Reacting $ZrCl_4$ with excess lithium borohydride in ether followed by vacuum distillation yields $Zr(BH_4)_4$. Treating the obtained $Zr(BH_4)_4$ with excess trimethylphosphine in ether and crystallizing with pentane produces an amber-colored solid. The byproduct of this reaction is a 1:1 Lewis acid-base adduct. X-ray diffraction results indicate that the amber-colored crystal is a dinuclear complex centered around zirconium atoms. Both zirconium atoms have an oxidation state of +4, with coordination numbers of 10 and 12, respectively. The ligands coordinated to zirconium are exclusively $H^-$, $BH_4^-$, and $P(CH_3)_3$. There are a total of three $H^-$ ligands, all serving as edge-bridging groups connecting the two zirconium atoms (with no other edge-bridging ligands). Additionally, the mass fraction of zirconium in this amber-colored solid was measured to be 44.3%. Based on the above information, which of the following options is incorrect?
将$ZrCl_4$和过量的硼氢化锂在乙醚中反应，并在真空中蒸馏，可制得$Zr(BH_4)_4$；将制得的 $Zr(BH_4)_4$同过量的三甲基膦在乙醚中反应，用戊烷结晶，可析出琥珀色固体。该反应的副产物为1:1型Lewis酸碱加合物。X-射线衍射结果表明，该琥珀色品体是以锆原子为中心的双核配合物；两个锆原子的氧化态均为+4，其配位原子数分别为10和12；与锆配位的配体仅有$H^-$、$BH_4^-$和$P(CH_3)_3$三种，$H^-$配体共有3个，它们均作为边桥基连接两个锆原子（无其它边桥基配体）。此外还测得，该琥珀色固体中，锆的质量分数为 44.3%。根据以上的信息可以得出下列选项中错误的是
The ratio of the number of $P$ atoms to $B$ atoms in this compound is 2:5.
In this compound, the $BH_4^-$ adopts two different coordination modes, with a ratio of 2:3 between the different coordination modes of this ligand.
All $P(CH_3)_3$ ligands are coordinated to the zirconium atom with a coordination number of 10.
The mass fraction of phosphorus in the by-products obtained from the preparation is 34.5%.
该化合物中的$P$原子与$B$原子数目比为2:5
该化合物中的$BH_4^-$采取2种不同的配位方式，且不同配位方式的该配体的比例为2:3
所有$P(CH_3)_3$配体都在配位数为10的锆原子上
制备所得的副产物中，磷元素的质量分数为34.5%
Based on charge balance, it can be deduced that the complex molecule contains a total of 3 $H^-$ and 5 $BH_4^-$.

<Checkpoint>[The complex molecule contains a total of 3 $H^-$ and 5 $BH_4^-$](0.5)</Checkpoint>

The molar mass of the complex is $M=91.22×2÷0.4430=411.8(g/mol)$.

The number of $P(CH_3)_3$ units in the complex is:

$[411.8-3×1.008-5×(4×1.008+10.81)-91.22×2]÷(30.97+12.01×3+1.008×9)=2$

This indicates that the complex contains 2 $P(CH_3)_3$ ligands.

<Checkpoint>[The complex contains 2 $P(CH_3)_3$ ligands](0.5)</Checkpoint>

Therefore, the chemical formula of the compound is $H_3Zr_2(P(CH_3)_3)_2(BH_4)_5$, and A is correct.

<Checkpoint>[The chemical formula of the compound is $H_3Zr_2(P(CH_3)_3)_2(BH_4)_5$](1)</Checkpoint>

From this, the chemical reaction equation can be written as:

$2Zr(BH_4)_4+5P(CH_3)_3=H_3Zr_2(P(CH_3)_3)_2(BH_4)_5+ 3P(CH_3)_3BH_3$

<Checkpoint>[The reaction equation is $2Zr(BH_4)_4+5P(CH_3)_3=H_3Zr_2(P(CH_3)_3)_2(BH_4)_5+ 3P(CH_3)_3BH_3$](1)</Checkpoint>

The mass percentage of phosphorus in $P(CH_3)_3BH_3$ is 34.5%, so D is correct.

<Checkpoint>[The mass percentage of phosphorus in $P(CH_3)_3BH_3$ is 34.5%](0.5)</Checkpoint>

The total coordination number is 12+10=22. Typically, hydrogen in the complex acts as a bridging ligand, with the 3 H atoms providing a total of 6 coordination numbers. Each $P(CH_3)_3$ provides 1 coordination number, so among the remaining 5 $BH_4^-$, 4 act as tridentate ligands and 1 as a bidentate ligand.

