$\mathrm{Na^+}$ ions and water molecules may exist in the cages of $\mathrm{Y}$- type zeolites, and their content varies with preparation conditions. The chemical formula of a certain artificially prepared $\mathrm{Y}$-type zeolite can be written as $\mathrm{Na_m[Al_mSi_nO_{2}] \cdot xH_2O}$. The following determination results are known:

(1) Complete dehydration of more than $100.0\ \mathrm{g}$ of zeolite sample yields $72.6\ \mathrm{g}$ of anhydrous solid $\mathrm{Na_m[Al_mSi_nO_{2}]}$;

(2) $\mathrm{^{29}Si}$-NMR measurements show that there are five silicons with different chemical shifts in the crystal, and the ratio of their contents is:

<MultiModal>![Silicon-oxygen tetrahedron connected to four silicon atoms, silicon-oxygen tetrahedron connected to three silicon atoms and one aluminum atom, silicon-oxygen tetrahedron connected to two silicon atoms and two aluminum atoms, silicon-oxygen tetrahedron connected to one silicon atom and three aluminum atoms, and silicon-oxygen tetrahedron connected to four aluminum atoms, from left to right.](/media/uploads/ac9c716b-32f8-4ddf-aac6-7327ad5f2929.png)</MultiModal>

$\mathrm{1.24 : 10.46 : 14.24 : 6.30 : 0.60}$

Which of the following options are incorrect:

1.\(\frac{m + x}{n} > 2\)  
   
2.\(m \times n < 0.3\) 

3.\(\frac{x - m}{n} < 1.4\)

4.\(\frac{m}{n} > 0.4\)  

5.\(\frac{n - m}{x} > 0.32\)  

6.\(\frac{m}{x} < 0.25\)  

7.\(n > \frac{x}{1.8}\)  

8.\(m \times x > 0.40\)
$\mathrm{Y}$型分子筛的笼中可能存在 $\mathrm{Na^+}$ 离子和水分子，其含量随制备条件而变化。某种人工制备的 $\mathrm{Y}$ 型分子筛，其化学式可记为 $\mathrm{Na_m[Al_mSi_nO_{2}] \cdot xH_2O}$，已知以下测定结果：

（1）将 $100.0\ \mathrm{g}$ 以上分子筛样品完全脱水，得到 $72.6\ \mathrm{g}$ 的无水固体 $\mathrm{Na_m[Al_mSi_nO_{2}]}$；

（2）$\mathrm{^{29}Si}$-NMR 测定表明该晶体中存在以下五种化学位移不同的硅，其含量比依次为：

<MultiModal>![从左到右依次为硅氧四面体连接四个硅原子、硅氧四面体连接三个硅原子和一个铝原子、硅氧四面体连接两个硅原子和两个铝原子、硅氧四面体连接一个硅原子和三个铝原子、硅氧四面体连接四个铝原子。](/media/uploads/ac9c716b-32f8-4ddf-aac6-7327ad5f2929.png)</MultiModal>

$\mathrm{1.24 : 10.46 : 14.24 : 6.30 : 0.60}$

下列选项错误的有：

1.\(\frac{m + x}{n} > 2\)  
   
2.\(m \times n < 0.3\) 

3.\(\frac{x - m}{n} < 1.4\)

4.\(\frac{m}{n} > 0.4\)  

5.\(\frac{n - m}{x} > 0.32\)  

6.\(\frac{m}{x} < 0.25\)  

7.\(n > \frac{x}{1.8}\)  

8.\(m \times x > 0.40\)
2,6
1,2,6,8
3,5,7
2,3,5,7
1,2,6
2,3,5,7,8
1,2,3,4,5,6,7,8
3,4,7
1,2,4,6
All of the above options are incorrect or the answer is incomplete.
2,6
1,2,6,8
3,5,7
2,3,5,7
1,2,6
2,3,5,7,8
1,2,3,4,5,6,7,8
3,4,7
1,2,4,6
以上选项均错误或答案不完全
Lowenstein's rule states that in aluminosilicates (such as zeolites), two aluminum-oxygen tetrahedra cannot be connected by a corner, that is, there is no $\mathrm{Al-O-Al}$ oxygen bridge.

