$\mathrm{AgO}$ reacts with $\mathrm{KIO}_{4}$ under alkaline conditions, undergoing a formal "disproportionation reaction" to obtain two silver-containing substances. The anion part of one of the compounds, $\bf{A}$, is a complex ion of $\mathrm{Ag}$ and the coordination number of $\mathrm{Ag}$ is 4. The mass fraction of $\mathrm{Ag}$ in this compound is $14.36\%$. Deduce the chemical formula of $\bf{A}$ and write the reaction equation.

There are the following statements, which of the statements are true?

1. In the anion of compound $\bf{A}$, the sum of the number of atoms in all ligand molecules of $\mathrm{Ag}$ is 16.
2. The product of the coefficients in front of each compound in the simplest reaction equation is 192.
3. Silver in compound $\bf{A}$ has a +1 valence.
4. The reaction also produces another silver compound with a molecular weight greater than 250.
$\mathrm{AgO}$在碱性条件下与$\mathrm{KIO}_{4}$作用, 发生形式上的 “歧化反应” 得到两种含银物质。其中一种化合物$\bf{A}$的阴离子部分为$\mathrm{Ag}$的配离子且$\mathrm{Ag}$的配位数为4。该化合物中$\mathrm{Ag}$的质量分数为 $14.36 \%$。请推理出$\bf{A}$的化学式，并写出反应的方程式。

有以下几个说法，其中正确的有？

1. 化合物$\bf{A}$的阴离子中，$\mathrm{Ag}$的所有配体分子中的原子数加和为16
2. 最简反应方程式各化合物前系数乘积为192
3. 化合物$\bf{A}$中银为+1价
4. 反应还生成了另外一种分子量大于250的银化合物
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
Coordination number 4 $\mathrm{Ag}$ is usually +2 or +3 valent. The problem mentions "disproportionation", so the coordination ion is likely to be +3 valent $\mathrm{Ag}$.

<Checkpoint>[The coordination ion is +3 valent $\mathrm{Ag}$](0.5)</Checkpoint>

Considering the size of an $\mathrm{Ag}$ atom, it is unlikely that there are four large-volume ligands around it. At the same time, $\mathrm{IO_6^{~5-}}$ is a common bidentate ligand, so it is speculated that the anion part is $\mathrm{Ag(IO_6)_2^{~7-}}$.

<Checkpoint>[The ion coordinated with $\mathrm{Ag}$ is the bidentate ligand $\mathrm{IO_6^{~5-}}$](1)</Checkpoint>

According to the mass fraction of $\mathrm{Ag}$, it can be calculated that $M_\bold{A}=M_{Ag}/14.36\%=751.4$. The molecular weight of the remaining part excluding silver and periodate is 197.7. The main cation in this reaction is $\mathrm{K^+}$. According to the molecular weight, it can be calculated that compound $\mathbf{A}$ contains five $\mathrm{K^+}$ and two $\mathrm{H^+}$ atoms. The chemical formula is written as $\mathrm{K}_{5} \mathrm{H}_{2} \mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}$. In fact, two hydrogen atoms are combined with $\mathrm{IO_6^{~5-}}$, which does not affect its coordination ability.

<Checkpoint>[$M_\bold{A}=751.4$](0.5)</Checkpoint>

<Checkpoint>[The chemical formula of $\mathbf{A}$ is written as $\mathrm{K}_{5} \mathrm{H}_{2} \mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}$](1)</Checkpoint>

In the anion of compound $\mathrm{A}$, the ligand molecules of $\mathrm{Ag}$ are two $\mathrm{HIO_6^{~4-}}$, and the sum of the number of atoms is 16, so statement 1 is correct.
The valence of silver is +3, so statement 3 is wrong.

