Question

Which of the following hydride is electron deficient?
A. $NaH$
B. $Ca{H_2}$
C. $C{H_4}$
D. ${B_2}{H_6}$

Answer 1

Hint: From the above option, the hydride which is electron deficient is a colorless, pyrophoric gas with sweet odor. Usually, hydrides of group 13 of periodic table are electron deficient because they have a lesser number of electrons. For example $LiH$ , $B{H_3}$ and so on.

Complete answer:
Generally, electron deficient compounds are those compounds which have less number of electrons to form normal covalent bonds. Usually, hydrides of group 13 of periodic table are electron deficient because they have a lesser number of electrons.
From the given option Boron belongs to group 13 that means ${B_2}{H_6}$ is an electron deficient hydride. ${B_2}{H_6}$ is the diborane. The chemical formula for borane molecules is $B{H_3}$ . It is surrounded by six electrons, so it is electron deficient due to its incomplete octet. Boron atom in diborane is $s{p^3}$ hybridized. ${B_2}{H_6}$​ molecule has a total of $8$ covalent bonds but only twelve electrons that is six pairs. So, it has two electron pairs less than the maximum required number for bonding, which makes it electron deficient.

Hence, option D is correct.

Note:
Diborane has the same number of electrons as ${C_2}{H_6}^{2 + }$ . This is due to the deprotonation of the planar molecule ethylene. Diborane is a compound with unusual bonding. The structure of diborane has ${D_{2h}}$ symmetry. Four hydrides are terminal, having two bridges between the boron centers. The lengths of the Boron hydrogen bridges bonds and the Boron hydrogen terminal bonds are $1.33$ and $1.19\mathop A\limits^0 $ , respectively. This difference in bond lengths reflects the difference in their strengths, the Boron hydrogen bridges bonds are relatively weaker. The weakness of the Boron hydrogen bridges vs Boron hydrogen terminal bonds is indicated by their vibrational signatures in the infrared spectrum, being $ \approx 2100$ and $2500c{m^{ - 1}}$ , respectively.