Trace element $\mathbf{X}$ is named after a Western European country. Extraction of $\mathbf{X}$ from minerals of elements from the same group requires a complex process, in which the key step is the formation of a precipitate $\mathbf{A}$ containing $\mathbf{X}$ and a precipitate $\mathbf{B}$ of the same type but without $\mathbf{Χ}$ to enrich $\mathbf{X}$.

The element $\mathbf{X}$ can react with a colorless irritating gas $\mathbf{C}$ to form a gray semi-metallic solid $\mathbf{D}$, and $\mathbf{D}$-type compounds have important industrial applications.

A monovalent anion $\mathbf{E}$ containing a mass fraction of $37.11\%$ of $\mathbf{X}$ has a $C_2$ axis and 4 bridging bonds, while the neutral compound $\mathbf{F}$ with the same elemental composition as $\mathbf{E}$ exhibits a $+2$ apparent valence for $\mathbf{X}$, but in reality, there is an anion-cation pair. $\mathbf{F}$ dissolves in diethyl ether to obtain a solvate $\mathbf{G}$, which also contains discrete anions and cations with the same coordination number for the anions and cations.

The reaction of $LiBH_4$ with a high-valent halide of $\mathbf{X}$ at $-45\degree C$ yields a ternary compound $\mathbf{H}$, where $\omega_\mathbf{X}=69.43\%$, which can be approximately considered to have $C_{2v}$ symmetry.

A binary compound $\mathbf{I}$ of $\mathbf{X}$ can be prepared below $308K$. Its monomer is a viscous liquid with slight polarity, while the polymer is a white solid. Both have strong reducing properties. The $1:1$ adduct formed by $\mathbf{I}$ with ligand $\mathbf{J}$ can be used to prepare $\mathbf{I}$ and can also form a $1:2$ complex $\mathbf{K}$, whose configuration is different from $\mathbf{H}$.

$\mathbf{C, J, D}$ contain the same element, and $\mathbf{J}$ can be prepared by complete methylation of $\mathbf{C}$.

Given the following options for the above process, let $x$ be the product of the serial numbers of all incorrect statements, and $y$ be the sum of the serial numbers of all correct statements. Find $\dfrac{\sqrt x}{y+1}$ and select the option with an absolute error $<0.001$, if there is no option that meets the condition, choose $A$.

1. The acidity of precipitate $\mathbf{A}$ is stronger than that of precipitate $\mathbf{B}$ partly due to the effect of electron shielding on the effective nuclear charge

2. Gas $\mathbf{C}$ dissolves in water to give an acidic solution

3. Compound $\mathbf{D}$ reacts violently with water at room temperature

4. One $C_2$ axis of anion $\mathbf{E}$ passes through a hexacoordinate atom and two ligands

5. Compound $\mathbf{F}$ has a molecular weight of $281.2$

6. The anion and cation of solvate $\mathbf{G}$ have the same stereoconfiguration

7. One molecule of compound $\mathbf{H}$ contains 13 atoms

8. Compound $\mathbf{I}$ is extremely prone to decompose into its elemental form at room temperature

9. The number of atoms of each element in ligand $\mathbf{J}$ constitutes a geometric sequence

10. The configuration of complex $\mathbf{K}$ is square pyramidal
稀散元素 $\mathbf{X}$ 名字来自西欧国家。从同族元素矿物中提取 $\mathbf{X}$ 需经过复杂的过程，其中关键步骤为形成含 $\mathbf{X}$ 的沉淀 $\mathbf{A}$ 与同类型不含 $\mathbf{Χ}$ 的沉淀 $\mathbf{B}$ 以富集 $\mathbf{X}$。

$\mathbf{X}$ 单质可与无色刺激性气体 $\mathbf{C}$ 反应生成灰色半金属状固体 $\mathbf{D}$，$\mathbf{D}$ 类型化合物具有重要工业用途。

