a simple metod valence electron in s are 6 and if nigbouring atoms are divalent or trivalent then dont count them and add the charge
6+2/2=4 so hybridisation is sp3 as there are 4 sp3 hybridised orbitals
Sanjeev Malik IIT BHU
Last Activity: 13 Years ago
S has 6 valence electrons . with divalent O it has S=O type bonding thus taking 2 elec. for each this type bonding.
2- charge on S act as lone pair i.e 2 elec for S again used for 4th O. thus it has 4 bondings total giving sp3 hybridisation.
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BABLU KUMAR
Last Activity: 13 Years ago
sp3 because it has 4 sigma bonds and no lone pair.so the total no of sigma bond + LP IS 4 .SO IT has 4 sp3 orbitals.
Trick:-> if sigma bond + lp is 2 then sp.
if sigma bond + lp is 3 then sp 2.
if sigma bond + lp is 4 then sp 3. and so on.
thanks pls approve it...............
Bhawna Singh
Last Activity: 3 Years ago
Formula to find hybrid orbitals = 1/2 ( V + M - C + A ) , where V = valence electron in central atom , M = Monovalent atoms in surroundings , C = Cationic Charge , A = Anionic Charge ..... Formula to find lone pair = No. Of hybrid orbitals - surrounding atoms ....... So here we have to find the Hybridisation of So42- as we only count no. Of monovalent atoms to find hybrid orbitals so oxygen would not be considered as it is not monovalent now configuration of S is 2, 8, 6 we would be taking 6 as valence electron .... Hybrid orbitals = 1/2 ( 6 - (-2)) = 4 that means it's hybridization would be sp3... Now lone pair = 4 - no.of oxygen atom = 4 - 4 = 0 so 0 lone pair .... that shows that it's geometry would be tetrahedral...... That was my answer thank you so much I hope you will understand it now
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