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Q.

Reaction of  3 moles of  BCl3    with 3 moles of  NH4Cl  at   1400Cproduces compound P(containing boron). Futher, P  reacts  NaBH4 to give colourless liquid Q as one of the product. If heated with water, Q hydrolyses slowly, Produces three compound R,S and T. Sum of the atomicity of R, S and T molecule is:

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answer is 13.

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Detailed Solution

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Boron Trichloride Reduction Reaction Sequence

Step 1: Initial Reaction
BCl3 + NH4Cl B3N3Cl3H3

Reactants: Boron trichloride (BCl3) and ammonium chloride (NH4Cl)

Product: B3N3Cl3H3 (Trichloroborazine - a boron-nitrogen heterocycle)

Step 2: Reduction with Sodium Borohydride
NaBH4 B3N3Cl3H3 B3N3H6

Reagent: Sodium borohydride (NaBH4) - a reducing agent

Product: B3N3H6 (Borazine - also known as "inorganic benzene")

This step replaces the chlorine atoms with hydrogen atoms through reduction.

Step 3: Hydrolysis
H2O B3N3H6 NH3 + H3BO3 + H2

Reagent: Water (H2O)

Products:

  • NH3 - Ammonia
  • H3BO3 - Boric acid
  • H2 - Hydrogen gas

This step involves the hydrolysis (reaction with water) of borazine, breaking down the B-N ring structure.

Overall Reaction Summary

This three-step reaction sequence demonstrates the synthesis and subsequent decomposition of boron-nitrogen compounds:

  1. Formation: BCl3 reacts with NH4Cl to form trichloroborazine (B3N3Cl3H3)
  2. Reduction: NaBH4 reduces trichloroborazine to borazine (B3N3H6), replacing Cl with H
  3. Hydrolysis: Water breaks down borazine into ammonia, boric acid, and hydrogen gas

Note: Borazine (B3N3H6) is structurally similar to benzene (C6H6) and is sometimes called "inorganic benzene" due to its analogous cyclic structure with alternating boron and nitrogen atoms.

 

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