A certain halide $\bf{X}$ of sulfur contains halogen $\bf{Z}$ 70.32%. $\bf{X}$ can react with acetone to generate $\bf{A}$ and the common gas $\bf{B}$ that causes acid rain. $\bf{X}$ can also combine with a pale yellow gas $\bf{C}$ to obtain $\bf{D}$, which contains 19.73% sulfur. $\bf{D}$ can also be formed by oxidation of $\bf{X}$ with element $\bf{E}$ in the presence of $\mathrm{CsF}$ (reaction 1). Irradiating $\bf{D}$ and reacting it with hydrogen produces $\bf{F}$. $\bf{F}$ reacts with ammonia to form $\bf{G}$ containing a threefold axis. $\bf{D}$ can also react with $\bf{B}$ to obtain sulfone compound $\bf{H}$.

$\bf{X}$ can be prepared by reacting a binary compound $\bf{I}$ containing 31.14% sulfur with a binary compound $\bf{J}$ containing 54.75% metal, while producing another sulfur-containing compound (reaction 2). $\bf{X}$ reacts with a boron-containing binary compound $\bf{K}$ to obtain substances such as $\bf{I}$ (reaction 3).

The following statements are correct:
硫的某种卤化物$\bf{X}$含卤素$\bf{Z}$ 70.32%。$\bf{X}$可以与丙酮作用生成$\bf{A}$与常见造成酸雨的气体$\bf{B}$。$\bf{X}$亦可以与淡黄色气体$\bf{C}$化合得到$\bf{D}$，$\bf{D}$含硫 19.73%，$\bf{D}$还可由$\bf{X}$在 $\mathrm{CsF}$ 的存在下被单质$\bf{E}$ 氧化而成（反应 1）。将$\bf{D}$光照，与氢气作用产生 $\bf{F}$，$\bf{F}$ 与氨反应生成含有三次轴的 $\bf{G}$，$\bf{D}$ 亦可与 $\bf{B}$反应得到砜类化合物$\bf{H}$。

$\bf{X}$可由一种含硫 31.14%的二元化合物$\bf{I}$与含金属 54.75%的二元化合物 $\bf{J}$反应制备，同时产生另一种含硫化合物（反应 2）。$\bf{X}$ 与含硼二元化合物$\bf{K}$ 反应，可以得到$\bf{I}$等物质（反应 3）。

下列说法正确的是：
All other options are incorrect
$\bf{X}$ and $\bf{G}$ have the same spatial configuration.
$\bf{A}$ is a nonpolar molecule
The oxidizing power of $\bf{C}$ is stronger than that of $\bf{E}$.
$\bf{G}$ contains no lone pairs of electrons.
A certain product of reaction 1 can serve as a reactant in reaction 2.
The central atoms of $\bf{B}$ and $\bf{I}$ have the same hybridization.
其他选项均不正确
$\bf{X}$与$\bf{G}$具有相同的的空间构型
$\bf{A}$是非极性分子
$\bf{C}$的氧化性强于$\bf{E}$
$\bf{G}$中没有孤电子对
反应1的某一种产物可以作为反应2的反应物
$\bf{B}$与$\bf{I}$的中心原子杂化方式相同
### Determine $\bf{X}$ and $\bf{Z}$

The mass fraction of halogen $\bf{Z}$ in $\bf{X}$ is 70.32%. This means the mass fraction of sulfur is 29.68%.
We can calculate the molar mass ratio of sulfur and halogen:

$$ \frac{n × M(Z) }{ a × M(S)}= \frac{70.32} {29.68} $$

Simplified to

$$ \frac{n}{a} × M(Z) = 75.96 g/mol$$

We try each halogen for $\bf{Z}$ one by one:
If $\bf{Z}$ is $\rm{F}$ ($M(F) = 19.00 g/mol$), then $\frac{n}{a} ≈ 75.96 / 19.00 ≈ 4.00$. The simplest integer ratio is n=4, a=1. The chemical formula is $SF_4$.

