In diborane, the Tau-bonds can be formed by the overlapping of orbitals :

s p^{3} - s - s p^{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s p^{3} - s p^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s p^{2} - s p^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s p^{2} - s - s p^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $s p^{3} - s - s p^{3}$
In diborane, ${s p}^{3}$ hybrid orbital from each boron overlaps with 1 $s$ orbital of $H$ atom to form 3 center-2 electron (3c-2e) bond.

Suggest Corrections

Similar questions

N O_{3}^{-}, N O_{2}^{+} a n d N H_{4}^{+}

C H_{4}, C_{2} H_{6}, a n d C_{2} H_{4}

C 3 - C 4

The hybridization of central atom in dimer of $B H_{3}$ and $B e H_{2}$ is:

C_{2} H_{2} - -- \to C_{2} H_{4} - -- \to C_{2} H_{6}

Join BYJU'S Learning Program