-->

Chemical Bonding And Molecular Structure

Question
CBSEENCH11005794

Give the reason for the following: Bond order in N2 is 3 whereas it is 2.5 in NO ?

 

Solution

Total number of electrons in  N2 = 7 + 7 = 14
The electronic configuration is:
   KK open square brackets open parentheses straight sigma subscript 2 straight S end subscript close parentheses squared space left parenthesis straight sigma asterisk times subscript 2 straight S end subscript right parenthesis squared space left parenthesis straight pi 2 straight p subscript straight X right parenthesis squared space left parenthesis straight pi 2 straight p subscript straight y right parenthesis squared space left parenthesis straight sigma 2 straight p subscript straight z right parenthesis squared close square brackets
Now space straight N subscript straight b space equals space 8 semicolon space space space straight N subscript straight a space equals space 2
Here space bond space order space equals space 1 half left square bracket straight N subscript straight b space minus space straight N subscript straight a right square bracket
space space space space space space space space space space space space space space space space space space space space space equals space 1 half left square bracket 8 minus 2 right square bracket
space space space space space space space space space space space space space space space space space space space space space space equals space 3
Total number of electrons in NO = 7 + 8 =  15. The 15th electron enters into left parenthesis straight pi asterisk times 2 straight p subscript straight x right parenthesis
Now Nb = 8 and Na = 3
Hence space Bond space order space equals space 1 half left square bracket straight N subscript straight b space minus space straight N subscript straight a right square bracket space equals space 1 half left square bracket 8 minus 3 right square bracket space equals space 2.5