The following chemical processes exist:

1.  Passing $\mathrm{AsH_3}$ into the $\mathrm{HF}-\mathrm{SbF_5}$ system can yield a highly symmetrical anion and cation.

2.  At $203 \mathrm{~K}$, $\mathrm{AlCl_3}$ can react with $\mathrm{N_2F_4}$ to produce a fluoride with one $\rm{C_2}$ axis and one plane of symmetry. This reaction can be understood as $\mathrm{N_2F_4}$ first undergoing disproportionation, and then reacting with $\mathrm{AlCl_3}$.

3.  At high temperature and in the presence of $\mathrm{H_2}$, chromium(III) oxide can react with graphite to obtain a compound with a chromium content of $86.7 \%$.

4.  At $242 \mathrm{~K}$, $\mathrm{[NMe_4]I}$ can react with $\mathrm{XeF_2}$ in a molar ratio of $1: 2$.

5.  The carbonyl compound $\mathrm{O_2NOMn(CO)_5}$ is a stable nitrate complex that decomposes upon heating to produce a common nitrate and release gas.

Regarding the chemical equations of the above reactions, assuming the coefficients are balanced to the simplest integer ratio, the following statements are correct:
有下列化学过程：

1. 将 $\mathrm{AsH_3}$ 通入 $\mathrm{HF}-\mathrm{SbF_5}$ 体系中, 可以得到一高对称性的阴离子和阳离子。

2. 在 $203 \mathrm{~K}$ 下, $\mathrm{AlCl_3}$ 可以与 $\mathrm{N_2F_4}$ 反应, 得到一种具有一条 $\rm{C_2}$ 轴和一个对称面的氟化物。该反应可理解为 $\mathrm{N_2F_4}$ 首先发生歧化, 再和 $\mathrm{AlCl_3}$ 发生反应。

3. 在高温、 $\mathrm{H_2}$ 存在下, 三氧化二铬可以与石墨反应, 得到一种铬含量为 $86.7 \%$ 的化合物。

4. 在 $242 \mathrm{~K}$ 下, $\mathrm{[NMe_4]I}$ 可以与 $\mathrm{XeF_2}$ 按摩尔比 $1: 2$ 发生反应。

5. 羰基化合物 $\mathrm{O_2NOMn(CO)_5}$ 是一种稳定的硝酸盐络合物, 在加热条件下发生分解生成一种常见的硝酸盐且放出气体。

关于上述反应的化学方程式，默认系数配平为最简整数比时，下列说法正确的是：
All other options are incorrect
The anion and cation produced in Process 1 possess the same symmetry.
Chemical equation for process 2, the sum of coefficients of all species is less than 10.
The chemical equation for Process 3, the sum of the coefficients of all species is less than 25.
The stoichiometric coefficients sum of all species in the chemical equation for Process 4 is a prime number.
The chemical equation for process 5, the sum of the coefficients of all species is a prime number.
其他选项均不正确
过程1中产生的阴离子和阳离子具有相同对称性
过程2的化学方程式，所有物种的系数和小于10
过程3的化学方程式，所有物种的系数和小于25
过程4的化学方程式，所有物种的系数和为质数
过程5的化学方程式，所有物种的系数和为质数
Process 1: The anion of the $\mathrm{HF}-\mathrm{SbF_5}$ system is easily conceived as $\rm{SbF_6^-}$; the cation has high symmetry. Considering the Lewis acid-base reaction of non-aqueous solvents, it is likely to be $\rm{AsH_4^+}$. Balancing the equation:

$\rm{AsH_3 + HF + SbF_5 = [AsH_4][SbF_6]}$

<Checkpoint>[$\rm{AsH_3 + HF + SbF_5 = [AsH_4][SbF_6]}$](1)</Checkpoint>

The anion and cation in the product do not have the same symmetry. Option B is incorrect.

