The second-period elements $\bf{X}$ and element $\bf{Y}$ react at $20 \degree \mathrm { C }$ to obtain the common Lewis acid molecule $\bf{A}$. When $\bf{A}$ passes through the crystal of $\bf{X}$ at $1850 \degree \mathrm { C }$ and a pressure lower than $133 \ \mathrm { P a }$, a reactive gas $\bf{B}$ is obtained, which is isoelectronic with $\rm{CO}$. $\bf{B}$ and $\bf{A}$ co-condense at $-196 \degree\mathrm { C }$ in a 2:1 stoichiometric ratio to obtain compound $\bf{C}$. $\bf{C}$ is unstable and disproportionates above $-50 \mathrm { \degree C } $ to form the yellow compound $\bf{D}$ and $\bf{E}$ in a ratio of 1:2. $\bf{E}$ exists in two conformations in the solid and gaseous states. $\bf{D}$ has a similar structure and bonding mode to the hydride of element $\bf{X}$. Reacting $\bf{D}$ with $\rm{CO}$ can cause symmetrical cleavage to obtain the adduct $\bf{F}$, and there is a $C_3$ axis in $\bf{F}$.

According to the above description, which of the following statements is incorrect (including calculations, error within 1%)?
第二周期元素$\bf{X}$和元素$\bf{Y}$在 $2 0  { \degree } \mathrm { C }$ 下反应可以得到常见路易斯酸分子$\bf{A}$。$\bf{A}$在 $1 8 5 0  \ { \degree } \mathrm { C }$ 和低于 $1 3 3 \ \mathrm { P a }$ 的压力下通过$\bf{X}$的晶体时可得一活泼的气体$\bf{B}$，与$\rm{CO}$互为等电子体。$\bf{B}$与$\bf{A}$在 $-1 9 6 \degree\mathrm { C }$ 按照2:1计量比共同凝聚可得到化合物$\bf{C}$，$\bf{C}$不稳定，在 $- 5 0 \mathrm { \degree C } $以上便歧化生成黄色化合物$\bf{D}$与$\bf{E}$，比例为1:2。$\bf{E}$在固态和气态中存在两种构象。$\bf{D}$与$\bf{X}$元素的氢化物有相似的结构与成键方式。将$\bf{D}$与$\rm{CO}$反应可发生对称裂解得到加合物$\bf{F}$，$\bf{F}$中存在一根 $C_3$ 轴。

依据上述描述，下列说法中错误的是(包含计算的，误差在1%以内)
All other options are incorrect
$\bf{Y}$ element is a pale yellow gas
The relative molecular mass of $\bf{C}$ is approximately $127$
There exist 3 chemical environments of $\bf{X}$ in $\bf{D}$
The simplest formula of $\bf{E}$ is $\bf{XY_2}$
The mass fraction of $\rm{C}$ in $\bf{F}$ is $6.5\%$
其他选项均不正确
$\bf{Y}$单质为淡黄色气体
$\bf{C}$的相对分子质量约为$127$
$\bf{D}$中存在3种化学环境的$\bf{X}$
$\bf{E}$的最简式为$\bf{XY_2}$
$\bf{F}$中$\rm{C}$的质量分数为$6.5\%$
For a second-period element to form a common Lewis acid, then $\bf{A}$ can only be $\rm{BF_3}$.

<Checkpoint>[$\bf{A}$ is $\rm{BF_3}$](1)</Checkpoint>

If $\bf{B}$ and $\rm{CO}$ are isoelectronic, then $\bf{B}$ can only be $\rm{BF}$. Therefore, $\bf{X}$ is $\rm{B}$, and $\bf{Y}$ is $\rm{F}$. The simple substance of $\bf{Y}$ is a pale yello gas, so B is correct.

<Checkpoint>[$\bf{B}$ is $\rm{BF}$](1)</Checkpoint>
<Checkpoint>[$\bf{X}$ is $\rm{B}$, $\bf{Y}$ is $\rm{F}$](1)</Checkpoint>

According to the question stem "$\bf{B}$ and $\bf{A}$ co-condense at $-1 9 6 \degree\mathrm { C }$ with a stoichiometric ratio of 2:1 to obtain compound $\bf{C}$", $\bf{C}$ should be $\rm{B_3F_5}$, with a relative molecular mass of approximately $127$, so C is correct.

<Checkpoint>[$\bf{C}$ is $\rm{B_3F_5}$](1)</Checkpoint>

Since "$\bf{E}$ exists in two conformations in the solid and gaseous states", it can be determined that $\bf{E}$ is $\rm{B_2F_4}$, and the two conformations are planar and trigonal pyramidal, respectively. The simplest formula is $\bf{XY_2}$, so E is correct.

<Checkpoint>[$\bf{E}$ is $\rm{B_2F_4}$](1)</Checkpoint>

According to the question stem "$\bf{D}$ has a similar structure and bonding mode to the hydride of element $\bf{X}$", there should be a three-center two-electron bond in $\bf{D}$. Under this condition, $\rm{B}$ has 3 forms of existence: terminal $\rm{BF_2}$, bridging $\rm{BF_2}$, and central $\rm{B}$. Therefore, for any structure $\rm{B_a(BF_2)_b}$, $\rm{F}$ must be an even number. Let there be $x$ terminal $\rm{BF_2}$ groups, which should satisfy $4a=x+2(b-x)$, i.e., $4a=2b-x$. It should be ensured that $0\lt 2a\lt b$ and $x\lt b$ and $a\lt b-x$, i.e., $0\lt 2a\lt b\lt 3a$.

