**X** and **Y** are short-period elements of the same main group.  

The reaction between the chloride of **X** and a compound containing approximately 65% of **Y** can prepare **A**; alternatively, the chloride of **X** and the hydride of **Y** in a 3:5 ratio can also yield compound **A** under high temperature. In **A**, the coordination number of **X** atoms is 4, while that of **Y** atoms is 2 and 3. Under high temperature and a nitrogen atmosphere, the reaction between **A** and salt **B**, or the 1:2 reaction between the lithium salt of **Y** and $\mathrm{LiXY_2}$, both produce compound **P**. The anion of **P** is isoelectronic with phosphorus pentoxide, while the anion of **B** is isoelectronic with the perchlorate ion.  

Which of the following statements best fits the context?
**X**和**Y**是同主族短周期元素。
**X**的氯化物和含**Y**约65%的化合物反应可以制备**A**；或**X**的氯化物和**Y**的氢化物按照比例为3：5反应在高温下也可制得化合物**A**。**A**中**X**原子的配位数均为4，而**Y**原子的配位数为2和3。在高温和氮气气氛下，**A**和盐**B**反应或**Y**的锂盐和$\mathrm{LiXY_2}$发生1：2反应均能够制得化合物**P**。**P**的阴离子是五氧化二磷的等电子体,而**B**的阴离子是高氯酸根的等电子体。


下列说法最符合题意的是：
The anion of **P** carries eight negative charges.
The anion of **P** possesses three fourfold rotation axes.
The equation for the preparation of **A** from the chloride of **X** and a compound containing approximately $\mathrm{65\%}$ of **Y**, with the coefficients in the simplest whole number ratio such that the sum of the coefficients of the reactants is greater than 20.
The lithium salt of **Y** reacts with $\mathrm{LiXY_2}$ to form compound **P**. When the coefficients in the chemical equation for this reaction are in their simplest whole-number ratio, the sum of the coefficients of the reactants is 9.
It is known that above 1100K, **A** decomposes through a two-step reaction, each step producing an elemental substance, and the decomposition proceeds via the intermediate **O**; considering the chemical formula of **O** in its simplest integer ratio, its relative molecular mass is greater than or equal to 45.
The ratio of **Y** atoms with two-coordination and three-coordination in **A** is 2:3.
The chemical equation for the reaction between **A** and salt **B**, with the coefficients in the simplest whole-number ratio, has a sum of the product coefficients equal to 7.
None of the above options is correct.
**P**的阴离子带8个负电荷。
**P**的阴离子具有3个四重旋转轴。
**X**的氯化物和含**Y**约$\mathrm{65\%}$的化合物制备**A**的方程式，系数为最简整数比时，反应物的系数和大于20。
**Y**的锂盐和$\mathrm{LiXY_2}$发生反应制得了化合物**P**，该反应的化学方程式，系数为最简整数比时，反应物的系数和为9。
已知在1100K以上，**A**通过两步反应分解，每步反应都生成单质，该分解经过中间体**O**；考虑**O**最简整数比的化学式，其相对分子质量大于等于45。
**A**中二配位和三配位的**Y**原子，个数比为2:3。
**A**和盐**B**反应的化学方程式，系数为最简整数比时，产物的系数和为7。
以上选项均不正确。
**X** and **Y** are short-period elements in the same main group. Since **Y** forms a lithium salt, alkali metals and alkaline earth metals are first excluded.  

<Checkpoint>[**Y** forms a lithium salt, excluding alkali metals and alkaline earth metals](1)</Checkpoint>  

The coordination numbers of the **Y** atom are 2 and 3, suggesting that **X** and **Y** are nitrogen and phosphorus.  

<Checkpoint>[**X** and **Y** are inferred to be nitrogen and phosphorus](1)</Checkpoint>  

Since the compound $\mathrm{LiXY_2}$ exists, **X** and **Y** cannot be chalcogens, halogens, or elements of Group 13, as no such compounds are known. Therefore, **X** and **Y** must belong to the nitrogen group, with **Y** having a lower coordination number, likely nitrogen, and **X** being phosphorus. In this case, $\mathrm{LiXY_2}$ corresponds to $\mathrm{LiPN_2}$, a known compound, confirming the deduction.  

<Checkpoint>[**Y** is nitrogen](0.5)</Checkpoint>  

<Checkpoint>[**X** is phosphorus](0.5)</Checkpoint>  

The chloride of **X** should be phosphorus trichloride or pentachloride. When reacting with a compound containing approximately $\mathrm{65\%}$ of **Y**, the mass fraction of nitrogen suggests the compound is sodium azide.  

