Adding $\rm NaNO_2$ to a concentrated aqueous solution of $\rm K_3IrBr_6$ and heating it, the color of the system gradually changes from green to yellow; then concentrated hydrobromic acid is added to the system, after sufficient reaction and cooling, it is stored at $5^\circ\rm C$ for several days, and a red solid $\bf X$ is obtained. Further studies have shown that $\bf X$ is a diamagnetic potassium salt, its anion is a mononuclear coordination anion of $\rm Ir$ with an integral number of crystal water molecules (and located outside the coordination anion); low-temperature drying can completely remove all crystal water molecules from $\bf X$, with a weight loss of 2.65%; the content (mass fraction) of some elements in $\bf X$ is: $\rm N$, 2.06%; $\rm Br$, 58.85%. The $\rm Ir-Br$ bond lengths in $\bf X$ are 242 pm and 248 pm.

Which of the following statements are correct:

1.The hybridization mode of $\rm Ir$ in the complex is $d^2sp^3$

2.Only one $\rm Ir-Br$ bond has a bond length of 248 pm

3.When $\bf X$ reacts with concentrated ammonia water, gas is released, and the difference in stoichiometric numbers between the left and right sides of the ionic equation for this reaction is 1

Studies have shown that when $\bf X$ is dissolved in water, if the system is strongly acidic, the anionic structure of $\bf X$ does not change; while when the system is weakly acidic ($\rm c(H^+)<0.5~mol/L$), the anion of $\bf X$ will be converted into a diamagnetic anion $\rm {\bf A}^{3-}$ with 3 negative charges; when the acidity of the system is increased, the reaction proceeds in the reverse direction, and $\rm {\bf A}^{3-}$ is converted back to the anion of $\bf X$. 

4.The molar mass of $\rm {\bf A}^{3-}$ is approximately 2.6% greater than that of the anion of $\bf X$ (error not exceeding 1.5%)
向$\rm K_3IrBr_6$的浓水溶液加入$\rm NaNO_2$并加热，体系颜色逐渐由绿色变为黄色；然后向体系中加入浓氢溴酸，充分反应后冷却，在$5^\circ\rm C$下保存数日，得到红色固体$\bf X$。进一步研究表明，$\bf X$是抗磁性钾盐，其阴离子是$\rm Ir$的单核配离子，带有整数个结晶水（且处于配离子的外界）；通过低温干燥可使得$\bf X$的所有结晶水完全脱除，失重2.65%；$\bf X$中部分元素的含量（质量分数）为：$\rm N$，2.06%；$\rm Br$，58.85%。$\bf X$中的$\rm Ir-Br$键长为242 pm和248 pm。

下列说法正确的有：

1.配合物中$\rm Ir$的杂化方式为$d^2sp^3$

2.仅有一根$\rm Ir-Br$键键长为248 pm

3.$\bf X$与浓氨水反应时放出气体，该反应的离子方程式左右两侧化学计量数之差为1

研究表明，将$\bf X$溶于水时，若体系酸性较强，则$\bf X$的阴离子结构不发生变化；而当体系为弱酸性体系（$\rm c(H^+)<0.5~mol/L$）时，$\bf X$的阴离子则会转化为带3个负电的抗磁性阴离子$\rm {\bf A}^{3-}$；增强体系酸性时，反应逆向进行，$\rm {\bf A}^{3-}$重新转化为$\bf X$的阴离子。

4.$\rm {\bf A}^{3-}$比$\bf X$的阴离子的摩尔质量大了约2.6%（误差不超过1.5%）
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
The number ratio of crystal water, $\rm N$ atoms, and $\rm Br$ atoms in the complex is:
$\rm (2.65 \div 18.02) : (2.06 \div 14.01) : (58.85 \div 79.90) = 1 : 1 : 5$

<Checkpoint>[The number ratio of crystal water, $\rm N$ atoms, and $\rm Br$ atoms in the complex is 1 : 1 : 5](1)</Checkpoint>

Assuming there is 1 $\rm Ir$ atom and 5 $\rm Br$ atoms in the coordination ion, then the formula weight of $\bf X$ is:

$\rm 79.90 \times 5 \div 0.5885 = 678.8$

The formula weight of the undetermined part is:

$\rm 678.8 - 79.90 \times 5 - 192.22 - 14.01 - 18.02 = 55.05$

$\bf X$ is a potassium salt, containing at least one $\rm K^+$, and the remaining part can only be one oxygen atom; therefore, the chemical formula of $\bf X$ is: $\rm K[Ir(NO)Br_5]\cdot H_2O$

<Checkpoint>[The chemical formula of $\bf X$ is: $\rm K[Ir(NO)Br_5]\cdot H_2O$](2)</Checkpoint>

In $\bf X$, $\rm Ir$ has a +3 valence, at which point $\rm NO^+$ is linearly coordinated, satisfying the 18-electron rule. $\bf X$ also has 6 d electrons and is a post-transition metal with a strong-field ligand $\rm NO^+$ present, so there are 2 remaining empty d orbitals, resulting in $d^2sp^3$ hybridization. Statement 1 is correct.

