Question

Hydride of boron occurs as $B_2{H}_6$ but $B_{2}C{l}_6$ does not exist. This is because:

• A. $p\pi -d\pi$ back bonding is possible in $B_2{H}_6$ bout not in $B_{2}C{l}_6$
• B. boron and hydrogen have almost equal values of electronegativity
• C. boron and chloride have almost equal atomic sizes
• D. small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.

Answer 1

The correct option is D small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.

Hydride of boron occurs as $B_2{H}_6$ but $B_{2}C{l}_6$ does not exist. This is because small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.

Note: Instead of $B_{2}C{l}_6$, the stable compound is $BC{l}_3$ which acts as a Lewis acid and accepts a lone pair of electrons from ammonia to form $BC{l}_3\leftarrow N{H}_3$.