<Checkpoint>[Among the 5 $BH_4^-$, 4 are tridentate ligands and 1 is a bidentate ligand](1)</Checkpoint>

For the 12-coordinate zirconium atom, in addition to the three hydrogen atoms participating in coordination, there are 9 remaining coordination numbers. Considering steric effects, if $P(CH_3)_3$ ligands are present, their monodentate nature would cause excessive steric repulsion around the central Zr. Therefore, all its ligands are $BH_4^-$, with each $BH_4^-$ acting as a tridentate ligand, totaling 3 such ligands.

<Checkpoint>[The 12-coordinate zirconium atom has 3 $BH_4^-$ ligands, each acting as a tridentate ligand](1)</Checkpoint>

For the 10-coordinate zirconium atom, it is surrounded by 2 $P(CH_3)_3$ ligands and the remaining 2 $BH_4^-$ ligands.

The coordination number provided by $BH_4^-$ is the total coordination number minus those from other ligands, i.e., $10-3-1×2=5$. Thus, these two $BH_4^-$ must consist of one tridentate and one bidentate ligand.

<Checkpoint>[The two $BH_4^-$ ligands around the 10-coordinate zirconium atom consist of one tridentate and one bidentate ligand.](1)</Checkpoint>

B is incorrect.

If both zirconium atoms had $P(CH_3)_3$ ligands, the steric hindrance for the 12-coordinate zirconium atom would be too large, ruling out this possibility. The two phosphine ligands must both coordinate to the 10-coordinate zirconium atom, so C is correct.

<Checkpoint>[Both phosphine ligands coordinate to the 10-coordinate zirconium atom](1)</Checkpoint>
根据电荷平衡可推知，配合物分子中共有3个$H^-$和5个$BH_4^-$

<Checkpoint>[配合物分子中共有3个$H^-$和5个$BH_4^-$](0.5)</Checkpoint>

配合物的摩尔质量为$M=91.22×2÷0.4430=411.8(g/mol)$，

配合物中含有的$P(CH_3)_3$的数目为：

$[411.8-3×1.008-5×(4×1.008+10.81)-91.22×2]÷(30.97+12.01×3+1.008×9)=2$

推得该配合物含有2个$P(CH_3)_3$配体

<Checkpoint>[配合物含有2个$P(CH_3)_3$配体](0.5)</Checkpoint>

因此，该化合物的化学式为 $H_3Zr_2(P(CH_3)_3)_2(BH_4)_5$，A正确。

<Checkpoint>[该化合物的化学式为 $H_3Zr_2(P(CH_3)_3)_2(BH_4)_5$](1)</Checkpoint>

由此可写出发生的化学反应方程式：

$2Zr(BH_4)_4+5P(CH_3)_3=H_3Zr_2(P(CH_3)_3)_2(BH_4)_5+ 3P(CH_3)_3BH_3$

<Checkpoint>[反应方程式$2Zr(BH_4)_4+5P(CH_3)_3=H_3Zr_2(P(CH_3)_3)_2(BH_4)_5+ 3P(CH_3)_3BH_3$](1)</Checkpoint>

$P(CH_3)_3BH_3$中磷的质量分数为34.5%，D正确。

<Checkpoint>[$P(CH_3)_3BH_3$中磷的质量分数为34.5%](0.5)</Checkpoint>

总共的配位数为12+10=22，通常来说H在配合物中作为氢桥配位，3个H一共提供6个配位数，1个$P(CH_3)_3$提供1个配位数，则剩余的5个$BH_4^-$中有4个为三齿配体，1个为双齿配体。

<Checkpoint>[5个$BH_4^-$中有4个为三齿配体，1个为双齿配体](1)</Checkpoint>

对于12配位的锆原子，除了已知的三个氢原子参与配位，还有9个配位数，再考虑位阻的影响，若存在$P(CH_3)_3$配体，由于其为单齿配体，会造成中心Zr周围的位阻斥力过大。因此其配体全为$BH_4^-$，则每个$BH_4^-$为三齿配体，共有3个该配体。

<Checkpoint>[12配位的锆原子，其配体为3个$BH_4^-$，且每个$BH_4^-$均为三齿配体](1)</Checkpoint>

对于10配位的锆原子，则其周围有2个$P(CH_3)_3$配体，以及剩余的2个$BH_4^-$。

$BH_4^-$提供的配位数为总配位数排除其它配体的配位数，即$10-3-1×2=5$，所以这两个$BH_4^-$只能为一个三齿和一个双齿配体。

<Checkpoint>[10配位的锆原子周围的两个$BH_4^-$为一个三齿和一个双齿配体。](1)</Checkpoint>

B错误。

若两个锆原子上都有$P(CH_3)_3$配体，对于12配位的锆原子来说位阻过大，排除。一定是两个膦全配在10配位数的锆原子上，C正确。

<Checkpoint>[两个膦配体全配在10配位数的锆原子上](1)</Checkpoint>