Process 2: First consider the disproportionation reaction of $\rm{N_2F_4}$; the +2 valence disproportionates to a stable valence state, generally +3 $\rm{NF_3}$ and 0 valence $\rm{N_2}$; then consider a fluoride with a $\rm{C_2}$ axis and a plane of symmetry, the only thing that comes to mind is $\rm{N_2F_2}$. Therefore, in the reaction of $\rm{NF_3}$ with $\rm{AlCl_3}$, nitrogen is reduced, so chlorine can only be oxidized to chlorine gas, while producing the fluorinated product $\rm{AlF_3}$. Therefore, the equation is balanced as follows:

<Checkpoint>[+2 valence $\mathrm{N}$ disproportionates to a stable valence state, generally +3 $\rm{NF_3}$ and 0 valence $\rm{N_2}$](1)</Checkpoint>

<Checkpoint>[A fluoride with a $\rm{C_2}$ axis and a plane of symmetry, the only thing that comes to mind is $\rm{N_2F_2}$](1)</Checkpoint>

$\rm{2AlCl_3 + 2N_2F_4 = N_2F_2 + 2AlF_3 + N_2 + 3Cl_2}$

<Checkpoint>[$\rm{2AlCl_3 + 2N_2F_4 = N_2F_2 + 2AlF_3 + N_2 + 3Cl_2}$](1)</Checkpoint>

The sum of the coefficients is 11, which is greater than 10, so option C is incorrect.

Process 3: The chromium content is $86.7 \%$ and can only contain carbon, hydrogen, and oxygen elements. It is easy to deduce that the compound is $\rm{Cr_3C_2}$ through the mass fraction; hydrogen is a reducing atmosphere, and the oxidation product of carbon should be $\rm{CO}$. Therefore, the equation is balanced:

<Checkpoint>[The compound is $\rm{Cr_3C_2}$](1)</Checkpoint>

<Checkpoint>[Hydrogen is a reducing atmosphere, and the oxidation product of carbon should be $\rm{CO}$](1)</Checkpoint>

$\rm{3Cr_2O_3 + 13C = 2Cr_3C_2 + 9CO}$

<Checkpoint>[$\rm{3Cr_2O_3 + 13C = 2Cr_3C_2 + 9CO}$](1)</Checkpoint>

The sum of the coefficients is 27, so option D is incorrect.

Process 4: The reaction occurs at a molar ratio of $1: 2$. Since $\mathrm{XeF_2}$ is a strong oxidizing/fluorinating agent, only -1 valent iodine can be oxidized; it can be known from the molar ratio that $\rm{IF_4^-}$ is produced, and $\mathrm{XeF_2}$ is reduced to $\mathrm{Xe}$. Balancing the equation:

<Checkpoint>[$\mathrm{XeF_2}$ is a strong oxidizing/fluorinating agent](1)</Checkpoint>

$\rm{[NMe_4]I +  2XeF_2 = [NMe_4]IF_4 + 2Xe}$

<Checkpoint>[$\rm{[NMe_4]I +  2XeF_2 = [NMe_4]IF_4 + 2Xe}$](1)</Checkpoint>

The sum of the coefficients is 6, so option E is incorrect.

Process 5: The common manganese nitrate is $\rm{Mn(NO_3)_2}$. Judging the valence shows that manganese is oxidized; since a decomposition reaction occurs and the nitrate radical still exists, it can only be a disproportionation reaction, so $\rm{Mn(0)}$ is produced, and a stable carbonyl compound $\rm{Mn_2(CO)_{10}}$ can be conceived; therefore, the equation is balanced:

<Checkpoint>[A decomposition reaction occurs and the nitrate radical still exists, it can only be a disproportionation reaction, so $\rm{Mn(0)}$ is produced, and a stable carbonyl compound $\rm{Mn_2(CO)_{10}}$ can be conceived](1)</Checkpoint>

$\rm{4O_2NO-Mn(CO)_5 = 2Mn(NO_3)_2 + Mn_2(CO)_{10} + 10CO}$

<Checkpoint>[$\rm{4O_2NO-Mn(CO)_5 = 2Mn(NO_3)_2 + Mn_2(CO)_{10} + 10CO}$](1)</Checkpoint>

The sum of the coefficients is 17, which is a prime number, so option F is correct.
过程1：$\mathrm{HF}-\mathrm{SbF_5}$ 体系的阴离子很容易想到为$\rm{SbF_6^-}$；阳离子具有高对称性，考虑非水溶剂的lewis酸碱反应，很可能为$\rm{AsH_4^+}$。配平方程式：

