Question

Hydride of boron occurs as $B_2H_6$ but $B_2Cl_6$ does not exist. This is because

• A. p$\pi$-d$\pi$ back bonding is possible in $B_2H_6$ but not in $B_2Cl_6$
• B. boron and hydrogen have almost equal values of electronegativity
• C. boron and chlorine have almost equal atomic sizes
• D. small hydrogen atoms can easily fit in between boron atoms but large chlorine atoms do not.

Answer 1

Answer: (D)

Hydride of boron occurs as $B _{2} H _{6}$ but $B _{2} Cl _{6}$ 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 $B _{2} Cl _{6}$, the stable compound is $BCl _{3}$ which acts as a Lewis acid and accepts a lone pair of electrons from ammonia to form $BCl _{3} \leftarrow NH _{3}$.