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Question: Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0\({A^0}\). Assuming density...

Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0A0{A^0}. Assuming density of the oxide as 4.0g−cm−3g - c{m^{ - 3}}, then the number of Fe2+F{e^{2 + }} and O2−{O^{2 - }}ions present in each unit cell will be
a.) Four Fe2+F{e^{2 + }}and two O2−{O^{2 - }}
b.) Four Fe2+F{e^{2 + }}and four O2−{O^{2 - }}
c.) Two Fe2+F{e^{2 + }}and four O2−{O^{2 - }}
d.) Two Fe2+F{e^{2 + }}and two O2−{O^{2 - }}

Explanation

Solution

In the Ferrous oxide structure, both the ions are arranged so that they show a cubic arrangement. The number of Fe2+F{e^{2 + }} ions in each unit cell is equal to the number of O2−{O^{2 - }} ions. Further, one can also say that there is one O2−{O^{2 - }} ion and one Fe2+F{e^{2 + }} ion behind every three vertices.

Complete step by step answer:
First let us understand the cubic structure and then we will move step by step on the answer.
So, the cubic structure is one in which atoms are arranged so that they form a cube-like appearance with all sides equal.
Ferrous oxide is expected to have a cubic structure.
Now, in the question we are given -
Edge of the unit cell = 5.0A0{A^0}= 5.0×10−85.0 \times {10^{ - 8}}cm
The unit cell is cubic.
Density of the oxide is also given, = 4.0g−cm−3g - c{m^{ - 3}}
What we have to find is the number of Fe2+F{e^{2 + }} and O2−{O^{2 - }} ions present in each unit cell.
Let us move step by step to answer.
First we can find volume from the edge. We know that volume of a cubic unit cell is as-
Volume of unit cell = (side)3{\left( {side} \right)^3}
Thus, Volume of unit cell = (5.0×10−8)3{\left( {5.0 \times {{10}^{ - 8}}} \right)^3}
Volume of unit cell = 1.25×10−22cm31.25 \times {10^{ - 22}}c{m^3}
We have density also. So, we can find the mass of one unit cell.
We have the formula, mass = density ×\times volume
Thus, mass of one unit cell = 4.0g−cm−3g - c{m^{ - 3}} ×\times 1.25×10−22cm31.25 \times {10^{ - 22}}c{m^3}
mass of one unit cell = 5×10−22g5 \times {10^{ - 22}}g
So, now we have the mass of the unit cell also.
Further, mass of one mole of Ferrous oxide = Mass of Fe + Mass of O
Thus, mass of one mole of Ferrous oxide = 55.8 + 16
mass of one mole of Ferrous oxide = 71.8g
Mass of one molecule of Ferrous oxide can also be found out as-
Mass of one molecule of Ferrous oxide = Mass of one mole of Ferrous oxide6.022×1023\dfrac{{Mass{\text{ }}of{\text{ }}one{\text{ }}mole{\text{ }}of{\text{ }}Ferrous{\text{ }}oxide}}{{6.022 \times {{10}^{23}}}}
Mass of one molecule of Ferrous oxide = 71.86.022×1023=1.192×10−22g\dfrac{{71.8}}{{6.022 \times {{10}^{23}}}} = 1.192 \times {10^{ - 22}}g
Now, we have the mass of the unit cell and mass of one molecule. So, by dividing the mass of the unit cell with the mass of one molecule of Ferrous oxide, we can find out the number of FeO formula units present.
Thus, Number of FeO formula units = Mass of unit cellMass of one molecule\dfrac{{Mass{\text{ of unit cell}}}}{{{\text{Mass of one molecule}}}}
Number of FeO formula units = 5×10−221.192×10−22≃4\dfrac{{5 \times {{10}^{ - 22}}}}{{1.192 \times {{10}^{ - 22}}}} \simeq 4
So, FeO formula units = 4.
Thus, number of Fe2+F{e^{2 + }}ions = 4
and number of O2−{O^{2 - }} ions = 4

So, option b.) is the correct answer.

Note: Always check that all the values have the same units as in the question the edge length is mentioned in the A0{A^0} while other values are in cm. So, we should convert and get one unit first. We know that 1 A0{A^0}= 10−10m{10^{ - 10}}m = 10−8cm{10^{ - 8}}cm.
Further, the term 6.022×10236.022 \times {10^{23}}corresponds to Avogadro number. It is denoted by NA{N_A}. We know that one mole of a substance always contains 6.022×10236.022 \times {10^{23}}number of molecules or ions or atoms.