The order of overlapping strength of orbitals is given as:
2pπ-2pπ>2pπ-3dπ>2pπ-3pπ>3pπ-3pπ.
The strength of π-bond is directly proportional to the overlapping strength of orbitals because on greater overlapping distance between two nuclei decreases and bond strength increase hence order of π-bond also given as:
2pπ-2pπ>2pπ-3dπ>2pπ-3pπ>3pπ-3pπ.
2pπ-2pπ, π-bonding is stronger than 2pπ-3dπ, π-bonding due to the lower energy gap (n=2) between them.
The importance comparison is in between 2pπ-3dπ and 2pπ-3pπ. The overlapping strength of 2pπ-3dπ is more than 2pπ-3pπ this is due to fact that , the 3d orbitals have two lobes which are more bent than 3p orbitals. That why overlapping strength increase.
is there any method to get the order quickly like we can check for the rest of the order other than 2p-3d and 2p-3p via the method of (n1+n2),if the value is more then strenght of sigma or pi bond is less . is there any method like this for the above mentioned two orbital bonds ??
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