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$d\pi - p\pi$ bond is present in :
A
$CO_3^{2-}$
B
$PO_4^{3-}$
C
$NO_3^-$
D
$NO_2^-$
Detailed Solution
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 $PO_4^{3-}$.
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 $PO_4^{3-}$.