if SiH 4 exists then why not PH 3 ?( 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?

Question

if SiH 4 exists then why not PH 3 ?( 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?

Umakant biswal · Answer

@ 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

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if SiH 4 exists then why not PH 3 ?( 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?

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?

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if SiH 4 exists then why not PH 3 ?( 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?

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?

1 Answers

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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?