Verify the fluorine content of $SF_4$: $w(F) = \frac{4 × 19.00} {32.06 + 4 × 19.00} = \frac{76}  {108.06} ≈ 0.7033 $. This is very consistent with the 70.32% given in the problem.

<Checkpoint>[$\bf{X}$ is $SF_4$, $\bf{Z}$ is F (fluorine)](2)</Checkpoint> 

$\bf{X}$ is $SF_4$, $\bf{Z}$ is F (fluorine).

### Determine $\bf{A}$ and $\bf{B}$

$\bf{X}$ ($SF_4$) reacts with acetone ($(CH_3)_2CO$) to produce $\bf{A}$ and $\bf{B}$. $\bf{B}$ is a common gas that causes acid rain.
The common sulfur-containing acid rain gas is $\bf{B}$ = $SO_2$ (sulfur dioxide).

<Checkpoint>[ $\bf{B}$ is $SO_2$ ](1)</Checkpoint>


$SF_4$ is a strong fluorinating agent that can replace carbonyl oxygen atoms with two fluorine atoms. Its complete reaction with acetone usually produces $SO_2$. The reaction formula is: $SF_4+ 2 (CH_3)_2CO \to 2(CH_3)_2CF_2 + SO_2$.
In this reaction, the organic product is 2,2-difluoropropane. So $\bf{A}$= $(CH_3)_2CF_2$.

<Checkpoint>[$\bf{A}$ is$(CH_3)_2CF_2$](2)</Checkpoint>

### Determine $\bf{C}$ ,$\bf{D}$，$\bf{E}$

$\bf{X}$($SF_4$ ) combines with the pale yellow gas $\bf{C}$ to obtain $\bf{D}$. Starting from substance $\bf{D}$ for derivation, $\bf{C}$ is pale yellow, and it is speculated to be a gas formed by halogen elements, then $\bf{D}$ should be a sulfur hexafluoride, where the number of S is 1, so its molecular weight is $ \frac {32.06} {0.1973}=162.5g/mol $ . Since it is obtained from the reaction of $SF_4$, it is speculated that there are at least 4 F atoms in it, so the mass sum of the remaining two halogen atoms is $ 162.5-32.06-4\times 19=54.43g/mol $, which happens to be one $\rm{F}$ atom and one $\rm{Cl}$ atom, so the pale yellow gas $\bf{C}$ is $ClF$ .

<Checkpoint>[ $\bf{C}$ is$ClF$](1)</Checkpoint>

The reaction is: $SF_4 + ClF \to SF_5Cl$. So $\bf{D}$= $SF_5Cl$.

<Checkpoint>[$\bf{D}$ is$SF_5Cl$](2)</Checkpoint>

$\bf{E}$ is a simple substance. From the perspective of atom conservation, $\bf{E}$ must provide $\rm{Cl}$ atoms, so E is $Cl_2$ (chlorine gas)

<Checkpoint>[$\bf{E}$ is $Cl_2$](1)</Checkpoint>

### Determine $\bf{F}$ and $\bf{G}$

$\bf{D}$ ($SF_5Cl$) reacts with hydrogen under light to produce $\bf{F}$. Under light, the S-Cl bond is easier to break than the S-F bond, and the S-S bond is easier to form than the S-H bond, so the reaction can occur: $2SF_5Cl + H_2 \to S_2F_{10} + 2HCl$. So $\bf{F}$ = $S_2F_{10}$ (disulfur decafluoride)

<Checkpoint>[F is$S_2F_{10}$ ](2)</Checkpoint>

$\bf{F}$ ($S_2F_{10}$) reacts with ammonia to produce $\bf{G}$. And it is known that one product has a threefold axis, and one known reaction is $S_2F_{10} + 4NH_3 \to NSF_3 + 3NH_4F + S$. So $\bf{G}$= $NSF_3$ (sulfur trifluoride nitride)

