Question
Question: Ferrous oxide has a cubic structure with an edge length of the unit cell equal to 5.0 A˚. Assuming t...
Ferrous oxide has a cubic structure with an edge length of the unit cell equal to 5.0 A˚. Assuming the density of ferrous oxide to be 3.84cm3g , the number of Fe2+ and O2− ions present in each unit cell will be respectively, (use NA= 6 × 1023)
A) 4 and 4
B) 2 and 2
C) 1 and 1
D) 3 and 4
Solution
In the ferrous oxide structure, both ions ( Fe2+ and O2− ) are arranged in such a way that they show a cubic arrangement. Since the edge length of the unit cell and the density of ferrous oxide is given in the question, and we know the expression of density, d=VNAnM. Substituting the values, we will calculate the number of Fe2+ and O2− ions present in each unit cell.
Complete step by step answer:
Given in the question is that ferrous oxide has cubic structure with all sides equal.
The expression for density (d) is as follows,
d=VNAnM ⇒n=MdVNA
Here, ‘a’ is the edge length of the unit cell, M is the molecular mass and V is the volume of a unit cell.
Given in the question are,
A=5A0=5×10−8cm,
d=3.84cm3g
Let the units of ferrous oxide in a unit cell =n
Molecular weight of Ferrous Oxide (FeO) = 56 + 16 = 72 molg
Volume of one unit = cube of length of corner =(5×10−8)3
Now, substituting all the values in the expression of density,
So, the number of 4 Fe2+ and 4 O2− ions present in each unit cell will be respectively.
Therefore, the correct answer is option (A).
Note: The term NA is the Avogadro number which corresponds to 6 × 1023 . We know that one mole of a substance contains NA i.e. 6 × 1023 number of molecules or ions or atoms.