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Q. $NaBH_{4} +I_{2}\to X + Y + Z $
$ BF_{3}+NaH $ $\xrightarrow{450\,K} X + P$
$BF_{3}+LiAlH_{4} \to X +Q + R$
$X, Y, Z, P, Q$ and $R$ in the reactions are
X Y Z P Q R
(a) $Na_{4}B_{4}O_{7} $ $NaI$ $HI$ $HF$ $LiF$ $AlF_{3} $
(b) $B_{2}H_{6}$ $NaI$ $H_{2}$ $NaF$ $LiF$ $AlF_{3}$
(c) $B_{2}H_{6}$ $BH_{3}$ $NaI$ $B_{3}N_{3}H_{6} $ $Al_{2}F_{6} $ $AlF_{3}$
(d) $BH_{3}$ $B_{2}H_{6}$ $H_{2}$ $B_{3}N_{3}H_{6}$ $LiF$ $AlF_{3}$

The p-Block Elements

Solution:

$2 NaBH _4+ I _2 \rightarrow \underset{(X)}{B _2 H _6} +2 \underset{\text { (Y) }}{ NaI }+ \underset{(Z)}{H _2}$
$2 BF _3+6 NaHH \xrightarrow{450 \,K } \underset{(X)}{B _2 H _6}+6 \underset{( P )}{ NaF )}$
$4 BF _3+3 LiAlH _4 \rightarrow 2 \underset{(X)}{B _2 H _6}+3 \underset{( Q)}{ LiF} +3 \underset{(R)}{AlF _3}$