Question

$NaB{H}_4+I_2\to X+Y+Z$
$B{F}_3+NaH\stackrel{450K}{\to }X+P$
$B{F}_3+LiAl{H}_4\to X+Q$
Identify $X,Y,Z,P\text{and}Q.$

• A. $X-N{a}_4{B}_4{O}_7,Y-NaI,Z-HI,P-HF,Q-3Li\left[AlF4\right]$
• B. $X-B_2{H}_6,Y-NaI,Z-H_2,P-NaF,Q-3Li\left[AlF4\right]$
• C. $X-B_2{H}_6,Y-B{H}_3,Z-NaI,P-B_3{N}_3{H}_6,Q-A{l}_2{F}_6$
• D. $X-B{H}_3,Y-B_2{H}_6,Z-H_2,P-B_2{N}_3{H}_6,Q-3Li\left[AlF4\right]$

Answer 1

The correct option is B $X-B_2{H}_6,Y-NaI,Z-H_2,P-NaF,Q-3Li\left[AlF4\right]$

$X=B_2{H}_6,Y=NaI,Z=H_2,P=NaF,Q=LiF,R=Al{F}_3$
Sodium borohydrate react with iodine to produce diborane, sodium iodide and hydrogen.
The reaction proceeds in the digilyme $(MeOC{H}_{2}C{H}_2{)}_{2}O.$
$2Na\left[B{H}_4\right]+I_2\stackrel{(MeOC{H}_{2}C{H}_2{)}_{2}O}{\to }{B}_2{H}_6+2NaI+H_2$
Thus, $X=B_2{H}_6,Y=NaI,Z=H_2$
Boron trifluoride reacts with sodium hydride to give diborane and sodium fluoride. This reaction takes place at a temperature around ${200}^{o}C.$
$2B{F}_3+6NaH\stackrel{\simeq {200}^{o}C}{\to }{B}_2{H}_6+6NaF$
Thus, $P=NaF$
Boron trifluoride on reaction with Lithium aluminium hydride produces diborane, lithium fluoride and aluminium fluoride.
$4B{F}_3+3LiAl{H}_4\to 2B_2{H}_6+3Li\left[Al{F}_4\right]$
Thus, $Q=3Li\left[Al{F}_4\right]$