Sodium borohydride upon treatment with iodine | vedclass

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Question

(B) The reaction of sodium borohydride with iodine produces diborane,which is a Lewis acid $(X = B_2H_6)$.$2 NaBH_4 + I_2 \xrightarrow{	ext{Diglyme}}

Vedclass · Accepted Answer

$X= B_2H_6; Y= B_3N_3H_6; Z= H_2$

Vedclass · Answer

(B) The reaction of sodium borohydride with iodine produces diborane,which is a Lewis acid $(X = B_2H_6)$.$2 NaBH_4 + I_2 \xrightarrow{	ext{Diglyme}} B_2H_6(X) + H_2 + 2 NaI$Heating diborane with ammonia in a $1:2$ molar ratio produces borazine,which is a cyclic compound $(Y = B_3N_3H_6)$ and hydrogen gas $(Z = H_2)$.$3 B_2H_6 + 6 NH_3 \xrightarrow{\Delta} 2 B_3N_3H_6(Y) + 12 H_2(Z) \uparrow$

Sodium borohydride upon treatment with iodine produces a Lewis acid $(X)$,which on heating with ammonia produces a cyclic compound $(Y)$ and a colourless gas $(Z)$. $X, Y$ and $Z$ are $....$

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