To balance the equation with a product stoichiometric ratio of 1:2, when the reactant stoichiometric number is 2, $\bf{D}$ is $\rm{B_2F_2}$, i.e., $\rm{B_0(BF_2)_2}$, which cannot be satisfied; when the reactant stoichiometric number is 4, $\bf{D}$ is $\rm{B_8F_{12}}$, i.e., $\rm{B_2(BF_2)_6}$, which is feasible; when the reactant stoichiometric number is 6, $\bf{D}$ is $\rm{B_{14}F_{22}}$, i.e., $\rm{B_3(BF_2)_{11}}$, which cannot be satisfied; when the reactant stoichiometric number is 8, $\bf{D}$ is $\rm{B_{20}F_{32}}$, i.e., $\rm{B_4(BF_2)_{16}}$, which cannot be satisfied; higher stoichiometric numbers also cannot satisfy the conditions. Therefore, $\bf{D}$ is $\rm{B_8F_{12}}$, i.e., $\rm{B_2(BF_2)_6}$, which has 2 central $\rm{B}$ atoms, 4 terminal $\rm{BF_2}$ groups, and 2 bridging $\rm{BF_2}$ groups, and there are 3 kinds of chemical environments of $\bf{X}$, so D is correct.

<Checkpoint>[$\bf{D}$ is $\rm{B_8F_{12}}$, i.e., $\rm{B_2(BF_2)_6}$, which has 2 central $\rm{B}$ atoms, 4 terminal $\rm{BF_2}$ groups, and 2 bridging $\rm{BF_2}$ groups, and there are 3 kinds of chemical environments of $\bf{X}$](2)</Checkpoint>

According to the question stem "$\bf{F}$ has a $C_3$ axis", it can be directly determined that $\bf{F}$ is $\rm{B(BF_2)_3CO}$, and the mass fraction of $\rm{C}$ is $6.5\%$, so F is correct. Therefore, choose A.

<Checkpoint>[$\bf{F}$ is $\rm{B(BF_2)_3CO}$](1)</Checkpoint>
第二周期元素要形成常见路易斯酸，则$\bf{A}$仅能为$\rm{BF_3}$。

<Checkpoint>[$\bf{A}$为$\rm{BF_3}$](1)</Checkpoint>

$\bf{B}$与$\rm{CO}$互为等电子体，则$\bf{B}$只能为$\rm{BF}$，因此$\bf{X}$为$\rm{B}$，$\bf{Y}$为$\rm{F}$。$\bf{Y}$单质为淡黄色气体，B正确。

<Checkpoint>[$\bf{B}$为$\rm{BF}$](1)</Checkpoint>
<Checkpoint>[$\bf{X}$为$\rm{B}$，$\bf{Y}$为$\rm{F}$](1)</Checkpoint>

根据题干“$\bf{B}$与$\bf{A}$在 $-1 9 6 \degree\mathrm { C }$ 按照2:1计量比共同凝聚可得到化合物$\bf{C}$”，$\bf{C}$应当为$\rm{B_3F_5}$，相对分子质量约为$127$，C正确。

<Checkpoint>[$\bf{C}$为$\rm{B_3F_5}$](1)</Checkpoint>

由于“$\bf{E}$在固态和气态中存在两种构象”，可以判断$\bf{E}$为$\rm{B_2F_4}$，两种构象分别为平面和三角锥，最简式为$\bf{XY_2}$，E正确。

<Checkpoint>[$\bf{E}$为$\rm{B_2F_4}$](1)</Checkpoint>

根据题干“$\bf{D}$与$\bf{X}$元素的氢化物有相似的结构与成键方式”，$\bf{D}$中应该存在三中心两电子键。这种条件下$\rm{B}$有3种存在形式：端基$\rm{BF_2}$，桥连$\rm{BF_2}$和中心$\rm{B}$。因此，对于任意结构$\rm{B_a(BF_2)_b}$，$\rm{F}$必须为偶数。设端基$\rm{BF_2}$有$x$个，应当满足$4a=x+2(b-x)$，即$4a=2b-x$，应当保证$0\lt 2a\lt b$且$x\lt b$且$a\lt b-x$，即$0\lt 2a\lt b\lt 3a$。

由1:2的产物计量比配平方程式，当反应物化学计量数为2时，$\bf{D}$为$\rm{B_2F_2}$，即$\rm{B_0(BF_2)_2}$,无法满足；当反应物化学计量数为4时，$\bf{D}$为$\rm{B_8F_{12}}$,即$\rm{B_2(BF_2)_6}$，有可行性；当反应物化学计量数为6时，$\bf{D}$为$\rm{B_{14}F_{22}}$,即$\rm{B_3(BF_2)_{11}}$，无法满足；当反应物化学计量数为8时,$\bf{D}$为$\rm{B_{20}F_{32}}$,即$\rm{B_4(BF_2)_{16}}$，无法满足;更高的化学计量数也无法满足条件。因此，$\bf{D}$为$\rm{B_8F_{12}}$，即$\rm{B_2(BF_2)_6}$，有2个中心$\rm{B}$，4个端基$\rm{BF_2}$，2个桥连$\rm{BF_2}$，存在3种化学环境的$\bf{X}$，D正确。

<Checkpoint>[$\bf{D}$为$\rm{B_8F_{12}}$，即$\rm{B_2(BF_2)_6}$，有2个中心$\rm{B}$，4个端基$\rm{BF_2}$，2个桥连$\rm{BF_2}$，存在3种化学环境的$\bf{X}$](2)</Checkpoint>

根据题干"$\bf{F}$中存在一根 $C_3$ 轴",可以直接判断$\bf{F}$为$\rm{B(BF_2)_3CO}$，$\rm{C}$的质量分数为$6.5\%$，F正确。因此选A。

<Checkpoint>[$\bf{F}$为$\rm{B(BF_2)_3CO}$](1)</Checkpoint>