There exists a binuclear iron complex with the chemical formula $\mathrm{Na}_{2}\left[\mathrm{Fe}_{2}(\mathrm{NO})_{4} \mathrm{S}_{2}\right]$. In this complex, both $\mathrm{Fe}$ atoms satisfy the 18-electron rule. The structure of the complex should be deduced solely based on the information provided in the problem, without referencing known complex structures.

This complex can be obtained by reacting $\mathrm{NH}_{4}\left[\mathrm{Fe}_{4} \mathrm{S}_{3}(\mathrm{NO})_{7}\right]$ and $\mathrm{NaOH}$ in a stoichiometric ratio of $1: 3$ under heating conditions, with the reaction releasing nitrous oxide.

The following statements are given:

1. The apparent oxidation state of the $\mathrm{Fe}$ atom is -1.  
2. The apparent oxidation state of the $\mathrm{Fe}$ atom is +1.  
3. In complex anions, the two atoms farthest apart are separated by at most 6 covalent or coordination bonds, as measured by bond count. 
4. In complex anions, the two atoms farthest apart are separated by at most 5 covalent or coordination bonds, as measured by bond count.
5. The sum of the coefficients on both sides of the chemical reaction equation in the problem is 17.  
6. The sum of the coefficients on both sides of the chemical reaction equation in the problem is 19.

Which of the following is correct?​​
现有一种双核铁配合物, 化学
式 $\mathrm{Na}_{2}\left[\mathrm{Fe}_{2}(\mathrm{NO})_{4} \mathrm{S}_{2}\right]$ 。在该配合物中两个 $\mathrm{Fe}$ 原子均满足 18 电子规则。仅根据题目给出的信息推断配合物结构，不要参考已知的配合物信息。

该配合物可由 $\mathrm{NH}_{4}\left[\mathrm{Fe}_{4} \mathrm{S}_{3}(\mathrm{NO})_{7}\right]$和 $\mathrm{NaOH}$ 以 $1: 3$ 的化学计量比在加热条件下反应得到, 反应放出了笑气。

有以下说法：

1. $\mathrm{Fe}$ 原子的表观氧化数为-1
2. $\mathrm{Fe}$ 原子的表观氧化数为+1
3. 按间隔的化学键计，配合物阴离子中相距最远的两个原子间最多间隔6根共价键或配位键
4. 按间隔的化学键计，配合物阴离子中相距最远的两个原子间最多间隔5根共价键或配位键
5. 题目中的化学反应方程式左右系数加和为17
6. 题目中的化学反应方程式左右系数加和为19


下列选项正确的为？
1,3,5
1,3,6
1,4,5
1,4,6
2,3,5
2,3,6
2,4,5
2,4,6
1,3,5
1,3,6
1,4,5
1,4,6
2,3,5
2,3,6
2,4,5
2,4,6
The number of valence electrons in $\mathrm{Fe}$ is 8, and the negative charge is 2. To satisfy the 18-electron rule, ligands must provide an additional $10-2/2=9$ electrons. In dinuclear $\mathrm{Fe}$ complexes containing $\mathrm{S}$, the $\mathrm{S}$ atom typically serves as a bridging group, donating one electron to each $\mathrm{Fe}$ atom. Therefore, each $\mathrm{Fe}$ in this complex still requires $18-8-1-2=7$ electrons. $\mathrm{NO}$ can act as either a one-electron or three-electron ligand. To provide 7 electrons, $\mathrm{NO}$ here should be a three-electron ligand, and the remaining 1 electron must be supplied by the $\mathrm{Fe-Fe}$ bond.

<Checkpoint>[In the complex, $\mathrm{NO}$ acts as a three-electron ligand, and the presence of an $\mathrm{Fe-Fe}$ bond ensures that each $\mathrm{Fe}$ satisfies the 18-electron rule](1)</Checkpoint>

Thus, the structure of the complex anion consists of one $\mathrm{Fe-Fe}$ bond, two bridging $\mathrm{S}$ atoms connecting the two $\mathrm{Fe}$ atoms, and four $\mathrm{NO}$ ligands acting as three-electron donors, evenly distributed between the two $\mathrm{Fe}$ atoms:
<MultiModal>![Fig. 1. The following SMILES representation corresponds to a complex anion with an overall charge of -2: O=N[Fe]1(S2)(N=O)[Fe]2(N=O)(N=O)S1](/media/uploads/976e2362-280c-464a-b846-49bda139e747.png)</MultiModal>

<Checkpoint>[By counting the intervening chemical bonds, the two farthest-apart atoms in the complex anion are separated by a maximum of 6 covalent or coordination bonds](0.5)</Checkpoint>

Based on electronegativity, the two negative charges contribute an oxidation state of $-2/2=-1$, the $\mathrm{Fe-Fe}$ bond contributes 0, the bridging $\mathrm{S}$ contributes $+1×2=+2$, and the three-electron $\mathrm{NO}$ ligands contribute $-1×2=-2$. Therefore, the apparent oxidation state of each $\mathrm{Fe}$ is $-1+0+2-2=-1$, confirming that statement 1 is correct.

