The edge length of face centered unit cubic cell is 508pm.if the radius of the cation is 110 pm,the radius of anion is
To find the radius of the anion in a face-centered unit cubic cell, we can use the relationship between the edge length of the unit cell and the radius of the cation and anion.
In a face-centered unit cubic cell, each edge of the cube (represented by the edge length) is equal to 4 times the radius of the atoms (ions) involved. This means that if the radius of the cation is known, we can find the radius of the anion by applying the following formula:
radius of anion = edge length / (4 * radius of cation)
Let's substitute the given values into the equation:
edge length = 508 pm
radius of cation = 110 pm
radius of anion = 508 pm / (4 * 110 pm)
Now, let's calculate the radius of the anion:
radius of anion = 508 pm / 440 pm
radius of anion ≈ 1.155 pm
Therefore, the radius of the anion in this face-centered unit cubic cell is approximately 1.155 pm.
If you construct a diagram of a fcc crystal, the atoms touch along the diagonal but they don't touch along the edges. So one edge is 508, the other edge is 508 and the diagonal is sin 45o = 508/x
x = length of the diagonal. It will be 4r in length.
So radius A + 2*radius B + radius A = x
You know radius A, solve for radius B. Post your work if you get stuck. I urge you to look a diagram, either in your text/notes or on the web to see how this fits.