含有 $\mathbf{X}$ 质量分数为 $ 3 7 . 1 1 \%$ 的一价阴离子 $\mathbf{E}$ 具有 $C_2$ 轴与4个桥键，而与 $\mathbf{E}$ 组成元素相同的电中性化合物 $\mathbf{F}$ 中 $\mathbf{X}$ 表观 $+2$ 价，实则存在一对阴阳离子。$\mathbf{F}$ 溶于乙醚得溶剂合物 $\mathbf{G}$，亦存在分立的阴阳离子且阴阳离子配位数相同。

将 $LiBH_4$ 与 $\mathbf{X}$ 的高价卤化物于 $-45\degree C$ 下反应得三元化合物 $\mathbf{H}$ ，其中 $\omega_\mathbf{X}=69.43\%$ ，可近似认为具有 $C_{2v}$ 对称性。

在低于 $308K$ 时可制得 $\mathbf{X}$ 的二元化合物 $\mathbf{I}$ ，其单体为粘稠状液体，略显极性，多聚体却为白色固体，二者均具有强还原性。$\mathbf{I}$ 与配体 $\mathbf{J}$ 形成的 $1:1$ 加合物可用于制备 $\mathbf{I}$，亦可形成 $1:2$ 型配合物 $\mathbf{K}$，其构型与 $\mathbf{H}$ 不同。

$\mathbf{C、J、D}$ 中含同种元素，且 $\mathbf{J}$ 可由 $\mathbf{C}$ 发生彻底甲基化反应制得。

针对上述过程给出如下选项，记所有错误叙述序号的乘积为 $x$，所有正确叙述序号的和为 $y$，求 $\dfrac{\sqrt x}{y+1}$，选出误差绝对值 $<0.001$ 的选项，如果没有符合条件的选项则选择 $A$。

1.沉淀 $\mathbf{A}$ 酸性强于沉淀 $\mathbf{B}$ 部分由于电子屏蔽效应对有效核电荷数的影响

2.气体 $\mathbf{C}$ 溶于水得到酸性溶液

3.化合物$\mathbf{D}$ 常温下与水剧烈反应

4.阴离子 $\mathbf{E}$ 的某一个 $C_2$ 轴穿过一个六配位的原子和两个配体

5.化合物 $\mathbf{F}$ 分子量 $281.2$ 

6.溶剂合物 $\mathbf{G}$ 的阴阳离子立体构型完全相同

7.一分子化合物 $\mathbf{H}$ 原子数为13

8.化合物$\mathbf{I}$ 室温下极易分解为单质

9.配体 $\mathbf{J}$ 的元素原子个数构成等比数列

10.配合物 $\mathbf{K}$ 的构型为四方锥形
All other options are incorrect
0.373
3.162
0.727
3.036
1.793
1.623
其他选项均不正确
0.373
3.162
0.727
3.036
1.793
1.623
$\mathbf{X}$ is both a rare element and named after a Western European country. Consider the three elements $Ga(\text{Gallia}),Ge(\text{Germenia}),Sc(\text{Scandia})$.

Consider the reaction of $LiBH_4$ with a high-valent halide of $\mathbf{X}$ to obtain a ternary compound $\mathbf{H}$. The strong reducing power of $LiBH_4$ completely replaces the halogen and provides two kinds of ions, $BH_4^-,H^-$. Therefore, the other two elements besides $\mathbf{X}$ should be $B,H$. Let the chemical formula of $\mathbf{H}$ be $\mathbf{X}H_x(BH_4)_y$, then $\omega_{\mathbf{X}}=\dfrac{Ar(\mathbf{X})}{Ar(\mathbf{X})+1.008x+14.84y}=69.43\%$.

Consider the element $\mathbf{X}=Ga$, substitute to get the equation $1.008x+14.84y=30.70$, the approximate non-negative integer solution is $x=1,y=2$, corresponding to $\mathbf{H}=GaH(BH_4)_2$, actually $\omega_{Ga}=\dfrac{69.72}{100.408}=69.43\%$.

Consider the element $\mathbf{X}=Ge$, substitute to get the equation $1.008x+14.84y=31.97$, the approximate non-negative integer solution is $x=2,y=2$, corresponding to $\mathbf{H}=GeH_2(BH_4)_2$, actually $\omega_{Ge}=\dfrac{72.61}{104.306}=69.61\%$.

