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A compound X, of boron reacts with NH_3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF_3 with lithium aluminum hydride. The compounds X and Y are represented by the formulas… |
Answer» <html><body><p>`B_2H_6, B_3N_3H_6`<br/>`B_2O_3 , B_3N_3H_6`<br/>`BF_3, B_3N_3H_6`<br/>`B_3N_3H_6, B_2H_6`</p>Solution :(i)`B_2H_6` reacts with ammonia and gives `B_2H_6 . 2NH_3`which is formulated as `[BH_2(NH_3)_2]^(+) [BH_4]^(-)`and on <a href="https://interviewquestions.tuteehub.com/tag/heating-1017297" style="font-weight:bold;" target="_blank" title="Click to know more about HEATING">HEATING</a> gives `B_3H_3H_6`borazine <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> called borazole. <br/> `underset"(X)"(3B_2H_6+6NH_3) to <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>[BH_2(NH_3)_2]^(+) [BH_4]^(-) oversetDeltato underset"(Y)"(2B_3N_3H_6) + 12H_2` <br/> (ii)`B_2H_6` can be prepared by reduction of `BF_3` with `LiAlH_4 . 4BF_3 + 3LiAlH_4 to underset"(Y)"(2B_2H_6) + 3LiF + 3AlF_3`</body></html> | |