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The type of hybridisation of boron in diborane is
$sp$ - hybridisation
$s{p^2}{\rm{ - }}$ hybridisation
$s{p^3}{\rm{ - }}$ hybridisation
$s{p^3}{d^2}{\rm{ - }}$ hybridisation
Solution
Boron has three valence electrons, so it is supposed to make $3$ bonds in a molecule with hybridization, $sp ^2$ as only s and two $p$ orbitals are used in hybridization and last $p$ orbitalis vacant.
But diborane, $B _2 H _6$ contains two electrons each, three centred bonds. Each Boron atom is in a link with four hydrogen atoms. This makes tetrahedral geometry.
Hence, each Boron atom is $sp ^3$ – hybridized.
Similar Questions
Match List $I$ with List $II$. Choose the correct answer from the options given below:
List $-I$ | List $-II$ |
$A.$ Melting point $[\mathrm{K}]$ |
$I.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$ |
$B.$ Ionic Radius $\left[\mathrm{M}^{+3} / \mathrm{pm}\right]$ |
$II.$ $\mathrm{B}>\mathrm{Tl}>\mathrm{Al} \approx \mathrm{Ga}>\mathrm{In}$ |
$C.$ $\Delta_{\mathrm{i}} \mathrm{H}_1 $ $ [\mathrm{~kJ} \mathrm{~mol}^{-1}]$ | $III.$ $\mathrm{Tl}>\mathrm{In}>\mathrm{Al}>\mathrm{Ga}>\mathrm{B}$ |
$D.$ Atomic Radius $[pm]$ | $IV.$ $\mathrm{B}>\mathrm{Al}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}$ |