<Checkpoint>[The compound containing ~$\mathrm{65\%}$ **Y** is sodium azide](1)</Checkpoint>  

The hydride of **Y** should be ammonia, which reacts with the chloride of **X** in a 5:3 ratio. Based on elemental knowledge and coordination number information, the product **A** is $\mathrm{P_3N_5}$, indicating the chloride is phosphorus pentachloride. The equations for the formation of **A** are:  

<Checkpoint>[**A** is $\mathrm{P_3N_5}$](2)</Checkpoint>  

$\mathrm{3PCl_5 + 15NaN_3 = 20N_2 + 15NaCl + P_3N_5}$  

The sum of the reactant coefficients is 18, making option C incorrect.  

<Checkpoint>[$\mathrm{3PCl_5 + 15NaN_3 = 20N_2 + 15NaCl + P_3N_5}$](1)</Checkpoint>  

$\mathrm{3PCl_5 + 5NH_3 = 15HCl + P_3N_5}$  

<Checkpoint>[$\mathrm{3PCl_5 + 5NH_3 = 15HCl + P_3N_5}$](1)</Checkpoint>  

**A** is $\mathrm{P_3N_5}$, where each phosphorus atom has a coordination number of 4, while nitrogen atoms have coordination numbers of 2 and 3. Let the number of 2-coordinate nitrogen atoms be *x*; the equation can be written as:  

$\mathrm{4 \times 3 = 2x + 3(5 - x)}$  

Solving gives *x* = 3, so the ratio of 2-coordinate to 3-coordinate nitrogen is 3:2, making option F incorrect.  

<Checkpoint>[In **A**, the ratio of 2-coordinate to 3-coordinate nitrogen is 3:2](1)</Checkpoint>  

$\mathrm{P_3N_5}$ decomposes at high temperatures into two elemental substances, nitrogen and phosphorus. The initially formed product is nitrogen, so the intermediate, in its simplest ratio, is phosphorus nitride ($\mathrm{PN}$). The combined relative molecular mass is 44.98, less than 45, making option E incorrect.  

<Checkpoint>[The intermediate decomposition product, phosphorus nitride, has the simplest ratio 1:1, i.e., $\mathrm{PN}$](1)</Checkpoint>  

The lithium salt of **Y** could be lithium nitride or lithium azide.  
The lithium salt of **Y** reacts with $\mathrm{LiPN_2}$ in a 1:2 ratio to form compound **P**. Since the anion of **P** is isoelectronic with $\mathrm{P_2O_5}$ and likely contains nitrogen, and $\mathrm{P_2O_5}$ typically exists as $\mathrm{P_4O_{10}}$, the well-known anion structure $\mathrm{P_4N_{10}^{10-}}$ is easily deduced. Thus, **P** is $\mathrm{Li_{10}P_4N_{10}}$. The anion has an adamantane-like structure with a $T_d$ point group, lacking a fourfold rotation axis, making options A and B incorrect.  

<Checkpoint>[**P** is $\mathrm{Li_{10}P_4N_{10}}$](2)</Checkpoint>  

<Checkpoint>[The $\mathrm{P_4N_{10}^{10-}}$ anion belongs to the $T_d$ point group and lacks a fourfold rotation axis](1)</Checkpoint>  

The reaction between the lithium salt of **Y** and $\mathrm{LiPN_2}$ to form $\mathrm{Li_{10}P_4N_{10}}$ suggests the lithium salt is lithium nitride. The equation is:  

$\mathrm{2Li_3N + 4LiPN_2 = Li_{10}P_4N_{10}}$  

<Checkpoint>[$\mathrm{2Li_3N + 4LiPN_2 = Li_{10}P_4N_{10}}$](1)</Checkpoint>  

The sum of reactant coefficients is 6, making option D incorrect.  

$\mathrm{P_3N_5}$ can also react with salt **B** to form $\mathrm{Li_{10}P_4N_{10}}$. Since the anion of **B** is isoelectronic with perchlorate, it is deduced to be $\mathrm{PN_4^{7-}}$, making **B** $\mathrm{Li_7PN_4}$. The reaction equation is:  

<Checkpoint>[Salt **B** is $\mathrm{Li_7PN_4}$](2)</Checkpoint>  

$\mathrm{10Li_7PN_4 + 6P_3N_5 = 7Li_{10}P_4N_{10}}$  

<Checkpoint>[$\mathrm{10Li_7PN_4 + 6P_3N_5 = 7Li_{10}P_4N_{10}}$](1)</Checkpoint>  

The sum of product coefficients is 7, making option G correct.  