Consider the element $\mathbf{X}=Sc$, substitute to get the equation $1.008x+14.84y=19.80$, the approximate non-negative integer solution is $x=5,y=1$, corresponding to $\mathbf{H}=ScH_5(BH_4)$, actually $\omega_{Sc}=\dfrac{44.96}{64.84}=69.34\%$.

Therefore, $\mathbf{X}=Ga,\mathbf{H}=GaH(BH_4)_2$, approximately has $C_{2v}$ symmetry. The number of atoms in one molecule of compound $\mathbf{H}$ is 12, statement 7 is wrong.

<Checkpoint>[$\mathbf{X}=Ga,\mathbf{H}=GaH(BH_4)_2$](1)</Checkpoint>

The homologous mineral is bauxite, then the two precipitates are $\mathbf{A}=Ga(OH)_3,\mathbf{B}=Al(OH)_3$. Acidity $Ga(OH)_3>Al(OH)_3$, because the poor shielding effect of 3d electrons and the lanthanide contraction make the effective nuclear charge of $Ga$ larger, and the ion potential of $Ga^{3+}$ is greater than $A1^{3+}$, statement 1 is correct.

<Checkpoint>[$\mathbf{A}=Ga(OH)_3,\mathbf{B}=Al(OH)_3$](1)</Checkpoint>

The strong reducing binary compound $\mathbf{I}=GaH_3$ is similar to $BH_3$, and the corresponding polymer $(GaH_3)_n$ is a white solid. The decomposition reaction $2GaH_3\rightarrow 2Ga + 3H_2$ proceeds rapidly above $243K$, statement 8 is correct.

<Checkpoint>[$\mathbf{I}=GaH_3$](0.5)</Checkpoint>

The colorless irritating gas $\mathbf{C}$ is considered $HCl,HBr,NH_3$, etc., because the ligand $\mathbf{J}$ can be prepared by complete methylation reaction of $\mathbf{C}$, consider $\mathbf{C}=NH_3,\mathbf{J}=N(CH_3)_3$.

<Checkpoint>[$\mathbf{C}=NH_3,\mathbf{J}=N(CH_3)_3$](1)</Checkpoint>

The gas $\mathbf{C}$ is dissolved in water to obtain weakly alkaline ammonia water, statement 2 is wrong. The number of element atoms in $\mathbf{J}$ constitutes a geometric sequence of $1:3:9$, statement 9 is correct.

The formed complex of $\mathbf{I}:\mathbf{J}=1:2$ is $\mathbf{K}=GaH_3\cdot2N(CH_3)_3$. The configuration is trigonal bipyramidal, and the two large ligands tend to be in the axial direction, statement 10 is wrong.

<Checkpoint>[$\mathbf{K}=GaH_3\cdot2N(CH_3)_3$](0.5)</Checkpoint>

Therefore, the gray semi-metallic solid $\mathbf{D}=GaN$ produced by $\mathbf{X}$ and $\mathbf{C}$ is widely used in industry as a semiconductor, and it is not easily soluble in water at room temperature and does not react violently with water, statement 3 is wrong.

<Checkpoint>[$\mathbf{D}=GaN$](0.5)</Checkpoint>

The common valence of $Ga$ is $+1,+3$, while the apparent valence in $\mathbf{F}$ is $+2$, indicating that the two common valences account for half each; the monovalent anion $\mathbf{E}$ with the same element composition has a $C_2$ axis and 4 bridging bonds, inferring that both contain halogen. Let the chemical formula of $\mathbf{E}$ be $GaY_a^-$, then $\omega_{Ga}=\dfrac{Ar(Ga)}{Ar(Ga)+aAr(Y)}=37.11\%$, simplified to $aAr(Y)=118.15$, there are solutions $6.22F、3.33Cl、1.48Br、0.93I$ respectively.

Combined with common valences, we get $\mathbf{E}= Ga_3Cl_{10}^-$. The corresponding $\mathbf{F}=Ga[GaCl_4]$, the solvate of $\mathbf{G}=[Ga(Et_2O)_4][GaCl_4]$ dissolved in ether, ensures that the coordination number of anions and cations is the same.

