Potassium salt $\bf A$ contains oxygen elements and has strong oxidizing properties. The mass fraction of potassium is 28.93%, the cation-anion ratio is 1:2, and the anion contains two elements with an atomic ratio of 1:4.

Adding pyridine and excess $\bf A$ to silver nitrate solution can produce an orange salt precipitate $\bf {X}$ with an oxygen mass fraction of 20.765%. Choose the option related to $\bf {X}$ below that has the closest numerical value and the most correct conclusion.
钾盐$\bf A$具有强氧化性，其中钾的质量分数为28.93%，阴阳离子比为1:2，阴离子中含有原子比为1:4的两种元素。

向硝酸银溶液中加入吡啶和过量$\bf A$，可以制得氧的质量分数为20.765%的橙色盐沉淀$\bf {X}$。选择下列与$\bf {X}$的有关选项中数值最接近且结论最正确的选项。
All other options are incorrect
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The cation of $\bf {X}$ does not obey the 18-electron rule.
The molar mass of the cation(s) contained in $\bf {X}$ is 537.3 g/mol
The paramagnetism of $\bf {X}$ is due to the presence of unpaired electrons in the contained anions.
The molar mass of the cation contained in $\bf {X}$ is 492.2 g/mol.
$\bf {X}$ The main reason for the color display is: The energy of charge transfer transitions (i.e., LMCT mechanism) from oxygen atoms to high-valent central atoms in anions is low, and the absorption of visible light displays color.
其他选项均不正确
$\bf {X}$所含阴离子和阳离子中，所有金属离子的配位构型构型均为四面体形。
$\bf {X}$的阳离子不符合18电子规则
$\bf {X}$所含阳离子的摩尔质量为537.3 g/mol。
$\bf {X}$的顺磁性是由于所含阴离子中存在不成对的电子。
$\bf {X}$所含阴离子的摩尔质量为492.2 g/mol。
$\bf {X}$显示颜色的主要原因是：阴离子中从氧原子到高价中心原子的荷移跃迁（即LMCT机理）的能量较低，吸收可见光显示颜色。
According to the cation-anion ratio of 1:2, first calculate the relative molar mass corresponding to 2 potassium ions as:

$$
M = 2 \times 39.10 \div 28.93\% = 270.3 (g/mol)
$$

Then the molar mass of the anion is $270.3-2\times39.10=192.1(g/mol)$

Common oxidizing salts are all oxoacid salts, so it is speculated that $\bf A$ contains oxygen. Since the oxygen atom generally does not act as the central atom in the anion, assume the anion is in the form of $XO_4$.

Calculate the molar mass of $X$:

$$
M(X) = 192.1 - 4\times16.00 = 128.1(g/mol)
$$

Close to the molar mass of tellurium, but the tellurate radical has only medium oxidizing power, which does not meet the condition of strong oxidizing power, so the possibility of the anion being in the form of $XO_4^{2-}$ is excluded.

Assume the anion is in the form of $X_2O_8^{2-}$.

Calculate the molar mass of $X$:

$$
M(X) = (192.1 - 4\times16.00) \div 2 = 32.05(g/mol)
$$

Close to the molar mass of sulfur. Potassium persulfate has strong oxidizing power, which meets the properties such as the cation-anion ratio of 1:2 and the atomic ratio of the two elements in the anion being 1:4. Therefore, $\bf A$ is potassium persulfate.

<Checkpoint>[$\bf A$ is potassium persulfate](2)</Checkpoint>

Potassium persulfate can oxidize Ag to +2 or +3 valence. Assuming oxidation to +2 valence, the chemical formula of the precipitate can be assumed as $Ag(Py)_xS_2O_8$

Calculate the molar mass of the compound:

$$
M = 8\times16.00 \div 20.765\% = 616.4 (g/mol)
$$

The number of pyridine is:

$$
x = (616.4-107.9-2\times32.06-8\times16.00)\div79.10=4.00
$$

Therefore, 4 pyridine molecules coordinate to silver, which is consistent with the typical coordination number of Ag(II), so the chemical formula of the orange salt precipitate is $[Ag(C_5H_5N)_4]S_2O_8$

<Checkpoint>[The chemical formula of the orange salt precipitate is $[Ag(C_5H_5N)_4]S_2O_8$](2)</Checkpoint>

The cation in the compound is $[Ag(C_5H_5N)_4]^{2+}$, which is coordinated by four pyridine molecules to $Ag^{2+}$ in a square planar configuration through the lone pair electrons of the nitrogen atom.

Therefore, option B is incorrect.

<Checkpoint>[The cation $[Ag(C_5H_5N)_4]^{2+}$ of the orange salt precipitate has a square planar configuration](1)</Checkpoint>

The molar mass of the cation:

$$
M = 107.9+20 \times 12.01 + 20 \times 1.008 + 4 \times 14.01 = 424.3 {(g/mol)}
$$

<Checkpoint>[The molar mass of the cation: $ M = 107.9+20 \times 12.01 + 20 \times 1.008 + 4 \times 14.01 = 424.3 {(g/mol)} $](1)</Checkpoint>

Therefore, option D is incorrect.

Since Ag(II) has a $d^9$ valence electron configuration, there are unpaired electrons, making the complex paramagnetic. All electrons in the anion are paired, with no paramagnetic moment. Therefore, option E is incorrect.

