$3 B_{2} H_{6} + 6 N H_{3} s l o w h e a t i n g - ------ \to l o w T 3 X H e a t - -- \to 2 Y + 12 H_{2}$ . In the anionic part of the intermediate $X$ covalency of the central atom is:

2

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4

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6

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8

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Solution

The correct option is B $4$
When $1$ mole of $B_{2} H_{6}$ reacts with $2$ moles of $N H_{3}$ by slow heating, an addition compound $B_{2} H_{6} .2 N H_{3}$ is formed which on further heating at $200^{o} C$ forms borazine $(B_{3} N_{3} H_{6})$ .
(Refer to Image)
Covalency of amionic part= $4$

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Similar questions

3 B_{2} H_{6} + 6 X \to 3 [B H_{2} (X)_{2}]^{+} [B H_{4}]^{-} H e a t - -- \to Y + 12 H_{2}

N a_{2} C r O F_{4}

C r (H_{2} O)_{6} C l_{3}

∠ X A X

A :

X :

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