<Checkpoint>[$\mathbf{E}= Ga_3Cl_{10}^-,\mathbf{F}=Ga[GaCl_4],\mathbf{G}=[Ga(Et_2O)_4][GaCl_4]$](1.5)</Checkpoint>

Therefore, the $C_2$ axis of $\mathbf{E}$ passes through a central six-coordinated $Ga$ atom and the upper and lower two $Cl$ ligands, statement 4 is correct.

The molecular weight of $\mathbf{F}$ is $2\times69.72+4\times35.45=281.24$, statement 5 is correct.

The anion of $\mathbf{G}$ has four-coordinated $T_d$ symmetry, and the cation has seesaw-shaped $C_{2v}$ symmetry, the stereoconfigurations are different, statement 6 is wrong.

The correct statements are $1,4,5,8,9$, and the wrong statements are $2,3,6,7,10$.

<Checkpoint>[正确的叙述为 $1,4,5,8,9$，错误的叙述为 $2,3,6,7,10$](1)</Checkpoint>

Calculate $x=2\times3\times6\times7\times10=2520,y=27,\dfrac{\sqrt x}{y+1}=1.793$.

<Checkpoint>[$x=2520,y=27,\dfrac{\sqrt x}{y+1}=1.793$](1)</Checkpoint>
$\mathbf{X}$ 既是稀散元素又取名自西欧国家，考虑 $Ga(\text{Gallia}),Ge(\text{Germenia}),Sc(\text{Scandia})$三种元素。

考虑 $LiBH_4$ 与 $\mathbf{X}$ 的高价卤化物反应得到三元化合物 $\mathbf{H}$ ， $LiBH_4$ 的强还原性将卤素完全取代，并提供了 $BH_4^-,H^-$ 两种离子，因此除 $\mathbf{X}$ 之外的另两种元素应分别是 $B,H$。记 $\mathbf{H}$ 的化学式 $\mathbf{X}H_x(BH_4)_y$，那么 $\omega_{\mathbf{X}}=\dfrac{Ar(\mathbf{X})}{Ar(\mathbf{X})+1.008x+14.84y}=69.43\%$。

考虑 $\mathbf{X}=Ga$ 元素，代入得到方程 $1.008x+14.84y=30.70$，近似非负整数解 $x=1,y=2$，对应 $\mathbf{H}=GaH(BH_4)_2$，实际 $\omega_{Ga}=\dfrac{69.72}{100.408}=69.43\%$。

考虑 $\mathbf{X}=Ge$ 元素，代入得到方程 $1.008x+14.84y=31.97$，近似非负整数解 $x=2,y=2$，对应 $\mathbf{H}=GeH_2(BH_4)_2$，实际 $\omega_{Ge}=\dfrac{72.61}{104.306}=69.61\%$。

考虑 $\mathbf{X}=Sc$ 元素，代入得到方程 $1.008x+14.84y=19.80$，近似非负整数解 $x=5,y=1$，对应 $\mathbf{H}=ScH_5(BH_4)$，实际 $\omega_{Sc}=\dfrac{44.96}{64.84}=69.34\%$。

因此 $\mathbf{X}=Ga,\mathbf{H}=GaH(BH_4)_2$ ，近似具有 $C_{2v}$ 对称性。一分子化合物 $\mathbf{H}$ 原子数为12，叙述7错误。

<Checkpoint>[$\mathbf{X}=Ga,\mathbf{H}=GaH(BH_4)_2$](1)</Checkpoint>

同族矿物为铝土矿，则两种沉淀 $\mathbf{A}=Ga(OH)_3,\mathbf{B}=Al(OH)_3$。酸性 $Ga(OH)_3>Al(OH)_3$，因为3d电子屏蔽效应较差和钪系收缩使得 $Ga$ 有效核电荷较大，$Ga^{3+}$ 离子势大于 $A1^{3+}$ ，叙述1正确。

