Chemical Bonding and Molecular Structure

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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

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Question
CBSEENCH11008417

Bond dissociation enthalpy of H2, Cl2 and HCl and 434, 242 and 431 kJ mol-1 respectively.  Enthalpy of formation of HCl is 

  • 93 kJ mol-1

  • -245 kJ mol-1

  • -39 kJ mol-1

  • 245  kJ mol-1

Solution

C.

-39 kJ mol-1

increment straight H subscript reaction space equals space begin inline style stack sum space Bond space space energy space of space reactant space with space below and space on top end style begin inline style stack sum Bond space space energy space of space product space with space below and space on top end style
Here comma space increment straight H subscript straight H minus straight H end subscript space equals space 434 space kJ space mol to the power of negative 1 end exponent
increment straight H subscript Cl minus space Cl end subscript space equals space 242 space kJ space mol to the power of minus
increment straight H subscript straight H minus Cl end subscript space equals space 431 space kJ space mol to the power of negative 1 end exponent
because space 1 half space straight H subscript 2 space plus 1 half Cl subscript 2 space rightwards arrow HCl
increment straight H subscript reaction space end subscript space equals space 1 half increment straight H subscript straight H minus straight H end subscript space plus space 1 half increment straight H subscript Cl space minus Cl end subscript space minus space increment straight H subscript straight H minus Cl end subscript
equals space 1 half space straight x space 434 space straight x space 1 half space straight x space 242 minus 431
equals space 217 space plus 121 minus 431
equals negative 93 space kJ space mol to the power of negative 1 end exponent

Question
CBSEENCH11008277

Bond order of 1.5 is shown by

  • O2+

  • O2-

  • O22-

  • O2

Solution

B.

O2-

MO configuration of O2+ (8+8-1=15)
straight sigma 1 straight s squared comma space straight sigma asterisk times 1 straight s squared comma space straight sigma 2 straight s squared comma straight sigma asterisk times 2 straight s squared comma straight sigma 2 straight p subscript straight z superscript 2 comma space straight pi 2 straight p subscript straight x superscript 2 almost equal to straight pi 2 straight p subscript straight y superscript 2 space comma space straight pi asterisk times 2 straight p subscript straight x superscript 1 almost equal to straight pi asterisk times 2 straight p subscript straight y superscript 0
Bond space order space space equals space fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction
left parenthesis where space straight N subscript straight b space equals space number space of space electrons space in space bonding space molecular space orbital
straight N subscript straight a space equals space number space of space electrons space in space anti minus bonding space molecular space orbital right parenthesis

therefore space BO space equals space fraction numerator 10 minus 5 over denominator 2 end fraction space equals space 2.5
Simiarly comma
straight b right parenthesis space straight O subscript 2 superscript minus space left parenthesis 8 plus 8 plus 1 space equals space 17 right parenthesis
so space Bond space order space equals space fraction numerator straight N subscript straight b space minus space straight N subscript straight a over denominator 2 end fraction space equals fraction numerator space 10 minus 7 over denominator 2 end fraction space equals space 1.5

straight c right parenthesis space straight O subscript 2 superscript 2 minus end superscript space left parenthesis 8 plus 8 plus 2 equals 18 right parenthesis
BO space equals space fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction space equals space fraction numerator 10 minus 8 over denominator 2 end fraction space equals 1

straight d right parenthesis space straight O subscript 2 space left parenthesis 8 plus 8 equals 16 right parenthesis
BO space equals space fraction numerator 10 minus 6 over denominator 2 end fraction space equals space 2
Thus space straight O subscript 2 superscript minus space shows space the space bond space order space 1.5. space

Question
CBSEENCH12011007

Consider the following compounds


Hyperconjugation occurs in 

  • I only

  • II only

  • III only

  • I and III

Solution

C.

III only

Hyperconjugation occurs through the H- atoms present on the carbon atom next to the double bond i.e alpha hydrogen atoms. There is no alpha -H in the structure I and II.
So, hyperconjugation occurs in structure III only ie. 

Question
CBSEENCH12011001

Consider the molecules CH4, NH3 and H2O. Which of the given statement is false?

  • The H-O-H bond angle in H2O  is larger than the H-C-H bond angle in CH4.

  • The H-O-H bond angle in H2O is smaller than the H-N-H bond angle in NH3.

  • The H-C-H bond angle in CH4 is larger than the H-N-H bond angle in NH3

  • The H-C-H bond angle in CH4, the H-N-H bond angle in H2O are all greater than 900.

Solution

A.

The H-O-H bond angle in H2O  is larger than the H-C-H bond angle in CH4.

According to VBT: lone pair-lone pair repulsion is more than a bond pair -lone pair repulsion.
As the number of lone pair of electrons on central element increases, repulsion between that lone pair of electrons increases and therefore, bond angle decreases.

Molecules

Bond angle

CH4 (no lone pair of electrons)

1090.5

NH3(one lone pair of electrons)

107.50

H2O (two lone pair of electrons)

104.450