Home
Class 12
CHEMISTRY
The intermolecular forces of attraction ...

The intermolecular forces of attraction (i.e., H-bonding and van der waal's forces) exist among polar and non-polar species which effect melting point, boiling point, solubility and viscosity of covalent compounds :
Q. Select the incorrect order of boiling point between the following compounds:

A

`N_(3)HltCH_(3) N_(3)`

B

`Me_(2)SO_(4)ltH_(2)SO_(4)`

C

`Me_(3)BO_(3)ltB(OH)_(3)`

D

`BF_(3)ltBI_(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the incorrect order of boiling points among the given compounds, we need to analyze the intermolecular forces present in each compound and how they influence boiling points. Here’s a step-by-step breakdown of the solution: ### Step 1: Identify the Compounds We have four pairs of compounds to analyze: 1. N3H (Hydrazoic acid) and CH3N3 (Methyl azide) 2. Me2SO4 (Dimethyl sulfate) and H2SO4 (Sulfuric acid) 3. Me3BO3 (Trimethyl borate) and BOH3 (Boric acid) 4. BF3 (Boron trifluoride) and Bi3 (Bismuth triiodide) ### Step 2: Analyze Intermolecular Forces - **N3H vs CH3N3**: - N3H has hydrogen bonding (H is attached to N). - CH3N3 has no hydrogen bonding (H is attached to C). - **Conclusion**: N3H has a higher boiling point than CH3N3. - **Me2SO4 vs H2SO4**: - Me2SO4 has dipole-dipole interactions. - H2SO4 has hydrogen bonding (H is attached to O). - **Conclusion**: H2SO4 has a higher boiling point than Me2SO4. - **Me3BO3 vs BOH3**: - Me3BO3 has dipole-dipole interactions. - BOH3 has hydrogen bonding (H is attached to O). - **Conclusion**: BOH3 has a higher boiling point than Me3BO3. - **BF3 vs Bi3**: - Both compounds exhibit dipole-dipole interactions. - Compare molecular masses: Bi3 has a greater molecular mass than BF3. - **Conclusion**: Bi3 has a higher boiling point than BF3. ### Step 3: Determine the Incorrect Order Now, let's summarize the expected boiling point orders based on our analysis: 1. N3H > CH3N3 (Correct) 2. H2SO4 > Me2SO4 (Correct) 3. BOH3 > Me3BO3 (Correct) 4. Bi3 > BF3 (Correct) Since all the analyzed pairs show the expected order, we need to find the incorrect order among the provided options. ### Final Conclusion Based on the analysis, if any of the options provided in the question contradicts our conclusions, that would be the incorrect order. In this case, the incorrect order of boiling points is likely to be the one that suggests a compound with hydrogen bonding has a lower boiling point than one without.

To determine the incorrect order of boiling points among the given compounds, we need to analyze the intermolecular forces present in each compound and how they influence boiling points. Here’s a step-by-step breakdown of the solution: ### Step 1: Identify the Compounds We have four pairs of compounds to analyze: 1. N3H (Hydrazoic acid) and CH3N3 (Methyl azide) 2. Me2SO4 (Dimethyl sulfate) and H2SO4 (Sulfuric acid) 3. Me3BO3 (Trimethyl borate) and BOH3 (Boric acid) 4. BF3 (Boron trifluoride) and Bi3 (Bismuth triiodide) ...
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL BONDING (ADVANCED)

    VK JAISWAL ENGLISH|Exercise ONE OR MORE ANSWER IS/ARE CORRECT|78 Videos
  • CHEMICAL BONDING (ADVANCED)

    VK JAISWAL ENGLISH|Exercise MATCH THE COLUMN|21 Videos
  • CHEMICAL BONDING (ADVANCED)

    VK JAISWAL ENGLISH|Exercise Level 2|150 Videos
  • CHEMICAL BONDING (BASIC)

    VK JAISWAL ENGLISH|Exercise SUBJECTIVE PROBLEMS|54 Videos
VK JAISWAL ENGLISH-CHEMICAL BONDING (ADVANCED)-Level 3
  1. The intermolecular forces of attraction (i.e., H-bonding and van der w...

    Text Solution

    |

  2. The intermolecular forces of attraction (i.e., H-bonding and van der w...

    Text Solution

    |

  3. The intermolecular forces of attraction (i.e., H-bonding and van der w...

    Text Solution

    |

  4. There are five species P, Q, R, S and T. Spectroscopical analysis show...

    Text Solution

    |

  5. In all expected compounds each central atom only uses its s and p-orbi...

    Text Solution

    |

  6. The comcept of redistribution of energy in different orbitals of an at...

    Text Solution

    |

  7. The comcept of redistribution of energy in different orbitals of an at...

    Text Solution

    |

  8. The comcept of redistribution of energy in different orbitals of an at...

    Text Solution

    |

  9. Drago suggested an emprical rule which is compatible with the energeti...

    Text Solution

    |

  10. Drago suggested an emprical rule which is compatible with the energeti...

    Text Solution

    |

  11. According to hybridisation theory, the % s-character in sp, sp^(2) and...

    Text Solution

    |

  12. According to hybridisation theory, the % s-character in sp, sp^(2) and...

    Text Solution

    |

  13. The correct statement is :

    Text Solution

    |

  14. PCl(5) is an example of a molecule having sp^(3)d-hybridisation. Three...

    Text Solution

    |

  15. PCl(5) is an example of a molecule having sp^(3)d-hybridisation. Three...

    Text Solution

    |

  16. The first compound of the noble gasees was made in 1962. Barlett and L...

    Text Solution

    |

  17. The first compound of the noble gasees was made in 1962. Barlett and L...

    Text Solution

    |

  18. The first compound of the noble gasees was made in 1962. Barlett and L...

    Text Solution

    |

  19. According to MOT, two atomic orbitals overlap resulting in the formati...

    Text Solution

    |

  20. According to MOT, two atomic orbitals overlap resulting in the formati...

    Text Solution

    |