Question

Hydride of Boron occurs as ${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ but ${{\text{B}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}$ does not exist because:
A.${p\pi - d\pi }$ back bonding is possible in ${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$but not in ${{\text{B}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}$
B.Boron and hydrogen have almost equal values of electro-negativity.
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 chemicals compounds having hydrogen atoms in them are known as hydrides. Water molecule is an example of hydrides. Hydrides are basically distinguished on the basis of their bonding between atoms.

Complete step by step answer:
All the chemical compounds containing hydrogen atoms are considered as hydrides but mostly the hydride word is used for ${{\text{H}}^{\text{ - }}}$ anion in chemistry. In chemistry hydrides are distinguished on the basis of bonding between the atoms i.e. ionic hydrides, metallic hydrides, covalent hydrides and polymeric hydrides.
Back bonding: The bonding between two atoms placed adjacent to each other in which one of the atoms has vacant orbitals and the other atom has lone pair is known as back bonding. Diborane (${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$) is one of the examples of back bonding.
Back bonding is not reason for no existence of ${{\text{B}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}$ in chemistry, the main reason behind it is its large size. As we all know that hydrogen atoms are very small in size as compared to chlorine atoms, so these hydrogen atoms are able to fit between the two boron atoms easily giving rise to the formation of diborane but due to large size of chlorine atoms, they didn’t fit in between the boron atoms, hence ${{\text{B}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}$ doesn’t exist. Instead of ${{\text{B}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{6}}}$ it forms ${\text{BC}}{{\text{l}}_{\text{3}}}$ which is stable compound and ${\text{BC}}{{\text{l}}_{\text{3}}}$acts as a Lewis acid also. It gains lone pairs and forms stable products.
Example: $BC{l_3} \leftarrow N{H_3}$ , ammonia provides a lone pair in this product.

Hence from the above discussions we can conclude that option (D) is correct.

Note: The binary compounds of boron with hydrogen are known as boron hydrides. Hydrides of boranes are also known as boranes and these are classified in various series such as nido boranes and arachno boranes.