How to find hybridisation of compounds like N2O4 and B2H6 ?

Dear anurag

The unorthodox nature of N2O4 makes this problem a bit more confusing. Basically, the structure of N2O4 is (O=N(O))2, where you have a N-N bond, both nitrogens bear a positive charge and the two single-bonded O are negative (overal charge: 0). Nitrogen has in its valency 1 s orbital and 3 p orbitals. Since each nitrogen must make one double (pi) bond with oxygen, it must use 1 p orbital for this. This then leaves 2p orbitals to mix with 1 s orbital. Total orbitals must be conserved, so nitrogen will have 3 sp^2 orbitals. One for N-N single(sigma) bond, two for single (sigma) N-O bonds.

 The hybridization of B2H6 (diborane) is sp^3 because each boron molecule has a tetrahedral geometry.

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Badiuddin

I didnot get regarding N₂O_{4..............}

using formula- 1/2(v+m-c+a).......v- valency of central atom, m- no. of monovalent atoms, c- nharge on cation, a- charge on anion.......