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Chemistry Question on p -Block Elements

The intermediate product (X)(X) formed in the following reaction is B2H6+6NH3→3X−>[Heat]2B3N3H6+12H2B_{2}H_{6}+6NH_{3} \to 3X { ->[Heat] } 2B_{3}N_{3}H_{6}+12H_{2}

A

[BH(NH3)3]+[BH4]−[BH(NH_{3})_{3}]^{+} [BH_{4}]^{-}

B

[BH2(NH3)4]+[BH4]−[BH_{2} (NH_{3})_{4}]^{+} [BH_{4}]^{-}

C

[BH(NH3)4]+[BH4]−[BH(NH_{3})_{4}]^{+} [BH_{4}]^{-}

D

[BH2(NH3)2]+[BH4]−[BH_{2} (NH_{3})_{2}]^{+} [BH_{4}]^{-}

Answer

[BH2(NH3)2]+[BH4]−[BH_{2} (NH_{3})_{2}]^{+} [BH_{4}]^{-}

Explanation

Solution

B2H6B_{2}H_{6} and NH3NH_{3} react in 1:21 : 2 to form borazine, B3N3H6B_{3}N_{3}H_{6}
The intermediate product 'X' is
[BH2(NH3)2]+[BH4]−[BH_{2}(NH_{3})_{2}]^{+} [BH_{4}]^{-}
3B2H6+6NH3→3[BH2(NH3)2]+[BH4−]3B_{2}H_{6}+6NH_{3} \to 3[BH_{2}(NH_{3})_{2}]^{+} [BH^{-}_{4}]
[BH4]−−>[Heat]2B3N3H6+12H2[BH_{4}]^{-} { ->[Heat] }2B_{3}N_{3}H_{6}+12H_{2}