BF3 reacts with NaH at 450 K to form NaF and X. When X reacts with LiH in diethyl ether. Y is formed. What is Y?

Question

BF3 reacts with NaH at 450 K to form NaF and X. When X reacts with LiH in diethyl ether. Y is formed. What is Y? (A) LiBO2 (B) Li2B4O7 (C) LiBH4 (D) B2H6⋅ LiH

Tardigrade Admin · Accepted Answer

2 BF 3+6 NaH stackrel450 K longrightarrow 6 NaF + B 2 H 6(X) underset textDiboraneB 2 H 6+2 LiH stackrel text Diethyl ether longrightarrow underset textLithium boron hydride2 LiBH 4(Y)

Q. $B F_{3}$ reacts with $N a H$ at $450 K$ to form $N a F$ and $X$ . When $X$ reacts with $L i H$ in diethyl ether. Y is formed. What is Y?

$L i B O_{2}$

$L i_{2} B_{4} O_{7}$

$L i B H_{4}$

$B_{2} H_{6} \cdot L i H$

$2 B F_{3} + 6 N a H ⟶ 450 K 6 N a F + B_{2} H_{6} (X)$

$Diborane B_{2} H_{6} + 2 L i H ⟶ Diethyl ether Lithium boron hydride 2 L i B H_{4} (Y)$

Q. BF3​ reacts with NaH at 450K to form NaF and X. When X reacts with LiH in diethyl ether. Y is formed. What is Y?

LiBO2​

Li2​B4​O7​

LiBH4​

B2​H6​⋅LiH