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H2O has two lone pair electron and H2S also have two lone pair electron but Bond Angle in H2O is 104.5° but H2S has 92.12°bond Angle .

Both H2O and H2S has same hybridization ,which is sp3 and i know that the four hybrid orbitals of sp3 hybridization are directed towards the four corner of a regular tetrahedron making Angle of 109.5° .

My teacher taught me that for every lone pair you have to subtract 2.5° from 109.5° so if that's the case then both of them should have same Angle as both of them have 2 lone pair electron ? If that's not the case then what is the exact way to determine Bond Angle ? And the way my teacher taught me is it ok ?

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marked as duplicate by Ivan Neretin, Tyberius, pentavalentcarbon, Mithoron, airhuff Feb 22 '18 at 16:24

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  • $\begingroup$ Again Bent's Rule does the job. $\endgroup$ – Abcd Feb 22 '18 at 13:34
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In H20 it is sp3 hybridisation, but in H2S it is purely p3 orbital , there is no hybridization in H2S, it cannot afford the cost of hybridization you may recall that all the p orbitals I.e. Px,Py, Pz are 90 degree to each other as along x y and z axis respectively and if u compare this with the Bond angle of H2S which is 92.12 degree( almost 90 degree) which clearly shows that it is formed by p orbitals only, hybridization is a costly process , when sulfur can easily bond with p orbitals itself , why will it go for expensive process of hybridization unnecessarily isn't it?

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    $\begingroup$ Hybridisation is a purely mathematical model; energy cannot be gained nor lost due to it. This is wrong. $\endgroup$ – Martin - マーチン Feb 22 '18 at 16:15

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