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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.
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
I didnot get regarding N2O4..............
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.......
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