Write chemical reaction to prepare inorganic benzene (B₃N₃Ht) .

English

Gujarati

Hindi

p-Block Elements - I

easy

Write chemical reaction to prepare inorganic benzene $(B_3N_3H_6)$.

Option A

Option B

Option C

Option D

Solution

$3 \mathrm{~B}_{2} \mathrm{H}_{6}+6 \mathrm{NH}_{3} \rightarrow 2\left[\mathrm{BH}_{2}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}\left[\mathrm{BH}_{4}\right]^{-} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}+12 \mathrm{H}_{2}$

Standard 11

Chemistry

The number of neutrons present in the more abundant isotope of boron is ' $\mathrm{x}$ '. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is ' $y$ '. The value of $x+y$ is …

medium

(JEE MAIN-2024)

Which is not obtained when metal carbides react with $H_2O$ ?

medium

Given below are two statements:

Statement $I$ : The decrease in first ionization enthalpy from $B$ to $Al$ is much larger than that from $Al$ to $Ga$.

Statement $II$ : The d orbitals in Ga are completely filled.

In the light of the above statements, choose the most appropriate answer from the options given below

hard

(JEE MAIN-2023)

When $BCl_3$, is treated with water, it hydrolyses and forms $[B(OH)_4]^-$ only whereas $AlCl_3$ in acidified aqueous solution forms $[Al(H_2O)_6]^{3+}$ ion. Explain what is the hybridizations of boron and aluminum in these species?

medium

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$

How can this reaction is made to proceed in forward direction?

normal

Write chemical reaction to prepare inorganic benzene $(B_3N_3H_6)$.

Solution

Similar Questions

The number of neutrons present in the more abundant isotope of boron is ' $\mathrm{x}$ '. Amorphous boron upon heating with air forms a product, in which the oxidation state of boron is ' $y$ '. The value of $x+y$ is …

Which is not obtained when metal carbides react with $H_2O$ ?

When $BCl_3$, is treated with water, it hydrolyses and forms $[B(OH)_4]^-$ only whereas $AlCl_3$ in acidified aqueous solution forms $[Al(H_2O)_6]^{3+}$ ion. Explain what is the hybridizations of boron and aluminum in these species?

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$ How can this reaction is made to proceed in forward direction?

Start a Free Trial Now

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$

How can this reaction is made to proceed in forward direction?