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Chemical Bonding And Molecular Structure

Question
CBSEENCH11008388

According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?

  • straight N subscript 2 superscript minus space less than space straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript
  • straight N subscript 2 superscript 2 minus end superscript less than space straight N subscript 2 superscript minus space less than space straight N subscript 2
  • straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript space less than straight N subscript 2 superscript minus
  • straight N subscript 2 superscript minus space less than thin space straight N subscript 2 superscript 2 minus end superscript space less than thin space straight N subscript 2

Solution

A.

straight N subscript 2 superscript minus space less than space straight N subscript 2 space less than space straight N subscript 2 superscript 2 minus end superscript

Bond Order = fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction
Where
Nb = number of electrons in bonding MO
Na = number of electrons in antibonding MO

straight N subscript 2 space left parenthesis space 7 plus 7 space equals space 14 right parenthesis space equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma straight sigma 2 straight s squared comma space straight sigma asterisk times 2 straight s squared comma space straight pi 2 straight p subscript straight x squared almost equal to straight pi 2 straight p subscript straight y squared comma straight sigma 2 straight p subscript straight z squared

BO space equals space fraction numerator 10 minus 4 over denominator 2 end fraction space equals space 3

straight N subscript 2 superscript minus space left parenthesis 7 plus 7 plus 1 space equals space 15 right parenthesis
equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma straight sigma 2 straight s squared comma space straight sigma asterisk times 2 straight s squared comma space straight pi 2 straight p subscript straight x squared almost equal to straight pi 2 straight p subscript straight y squared comma straight pi asterisk times 2 straight p subscript straight x to the power of 1

BO space equals space fraction numerator 10 minus 5 over denominator 2 end fraction space equals space 2.5
straight N subscript 2 superscript 2 minus end superscript space left parenthesis space 7 plus 7 plus 2 space equals space 16 right parenthesis
equals space straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma straight sigma 2 straight s squared comma space straight sigma asterisk times 2 straight s squared space σp subscript straight z squared comma space straight pi 2 straight p subscript straight x squared almost equal to straight pi 2 straight p subscript straight y squared comma straight sigma 2 straight p subscript straight x to the power of 1 space almost equal to straight pi asterisk times 2 straight p subscript straight y to the power of 1
BO space equals space fraction numerator 10 minus 6 over denominator 2 end fraction space equals space 2
Hence space comma space the space increasing space order space of space straight B. straight O space is comma
straight N subscript 2 superscript 2 minus end superscript space less than space straight N subscript 2 superscript minus space less than straight N subscript 2