<Checkpoint>[there is no $\mathrm{Al-O-Al}$ oxygen bridge](1)</Checkpoint>

The relative content of $\mathrm{Si}$ in the crystal is: $1.24 + 10.46 + 14.24 + 6.30 + 0.60 = 32.84$

<Checkpoint>[The relative content of $\mathrm{Si}$ is: $32.84$](1)</Checkpoint>

The relative content of $\mathrm{Al}$ in the crystal is: $\frac{10.46 + 2\times14.24 + 3\times6.30 + 4\times0.60}{4} = 15.06$

<Checkpoint>[The relative content of $\mathrm{Al}$ is: $15.06$](1)</Checkpoint>

According to charge conservation, $m+n = 1$

<Checkpoint>[$m+n = 1$](1)</Checkpoint>

Therefore: $n = \frac{32.84}{32.84 + 15.06} = 0.686$

<Checkpoint>[$n = 0.686$](1)</Checkpoint>

$m = 1 - n = 0.314$

<Checkpoint>[$m = 0.314$](1)</Checkpoint>

From: $\frac{M(\mathrm{Na}_{0.314}[\mathrm{Al}_{0.314}\mathrm{Si}_{0.686}\mathrm{O}_2])}{M(\mathrm{Na}_{0.314}[\mathrm{Al}_{0.314}\mathrm{Si}_{0.686}\mathrm{O}_2]\cdot x\mathrm{H}_2\mathrm{O})} = \frac{72.6}{100.0}$

We get: $x = 1.40$

<Checkpoint>[$x = 1.40$](1)</Checkpoint>

Substituting into the calculation:

Option 1, $\frac{m+x}{n}=2.50>2$, correct

Option 2, $mn=0.215<0.3$, correct

Option 3, $\frac{x-m}{n}=1.58>1.4$, incorrect

Option 4, $\frac{m}{n}=0.458>0.4$, correct

Option 5, $\frac{n-m}{x}=0.266<0.32$, incorrect

Option 6, $\frac{m}{x}=0.224<0.25$, correct

Option 7, $\frac{x}{1.8}=0.778>n$, incorrect

Option 8, $mx=0.440>0.4$, correct

In summary, options 3, 5, and 7 are incorrect, so the correct answer is C.

<Checkpoint>[Options 3, 5, and 7 are incorrect, so the correct answer is C](1)</Checkpoint>
Lowenstein规则指出，在铝硅酸盐（如沸石）中，两个铝氧四面体不能共角相连，即不存在 $\mathrm{Al-O-Al}$ 氧桥。

<Checkpoint>[不存在 $\mathrm{Al-O-Al}$ 氧桥](1)</Checkpoint>

晶体中 $\mathrm{Si}$ 的相对含量为：$1.24 + 10.46 + 14.24 + 6.30 + 0.60 = 32.84$

<Checkpoint>[$\mathrm{Si}$ 的相对含量为：$32.84$](1)</Checkpoint>

晶体中 $\mathrm{Al}$ 的相对含量为：$\frac{10.46 + 2\times14.24 + 3\times6.30 + 4\times0.60}{4} = 15.06$

<Checkpoint>[$\mathrm{Al}$ 的相对含量为：$15.06$](1)</Checkpoint>

由电荷守恒，$m+n = 1$

<Checkpoint>[$m+n = 1$](1)</Checkpoint>

故：$n = \frac{32.84}{32.84 + 15.06} = 0.686$

<Checkpoint>[$n = 0.686$](1)</Checkpoint>  

$m = 1 - n = 0.314$

<Checkpoint>[$m = 0.314$](1)</Checkpoint>

由：$\frac{M(\mathrm{Na}_{0.314}[\mathrm{Al}_{0.314}\mathrm{Si}_{0.686}\mathrm{O}_2])}{M(\mathrm{Na}_{0.314}[\mathrm{Al}_{0.314}\mathrm{Si}_{0.686}\mathrm{O}_2]\cdot x\mathrm{H}_2\mathrm{O})} = \frac{72.6}{100.0}$

得：$x = 1.40$

<Checkpoint>[$x = 1.40$](1)</Checkpoint>

代入计算：

选项1，$\frac{m+x}{n}=2.50>2$，正确

选项2，$mn=0.215<0.3$，正确

选项3，$\frac{x-m}{n}=1.58>1.4$，错误

选项4，$\frac{m}{n}=0.458>0.4$，正确

选项5，$\frac{n-m}{x}=0.266<0.32$，错误

选项6，$\frac{m}{x}=0.224<0.25$，正确

选项7，$\frac{x}{1.8}=0.778>n$，错误

选项8，$mx=0.440>0.4$，正确

综上，选项3、5、7错误，故正确答案为C。

<Checkpoint>[选项3、5、7错误，正确答案为C](1)</Checkpoint>