<Checkpoint>[$\bf{G}$ is $NSF_3$ ](2)</Checkpoint>

### Determine $\bf{H}$
$\bf{D}$ ($SF_5Cl$) reacts with $\bf{B}$ ($SO_2$) to obtain the sulfone compound $\bf{H}$. One possible reaction pathway is the insertion reaction of $\cdot SF_5$ radicals:

$SF_5Cl  → \cdot SF_5 + \cdot Cl$

 $\cdot SF_5 + SO_2 → SF_5SO_2\cdot$

$SF_5SO_2\cdot + SF_5Cl → SF_5SO_2Cl + \cdot SF_5$

The product $\bf{H}$ is $SF_5SO_2Cl$ (pentafluorosulfanyl sulfuryl chloride), which is a sulfone compound (sulfonyl chloride is a derivative of sulfone). So

<Checkpoint>[$\bf{H}$ is$SF_5SO_2Cl$](2)</Checkpoint>

### Determine $\bf{I}$，$\bf{J}$，$\bf{K}$

$\bf{X}$ ($SF_4$) can be prepared by the reaction of $\bf{I}$ and $\bf{J}$ (reaction 2). $\bf{I}$ is a binary compound containing 31.14% sulfur. Assuming that there is only one S atom in $\bf{I}$, then:

$$w(S) = \frac{32.06}  {M(I)} = 0.3114 $$


That is $$M(I) ≈ \frac{32.06}  {0.3114} ≈ 102.95 g/mol$$

 $\bf{I}$ is a binary compound, set as $SY_y$.
If it is a sulfur oxide ($SO_3$, S=40%), ($SO_2$,S=50%). Not consistent.
If it is a sulfur chloride ($SCl_2$, S=31.14%). $M(SCl_2) = 32.06 + 2×35.45 = 102.96 g/mol$. Perfectly consistent. So $\bf{I}$ = $SCl_2$ (sulfur dichloride).

<Checkpoint>[$\bf{I}$ is$SCl_2$](2)</Checkpoint>

$\bf{J}$ is a binary compound containing 54.75% metal. $SCl_2 + \mathbf{J} → SF_4 + another sulfur-containing compound$. This reaction is a fluorination reaction, and $\bf{J}$ must be a metal fluoride. Let $\bf{J}$ be $M_aF_n$. 

$$w(M) =\frac{ a×M(M) }{ a×M(M) + n×M(F)}=0.5475$$。

Solving the indeterminate equation, we get that if $ n/a = 1$ (such as NaF), then $M(M) ≈ 22.99$, which is element Na. So $\bf{J}$ = NaF (sodium fluoride). Verify the Na content in NaF: $w(Na) = \frac{22.99}  {22.99+19.00} ≈ 0.5475$. Consistent.

<Checkpoint>[$\bf{J}$ is $NaF$](1)</Checkpoint>

$\bf{X}$ ($SF_4$) reacts with $\bf{K}$ to obtain $\bf{I}$ ($SCl_2$). $\bf{K}$ is a boron-containing binary compound, and from the conservation of elements, we know that $\bf{K}$ should also have Cl element. Therefore, there is reaction 3: $3SF_4 + 4BCl_3 \to 4BF_3 + 3SCl_2 + 3Cl_2$. This reaction can generate $SCl_2$. So $\bf{K}$ = $BCl_3$ (boron trichloride).