In conclusion, option G is correct.
**X**和**Y**是同主族短周期元素，**Y**有锂盐，首先排除碱金属和碱土金属；

<Checkpoint>[**Y**有锂盐，排除碱金属和碱土金属](1)</Checkpoint>

**Y**原子的配位数为2和3，可以推测**X**和**Y**为氮元素和磷元素；

<Checkpoint>[推测**X**和**Y**为氮元素和磷元素](1)</Checkpoint>

由于存在$\mathrm{LiXY_2}$这种物质，如果**X,Y**为氧族或者卤素或者第三主族，不存在这样的化合物；因此**X,Y**只能为氮族元素，且**Y**配位数更低，应当为氮元素，**X**为磷元素。此时$\mathrm{LiXY_2}$化学式为$\mathrm{LiPN_2}$，为一种常见化合物，印证了猜测正确。

<Checkpoint>[**Y**是氮元素](0.5)</Checkpoint>

<Checkpoint>[**X**是磷元素](0.5)</Checkpoint>

**X**的氯化物应当为三氯化磷或者五氯化磷，和含**Y**约$\mathrm{65\%}$的化合物反应，根据氮元素的质量分数可以推出该化合物为叠氮化钠；

<Checkpoint>[含**Y**约$\mathrm{65\%}$的化合物为叠氮化钠](1)</Checkpoint>

**Y**的氢化物应当为氨气，氨气与**X**的氯化物按照5：3反应，根据元素知识和配位数信息，生成产物**A**为$\mathrm{P_3N_5}$，此时**X**的氯化物为五氯化磷。从而生成**A**的方程式分别为

<Checkpoint>[**A**为$\mathrm{P_3N_5}$](2)</Checkpoint>

$\mathrm{3PCl_5 + 15NaN_3 = 20N_2 + 15NaCl + P_3N_5}$

反应反应物系数和为18，选项C错误。

<Checkpoint>[$\mathrm{3PCl_5 + 15NaN_3 = 20N_2 + 15NaCl + P_3N_5}$](1)</Checkpoint>

$\mathrm{3PCl_5 + 5NH_3 = 15HCl + P_3N_5}$

<Checkpoint>[$\mathrm{3PCl_5 + 5NH_3 = 15HCl + P_3N_5}$](1)</Checkpoint>


**A**为$\mathrm{P_3N_5}$，该结构中磷原子的配位数均为4，而氮原子的配位数为2和3；设二配位的氮原子有x个，可列方程：

$\mathrm{4\times3=2x+3(5-x)}$

得到x为3，从而二配位：三配位为3：2，选项F错误。

<Checkpoint>[**A**中氮原子二配位：三配位为3：2](1)</Checkpoint>

$\mathrm{P_3N_5}$高温分解为两种单质，应当为氮气和磷单质；先生成的单质应当为氮气，故此时中间体在化简为最简整数比应当为氮化磷$\mathrm{PN}$，相对分子质量之和为44.98，小于45，选项E错误。

<Checkpoint>[分解中间体氮化磷的最简整数比应当为1：1亦即$\mathrm{PN}$](1)</Checkpoint>

**Y**的锂盐可能为氮化锂或叠氮化锂；
**Y**的锂盐和$\mathrm{LiPN_2}$发生1：2反应制得了化合物**P**，由于**P**阴离子为$\mathrm{P_2O_5}$的等电子体且大概率含有氮元素，$\mathrm{P_2O_5}$一般表现为$\mathrm{P_4O_{10}}$，从而很容易联想到著名的阴离子结构$\mathrm{P_4N_{10}^{10-}}$，从而**P**为$\mathrm{Li_{10}P_4N_{10}}$。该结构阴离子为金刚烷式结构，点群为$T_d$，不具有四重旋转轴，选项A,B均错误。

<Checkpoint>[**P**为$\mathrm{Li_{10}P_4N_{10}}$](2)</Checkpoint>

<Checkpoint>[$\mathrm{P_4N_{10}^{10-}}$阴离子属于$T_d$点群，不具有四重旋转轴](1)</Checkpoint>

**Y**的锂盐和$\mathrm{LiPN_2}$发生反应制得$\mathrm{Li_{10}P_4N_{10}}$，根据比例可得**Y**的锂盐为氮化锂，方程式为

$\mathrm{2Li_3N + 4LiPN_2 = Li_{10}P_4N_{10}}$

<Checkpoint>[$\mathrm{2Li_3N + 4LiPN_2 = Li_{10}P_4N_{10}}$](1)</Checkpoint>

反应物系数和为6，选项D错误。

$\mathrm{P_3N_5}$和盐**B**反应也可制$\mathrm{Li_{10}P_4N_{10}}$，且盐**B**阴离子与高氯酸根为等电子体，可推出该阴离子为$\mathrm{PN_4^{7-}}$，盐**B**为$\mathrm{Li_7PN_4}$，反应方程式为

<Checkpoint>[盐**B**为$\mathrm{Li_7PN_4}$](2)</Checkpoint>

$\mathrm{10Li_7PN_4 + 6P_3N_5 = 7Li_{10}P_4N_{10}}$

<Checkpoint>[$\mathrm{10Li_7PN_4 + 6P_3N_5 = 7Li_{10}P_4N_{10}}$](1)</Checkpoint>

产物系数和为7，选项G正确。

综上，选项G正确。