Looking for classes? Ksquare Career Institute, Bengaluru →
Which of the following has $p\pi - d\pi$ bonding
A
$NO_3^-$
B
$SO_3^{2-}$
C
$BO_3^{3-}$
D
$CO_3^{2-}$
Detailed Solution
A $p\pi - d\pi$ 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.
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.