Question
Question: \(NaBH_{4} + I_{2} \rightarrow X + Y + Z\) \[BF_{3} + NaH\overset{\quad 450K\quad}{\rightarrow}X + ...
NaBH4+I2→X+Y+Z
BF3+NaH→450KX+P
BF3+LiAlH4→X+Q+R
X,Y, Z, P, Q and R in the reactions are
X | Y | Z | P | Q | R | ||
---|---|---|---|---|---|---|---|
A
| Na4B4O7 | NaI | | HI | HF | LiF | AlF3 | |
B
| B2H6 | NaI | | H2 | NaF | LiF | AlF3 | |
C
| B2H6 | BH3 | | NaI | B3N3H6 | Al2F6 | AlF3 | |
D
| BH3 | B2H6 | | H2 | B3N3H6 | LiF | AlF3 | |
Explanation
Solution
: 2NaBH4+I2→(X)B2H6+(Y)2NaI+(Z)H2
2BF3+6NaH→450K(X)B2H6+(P)6NaF
4BF3+3LiAlH4→2(X)B2H6+(Q)3LiF+(R)3AlF3