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Question: A mixture of boron trichloride and hydrogen is subjected to silent electric discharge to form 'A' an...

A mixture of boron trichloride and hydrogen is subjected to silent electric discharge to form 'A' and HCl{\text{HCl}} .' A' is mixed with NH3{\text{N}}{{\text{H}}_{\text{3}}} and heated to 200C200^\circ {\text{C}} to form 'B’. The formula of 'B' is:
A. H3BO3{{\text{H}}_{\text{3}}}{\text{B}}{{\text{O}}_3}
B. B2O3{{\text{B}}_{\text{2}}}{{\text{O}}_3}
C. B2H6{{\text{B}}_{\text{2}}}{{\text{H}}_6}
D. B3N3H6{{\text{B}}_{\text{3}}}{{\text{N}}_3}{{\text{H}}_6}

Explanation

Solution

In the first reaction, you can replace all boron-chlorine bonds with boron-hydrogen bonds. In the second reaction, two reagent molecules will add to form a single product which has structure similar to benzene.

Complete answer:
A mixture of boron trichloride and hydrogen is subjected to silent electric discharge to form 'A' and HCl{\text{HCl}} .
Write the balanced chemical equation for the above process as shown below:

2 BCl3 + 3 H2 dischargeelectric B2H6                      +6                     HCl                                                                        Compound ’A’  {\text{2 BC}}{{\text{l}}_{\text{3}}}{\text{ + 3 }}{{\text{H}}_2}{\text{ }}\xrightarrow[{{\text{discharge}}}]{{{\text{electric}}}}{\text{ }}{{\text{B}}_2}{{\text{H}}_6}{\;\;\;\;\;\;\;\;\;\;\;\text{+6}\;\;\;\;\;\;\;\;\;\;\text{ HCl}} \\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ Compound 'A' }}

In the above reaction all the boron-chlorine bonds are replaced with boron-hydrogen bonds. New hydrogen-chlorine bonds are formed.
Thus, the compound ‘A’ is diborane with chemical formula B2H6{{\text{B}}_2}{{\text{H}}_6} .
' A' is mixed with NH3{\text{N}}{{\text{H}}_{\text{3}}} and heated to 200C200^\circ {\text{C}} to form 'B’.
Write the balanced chemical equation for the above process as shown below:

B2H6 + 2 NH3 B2H6NH3 200oC B3N3H6                                                                                                                  Compound B {{\text{B}}_2}{{\text{H}}_6}{\text{ + 2 N}}{{\text{H}}_3}{\text{ }} \to {{\text{B}}_2}{{\text{H}}_6} \cdot {\text{N}}{{\text{H}}_3}{\text{ }}\xrightarrow{{{{200}^o}{\text{C}}}}{\text{ }}{{\text{B}}_3}{{\text{N}}_3}{{\text{H}}_6} \\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\text{ Compound B}}

The compound ‘B’ is B3N3H6{{\text{B}}_{\text{3}}}{{\text{N}}_3}{{\text{H}}_6}

**Hence, the correct option is the option D.

Note:**
The chemical formula B3N3H6{{\text{B}}_{\text{3}}}{{\text{N}}_3}{{\text{H}}_6}represents the compound borazine which is commonly called inorganic benzene. It contains 12 sigma bonds and 3 pi bonds, similar to those present in benzene. Alternate three BH{\text{BH}} units and three NH{\text{NH}} units are present.