Question

A compound XY crystalizes in BCC lattice with unit cell edge length of \[480pm\]. If the radius of \[{Y^ - }\] is \[225pm\] , then the radius of \[{X^ + }\] is:
A.\[127.5pm\]
B.\[190.68pm\]
C.\[225pm\]
D.\[255pm\]

Answer 1

Hint:The sum of the ionic radii is the nearest neighbour distance between the two atoms. In a body centred cubic cell the distance can be determined from the formula. By substituting the ionic radius of anion in the above distance the value of ionic radius of cation can be determined.

Complete answer:
Given that a compound XY crystalizes in a BCC lattice. BCC lattice is a body centred cubic cell. The unit cell is the smallest representing unit to make a crystal. Body centred cubic cell is one type of unit cell.
The sum of the ionic radii of the two ions i.e., cation and anion give the neighbouring distance.
The formula for the distance in BCC will be \[d = \dfrac{{\sqrt 3 }}{2}a\]
The edge length given is \[480pm\] , substitute the edge length in the above formula,
\[d = \dfrac{{\sqrt 3 }}{2}480 = 415.68pm\]
Given that the ionic radius of \[{Y^ - }\] is \[225pm\]
The given compound has two ions namely cation and anion.
\[{X^ + } + {Y^ - } = 415.68pm\]
Thus, the value of \[{X^ + }\] will be determined by substituting the value of \[{Y^ - }\]
\[{X^ + } = 415.68pm - 225pm\]
The value will be \[190.68pm\]

Option B is the correct one.

Note:
The ionic radius is the distance between the nucleus and outermost orbit of electrons. The sum of the ionic radius of two ions will be equal to the internuclear distance between the two ions in a crystal. The body centred cubic distance formula must be accurate and length of unit cell must be in picometers.