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 ..