Borazole Preparation Chemistry
Borazole, often called inorganic benzene due to its structural similarity to benzene ($C_6H_6$), has the formula $B_3N_3H_6$. Its preparation involves specific reactions between boron-containing compounds and nitrogen-containing compounds.
Reaction Pathway for Borazole
The standard laboratory synthesis of borazole involves the reaction between diborane ($B_2H_6$) and ammonia ($NH_3$). This reaction proceeds through several intermediate steps. Heating the reaction mixture is crucial to complete the formation of the cyclic borazole structure.
- Reactants: Diborane ($B_2H_6$) and Ammonia ($NH_3$).
- Process: Reaction followed by heating.
- Product: Borazole ($B_3N_3H_6$).
The overall simplified reaction can be represented as:
$ 3B_2H_6 + 6NH_3 \xrightarrow{Heat} 2B_3N_3H_6 + 12H_2 $Therefore, borazole is prepared by heating the product isolated from reacting diborane with ammonia.
Correct Option Identification
Based on the established chemical synthesis route:
- Option 1 is incorrect as boron and dinitrogen form boron nitride.
- Option 2 is incorrect; ammonium nitrate is not the standard reactant.
- Option 3 is correct as it accurately describes the reaction of diborane with ammonia.
- Option 4 is incorrect as boron and ammonia typically yield boron nitride at high temperatures.
Similar Questions
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Important Questions from Inorganic Chemistry
- A chromium complex with a formula $\text{CrCl}_3\cdot 6\text{H}_2\text{O}$ has a spin only magnetic moment value of $3.87 \text{ BM}$ and its solution conductivity corresponds to $1 : 2$ electrolyte. $2.75 \text{ g}$ of the complex solution was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of $\text{AgNO}_3$. The amount of $\text{AgCl}$ formed in the above process is _____ g. (Nearest integer)
[Given: Molar mass in $\text{g mol}^{-1}$ $\text{Cr} : 52; \text{Cl}: 35.5, \text{Ag}: 108, \text{O}: 16, \text{H}: 1$] - Consider the following reactions :
$\text{NaCl} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{A} + \text{KHSO}_4 + \text{NaHSO}_4 + \text{H}_2\text{O}$
$\text{A} + \text{NaOH} \rightarrow \text{B} + \text{NaCl} + \text{H}_2\text{O}$
$\text{B} + \text{H}_2\text{SO}_4 + \text{H}_2\text{O}_2 \rightarrow \text{C} + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$
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