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if SiH4 exists then why not PH3?( However,one electron is only present in d-orbital in excited state, it not the part of hybridised orbitals) SO, how to check validity of drago rule?
@ pranav the hybridisation of ph3 is sp3 and the bond angle is 90 degree and ph3 do exist , as phosphine in nature , this is beacuse phosphorous have 3 bond pairs and 1 loan pairs , that gives it a sp3 hybridisation . and acc to drago rule ph3 donot have any hybridisation , in ph3 the bonds are made by pure p orbitals and not by hybridised orbitals . drago rule is not valid here beacuse 1- the central atom and the surrounded atom has moderate electronegativity diffrence 2- absence of partial +ve charge and -ve charge HOPE IT CLEARS YOUR DOUBT ALL THE BEST ..
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