InterviewSolution
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InterviewSolution
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Ferrous oxide has cubes structure and each edge of the unit cell is `5.0 Å` .Assuming of the oxide as `4.0g//cm^(3)` then the number of `Fe^(2+) and O^(2)` inos present in each unit cell will beA. `Four Fe^(2+) and four O^(2-)`B. `Two Fe^(2+) and four O^(2-)`C. `Four Fe^(2+) and two O^(2-)`D. ` Three Fe^(2+) and three O^(2-)` |
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Answer» Correct Answer - a Let the units of ferrous oxide in a unit cell `= n`, molecular weight of ferrous `(FeO) = 56 + 16 = 72 g mol^(-1)` Weight of n unit`= (72 xx n)/(6.025 xxx 10^(23))` Volume of one unit `= ("length of corner")^(3)` `= (5 Å)^(3) = 125 xx 10^(-26) cm^(3)` Density `= ("wt, of cell" )/("volume")` `4.09 = (72 xx n)/(6.023 xx 10^(23) xx 125 xx 10^(-24))` `n = (3079.2 xx 10^(-1))/(72)` `= 42.7 xx 10^(-1) = 4.27 = 4` |
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