Question:
The intermediate product $ (X) $ formed in the following reaction is
$ B_{2}H_{6}+6NH_{3} \to 3X { ->[Heat] } 2B_{3}N_{3}H_{6}+12H_{2} $
The intermediate product $ (X) $ formed in the following reaction is
$ B_{2}H_{6}+6NH_{3} \to 3X { ->[Heat] } 2B_{3}N_{3}H_{6}+12H_{2} $
Updated On: Jun 14, 2022
- $ [BH(NH_{3})_{3}]^{+} [BH_{4}]^{-} $
- $ [BH_{2} (NH_{3})_{4}]^{+} [BH_{4}]^{-} $
- $ [BH(NH_{3})_{4}]^{+} [BH_{4}]^{-} $
- $ [BH_{2} (NH_{3})_{2}]^{+} [BH_{4}]^{-} $
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The Correct Option is D
Solution and Explanation
$B_{2}H_{6}$ and $NH_{3}$ react in $1 : 2$ to form borazine, $B_{3}N_{3}H_{6}$
The intermediate product 'X' is
$[BH_{2}(NH_{3})_{2}]^{+} [BH_{4}]^{-}$
$3B_{2}H_{6}+6NH_{3} \to 3[BH_{2}(NH_{3})_{2}]^{+} [BH^{-}_{4}]$
$[BH_{4}]^{-} { ->[Heat] }2B_{3}N_{3}H_{6}+12H_{2}$
The intermediate product 'X' is
$[BH_{2}(NH_{3})_{2}]^{+} [BH_{4}]^{-}$
$3B_{2}H_{6}+6NH_{3} \to 3[BH_{2}(NH_{3})_{2}]^{+} [BH^{-}_{4}]$
$[BH_{4}]^{-} { ->[Heat] }2B_{3}N_{3}H_{6}+12H_{2}$
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Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
- P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.
P block elements consist of:
- Group 13 Elements: Boron family
- Group 14 Elements: Carbon family
- Group 15 Elements: Nitrogen family
- Group 16 Elements: Oxygen family
- Group 17 Elements: Fluorine family
- Group 18 Elements: Neon family