Write the molecular-orbital valence electron-configuration of the CN radical and determine its bond order. Does the bond order match the type of bond in the Lewis structure? What does your calculated bond order suggest about the stability of the cyanide radical?
The answer states that the bond order doesn't match the Lewis structure. Why doesn't it match the Lewis structure?
Fall 2015 Final Exam #5
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Re: Fall 2015 Final Exam #5
The lewis structure should look like this
https://www.google.com/url?sa=i&rct=j&q ... 5690040006
The structure implies that the bond between C and N is a triple bond. However, if you do the MO electron-configuration out you get^{2} (\sigma_{2s}*)^{2} (\pi _{2px})^{2} (\pi _{2py})^{2} \left ( \sigma _{2pz})^{1})
from this you can calculate the bond order to be 2.5. Since bond order is usually indicative of the bond type (i.e. usually a triple bond means a bond order of 3), the fact that the bond order is less than the expected bond type indicates that the radical is less stable and more reactive.
https://www.google.com/url?sa=i&rct=j&q ... 5690040006
The structure implies that the bond between C and N is a triple bond. However, if you do the MO electron-configuration out you get
from this you can calculate the bond order to be 2.5. Since bond order is usually indicative of the bond type (i.e. usually a triple bond means a bond order of 3), the fact that the bond order is less than the expected bond type indicates that the radical is less stable and more reactive.
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