First slide
13th group or boron group
Question

When BCl3is strongly heated with ammonia the product formed is

Moderate
Solution

BCl3+6NH3BNH23+3NH4Cl

                          Boron amide

2BNH23ΔB2NH3Boron  imide+3NH3

B2NH3Δ2BN+NH3

                       Boron Nitride 

 

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