Analyze the following chemical reactions:

1. A small amount of methane reduces manganese(III) oxide, forming manganese(II,III) oxide.

2. Nitroethane is added to aqueous sulfuric acid, and the mixture is heated to produce a monobasic acid and a 1:1 salt with nitrogen in the -1 oxidation state.

3. Iron is added to molten sodium hydroxide, yielding two elemental substances, with the iron transforming into a solid with a spinel structure.

4. Phenylhydrazine and $\mathrm{AtI}$ are dissolved in a $\mathrm { K I - I _ { 2 } }$ solution to produce astatobenzene.

5. Sulfur dioxide is bubbled into a black suspension of $\mathrm { C u ( S C N } ) _ { 2 }$, converting the system into a white suspension.

6. Arsenic-bearing beudantite $\mathrm { P b F e _ { 3 } ( A s O _ { 4 } ) ( S O _ { 4 } ) ( O H ) _ { 6 } }$ reacts with calcium bicarbonate dissolved in mine water, transforming into four minerals: scorodite ($\mathrm { F e A s O _ { 4 }\cdot 2 H _ { 2 } O }$), goethite ($\mathrm{FeOOH}$), cerussite ($\mathrm{PbCO_3}$), and gypsum ($\mathrm{CaSO_4\cdot 2H_2O}$).

Among these, which reactions have chemical equations (with coefficients in the simplest integer ratio) where the sum of the product-side coefficients is even and the sum of the reactant-side coefficients is odd? Select the option that includes **all** such reactions.
分析以下化学反应：

1.少量甲烷还原三氧化二锰，生成四氧化三锰。

2.将硝基乙烷投入含水硫酸中，加热反应，生成一种一元酸以及一种氮为-1价态的1:1型盐。

3.铁投入到熔融的氢氧化钠中，生成2种单质，铁均转化为尖晶石结构的固体。

4.将苯肼和$\mathrm{AtI}$溶于 $\mathrm { K I - I _ { 2 } }$ 溶液中反应，制得砹苯。

5.向黑色 $\mathrm { C u ( S C N } ) _ { 2 }$ 的悬浊液中通入二氧化硫，体系转变为白色悬浊液。

6.砷硫铅铁矿 $\mathrm { P b F e _ { 3 } ( A s O _ { 4 } ) ( S O _ { 4 } ) ( O H ) _ { 6 } }$ 与矿井水中溶解的碳酸氢钙反应，转化为四种矿物：臭葱石（$\mathrm { F e A s O _ { 4 }\cdot 2 H _ { 2 } O }$ ），针铁矿（$\mathrm{FeOOH}$），白铅矿（$\mathrm{PbCO_3}$）和生石膏（$\mathrm{CaSO_4\cdot 2H_2O}$）。

其中哪些反应的化学方程式（系数为最简整数比）中产物一侧系数和为偶数且反应物一侧系数为奇数，选出包含**全部**这些反应的选项？
2
4
6
3,5
1,4
2,6
1,3,4
1,2,5
3,4,6
2,5,6
All of the above options are incorrect.
2
4
6
3,5
1,4
2,6
1,3,4
1,2,5
3,4,6
2,5,6
以上选项均不对
Reaction 1: The common oxidation product of carbon is carbon dioxide, with the equation $\mathrm{12Mn_2O_3+CH_4\rightarrow8Mn_3O_4+CO_2+2H_2O}$  

<Checkpoint>[$\mathrm{12Mn_2O_3+CH_4\rightarrow8Mn_3O_4+CO_2+2H_2O}$](1)</Checkpoint>  

Reaction 2: The cation of a 1:1-type salt with nitrogen at -1 oxidation state should be hydroxylamine, forming $\mathrm{[NH_3OH][HSO_4]}$ in a strongly acidic environment. The equation is $\mathrm{C_2H_5NO_2+H_2SO_4+H_2O\rightarrow CH_3COOH+[NH_3OH][HSO_4]}$  

<Checkpoint>[A 1:1-type salt with nitrogen at -1 oxidation state forms $\mathrm{[NH_3OH][HSO_4]}$ in a strongly acidic environment](1)</Checkpoint>  

<Checkpoint>[$\mathrm{C_2H_5NO_2+H_2SO_4+H_2O\rightarrow CH_3COOH+[NH_3OH][HSO_4]}$](1)</Checkpoint>  

Reaction 3: From the spinel structure, the iron-containing species has the formula $\mathrm{AB_2O_4}$ (where $\mathrm{A,B}$ are +2 and +3 oxidation states, respectively), indicating it is $\mathrm{Fe_3O_4}$. In a reducing environment, oxygen cannot be generated, and the resulting elemental products are sodium and hydrogen. The equation is $\mathrm{3Fe+4NaOH\rightarrow Fe_3O_4+4Na+2H_2}$  

<Checkpoint>[In a reducing environment, oxygen cannot be generated, and the elemental products are sodium and hydrogen](1)</Checkpoint>  

<Checkpoint>[$\mathrm{3Fe+4NaOH\rightarrow Fe_3O_4+4Na+2H_2}$](1)</Checkpoint>  

Reaction 4: The nitrogen in phenylhydrazine converts to nitrogen gas. The equation is $\mathrm{C_6H_5NHNH_2+AtI+I_2\rightarrow C_6H_5At+3HI+N_2}$  