<Checkpoint>[The ligand of $\mathrm{Ag}$ is $\mathrm{HIO_6^{4-}}$](1)</Checkpoint>

According to the "disproportionation reaction", the other substance is +1 valent $\mathrm{Ag}$, which is most reasonably written as $\mathrm{Ag_2 O}$ under alkaline conditions, so the equation is

$$
4 \mathrm{AgO}+6 \mathrm{KOH}+4 \mathrm{KIO}_{4} \rightarrow 2 \mathrm{~K}_{5} \mathrm{H}_{2}\left[\mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}\right]+\mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}
$$

<Checkpoint>[The other product is $\mathrm{Ag_2O}$](0.5)</Checkpoint>

<Checkpoint>[The equation is  $$ 4 \mathrm{AgO}+6 \mathrm{KOH}+4 \mathrm{KIO}_{4} \rightarrow 2 \mathrm{~K}_{5} \mathrm{H}_{2}\left[\mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}\right]+\mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} $$](1)</Checkpoint>

Hydrogen can also be written on periodate.

The product of the coefficients of the simplest reaction equation is 192, so statement 2 is correct. The other silver compound produced by the reaction is $\mathrm{Ag_2O}$, and the molecular weight is 231.8, so statement 4 is wrong.
配位数为4的$\mathrm{Ag}$通常为+2或者+3价,题目里提到了“歧化”，所以该配离子大概率为+3价的$\mathrm{Ag}$。

<Checkpoint>[配离子为+3价的$\mathrm{Ag}$](0.5)</Checkpoint>

考虑到一个$\mathrm{Ag}$原子的大小，周围不太可能有四个大体积配体。同时，$\mathrm{IO_6^{~5-}}$是常见的双齿配体，所以推测阴离子部分为$\mathrm{Ag(IO_6)_2^{~7-}}$。

<Checkpoint>[与$\mathrm{Ag}$配位的离子为双齿配体$\mathrm{IO_6^{~5-}}$](1)</Checkpoint>

根据$\mathrm{Ag}$的质量分数可以计算得到$M_\bold{A}=M_{Ag}/14.36\%=751.4$，除去银和高碘酸根剩余部分的分子量为197.7，该反应中正离子主要是$\mathrm{K^+}$，根据分子量可以算得化合物$\mathbf{A}$中含$\mathrm{K^+}$五个和$\mathrm{H^+}$两个，化学式写作$\mathrm{K}_{5} \mathrm{H}_{2} \mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}$。实际上两个氢原子与$\mathrm{IO_6^{~5-}}$结合，不影响其配位能力。

<Checkpoint>[$M_\bold{A}=751.4$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{A}$的化学式写作$\mathrm{K}_{5} \mathrm{H}_{2} \mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}$](1)</Checkpoint>

化合物$\mathrm{A}$ 的阴离子中，$\mathrm{Ag}$的配体分子为两个$\mathrm{HIO_6^{~4-}}$，原子数加和为16，说法1正确。
银的价态为+3，说法3错误。

<Checkpoint>[$\mathrm{Ag}$的配体是$\mathrm{HIO_6^{4-}}$](1)</Checkpoint>

根据“歧化反应”，另一种物质为+1价 $\mathrm{Ag}$, 碱性条件写成 $\mathrm{Ag_2 O}$ 最合理, 故方程式为

$$
4 \mathrm{AgO}+6 \mathrm{KOH}+4 \mathrm{KIO}_{4} \rightarrow 2 \mathrm{~K}_{5} \mathrm{H}_{2}\left[\mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}\right]+\mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}
$$

<Checkpoint>[另一产物是$\mathrm{Ag_2O}$](0.5)</Checkpoint>

<Checkpoint>[方程式为  $$ 4 \mathrm{AgO}+6 \mathrm{KOH}+4 \mathrm{KIO}_{4} \rightarrow 2 \mathrm{~K}_{5} \mathrm{H}_{2}\left[\mathrm{Ag}\left(\mathrm{IO}_{6}\right)_{2}\right]+\mathrm{Ag}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} $$](1)</Checkpoint>

氢写到高碘酸根上亦可以。

最简反应方程式系数乘积为192，说法2正确。反应生成的另一种银化合物为$\mathrm{Ag_2O}$，分子量为231.8，说法4错误。