Check whether the cation conforms to the 18-electron rule:

$EAN=9+4 \times 2 =17 \neq 18$

Therefore, it does not conform to the 18-electron rule, and option C is correct.

<Checkpoint>[$EAN=9+4 \times 2 =17 \neq 18$ Therefore, it does not conform to the 18-electron rule, and option C is correct.](1)</Checkpoint>

<Checkpoint>[Ag(II) has a $d^9$ valence electron configuration, so there are unpaired electrons, making the complex paramagnetic](1)</Checkpoint>

Since the 4d orbital of Ag(II) is partially filled, the complex exhibits color through the d-d transition mechanism, which is not related to the LMCT transition that may exist in the anion, so option G is incorrect.

<Checkpoint>[The 4d orbital of Ag(II) is partially filled, and the complex exhibits color through the d-d transition mechanism](1)</Checkpoint>

The molar mass of the anion:

$$
M = 2 \times 32.06 + 8 \times 16.00 = 192.1 {(g/mol)}
$$

<Checkpoint>[The molar mass of the anion: $ M = 2 \times 32.06 + 8 \times 16.00 = 192.1 {(g/mol)} $](1)</Checkpoint>

Therefore, option F is incorrect.

So choose option C.
根据阴阳离子比为1:2，先计算2个钾离子对应的相对摩尔质量为：

$$
M = 2 \times 39.10 \div 28.93\% = 270.3 (g/mol)
$$

则阴离子摩尔质量为$270.3-2\times39.10=192.1(g/mol)$

常见氧化性盐都是含氧酸盐，因此猜测$\bf A$含氧。由于氧原子在阴离子中一般不做中心原子，故设阴离子为$XO_4$形式。

计算$X$的摩尔质量：

$$
M(X) = 192.1 - 4\times16.00 = 128.1(g/mol)
$$

与碲的摩尔质量接近，但碲酸根仅有中等强度的氧化性，不符合强氧化性的条件，故排除阴离子为$XO_4^{2-}$形式的可能性。

设阴离子为$X_2O_8^{2-}$形式。

计算$X$的摩尔质量：

$$
M(X) = (192.1 - 4\times16.00) \div 2 = 32.05(g/mol)
$$

与硫的摩尔质量接近。过二硫酸钾有强氧化性，符合阴阳离子比1:2，阴离子中两种元素原子比1:4等性质。故$\bf A$为过二硫酸钾。

<Checkpoint>[$\bf A$为过二硫酸钾](2)</Checkpoint>

过二硫酸钾可将Ag氧化至+2价或+3价。设氧化至+2价，则可设沉淀化学式为$Ag(Py)_xS_2O_8$

计算化合物摩尔质量：

$$
M = 8\times16.00 \div 20.765\% = 616.4 (g/mol)
$$

吡啶个数为：

$$
x = (616.4-107.9-2\times32.06-8\times16.00)\div79.10=4.00
$$

故有4个吡啶分子对银配位，符合Ag(II)的典型配位数，因此橙色盐沉淀的化学式为 $[Ag(C_5H_5N)_4]S_2O_8$

<Checkpoint>[橙色盐沉淀的化学式为 $[Ag(C_5H_5N)_4]S_2O_8$](2)</Checkpoint>

化合物中阳离子为$[Ag(C_5H_5N)_4]^{2+}$，由四个吡啶分子以氮原子孤对电子对$Ag^{2+}$以平面正方形构型进行配位。

故选项B错误。

<Checkpoint>[橙色盐沉淀的阳离子$[Ag(C_5H_5N)_4]^{2+}$为平面正方形构型](1)</Checkpoint>

阳离子的摩尔质量：

$$
M = 107.9+20 \times 12.01 + 20 \times 1.008 + 4 \times 14.01 = 424.3 {(g/mol)}
$$

<Checkpoint>[阳离子的摩尔质量：$ M = 107.9+20 \times 12.01 + 20 \times 1.008 + 4 \times 14.01 = 424.3 {(g/mol)} $](1)</Checkpoint>

故选项D错误。

由于Ag(II)为$d^9$价电子组态，因此存在不成对电子，使配合物呈顺磁性。阴离子中电子全部成对，无顺磁磁矩。故选项E错误。

检验阳离子是否符合18电子规则：

$EAN=9+4 \times 2 =17 \neq 18$

故不符合18电子规则，选项C正确。

<Checkpoint>[$EAN=9+4 \times 2 =17 \neq 18$  故不符合18电子规则，选项C正确。](1)</Checkpoint>

<Checkpoint>[Ag(II)为$d^9$价电子组态，因此存在不成对电子，使配合物呈顺磁性](1)</Checkpoint>

由于Ag(II)的4d轨道部分填充，配合物通过d-d跃迁机理显色，与阴离子中可能存在的LMCT跃迁无关，故选项G错误。

<Checkpoint>[Ag(II)的4d轨道部分填充，配合物通过d-d跃迁机理显色](1)</Checkpoint>

阴离子的摩尔质量：

$$
M = 2 \times 32.06 + 8 \times 16.00 = 192.1 {(g/mol)}
$$

<Checkpoint>[阴离子的摩尔质量：$ M = 2 \times 32.06 + 8 \times 16.00 = 192.1 {(g/mol)} $](1)</Checkpoint>

故选项F错误。

故选择C选项。