<Checkpoint>[$\mathrm{C_6H_5NHNH_2+AtI+I_2\rightarrow C_6H_5At+3HI+N_2}$](1)</Checkpoint>  

Reaction 5: The solid in the white suspension is $\mathrm{Cu(SCN)}$, and sulfur dioxide is converted to sulfuric acid. The equation is $\mathrm{2Cu(SCN)_2+SO_2+2H_2O\rightarrow 2CuSCN+2HSCN+H_2SO_4}$  

<Checkpoint>[The solid in the white suspension is $\mathrm{Cu(SCN)}$](1)</Checkpoint>  

<Checkpoint>[$\mathrm{2Cu(SCN)_2+SO_2+2H_2O\rightarrow 2CuSCN+2HSCN+H_2SO_4}$](1)</Checkpoint>  

Reaction 6: Using elemental conservation, the equation is directly written as: $\mathrm{PbFe_3(AsO_4)(SO_4)(OH)_6+Ca(HCO_3)_2+H_2O\rightarrow FeAsO_4\cdot 2H_2O+2FeOOH+PbCO_3+CaSO_4\cdot 2H_2O+CO_2}$  

<Checkpoint>[$\mathrm{PbFe_3(AsO_4)(SO_4)(OH)_6+Ca(HCO_3)_2+H_2O\rightarrow FeAsO_4\cdot 2H_2O+2FeOOH+PbCO_3+CaSO_4\cdot 2H_2O+CO_2}$](1)</Checkpoint>  

Reactions 2 and 6 have an even sum of product coefficients and an odd sum of reactant coefficients. The answer is $\mathrm{F}$.  

<Checkpoint>[Reactions 2 and 6 have an even sum of product coefficients and an odd sum of reactant coefficients. The answer is $\mathrm{F}$](1)</Checkpoint>
反应1：碳元素的常见氧化产物为二氧化碳，方程式为$\mathrm{12Mn_2O_3+CH_4\rightarrow8Mn_3O_4+CO_2+2H_2O}$

<Checkpoint>[$\mathrm{12Mn_2O_3+CH_4\rightarrow8Mn_3O_4+CO_2+2H_2O}$](1)</Checkpoint>

反应2：氮为-1价的1：1型盐的阳离子应为羟胺，在强酸性环境为$\mathrm{[NH_3OH][HSO_4]}$。方程式为$\mathrm{C_2H_5NO_2+H_2SO_4+H_2O\rightarrow CH_3COOH+[NH_3OH][HSO_4]}$

<Checkpoint>[氮为-1价的1：1型盐在强酸性环境为$\mathrm{[NH_3OH][HSO_4]}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{C_2H_5NO_2+H_2SO_4+H_2O\rightarrow CH_3COOH+[NH_3OH][HSO_4]}$](1)</Checkpoint>

反应3：由尖晶石结构可知含铁物种化学式为$\mathrm{AB_2O_4}$（$\mathrm{A,B}$分别为+2和+3价），推断其为$\mathrm{Fe_3O_4}$。还原性环境下不能生成氧气，生成的单质为钠和氢气。方程式为$\mathrm{3Fe+4NaOH\rightarrow Fe_3O_4+4Na+2H_2}$

<Checkpoint>[还原性环境下不能生成氧气，生成的单质为钠和氢气](1)</Checkpoint>

<Checkpoint>[$\mathrm{3Fe+4NaOH\rightarrow Fe_3O_4+4Na+2H_2}$](1)</Checkpoint>

反应4：苯肼中的氮变为氮气。方程式为$\mathrm{C_6H_5NHNH_2+AtI+I_2\rightarrow C_6H_5At+3HI+N_2}$

<Checkpoint>[$\mathrm{C_6H_5NHNH_2+AtI+I_2\rightarrow C_6H_5At+3HI+N_2}$](1)</Checkpoint>

反应5：白色悬浊液中的固体为$\mathrm{Cu(SCN)}$，二氧化硫变为硫酸。方程式为$\mathrm{2Cu(SCN)_2+SO_2+2H_2O\rightarrow 2CuSCN+2HSCN+H_2SO_4}$

<Checkpoint>[白色悬浊液中的固体为$\mathrm{Cu(SCN)}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{2Cu(SCN)_2+SO_2+2H_2O\rightarrow 2CuSCN+2HSCN+H_2SO_4}$](1)</Checkpoint>

反应6：利用元素守恒直接写出方程式：$\mathrm{PbFe_3(AsO_4)(SO_4)(OH)_6+Ca(HCO_3)_2+H_2O\rightarrow FeAsO_4\cdot 2H_2O+2FeOOH+PbCO_3+CaSO_4\cdot 2H_2O+CO_2}$

<Checkpoint>[$\mathrm{PbFe_3(AsO_4)(SO_4)(OH)_6+Ca(HCO_3)_2+H_2O\rightarrow FeAsO_4\cdot 2H_2O+2FeOOH+PbCO_3+CaSO_4\cdot 2H_2O+CO_2}$](1)</Checkpoint>

产物一侧系数和为偶数且反应物一侧系数和为奇数的有反应2和6，选$\mathrm{F}$

<Checkpoint>[产物一侧系数和为偶数且反应物一侧系数和为奇数的有反应2和6，选$\mathrm{F}$](1)</Checkpoint>