Question

$LiH+{\underset{–}{B}}_2{H}_6\to 2Li\underset{–}{B}{H}_4$, find the change in hybridization of underlined atom.

• A. $s{p}^2\to s{p}^3$
• B. $s{p}^3\to s{p}^2$
• C. $sp\to s{p}^3$
• D. None

Answer 1

The correct option is D None

In diborane, each boron atom is bonded to 4 hydrogen atoms - two of them
terminal hydrogens plus two bridging hydrogen atoms. The bridging B-H bonds
are longer than the terminal B-H bonds but they are also stronger owing to
$3c-2e$ banana bonds. Thus each boron atom is $s{p}^3$ hybridized.

If we look at the borohydride anion, once again the central boron is bonded to
4 hydrogen atoms. Thus the boron atom does have a complete octet. Hence,
the B in $[B{H}_4{]}^{-}$ is also $s{p}^3$ hybridized!