<Checkpoint>[The hybridization mode of $\rm Ir$ in the complex is $d^2sp^3$](1)</Checkpoint>

As shown in the figure below, the $\rm NO^+$ ligand and the $\rm Br$ ligand in the para position can form two sets of vertical, linearly oriented conjugated systems through the donor-acceptor synergistic effect of $\pi$ electrons, shortening the $\rm Ir-Br$ bond length, while the ligands in the ortho position have only one set of similar effects, so the $\rm Ir-Br$ bond length is not significantly shortened. Therefore, there are 4 $\rm Ir-Br$ bonds of 248 pm, and statement 2 is incorrect.

<MultiModal>![This image is a schematic diagram of orbital overlap, showing two independent molecular orbital structures side-by-side on a pure white background. All atomic labels and connecting lines in the figure are red. The structure on the left shows 4 atoms arranged horizontally (from left to right: $\rm Br$, $\rm Ir$, $\rm N$, and $\rm O$), where $\rm Br-Ir$ and $\rm Ir-N$ are connected by single bonds, and $\rm N-O$ is connected by a triple bond. The vertical $p$ orbitals of $\rm Br$, $\rm N$, and $\rm O$ match the wave phase of the $d$ orbital of $\rm Ir$ at a 45-degree angle to form $\rm d$-$p$ $\pi$ bonds. The structure on the right shows $\rm Ir$, $\rm N$, and $\rm O$ arranged horizontally with $\rm Br$ located directly below $\rm Ir$, where $\rm Br-Ir$ and $\rm Ir-N$ are connected by single bonds, and $\rm N-O$ is connected by a triple bond. The horizontal $p$ orbital of $\rm Br$ matches the wave phase of the vertical $p$ orbitals of $\rm N$ and $\rm O$ and the $d$ orbital of $\rm Ir$ at a 45-degree angle to form $\rm d$-$p$ $\pi$ bonds.](/media/uploads/ca45f56d-1403-48f7-bb34-9eeb132a5adf.png)</MultiModal>

<Checkpoint>[The $\rm NO^+$ ligand and the $\rm Br$ ligand in the para position can form two sets of vertical, linearly oriented conjugated systems, while the $\rm Br$ ligand in the ortho position has only one set of similar effects](2)</Checkpoint>

When $\bf X$ reacts with concentrated ammonia, gas is released, and the gas can only be $\rm N_2$. This can be regarded as the reaction of $\rm NH_3$ with $\rm NO^+$ to obtain $\rm N_2$, $\rm H_2O$, and $\rm H^+$. $\rm H^+$ reacts with $\rm NH_3$ in the system to obtain $\rm NH_4^+$: $\rm [Ir(NO)Br_5]^- + 3NH_3 \rightarrow [Ir(NH_3)Br_5]^{2-} + H_2O + NH_4^+ + N_2$

<Checkpoint>[The reaction equation of $\bf X$ with concentrated ammonia is: $\rm [Ir(NO)Br_5]^- + 3NH_3 \rightarrow [Ir(NH_3)Br_5]^{2-} + H_2O + NH_4^+ + N_2$](1)</Checkpoint>

The sum of the stoichiometric numbers on the left and right sides are equal, so statement 3 is incorrect.

The reaction that occurs in statement 4 should be a non-redox reaction, so $\rm NO^+$ in the coordination ion becomes an anion with a single negative charge. According to Le Chatelier's principle, it can be seen from the reaction conditions that the conversion of the anion of $\bf X$ to $\rm {\bf A}^{3-}$ may produce $\rm H^+$, then it can only be: $\rm [Ir(NO)Br_5]^- + H_2O \rightarrow [Ir(NO_2)Br_5]^{3-} + 2H^+$, statement 4 is correct.

