Respuesta :
Answer:
73.2 pm
Explanation:
"The anions make contact along the diagonal of the face of the unit cell, so the diagonal of the face is equal to 4 ionic anion radii. By letting r represent the ionic radius of the anion, d represent the diagonal of the face of the unit cell, and x represent the edge length, then d=4r and
d^2=x^2+x^2= 2x^2
By substituting 4r for d, the second equation becomes
(4r)^2=(2)x^2
Solving this equation for r shows that
r=(x–√2)/4
Since the edge length x is 207 pm, the ionic radius of the anion is
r= ((207 pm) X [tex]\sqrt{2}[/tex])/4=73.2 pm
Notice that the formulae used to describe atomic radii include constants with exact values. Therefore, these values do not constrain the number of significant figures."
The ionic radius of the anion in the FCC unit cell where anions are in contact along the diagonal of the face of the unit cell is 73.1pm.
What is ionic radius?
Ionic radius is the distance between the center of the atom and the outermost part.
As in the question it is given that anions are in contact along the diagonal of the face of the unit cell, so the length of the diagonal is represented as: d = 4r, where
r = ionic radius
Length of edge i.e. a = 207 pm (given)
Two adjacent edges and diagonal of a FCC unit cell makes a right angle triangle, by using pythagoras theorem we can write as:
a² + a² = d²
2a² = (4r)²
2a²/16 = r²
2(207)²/16 = r²
85,698/16 = r²
0r r = 73.18 pm
Hence, 73.18 pm is the ionic radius of the anion.
To learn more about ionic radius, visit the below link:
https://brainly.com/question/8137711
