The platinum-containing salt $\mathbf{A}$ with $40.14\%$ platinum decomposes upon heating to produce elemental platinum, a salt $\mathbf{B}$, and a yellowish-green gas $\mathbf{C}$. The anion in $\mathbf{A}$ has an octahedral structure, and its anion reacts with $\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ to produce a square planar anion $\mathbf{D}$. $\mathbf{D}$ reacts with a binary compound to form an anion $\mathbf{E}$. The mass fraction of $\mathrm{Pt}$ in $\mathbf{E}$ is $59.22\%$, and it contains a ligand with more than 3 atoms. $\mathbf{E}$ can dimerize to form a neutral compound $\mathbf{F}$.

Which of the following statements is incorrect?
含铂 $40.14 \%$ 的盐$\mathbf{A}$受热分解得到单质铂，一种盐$\mathbf{B}$和一种黄绿色气体$\mathbf{C}$。$\mathbf{A}$中阴离子呈八面体结构，其阴离子与 $\mathrm {K_2C_2O_4}$ 作用生成一种呈平面四边形的阴离子$\mathbf{D}$。$\mathbf{D}$与一种二元化合物反应生成阴离子$\mathbf{E}$。$\mathbf{E}$中$\mathrm{Pt}$的质量分数为 $59.22\%$，含有原子数大于3的配体。$\mathbf{E}$可双聚形成中性化合物$\mathbf{F}$。

下列说法中，错误的是
The cation in $\mathbf{A}$ corresponds to an element that reacts violently with water.
The sum of the simplest integer ratios after balancing the chemical equation for the decomposition of $\mathbf{A}$ is 6.
Potassium salt of $\mathbf{D}$ is a red crystal.
The mass fraction of $\mathrm{Pt}$ in $\mathbf{F}$ is $66\%$ (correct within a 1% error margin).
All options A-D are correct
At most three of options A-D are correct.
$\mathbf{A}$中阳离子对应单质会和水剧烈反应
$\mathbf{A}$分解的化学方程式配平后的最简整数比之和为6
$\mathbf{D}$的钾盐为一种红色晶体
$\mathbf{F}$中$\mathrm{Pt}$的质量分数为$66\%$（误差在1%内即正确）
选项A-D均正确
选项A-D中至多有三个正确
From the description "yellow-green gas" in the question stem, it can be determined that $\mathbf{C}$ is $\mathrm{Cl_2}$.

<Checkpoint>[$\mathbf{C}$ is $\mathrm{Cl_2}$](0.5)</Checkpoint>

From the description "the anion in $\mathbf{A}$ has an octahedral structure" in the question stem, it can be inferred that the anion is $\mathrm{PtCl_6^{2-}}$. Assuming $\mathbf{A}$ contains 1 $\mathrm{Pt}$, the molecular weight of $\mathbf{A}$ can be calculated as $486.05$. Subtracting the anion, the cation can be determined to be 2 $\mathrm{K^+}$. Therefore, salt $\mathbf{A}$ is $\mathrm{K_2PtCl_6}$, and salt $\mathbf{B}$ is $\mathrm{KCl}$.

<Checkpoint>[$\mathbf{A}$ is $\mathrm{K_2PtCl_6}$, $\mathbf{B}$ is $\mathrm{KCl}$](1)</Checkpoint>

From the description "the planar quadrilateral anion $\mathbf{D}$" in the question stem, it can be inferred that $\mathbf{D}$ should be $\mathrm{PtCl_4^{2-}}$.

<Checkpoint>[$\mathbf{D}$ should be $\mathrm{PtCl_4^{2-}}$](1)</Checkpoint>

Since $\mathrm{Cl}$ in $\mathrm{PtCl_4^{2-}}$ is easily substituted, it is highly likely that during the formation of $\mathbf{E}$, one $\mathrm{Cl^{-}}$ is replaced by a binary compound or ions from a binary compound. Assuming $\mathbf{E}$ contains only one $\mathrm{Pt}$, the molecular weight of $\mathbf{E}$ can be calculated as $329.45$. Subtracting one $\mathrm{Pt}$ and three $\mathrm{Cl}$, the remaining mass is $28$, which should correspond to ethylene. Therefore, $\mathbf{E}$ is $\mathrm{PtCl_3(C_2H_4)^-}$, and the dimerized $\mathbf{F}$ is $\mathrm{[PtCl_2(C_2H_4)]_2}$.