<Checkpoint>[$\bf{K}$ is $BCl_3$](2)</Checkpoint>

### Summarize all substances

$\bf{X}$ = $SF_4$ (sulfur tetrafluoride)

$\bf{Z}$ = $F$ (fluorine)

$\bf{A}$= $(CH_3)_2CF_2$ (2,2-difluoropropane)

$\bf{B}$ = $SO_2$ (sulfur dioxide)

$\bf{C}$ = $ClF$ (chlorine monofluoride)

$\bf{D}$ = $SF_5Cl$ (sulfur pentafluoride chloride)

$\bf{E}$= $Cl_2$ (chlorine gas)

$\bf{F}$ = $S_2F_{10}$ (disulfur decafluoride)

$\bf{G}$ = $NSF_3$ (sulfur trifluoride nitride)

$\bf{H}$ = $SF_5SO_2Cl$ (pentafluorosulfanyl sulfuryl chloride)

$\bf{I}$= $SCl_2$ (sulfur dichloride)

$\bf{J}$ = $NaF$ (sodium fluoride)

$\bf{K}$ = $BCl_3$ (boron trichloride)

### Analyze options

Option B: 

<Checkpoint>[$SF_4$ is seesaw](1)</Checkpoint>


<Checkpoint>[$SNF_3$ is tetrahedral](1)</Checkpoint>


$SF_4$ (seesaw) and $SNF_3$ (tetrahedral) are different, incorrect.

Option C:  $(CH_3)_2CF_2$ is a polar molecule. Incorrect.

Option D: $ClF$ oxidizing property \>$ Cl_2$. Correct.

Option E:

<Checkpoint>[$SNF_3$ has 10 lone pairs of electrons](1)</Checkpoint>

The 10 lone pairs of electrons in $SNF_3$ exist on nitrogen and fluorine atoms. Incorrect.

Option F: Reaction 1: $SF_4 + Cl_2 +CsF→ SF_5Cl +CsCl$
  Reaction 2: $3SCl_2 + 4NaF → SF_4 + S_2Cl_2 + 4NaCl$. Incorrect.

Option G: $SO_2$ ($sp^2$) vs $SCl_2$ ($sp^3$). Different. Incorrect.
### 确定 $\bf{X}$ 和 $\bf{Z}$

$\bf{X}$ 中卤素 $\bf{Z}$ 的质量分数为 70.32%。这意味着硫的质量分数为 29.68%。
我们可以计算硫和卤素的摩尔质量比：

$$ \frac{n × M(Z) }{ a × M(S)}= \frac{70.32} {29.68} $$

化简后得到

$$ \frac{n}{a} × M(Z) = 75.96 g/mol$$

我们逐一尝试 $\bf{Z}$ 为何种卤素：
若 $\bf{Z}$ 为 $\rm{F}$ ($M(F) = 19.00 g/mol$)，则 $\frac{n}{a} ≈ 75.96 / 19.00 ≈ 4.00$。最简整数比为 n=4, a=1。化学式为 $SF_4$。

验证 $SF_4$ 的氟含量：$w(F) = \frac{4 × 19.00} {32.06 + 4 × 19.00} = \frac{76}  {108.06} ≈ 0.7033 $。这与题目给出的 70.32% 非常吻合。

<Checkpoint>[$\bf{X}$ 是 $SF_4$，$\bf{Z}$ 是 F (氟) ](2)</Checkpoint> 

$\bf{X}$ 是 $SF_4$，$\bf{Z}$ 是 F (氟)。

### 确定 $\bf{A}$ 和 $\bf{B}$

$\bf{X}$ ($SF_4$) 与丙酮 ($(CH_3)_2CO$) 作用生成 $\bf{A}$ 和 $\bf{B}$。$\bf{B}$ 是常见的造成酸雨的气体。
含硫的常见酸雨气体是 $\bf{B}$ = $SO_2$ (二氧化硫)。

<Checkpoint>[ $\bf{B}$ 是 $SO_2$ ](1)</Checkpoint>


$SF_4$是一种强氟化剂，能将羰基氧原子置换为两个氟原子。其与丙酮充分反应通常会生成 $SO_2$。反应式为：$SF_4+ 2 (CH_3)_2CO \to 2(CH_3)_2CF_2 + SO_2$。
在此反应中，有机产物是 2,2-二氟丙烷。所以 $\bf{A}$= $(CH_3)_2CF_2$。

