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Coulomb’s Law and Lattice Energy

Atoms involved in bonds are usually more stable than they would be otherwise. This means that energy is released in the formation of the bond. For ionic bonds, Coulomb’s law is used to calculate this energy:

E=kq1q2d

where E is energy in joules (J), k is a constant equal to 2.31×10−28 J⋅m, q1 and q2 are the charges of the two ions, and d is the distance between the nuclei of the two ions in meters (m).

 

Whereas the bond energy calculated by Coulomb’s law pertains to a single ionic bond, lattice energypertains to all the ionic bonding among a group of atoms arranged in a crystal lattice. Lattice energy is calculated by using the Coulomb’s law equation, but with a different constant (unique to each substance) that takes into account the crystalline structure of the substance.

By convention, lattice energy is defined as the amount required to either break an ionic solid into individual gaseous ions, or to form an ionic solid from gaseous ions.

Calculate the amount of energy released in the formation of one mole of SrO bonds (not lattice energy). The radius of the strontium ion is 1.13 Å, and the radius of the oxide ion is 1.40 Å. Note that 1Å=10−10m.

Arrange the following ionic compounds in order of decreasing amount of energy released in lattice formation: NaF, MgS, TlN, and CsI.

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Posted on October 21, 2013, in Question and tagged , , , , , , , . Bookmark the permalink. 1 Comment.

  1. 2.20*10^3 kJ/mol
    TIN, MgS, NaF, CsI

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