Home
Class 12
CHEMISTRY
a. "Stability of a crystal is reflected ...

a. "Stability of a crystal is reflected is reflected in the magnitude of its melting points" Comment.
b. Melting points of some compounds are given below water `= 273 K`, ethyl alcohol `= 153.7 K`, diethyl ether `= 156.8 K`, methane `= 90.5 K`. What can you say about the intermolecular forces between the molecules of these compounds?

Text Solution

Verified by Experts

The melting point of a crystal depends upon the magnitude of forces holding the constituent particles together, which determine the stability. Higher the melting point, greater are the forces holding the constituent particles together and hence greater is hte stability. For example , ionic crystals such as `NaCl, KNO_3`, etc. have very high melting points and are stable. On the other hand, molecular solids such as maphthalene, iodine, etc, are less stable because they have low values of melting points.
The melting points of some compounds are :
Water = 273 K, ethyl alcohol = 155.8 K, diethyl ether = 156.8 K and methane = 90.5 K.
The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding The higher melting point of water than ethyl alcohol indicates that the hydrogen bonding in water is stronger than in ethyl alcohol. Diethyl ether is a polar molecule and, therefore, the intermolecular forces in diethyl ether are dipole-dipole interactiosn. On the other hand, methane is a non-polar molecular and the only forces present in them are the weak van der Waals forces (London dispersion forces). These are weaker than dipole-dipole interaction and hence methane has very low melting point than diethyl ether.
Promotional Banner

Topper's Solved these Questions

  • SOLID STATE

    MODERN PUBLICATION|Exercise NCERT EXEMPLAR PROBLEMS (MULTIPLE CHOICE QUESTIONS) (TYPE I)|37 Videos
  • SOLID STATE

    MODERN PUBLICATION|Exercise NCERT EXEMPLAR PROBLEMS (MULTIPLE CHOICE QUESTIONS) (TYPE II)|16 Videos
  • SOLID STATE

    MODERN PUBLICATION|Exercise NCERT FILE SOLVED (NCERT) (IN TEST QUESTIONS )|24 Videos
  • POLYMERS

    MODERN PUBLICATION|Exercise Competition file (OBJECTIVE TYPE QUESTIONS) (C. MULTIPLE CHOICE QUESTIONS)(Integer Type Questions)|6 Videos
  • SOLUTIONS

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|14 Videos

Similar Questions

Explore conceptually related problems

(a) ‘Stability of a crystal is reflected in the magnitude of its melting point’. Comment. (b) The melting points of some compounds are given below : Water = 273 K, Ethyl alcohol = 155.7 K, Diethyl ether = 156.8 K, Methane = 90.5 K. What can you say about the intermolecular forces between these molecules ?

Use the information and data given below to answer the question (a) to (c), Stronger intermolecular forces result in higher boiling point. Strength of London forces increases with the number of electrons in the molecule. Boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively. (a) which type of intermolecular forces are present in the molecules HF, HCl, HBr and HI ? (b) Looking at the trend of boiling points of HCl, HBr and HI , explain out of dipole-dipole interaction and London interaction, which one is predominant here. (c) Why is boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest ?

A compound has the empirical formula C_(10)H_(8)Fe . A solution of 0.26 g of the compound in 11.2 g of benzene ( C_(6)H_(6) )boils at 80.26^(@)C . The boiling point of benzene is 80.10^(@)C , the K_(b) is 2.53^(@)C /molal. What is the molecules formula of the compound?

A special metal S conducts electricity without any resistance. A closed wire loop made of S does not allow any change in flux through itself by inducting a suitable current to generate a compensating flux. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius a, with its center at the origin. A magnetic dipole of moment m is brought along the axis of this loop from infinity to a point at distance r(>> a) from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole m at a point on its axis at distance r is (mu_0 m)/(2 pi r^3) . The magnitude of the force between two magnetic dipoles with moments m_1 and m_2 separated by a distance r on common axis with their north pole facing each other is (k m_1 m_2)/r^4 where k is a constant of appropriate dimensions. The direction of this force is along line joining two dipoles. Work done in bringing dipole from infinity to a distance r from center of the loop by given process is proportional to

Structures of molecules of two compounds are given below. a) Which of the two compounds will have intermolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding? b) The melting point of compound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point? c) Solubility of compounds in water depends on powers to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with easily and be more stable in it?

A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given: Freezing point depression constant of water (K_(f)^(water)=1.86 K kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=2.0 K kg mol^(-1)) Boiling point elevation constant of water (K_(b)^(water))=0.52 K kg mol^(-1)) Boiling point elevation constant of ethanol (K_(b)^(ethanol))=1.2 K kg mol^(-1)) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9 . The boiling point of this solution is

MODERN PUBLICATION-SOLID STATE-NCERT FILE SOLVED (NCERT) (TEXTBOOK EXERCISES )
  1. What is meant by the term "coordination number"? b. What is the coor...

    Text Solution

    |

  2. How can you determine the atoic mass of an unknown metal if you know i...

    Text Solution

    |

  3. a. "Stability of a crystal is reflected is reflected in the magnitude ...

    Text Solution

    |

  4. How will you distinguish between the following pairs of terms? a. He...

    Text Solution

    |

  5. How many lattice points are there in one unit cell of each of the foll...

    Text Solution

    |

  6. Explain a. The basic of sumilarities and differences between metalli...

    Text Solution

    |

  7. Calculate the efficiency of packing in case of a metal crystal for a...

    Text Solution

    |

  8. Silver crystallizes in fcc lattic. If the edge length of the cell is 4...

    Text Solution

    |

  9. A cubic solid is made of two elements P and Q . Atoms of Q are at the ...

    Text Solution

    |

  10. Niobium crystallizes in body-centred cubic structure. If the density i...

    Text Solution

    |

  11. If the radius of the octaheral void is r and the radius of the atoms i...

    Text Solution

    |

  12. Copper crystallizer into an fcc lattice with edge length 3.61 xx 10^(8...

    Text Solution

    |

  13. Analysis shows that nickel oxide has the formula Ni(0.98)O(1.00). What...

    Text Solution

    |

  14. What is a semiconductor? Describe the two main types of semiconductor ...

    Text Solution

    |

  15. Non-stoichiometric cuprous oxide. Cu(2)O can be perpared in laboratory...

    Text Solution

    |

  16. Ferric oxide crystalliizes in a hexagonal close-packed array of oxide ...

    Text Solution

    |

  17. Classify each of the following as being either a p-type or an n-type s...

    Text Solution

    |

  18. Gold (atomic radius = 0.154 nm ) crystallizes in a face centred unit c...

    Text Solution

    |

  19. In terms of band theory, what is the difference between a. a condcut...

    Text Solution

    |

  20. Explain the following terms with suitable example: a. Schottky defec...

    Text Solution

    |