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`FeO` crystal has a simple cubic structure and each edge of the unit cell is `5 A^(@)`. Taking density of crystal as `5gm//"cc"` the number of `Fe^(2+)` and `O^(2-)` ions present in each unit cell are:A. `4Fe^(2+)` and `4O^(2-)`B. `6Fe^(2+)` and `6O^(2-)`C. `2Fe^(2+)` and `2O^(2-)`D. `1Fe^(2+)` and `1O^(2-)` |
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Answer» Correct Answer - A Volume of unit cell `=1.25xx10^(-22)"cc"` Mass of unit cell `=1.25xx10^(-22)xx4` `=5xx10^(-22)` gm For one molecle `=72/(6.022xx10^(23))=1.195xx10^(-22)g` Hence number of `FeO` molecules per unit cell `=(5xx10^(-22))/(1.195xx10^(-22))=4.19~~4` Hence there are `4Fe^(2+)` and `4O^(2-)` in each unit cell. |
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