<Checkpoint>[$\mathbf{A}=Ga(OH)_3,\mathbf{B}=Al(OH)_3$](1)</Checkpoint>

强还原性二元化合物 $\mathbf{I}=GaH_3$ 与 $BH_3$类似，对应多聚体 $(GaH_3)_n$ 是白色固体。分解反应 $2GaH_3\rightarrow 2Ga + 3H_2$ 在高于 $243K$ 时快速进行，叙述8正确。

<Checkpoint>[$\mathbf{I}=GaH_3$](0.5)</Checkpoint>

无色刺激性气体 $\mathbf{C}$ 考虑 $HCl,HBr,NH_3$ 等，因为配体 $\mathbf{J}$ 可由 $\mathbf{C}$ 发生彻底甲基化反应制得，考虑 $\mathbf{C}=NH_3,\mathbf{J}=N(CH_3)_3$。

<Checkpoint>[$\mathbf{C}=NH_3,\mathbf{J}=N(CH_3)_3$](1)</Checkpoint>

气体 $\mathbf{C}$ 溶于水得到弱碱性氨水，叙述2错误。 $\mathbf{J}$ 的元素原子个数构成 $1:3:9$ 的等比数列，叙述9正确。

生成的 $\mathbf{I}:\mathbf{J}=1:2$ 的配合物 $\mathbf{K}=GaH_3\cdot2N(CH_3)_3$。构型为三角双锥，两个大配体倾向于处于轴向，叙述10错误。

<Checkpoint>[$\mathbf{K}=GaH_3\cdot2N(CH_3)_3$](0.5)</Checkpoint>

从而 $\mathbf{X}$ 与 $\mathbf{C}$ 生成的灰色半金属状固体 $\mathbf{D}=GaN$，作为半导体广泛运用于工业，常温下不易溶于水、不与水剧烈反应，叙述3错误。

<Checkpoint>[$\mathbf{D}=GaN$](0.5)</Checkpoint>

$Ga$ 常见化合价为 $+1,+3$ 价，而 $\mathbf{F}$ 中表观 $+2$ 价，说明两种常见化合价各占一半；与之元素组成的一价阴离子 $\mathbf{E}$ 具有 $C_2$ 轴与4个桥键，推断二者含卤素。记 $\mathbf{E}$ 化学式 $GaY_a^-$，则 $\omega_{Ga}=\dfrac{Ar(Ga)}{Ar(Ga)+aAr(Y)}=37.11\%$，化简为 $aAr(Y)=118.15$，分别有解 $6.22F、3.33Cl、1.48Br、0.93I$。

结合常见化合价得到 $\mathbf{E}= Ga_3Cl_{10}^-$。对应的 $\mathbf{F}=Ga[GaCl_4]$，溶于乙醚的溶剂合物 $\mathbf{G}=[Ga(Et_2O)_4][GaCl_4]$，保证了阴阳离子配位数相同。

<Checkpoint>[$\mathbf{E}= Ga_3Cl_{10}^-,\mathbf{F}=Ga[GaCl_4],\mathbf{G}=[Ga(Et_2O)_4][GaCl_4]$](1.5)</Checkpoint>

故 $\mathbf{E}$ 的 $C_2$ 轴穿过一个中心六配位的 $Ga$ 原子和上下两个 $Cl$ 配体，叙述4正确。

$\mathbf{F}$ 的分子量 $2\times69.72+4\times35.45=281.24$，叙述5正确。

$\mathbf{G}$ 的阴离子为四配位 $T_d$ 对称性，阳离子为跷跷板形 $C_{2v}$对称性，立体构型不同，叙述6错误。

正确的叙述为 $1,4,5,8,9$，错误的叙述为 $2,3,6,7,10$。

<Checkpoint>[正确的叙述为 $1,4,5,8,9$，错误的叙述为 $2,3,6,7,10$](1)</Checkpoint>

计算 $x=2\times3\times6\times7\times10=2520,y=27,\dfrac{\sqrt x}{y+1}=1.793$。

<Checkpoint>[$x=2520,y=27,\dfrac{\sqrt x}{y+1}=1.793$](1)</Checkpoint>