Halogen chemistry is an important part of elemental chemistry, and $\mathbf{X}$ element is a halogen. The simple ion $\mathbf{X}^+$ is often considered unable to exist independently, but it was later discovered that compound A containing a monovalent positive X element can be obtained under the following conditions: $\mathbf{X}_2+\mathbf{B}+\mathbf{C}\rightarrow\mathbf{A}$. Known: $\mathbf{B}$ is a common silver source, and its anion does not contain halogens and is an ionic cluster; $\mathbf{C}$ is a common base in organic synthesis and has a certain nucleophilicity, and is considered to be used as a nucleophilic catalyst. The molecular structure of $\mathbf{C}$ has multi-center delocalized bonds and contains only one $N$ atom; oxidation-reduction and acid-base complexation reactions occur during the reaction to generate $\mathbf{A}$.

Based on the above information, determine the correct option from the following statements.
卤素化学是元素化学中的重要部分，$\mathbf{X}$元素是一种卤素。$\mathbf{X}^+$ 简单离子往往被认为不能单独存在，但人们后来发现可以在以下条件获得包含正一价X元素的化合物A：$\mathbf{X}_2+\mathbf{B}+\mathbf{C}\rightarrow\mathbf{A}$。已知：$\mathbf{B}$为常见银源，且其阴离子不含卤素，为离子团；$\mathbf{C}$为有机合成中常见的碱，且具有一定亲核性，被认为能作为亲核性催化剂使用 ，$\mathbf{C}$的分子结构中有多中心离域键，只含有一个$N$原子；生成$\mathbf{A}$的反应过程中发生了氧化还原和酸碱络合反应。

根据上述信息，判断下列说法正确的选项。
$\mathbf{X}$ is the element bromine, $\mathbf{C}$ reacts with elemental bromine, $Br_2$ acts as an electrophile and is attacked by the nitrogen atom of $\mathbf{C}$.
$\mathbf{A}$'s monovalent $\mathbf{X}$ combines with one molecule of $\mathbf{C}$, thereby obtaining a stable anion which is nitrate.
$\mathbf{X}$ is an iodine atom, because the electronegativity of the iodine atom is relatively small, and it combines with $\mathbf{C}$ in a 1:1 ratio.
$\mathbf{B}$ is silver nitrate, which is often used to test for halide anions. In the synthesis of $\mathbf{A}$, the nitrate acts as an oxidizing agent.
The ratio of $\mathbf{X}$ atoms to $\mathbf{C}$ molecules in $\mathbf{A}$ is 1 to 2, because each $\mathbf{C}$ molecule contains one $N$ atom, so the ratio of $\mathbf{X}$ atoms to $N$ atoms is also 1 to 2. In addition to $\mathbf{A}$, the reaction also produces a product containing monovalent anionic $\mathbf{X}$.
Hypervalent bonding exists in $\mathbf{A}$.
$\mathbf{X}$为溴元素，$\mathbf{C}$与溴单质发生反应，$Br_2$作为亲电试剂被$\mathbf{C}$的氮原子进攻。
$\mathbf{A}$中正一价的$\mathbf{X}$与一分子$\mathbf{C}$结合，从而得到稳定，阴离子为硝酸根。
$\mathbf{X}$为碘原子，因为碘原子电负性相对小，与$\mathbf{C}$以1比1的比例结合。
$\mathbf{B}$为硝酸银，其常被用于检验卤素阴离子，在合成$\mathbf{A}$的过程中硝酸根起到氧化剂的作用。
$\mathbf{A}$中$\mathbf{X}$原子与$\mathbf{C}$分子的比例为1比2，因为每个$\mathbf{C}$分子中含有一个$N$原子，所以$\mathbf{X}$原子与$N$原子的比例也是为1比2。除了$\mathbf{A}外，该反应还生成了含负一价$\mathbf{X}$的产物。
$\mathbf{A}$中存在超价键作用。
Halogen elements have high electronegativity and are difficult to form $\mathbf{X}^+$. Therefore, $\mathbf{X}$ should be the $I$ element with the largest radius and the smallest electronegativity;

<Checkpoint>[$\mathbf{X}: I$](1)</Checkpoint>

$\mathbf{B}$ is a common silver source, and the anion does not contain halogens, so it should be silver nitrate ($AgNO_3$). Silver nitrite is uncommon;

<Checkpoint>[$\mathbf{B}$: $AgNO_3$](1)</Checkpoint>

$\mathbf{C}$ is a common base in organic chemistry that can be used as a nucleophilic catalyst and has only one nitrogen atom. It can be determined as pyridine ($C_5H_5N$);

<Checkpoint>[$\mathbf{C}$: $C_5H_5N$](1)</Checkpoint>

$I^+$ has one empty orbital and three lone pairs of electrons. If it is 1:1 coordination, it will form an N-I single bond, which is unstable and has too high energy. If it is 1:2 coordination, the empty orbital of I forms a 3-center 4-electron bond with the lone pair electrons of N on two pyridines, which is more stable than a single bond. So $\mathbf{A}$ is $[I(C_5H_5N)_2]NO_3$

<Checkpoint>[$\mathbf{A}$: $[I(C_5H_5N)_2]NO_3$](1)</Checkpoint>

According to the above analysis, the correct statement in each option should be F.

The element in A is wrong, and the reaction mechanism is complex, not a simple nucleophilic attack reaction, which is wrong.

The ratio of X to pyridine in B is wrong.

The ratio of X to pyridine in C is wrong.

In D, nitrate does not act as an oxidant, which is wrong.

In E, nitrate also has $N$ atoms, so the ratio of $\mathbf{X}$ to $N$ is 1:3.
卤族元素电负性大，难以形成$\mathbf{X}^+$，因此，$\mathbf{X}$应该为半径最大，电负性最小的$I$元素； 


<Checkpoint>[$\mathbf{X}: I$](1)</Checkpoint>


$\mathbf{B}$为常见银源，且阴离子不含有卤素，所以应该为硝酸银($AgNO_3$)，亚硝酸银不常见；


<Checkpoint>[$\mathbf{B}$: $AgNO_3$](1)</Checkpoint>


$\mathbf{C}$为有机化学中常见碱，能作为亲核催化剂，且只有一个氮原子，可以确定为吡啶($C_5H_5N$)；


<Checkpoint>[$\mathbf{C}$: $C_5H_5N$](1)</Checkpoint>


$I^+$有一个空轨道和三个孤对电子，若为1比1配位则形成N-I单键，不稳定，能量太高；若为1比2配位，I的空轨道与两个吡啶上N的孤对电子形成3中心4电子键，相比一根单键更稳定。所以$\mathbf{A}$为$[I(C_5H_5N)_2]NO_3$


<Checkpoint>[$\mathbf{A}$: $[I(C_5H_5N)_2]NO_3$](1)</Checkpoint>


根据上述分析，各选项中说法正确的应该为F。

A中元素错误，且该反应历程复杂，不是简单的亲核进攻反应，错误。

B中X与吡啶比例错误。

C中X与吡啶比例错误。

D中硝酸根不作为氧化剂，错误。

E中硝酸根也有$N$原子，所以$\mathbf{X}$与$N$比例为1比3。