If the ions are removed from a single body diagonal in above case after doping, then the molecular formula of the unit cell would be

Options

A. `Mg_(2)Be_(3.5)O_(2.5)`
B. `Mg_(3)Be_(3)O_(3.75)`
C. `Mg_(3)Be_(3)O_(3.5)`
D. `Mg_(4)Be_(4)O_(2.5)`

Answer

B

Explanation

i. Number of `O^(2-) = 4`
Number of `Be^(2+) = 4`
Number of `Mg^(2+) = 4`
ii. `Be^(2+)` and `Mg^(2+)` are present in `TVs` and they are present on each body diagonal of the cube.
When single body diagoanl is removed,
Number of corner ions `(O^(2-))` removed
`= 2 xx (1)/(8)` per corner share `= (1)/(4) = 0.25` Number of `O^(2-)` left `= 4 - 0.25 = 3.75`
iii. Number of `Be^(2+)` removed `= 1`
Number of `Be^(2+)` left `= 4 - 1 = 3`
iv. Number of `Mg^(2+)` removed `= 1`
Number of `Mg^(2+)` left `= 3`
Thus, formula is: `Mg_(3)Be_(3)O_(3.75)`.