- #1

izzy93

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## Homework Statement

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.

## Homework Equations

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

## 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