<Checkpoint>[$\mathbf{E}$ is $\mathrm{PtCl_3(C_2H_4)^-}$](1)</Checkpoint>

<Checkpoint>[The dimerized $\mathbf{F}$ is $\mathrm{[PtCl_2(C_2H_4)]_2}$](1)</Checkpoint>

Analyzing the options:

A: Potassium reacts violently with water to release $\mathrm{H_2}$, which is correct;

<Checkpoint>[Potassium reacts violently with water](0.5)</Checkpoint>

B: The balanced equation is $\mathrm{K_2PtCl_6=2KCl+2Cl_2+Pt}$, with a sum of coefficients of 6, which is correct;

<Checkpoint>[The balanced equation for the decomposition of $\mathbf{A}$ is $\mathrm{K_2PtCl_6=2KCl+2Cl_2+Pt}$, with a sum of coefficients of 6](1)</Checkpoint>

C: The potassium salt of $\mathbf{D}$ is $\mathrm{K_2PtCl_4}$, which is a red crystal;

<Checkpoint>[$\mathrm{K_2PtCl_4}$ is a red crystal](1)</Checkpoint>

D: The dimerized $\mathbf{F}$ is $\mathrm{[PtCl_2(C_2H_4)]_2}$, with a mass percentage of $\mathrm{Pt}$ at $66\%$, which is correct.

<Checkpoint>[The mass percentage of $\mathrm{Pt}$ in $\mathrm{[PtCl_2(C_2H_4)]_2}$ is $66\%$](0.5)</Checkpoint>

Options A, B, C, and D are all correct, so option E (A-D are all correct) is reasonable.

The question asks for the incorrect statement, so option F is incorrect, and the answer is F.
由题干“黄绿色气体”可以判断，$\mathbf{C}$为$\mathrm{Cl_2}$。

<Checkpoint>[$\mathbf{C}$为$\mathrm{Cl_2}$](0.5)</Checkpoint>

由题干“$\mathbf{A}$中阴离子呈八面体结构”可以判断，阴离子为$\mathrm{PtCl_6^{2-}}$。不妨先假设$\mathbf{A}$中含1个$\mathrm{Pt}$，可以计算$\mathbf{A}$的分子量为$486.05$。除去阴离子，可以算出阳离子为2个$\mathrm{K^+}$。因此，盐$\mathbf{A}$为$\mathrm{K_2PtCl_6}$，盐$\mathbf{B}$为$\mathrm{KCl}$。

<Checkpoint>[$\mathbf{A}$为$\mathrm{K_2PtCl_6}$，$\mathbf{B}$为$\mathrm{KCl}$](1)</Checkpoint>

由题干“平面四边形的阴离子$\mathbf{D}$”，可以判断$\mathbf{D}$应当为$\mathrm{PtCl_4^{2-}}$。

<Checkpoint>[$\mathbf{D}$应当为$\mathrm{PtCl_4^{2-}}$](1)</Checkpoint>

由于$\mathrm{PtCl_4^{2-}}$中的$\mathrm{Cl}$容易被取代，$\mathbf{E}$的生成过程中很有可能发生了一个$\mathrm{Cl^{-}}$被二元化合物取代或二元化合物中的离子取代。不妨先假设$\mathbf{E}$中仅有一个$\mathrm{Pt}$，可以计算$\mathbf{E}$的分子量为$329.45$。除去一个$\mathrm{Pt}$和三个$\mathrm{Cl}$，发现还剩$28$，应当为乙烯。因此，$\mathbf{E}$为$\mathrm{PtCl_3(C_2H_4)^-}$，二聚后的$\mathbf{F}$为$\mathrm{[PtCl_2(C_2H_4)]_2}$。

<Checkpoint>[$\mathbf{E}$为$\mathrm{PtCl_3(C_2H_4)^-}$](1)</Checkpoint>

<Checkpoint>[二聚后的$\mathbf{F}$为$\mathrm{[PtCl_2(C_2H_4)]_2}$](1)</Checkpoint>

对选项进行分析：

A中，钾会和水剧烈反应放$\mathrm{H_2}$，说法正确；

<Checkpoint>[钾会和水剧烈反应](0.5)</Checkpoint>

B中，配平的方程式为$\mathrm{K_2PtCl_6=2KCl+2Cl_2+Pt}$，系数和为6，正确；

<Checkpoint>[$\mathbf{A}$分解过程配平的方程式为$\mathrm{K_2PtCl_6=2KCl+2Cl_2+Pt}$，系数和为6](1)</Checkpoint>

C中，$\mathbf{D}$的钾盐为$\mathrm{K_2PtCl_4}$，为红色晶体；

<Checkpoint>[$\mathrm{K_2PtCl_4}$为红色晶体](1)</Checkpoint>


D中，二聚后的$\mathbf{F}$为$\mathrm{[PtCl_2(C_2H_4)]_2}$，$\mathrm{Pt}$的质量分数为$66\%$，正确。

<Checkpoint>[$\mathrm{[PtCl_2(C_2H_4)]_2}$中$\mathrm{Pt}$的质量分数为$66\%$](0.5)</Checkpoint>

A、B、C、D选项均正确，所以E选项（A-D均正确）合理。

本题所选择的是错误的表述，故选项F表述错误，答案是F.