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p-Block Elements - I
hard
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 $....$
A$X= BH _{3} ; Y= BH _{3} \cdot NH _{3} ; Z= N _{2}$
B$X= B _{2} H _{6} ; Y= B _{3} N _{3} H _{6} ; Z= H _{2}$
C$X= B _{2} H _{6} ; Y= B _{6} H _{6} ; Z= H _{2}$
D$X= B _{2} H _{6} ; Y= B _{3} N _{3} H _{6} ; Z= N _{2}$
(KVPY-2020)
Solution
(b) Sodium borohydride upon treatment with iodine preduces a Lewis acid, $B _{2} H _{6}(X)$, which on heating with ammonia, produces a cyclic compound, $B _{3} N _{3} H _{6}(Y)$ and a colourless gas $H _{(Z)}$. Reactions involved are as follows
$2 NaBH _{4}+ I _{2} \underset{\text { Diglyme }}{\longrightarrow} \underset{(X)}{ B _{2} H _{6}}+ H _{2}+2 NaI$
$3 B _{2} H _{6}+6 NH _{3} \stackrel{\Delta}{\longrightarrow} \underset{( Y )}{2 B _{3}} N _{3} H _{6}+\underset{(Z)}{12 H _{2} \uparrow}$
$2 NaBH _{4}+ I _{2} \underset{\text { Diglyme }}{\longrightarrow} \underset{(X)}{ B _{2} H _{6}}+ H _{2}+2 NaI$
$3 B _{2} H _{6}+6 NH _{3} \stackrel{\Delta}{\longrightarrow} \underset{( Y )}{2 B _{3}} N _{3} H _{6}+\underset{(Z)}{12 H _{2} \uparrow}$
Standard 11
Chemistry
Similar Questions
Match List$-I$ with List$-II :$
| List$-I$ | List$-II$ |
| $a$ Borax | $i$ $NaBO _2$ |
| $b$ Kernite | $ii$ $Na _2 B _4 O _7 .4 H _2 O$ |
| $c$ Orthoboric acid | $iii$ $H _3 BO _3$ |
| $d$ Borax bead | $iv$ $Na _2 B _4 O _7 .10 H _2 O$ |