$\rm{AsH_3 + HF + SbF_5 = [AsH_4][SbF_6]}$

<Checkpoint>[$\rm{AsH_3 + HF + SbF_5 = [AsH_4][SbF_6]}$](1)</Checkpoint>

产物中阴离子与阳离子不具有相同对称性。选项B错误。

过程2：首先考虑$\rm{N_2F_4}$的歧化反应；+2价歧化为稳定价态，一般为+3的$\rm{NF_3}$和0价的$\rm{N_2}$；之后考虑具有一条 $\rm{C_2}$ 轴和一个对称面的氟化物，只能想到是$\rm{N_2F_2}$。因此$\rm{NF_3}$与$\rm{AlCl_3}$的反应中氮被还原，则氯只能被氧化为氯气，同时生成氟化产物$\rm{AlF_3}$。因此方程式配平如下：

<Checkpoint>[+2价$\mathrm{N}$歧化为稳定价态，一般为+3的$\rm{NF_3}$和0价的$\rm{N_2}$](1)</Checkpoint>

<Checkpoint>[具有一条 $\rm{C_2}$ 轴和一个对称面的氟化物，只能想到是$\rm{N_2F_2}$](1)</Checkpoint>

$\rm{2AlCl_3 + 2N_2F_4 = N_2F_2 + 2AlF_3 + N_2 + 3Cl_2}$

<Checkpoint>[$\rm{2AlCl_3 + 2N_2F_4 = N_2F_2 + 2AlF_3 + N_2 + 3Cl_2}$](1)</Checkpoint>

系数和为11，大于10，选项C错误。

过程3：铬含量为 $86.7 \%$且只能含有碳氢氧元素，很容易通过质量分数推断出该化合物为$\rm{Cr_3C_2}$；氢气为还原性气氛，碳的氧化产物应当为$\rm{CO}$。因此配平方程式：

<Checkpoint>[化合物为$\rm{Cr_3C_2}$](1)</Checkpoint>

<Checkpoint>[氢气为还原性气氛，碳的氧化产物应当为$\rm{CO}$](1)</Checkpoint>

$\rm{3Cr_2O_3 + 13C = 2Cr_3C_2 + 9CO}$

<Checkpoint>[$\rm{3Cr_2O_3 + 13C = 2Cr_3C_2 + 9CO}$](1)</Checkpoint>

系数和为27，选项D错误。

过程4：摩尔比 $1: 2$ 发生反应，由于$\mathrm{XeF_2}$为强氧化/氟化剂，只有-1价的碘能被氧化；通过摩尔比可知生成了$\rm{IF_4^-}$，$\mathrm{XeF_2}$还原为$\mathrm{Xe}$。配平方程式：

<Checkpoint>[$\mathrm{XeF_2}$为强氧化/氟化剂](1)</Checkpoint>

$\rm{[NMe_4]I +  2XeF_2 = [NMe_4]IF_4 + 2Xe}$

<Checkpoint>[$\rm{[NMe_4]I +  2XeF_2 = [NMe_4]IF_4 + 2Xe}$](1)</Checkpoint>

系数和为6，选项E错误。

过程5：锰常见的硝酸盐为$\rm{Mn(NO_3)_2}$，判断化合价可知锰被氧化；由于发生分解反应且硝酸根仍旧存在，只能为歧化反应，故生成$\rm{Mn(0)}$，可以想到是稳定的羰基化合物$\rm{Mn_2(CO)_{10}}$；因此配平方程式：

<Checkpoint>[发生分解反应且硝酸根仍旧存在，只能为歧化反应，故生成$\rm{Mn(0)}$，可以想到是稳定的羰基化合物$\rm{Mn_2(CO)_{10}}$](1)</Checkpoint>

$\rm{4O_2NO-Mn(CO)_5 = 2Mn(NO_3)_2 + Mn_2(CO)_{10} + 10CO}$

<Checkpoint>[$\rm{4O_2NO-Mn(CO)_5 = 2Mn(NO_3)_2 + Mn_2(CO)_{10} + 10CO}$](1)</Checkpoint>

系数和为17，为质数，选项F正确。