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Solid state physics, lattice constants, ionic radii, nacl

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data

    NaCl (a0 = 5.64A° ), NaBr (a0 = 5.98A° ) and KCl (a0 = 6.30A° ) all have the same structure, which
    is the NaCl structure.

    (a) Assuming the spacings are determined by the ionic radii of the relevant ions, what would value
    would you expect for the lattice constant of KBr?
    (b) If the Na+ ion has a radius of 0.97A° , determine the radii of the other ions.

    2. Relevant equations

    lattice constant for nacl type structure =2cl-radius + 2 na+ radius ??

    3. The attempt at a solution

    a, I used the interionic distance of NaBr= 5.98A°/2 = 2.99A°

    and then pythagorous to finde the diameter of Br- =4.228A°

    do the same with KCl, I got the cl- diameter to be 4.455A°

    and then as the lattice parameter = the length of one side of the cubic structure;
    2cl-radius + 2 k+ radius = 6.30A°
    k+ radius =0.9225A°

    therefore the lattice constant of kbr = 2(0.9225A°) + 2(4.228/2) = 6.07 A°

    The answer given is 6.64 A°. Is there another way to do this or have I got the definition of the lattice parameter entirely wrong?

    b, got this part right r_Cl:1.85A° r_Br:2.02A° r_K:1.30A°

    Any help would be much appreciated
  2. jcsd
  3. Feb 10, 2014 #2


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    Homework Helper

    The distance between the nearest positive and negative ions is approximately equal to the sum of the ionic radii, but the distance between the negative ions, a0/√2 is not quite equal to their diameter. See picture. The opposite can be also true, in case of big negative ions they touch each other, but the positive ions do not touch the negative ones. Use only the lattice parameters and the given radius of the Na ion to get the radii of the other ions and the lattice parameter of KBr. You got rK and rBr: the lattice parameter of KBr is equal to twice the sum of these radii.


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    Last edited: Feb 10, 2014
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