

<Checkpoint>[$\bf{A^-}$应为$\mathrm{AuCl_4^-}$](1)</Checkpoint>

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

因此，2错误

其中$\bf{C}$中$\bf{X}$的价态与$\bf{A^-}$相同，含$\bf{X}$质量分数31.63%，根据题目可以推测出其阳离子应为$\mathrm{NBu_4^+}$，余下的质量对应阴离子为$\mathrm{AuCl_3Ph^-}$

1错误。

<Checkpoint>[$\bf{C}$为$\mathrm{NBu_4^+}\mathrm{AuCl_3Ph^-}$](1)</Checkpoint>

那么$\bf{B}$应为$Au$被还原的产物，为$\mathrm{NBu_4}\mathrm{AuCl_2}$

<Checkpoint>[$\bf{B}$应为$\mathrm{NBu_4}\mathrm{AuCl_2}$](1)</Checkpoint>


此处发生的化学反应方程式为
$\mathrm{2[NBu_4]AuCl_4+PhNHNH_3Cl==[NBu_4]AuCl_2+[NBu_4][PhAuCl_3]+N_2+4HCl}$

<Checkpoint>[$\mathrm{2[NBu_4]AuCl_4+PhNHNH_3Cl==[NBu_4]AuCl_2+[NBu_4][PhAuCl_3]+N_2+4HCl}$](1)</Checkpoint>

二者的化学计量比为1：1，同时反应产生了氮气，3.4正确。

<Checkpoint>[生成B和C的转换中，B，C两种产物计量比为1:1，并且还产生了氮气](1)</Checkpoint>

有机小分子$\bf{D}$为丙酮被1：1氧化的产物，应为氯代丙酮。

<Checkpoint>[$\bf{D}$为氯代丙酮$\mathrm{ClCH_2COCH_3}$](1)</Checkpoint>

可以被水解取代氯，5正确

综上，3，4，5正确，选F。