<Checkpoint>[The apparent oxidation state of $\mathrm{Fe}$ is -1](1)</Checkpoint>

According to the problem, this compound can be obtained by reacting $\mathrm{NH}_{4}\left[\mathrm{Fe}_{4} \mathrm{~S}_{3}(\mathrm{NO})_{7}\right]$ with $\mathrm{NaOH}$ in a 1:3 stoichiometric ratio under heating conditions, releasing nitrous oxide. First, draft the preliminary reaction equation:
$\mathrm{{NH}_4\left[Fe_4 S_3({NO_7})\right]+3NaOH\xrightarrow{\Delta} Na_2\left[{Fe}_{2}({NO})_{4} {S}_{2}\right]+N_2O+other products}$
The right side of the equation lacks $\mathrm{Fe}$ and $\mathrm{N}$. Under $\mathrm{NaOH}$, $\mathrm{NH_4^+}$ should be neutralized directly to $\mathrm{NH_3}$, and the $\mathrm{N}$ atoms in $\mathrm{N_2O}$ are partially reduced. Therefore, the oxidized species should be $\mathrm{Fe}$, which in alkaline conditions should oxidize to the stable $\mathrm{Fe(OH)_3}$. Balancing the equation based on electron transfer and material conservation:
$\mathrm{2{NH}_4\left[Fe_4 S_3({NO_7})\right]+6NaOH\xrightarrow{\Delta} 3Na_2\left[{Fe}_{2}({NO})_{4} {S}_{2}\right]+2Fe(OH)_3+N_2O+2NH_3+H_2O}$. The sum of coefficients on both sides of the reaction equation is 17, confirming that statement 5 is correct.

<Checkpoint>[$\mathrm{2{NH}_4\left[Fe_4 S_3({NO_7})\right]+6NaOH\xrightarrow{\Delta} 3Na_2\left[{Fe}_{2}({NO})_{4} {S}_{2}\right]+2Fe(OH)_3+N_2O+2NH_3+H_2O}$](1)</Checkpoint>
$\mathrm{Fe}$的价电子数为8，负电荷数为2，满足18电子规则需要配体提供额外$10-2/2=9$个电子。在含$\mathrm{S}$双核$\mathrm{Fe}$配合物中，$S$原子通常作为桥联基团为每个$\mathrm{Fe}$原子提供一个电子，因此该配合物中每个$\mathrm{Fe}$还需要$18-8-1-2=7$个电子。$\mathrm{NO}$可以作为单电子或者3电子配体，为提供7电子此处$\mathrm{NO}$应该为3电子配体，其余1电子需要由$\mathrm{Fe-Fe}$键提供。

<Checkpoint>[配合物中$\mathrm{NO}$为三电子配体，而且存在一根$\mathrm{Fe-Fe}$键，使得每个$\mathrm{Fe}$满足18电子规则](1)</Checkpoint>

因此配合物阴离子结构为一根$\mathrm{Fe-Fe}$键、两个桥联$S$连接两个$\mathrm{Fe}$原子、四个$\mathrm{NO}$作为三电子配体平均配位到两个$\mathrm{Fe}$原子上：
<MultiModal>![Fig. 1，以下为SMILES表示是一个配合物阴离子，整体带电荷为-2：O=N[Fe]1(S2)(N=O)[Fe]2(N=O)(N=O)S1](/media/uploads/976e2362-280c-464a-b846-49bda139e747.png)</MultiModal>

<Checkpoint>[按间隔的化学键计，配合物阴离子中相距最远的两个原子间最多间隔6根共价键或配位键](0.5)</Checkpoint>

根据电负性，两个负电荷贡献氧化数为$-2/2=-1$，$\mathrm{Fe-Fe}$键贡献氧化数为0，桥接$\mathrm{S}$贡献氧化数为$+1×2=+2$，三电子配体$\mathrm{NO}$贡献氧化数为$-1×2=-2$，因此每个$\mathrm{Fe}$的表观氧化数为$-1+0+2-2=-1$，说法1正确。

<Checkpoint>[$\mathrm{Fe}$的表观氧化数为-1](1)</Checkpoint>

根据题目，该化合物可由$\mathrm{NH}_{4}\left[\mathrm{Fe}_{4} \mathrm{~S}_{3}(\mathrm{NO})_{7}\right]$和 $\mathrm{NaOH}$ 以 $1: 3$ 的化学计量比在加热条件下反应得到, 反应放出了笑气。先写出初步的反应方程式：
$\mathrm{{NH}_4\left[Fe_4 S_3({NO_7})\right]+3NaOH\xrightarrow{\Delta} Na_2\left[{Fe}_{2}({NO})_{4} {S}_{2}\right]+N_2O+其他产物}$
反应式右侧缺少$\mathrm{Fe}$和$\mathrm{N}$，$\mathrm{NH_4^+}$在$\mathrm{NaOH}$下应该直接中和为$\mathrm{NH_3}$，$\mathrm{N_2O}$中$\mathrm{N}$原子被部分还原，那么被氧化的应该是$\mathrm{Fe}$，且在碱性下应该氧化至稳定的三价$\mathrm{Fe(OH)_3}$。根据电子转移数和物料守恒配平方程式：
$\mathrm{2{NH}_4\left[Fe_4 S_3({NO_7})\right]+6NaOH\xrightarrow{\Delta} 3Na_2\left[{Fe}_{2}({NO})_{4} {S}_{2}\right]+2Fe(OH)_3+N_2O+2NH_3+H_2O}$，化学反应式左右系数加和为17，说法5正确。

<Checkpoint>[$\mathrm{2{NH}_4\left[Fe_4 S_3({NO_7})\right]+6NaOH\xrightarrow{\Delta} 3Na_2\left[{Fe}_{2}({NO})_{4} {S}_{2}\right]+2Fe(OH)_3+N_2O+2NH_3+H_2O}$](1)</Checkpoint>