Question

$B{H_3}$ is a lewis acid because:
A. it has empty d-orbitals to accept lone pair of electrons
B. it has empty p-orbitals to accept lone pair of electrons
C. it has 2p electrons to donate
D. none of these

Answer 1

Boron is non-metal, and always forms covalent bonds. Normally it forms three covalent bonds at ${120^ \circ }$ using $s{p^2}$ hybrid orbitals. There is no tendency to form univalent compounds. All $B{X_3}$ compounds are electron deficient and may accept an electron from another atom, thus forming a coordinate bond.

Step by step answer: Boron has an unusual crystal structure which results in the melting point being very high. Boron has insufficient electrons to fill the valence shell even after forming bonds.
From lewis theory we know that acids are materials which accept electron pairs, and bases as substances that donate electron pairs. Thus a proton is a lewis acid. Boron is hence an acceptor atom with vacant p orbital. $B{H_3}$is not observed to exist in monomeric form at ordinary conditions, and undergo dimerization as readily as possible. Boron momentarily completes its octet, vacant orbitals of boron participate in hybridization. Diborane thus formed by banana bonds is still an electron deficient compound in nature.
Therefore from above we can conclude that $B{H_3}$is a lewis acid as it has empty p orbitals to accept lone pairs of electrons.

Hence the correct option is option B.

Note: Both the atoms in diborane are $s{p^3}$ hybridised. The electron distribution of bridge bond(B-H-B) has a banana-like appearance and has (3c-2e) bond as only two electrons are responsible for attracting three positive centres.