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Pi Bonding with d Orbitals
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Which of the following has $p\pi - d\pi$ bondingA $p\pi - d\pi$ bond is formed by sideways overlap of a filled p orbital of one atom (here O) with an empty d orbital of the other atom.
N, B and C are second period elements with no d orbitals in their valence shell. In $NO_3^-$, $BO_3^{3-}$ and $CO_3^{2-}$, the $\pi$ bonding is of $p\pi - p\pi$ type.
S is a third period element with vacant 3d orbitals. In $SO_3^{2-}$, the S=O $\pi$ bond is formed by overlap of a 2p orbital of O with a 3d orbital of S ($p\pi - d\pi$ bonding).
So $SO_3^{2-}$ has $p\pi - d\pi$ bonding. -
$d\pi - p\pi$ bond is present in :A $d\pi - p\pi$ bond is formed by sideways overlap of a d orbital of one atom with a p orbital of another.
C and N are second period elements without d orbitals, so in $CO_3^{2-}$, $NO_3^-$ and $NO_2^-$ the $\pi$ bonds are $p\pi - p\pi$.
P is a third period element with vacant 3d orbitals. In $PO_4^{3-}$, the P=O $\pi$ bond is formed by overlap of a filled 2p orbital of oxygen with a vacant 3d orbital of phosphorus ($d\pi - p\pi$).
So the $d\pi - p\pi$ bond is present in $PO_4^{3-}$.