I read that diphosphorus has 2 phosphorus atoms joined by a triple bond (ie) it has one sigma bond and 2 pi bonds with 2 lone pairs ( 1 lone pair on each atom). I'm confused about the hybridization of the orbitals. Phosphorus in excited state has one 3s orbital, three 3p orbitals and one d orbital .

If sp2 hybridization takes place, it will form one sigma bond and what about the other 2 hybridized orbitals what would become of them? I'm confused on the lone pair part :/

If the above even took place, would the unhybridized p orbital form pi bond with the unhybridized p orbital of the other phosphorus. And the same would go for the 2 d orbitals. Am I correct on this??

note :https://socratic.org/questions/does-nitrogen-n2-and-oxygen-o2-goes-through-hybridization-how-can-i-know-when-hy

Here it says that "Diatomic molecules don't need hybridisation that is phosphorus would be using p pi p pi bonding which it could never do cuz of large diffuse orbitals !!

  • $\begingroup$ Note: diphosphorus is unstable. Natural phosphorus will instead exist in alltropes like white, red, or black phosophorus, which have entirely different structures. $\endgroup$ – Gaurang Tandon Feb 17 '18 at 4:54
  • $\begingroup$ Yes diphosphorus is unstable but I'm curious about it's hybridization . If the phosphorus undergoes sp hybridisation then 2 p orbitals and a d orbital will be left behind . $\endgroup$ – Jacob P.J Feb 17 '18 at 4:59
  • $\begingroup$ You can leave orbitals even $\ce{N2}$ has extra $p$ orbitals. $\endgroup$ – Avnish Kabaj Feb 17 '18 at 5:46
  • $\begingroup$ @AvatarShiny which orbitals ? $\endgroup$ – Jacob P.J Feb 17 '18 at 7:29
  • $\begingroup$ My bad we don't leave orbitals in N2 but you can leave put orbitals $\endgroup$ – Avnish Kabaj Feb 17 '18 at 8:27

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