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Home > Chemical Bonding And Molecular Structure

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Chemical Bonding And Molecular Structure

Question
CBSEENCH11008388
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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
Multi-choise Question

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


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Some More Questions From Chemical Bonding and Molecular Structure Chapter

What bond is present in MgCl2 molecules?

Write Lewis dot symbols for atoms of the elements Mg, Na, B, O, N, Br.

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