Question

Which of the following hydride is electron-precise hydride?
A.${{B}_{2}}{{H}_{6}}$
B.$N{{H}_{3}}$
C.${{H}_{2}}O$
D.$C{{H}_{4}}$

Answer 1

Hint:To solve this question, first we should have prior knowledge about the octet configuration. According to the octet configuration as the name suggests Oct means eight the atom has eight electrons in the outermost orbit.

Complete answer:
In the year 1904, Richard Abegg proposed the concept of coordination number and said that the atoms behave as donors or acceptors of electrons. The octet rule uses this idea and states that every atom binds with another atom to have eight electrons in its outermost shell.
The octet rule is able to explain the formation of a number of compounds and also able to explain the stability of the compound.
Electron deficient compounds have a lesser number of electrons which are generally required for writing the Lewis structures. The hydrides of group 13 generally form electron deficient species. The most common example of electron deficient species is ${{B}_{2}}{{H}_{6}}$.
In ${{B}_{2}}{{H}_{6}}$ ,Valence electrons in boron = 3
Boron requires 5 electrons to complete its octet. Each boron atom uses 2 of its electrons to form bonds with the terminal hydrogens but still the octet of boron is incomplete.
The structure of ${{B}_{2}}{{H}_{6}}$ is mentioned below:

Hence, the correct answer is option (A).

Note:
The octet rule is based on the concept that electronic configuration of the noble gases are most stable and they have eight electrons in their outermost orbital so every element should attain the similar electronic configuration to become stable.