<Checkpoint>[$\bf{A}$ 是$(CH_3)_2CF_2$](2)</Checkpoint>

### 确定 $\bf{C}$ ，$\bf{D}$，$\bf{E}$

$\bf{X}$($SF_4$ ) 与淡黄色气体 $\bf{C}$ 化合得到 $\bf{D}$。从物质$\bf{D}$入手进行推导，$\bf{C}$为淡黄色，推测为卤族元素形成的气体，那么$\bf{D}$应该为S的六卤化物，其中S的个数为1，因此其分子量为 $ \frac {32.06} {0.1973}=162.5g/mol $  ,由于其从$SF_4$ 反应得到，因此推测其中至少有4个F原子，因此剩下的两个卤素原子质量和为$ 162.5-32.06-4\times 19=54.43g/mol $,恰好为一个$\rm{F}$原子和一个$\rm{Cl}$原子，所以，淡黄色气体$\bf{C}$是$ClF$ 。

<Checkpoint>[ $\bf{C}$是$ClF$](1)</Checkpoint>

反应为：$SF_4 + ClF \to SF_5Cl$。所以 $\bf{D}$= $SF_5Cl$。

<Checkpoint>[$\bf{D}$是$SF_5Cl$](2)</Checkpoint>

$\bf{E}$是一种单质。从原子守恒看，$\bf{E}$必须提供$\rm{Cl}$原子，因此 E 是 $Cl_2$ (氯气)

<Checkpoint>[$\bf{E}$是 $Cl_2$](1)</Checkpoint>

### 确定$\bf{F}$和$\bf{G}$

$\bf{D}$ ($SF_5Cl$) 光照与氢气作用产生$\bf{F}$。光照下 S-Cl 键比 S-F 键更容易断裂，并且S-S键比S-H键更易生成，因此可以发生反应：$2SF_5Cl + H_2 \to S_2F_{10} + 2HCl$。这样$\bf{F}$ = $S_2F_{10}$ (十氟化二硫)

<Checkpoint>[F 是$S_2F_{10}$ ](2)</Checkpoint>

$\bf{F}$ ($S_2F_{10}$) 与氨反应生成 $\bf{G}$。且已知有一种产物具有三次轴，一个已知的反应是 $S_2F_{10} + 4NH_3 \to NSF_3 + 3NH_4F + S$。因此$\bf{G}$= $NSF_3$ (三氟化硫氮)

<Checkpoint>[$\bf{G}$ 是 $NSF_3$ ](2)</Checkpoint>

### 确定$\bf{H}$
$\bf{D}$ ($SF_5Cl$) 与 $\bf{B}$ ($SO_2$) 反应得到砜类化合物 $\bf{H}$。可能的一种反应途径是 $\cdot SF_5$ 自由基的插入反应：

$SF_5Cl  → \cdot SF_5 + \cdot Cl$

 $\cdot SF_5 + SO_2 → SF_5SO_2\cdot$

$SF_5SO_2\cdot + SF_5Cl → SF_5SO_2Cl + \cdot SF_5$

产物 $\bf{H}$ 是 $SF_5SO_2Cl$ (五氟硫基磺酰氯)，这是一个砜类化合物（磺酰氯是砜的衍生物）。所以 

<Checkpoint>[$\bf{H}$ 是$SF_5SO_2Cl$](2)</Checkpoint>

### 确定$\bf{I}$，$\bf{J}$，$\bf{K}$

$\bf{X}$ ($SF_4$) 可由 $\bf{I}$ 和 $\bf{J}$ 反应制备（反应 2）。$\bf{I}$ 是含硫 31.14% 的二元化合物。假设$\bf{I}$中只有一个S原子，则有：

$$w(S) = \frac{32.06}  {M(I)} = 0.3114 $$


即$$M(I) ≈ \frac{32.06}  {0.3114} ≈ 102.95 g/mol$$

 $\bf{I}$是二元化合物，设为 $SY_y$。
若为硫的氧化物 ($SO_3$, S=40%)、($SO_2$,S=50%)。不符。
若为硫的氯化物 ($SCl_2$, S=31.14%)。$M(SCl_2) = 32.06 + 2×35.45 = 102.96 g/mol$。完全吻合。所以$\bf{I}$ = $SCl_2$ (二氯化硫)。