<Checkpoint>[$\rm {\bf A}^{3-}$ is $\rm [Ir(NO_2)Br_5]^{3-}$](1)</Checkpoint>

Choose H
配合物中结晶水、$\rm N$原子、$\rm Br$原子的数目比为：  
$\rm (2.65 \div 18.02) : (2.06 \div 14.01) : (58.85 \div 79.90) = 1 : 1 : 5$

<Checkpoint>[配合物中结晶水、$\rm N$原子、$\rm Br$原子的数目比为1 : 1 : 5](1)</Checkpoint>

设配离子中有1个$\rm Ir$原子和5个$\rm Br$原子，则$\bf X$的式量为：  

$\rm 79.90 \times 5 \div 0.5885 = 678.8$  

未确定部分的式量为：  

$\rm 678.8 - 79.90 \times 5 - 192.22 - 14.01 - 18.02 = 55.05$  

$\bf X$为钾盐，至少含有1个$\rm K^+$，剩余部分只能为1个氧原子；因此$\bf X$的化学式为：$\rm K[Ir(NO)Br_5]\cdot H_2O$

<Checkpoint>[$\bf X$的化学式为：$\rm K[Ir(NO)Br_5]\cdot H_2O$](2)</Checkpoint>

$\bf X$中$\rm Ir$为+3价，此时$\rm NO^+$直线型配位，满足18电子规则。$\bf X$还有6个d电子，自身为后过渡金属，且有强场配体$\rm NO^+$存在，故剩余2个空的d轨道，为$d^2sp^3$杂化，说法1正确

<Checkpoint>[配合物中$\rm Ir$的杂化方式为$d^2sp^3$](1)</Checkpoint>

如下图所示，$\rm NO^+$配体与其对位的$\rm Br$配体能通过$\pi$电子的给体-受体协同作用形成2套垂直的、沿直线方向的共轭体系，使$\rm Ir-Br$键长缩短，而与其邻位的配体只有一组类似的作用，因此$\rm Ir-Br$键长缩短不明显。故有4根248 pm的$\rm Ir-Br$键，说法2错误

<MultiModal>![这张图片是一幅轨道重叠示意图，在纯白背景上并列展示了两个独立的分子轨道结构，图中所有原子标签和连接线均为红色。左侧结构为4个原子水平排布（从左到右依次为$\rm Br$、$\rm Ir$、$\rm N$和$\rm O$），其中$\rm Br-Ir$和$\rm Ir-N$以单键连接、$\rm N-O$以三键连接，$\rm Br$、$\rm N$、$\rm O$的竖直$p$轨道与$\rm Ir$的45度夹角$d$轨道波相匹配形成$\rm d$-$p$ $\pi$键；右侧结构为$\rm Ir$、$\rm N$、$\rm O$水平排布且$\rm Br$位于$\rm Ir$正下方，其中$\rm Br-Ir$和$\rm Ir-N$以单键连接、$\rm N-O$以三键连接，$\rm Br$的水平$p$轨道与$\rm N$、$\rm O$的竖直$p$轨道及$\rm Ir$的45度夹角$d$轨道波相匹配形成$\rm d$-$p$ $\pi$键。](/media/uploads/ca45f56d-1403-48f7-bb34-9eeb132a5adf.png)</MultiModal>

<Checkpoint>[$\rm NO^+$配体与其对位的$\rm Br$配体能形成2套垂直的、沿直线方向的共轭体系，而与其邻位的$\rm Br$配体只有一组类似的作用](2)</Checkpoint>

$\bf X$与浓氨水反应时放出气体，气体只可能为$\rm N_2$，可以看成$\rm NH_3$与$\rm NO^+$反应得到$\rm N_2$、$\rm H_2O$和$\rm H^+$，$\rm H^+$与体系中的$\rm NH_3$反应得到$\rm NH_4^+$：$\rm [Ir(NO)Br_5]^- + 3NH_3 \rightarrow [Ir(NH_3)Br_5]^{2-} + H_2O + NH_4^+ + N_2$

<Checkpoint>[$\bf X$与浓氨水反应方程式为：$\rm [Ir(NO)Br_5]^- + 3NH_3 \rightarrow [Ir(NH_3)Br_5]^{2-} + H_2O + NH_4^+ + N_2$](1)</Checkpoint>

左右化学计量数之和相等，说法3错误



说法4中发生的反应应该是一个非氧化还原反应，因此配离子中$\rm NO^+$变成了一个带一个负电荷的阴离子。根据勒夏特列原理，由反应条件可知$\bf X$的阴离子转化为$\rm {\bf A}^{3-}$可能产生$\rm H^+$，则只可能是：$\rm [Ir(NO)Br_5]^- + H_2O \rightarrow [Ir(NO_2)Br_5]^{3-} + 2H^+$，说法4正确

<Checkpoint>[$\rm {\bf A}^{3-}$为$\rm [Ir(NO_2)Br_5]^{3-}$](1)</Checkpoint>

选H