Consider the structure of a diborane molecule ( B₂H₆ ). Let: N₁ be the number of 2 -centre- 2 -electron bonds. N₂ be the number of 3 -centre- 2 -electron bonds. N₃ be the maximum number of atoms that can lie in the same plane. N₄ be the number of sp^3 hybridized atoms. The value of (N₁ + N₂ + N₃ + N₄) is:

Step-by-step solution

In the structure of diborane ( B₂H₆ ): - There are four terminal B-H bonds which are normal 2 -centre- 2 -electron ( 2c-2e ) bonds. Thus, N₁ = 4 . - There are two bridging B-H-B bonds which are 3 -centre- 2 -electron ( 3c-2e ) bonds. Thus, N₂ = 2 . - The two boron atoms and the four terminal hydrogen atoms lie in the same plane. The two bridging hydrogen atoms lie above and below this plane. Therefore, the maximum number of coplanar atoms is 6 . Thus, N₃ = 6 . - Both the boron atoms undergo sp^3 hybridization to fo