<Checkpoint>[$\bf{I}$是$SCl_2$](2)</Checkpoint>

$\bf{J}$ 是含金属 54.75% 的二元化合物。 $SCl_2 + \mathbf{J} → SF_4 + 另一种含硫化合物$。该反应是氟化反应，$\bf{J}$ 必然是金属氟化物。设 $\bf{J}$ 为 $M_aF_n$。 

$$w(M) =\frac{ a×M(M) }{ a×M(M) + n×M(F)}=0.5475$$。

解该不定方程，得到若$ n/a = 1$ (如 NaF)，则$M(M) ≈ 22.99$，为元素Na。所以 $\bf{J}$ = NaF (氟化钠)。 验证 NaF 中 Na 含量：$w(Na) = \frac{22.99}  {22.99+19.00} ≈ 0.5475$。吻合。

<Checkpoint>[$\bf{J}$是 $NaF$](1)</Checkpoint>

$\bf{X}$ ($SF_4$) 与$\bf{K}$ 反应得到 $\bf{I}$ ($SCl_2$)。$\bf{K}$ 是含硼二元化合物，由元素守恒可以知道$\bf{K}$还应该有Cl元素。 因此有反应 3: $3SF_4 + 4BCl_3 \to 4BF_3 + 3SCl_2 + 3Cl_2$。这个反应可以生成 $SCl_2$。所以 $\bf{K}$ = $BCl_3$ (三氯化硼)。

<Checkpoint>[$\bf{K}$ 是 $BCl_3$](2)</Checkpoint>

### 总结所有物质

$\bf{X}$ = $SF_4$ (四氟化硫)

$\bf{Z}$ = $F$ (氟)

$\bf{A}$= $(CH_3)_2CF_2$ (2,2-二氟丙烷)

$\bf{B}$ = $SO_2$ (二氧化硫)

$\bf{C}$ = $ClF$ (氟化氯)

$\bf{D}$ = $SF_5Cl$ (一氯五氟化硫)

$\bf{E}$= $Cl_2$ (氯气)

$\bf{F}$ = $S_2F_{10}$ (十氟化二硫)

$\bf{G}$ = $NSF_3 $(三氟化硫氮)

$\bf{H}$ = $SF_5SO_2Cl $(五氟硫基磺酰氯)

$\bf{I}$= $SCl_2 $(二氯化硫)

$\bf{J}$ = $NaF$ (氟化钠)

$\bf{K}$ = $BCl_3$ (三氯化硼)

### 分析选项

选项 B: 

<Checkpoint>[$SF_4$为跷跷板](1)</Checkpoint>


<Checkpoint>[$SNF_3$为四面体](1)</Checkpoint>


$SF_4$(跷跷板) 和 $SNF_3$(四面体)不同，错误。

选项 C:  $(CH_3)_2CF_2$ 是极性分子。错误。

选项 D: $ClF$ 氧化性 \>$ Cl_2$。正确。

选项 E:

<Checkpoint>[$SNF_3$ 中有 10 个孤电子对](1)</Checkpoint>

$SNF_3$ 中的 10 个孤电子对存在于氮和氟原子上。错误。

选项 F：反应 1: $SF_4 + Cl_2 +CsF→ SF_5Cl +CsCl$
  反应 2: $3SCl_2 + 4NaF → SF_4 + S_2Cl_2 + 4NaCl$。错误。

选项 G: $SO_2$ ($sp^2$) vs $SCl_2$ ($sp^3$)。不同。错误。