The state of hybridisation of central atom in dimer form of both BH₃ and BeH₂ is :

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sp², sp³

sp³, sp³

sp³, sp²

sp², sp

answer is B.

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Detailed Solution

The state of hybridisation of the central atom in the dimer forms of BH₃ and BeH₂ is as follows:

In the dimer of BH₃ (which is B₂H₆, diborane), each boron atom is sp³ hybridised. This allows boron to form four bonds—three with hydrogen atoms and one with another boron atom—by utilizing three regular two-center bonds and two three-center two-electron bonds (bridge hydrogens) as seen in the structure.
In the dimer of BeH₂ (which is Be₂H₄), each beryllium atom is sp² hybridised. The beryllium atom forms bonds with two terminal hydrogens and participates in bonding with bridging hydrogens using sp² hybrid orbitals.

Summary:

Hybridisation in B₂H₆ (BH₃ dimer): sp³
Hybridisation in Be₂H₄ (BeH₂ dimer): sp²In the dimer form of BH₃ (diborane, B₂H₆), the central atom (boron) exhibits sp³ hybridisation.
In the dimer form of BeH₂ (beryllium hydride, Be₂H₄), the central atom (beryllium) exhibits sp² hybridisation.

This occurs because, in diborane, boron forms four bonds (three with hydrogen and one with another boron), requiring sp³ hybrid orbitals. In beryllium hydride dimer, each Be atom forms bonds via sp² hybridisation due to its geometry and bond formation with bridging hydrogens.