InterviewSolution
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InterviewSolution
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If the unit cell of a mineral has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by magnesium ions , m and n, respectively areA. `1/2,1/8`B. `1,1/4`C. `1/2,1/2`D. `1/4,1/8` |
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Answer» Correct Answer - A In ccp lattice, Z=4 . `therefore` No. of O-atoms per unit cell= `4(O^(2-))` . `therefore` No. of octahedral voids = 4 and No. of tetrahedral voids=8 As m fraction of octahedral voids is occupied by `Al^(3+)` ions, therefore , `Al^(3+)` ions present =4 m . Similarly, `Mg^(2+)` ions =8 n. Hence, formula of the mineral is `Al_(4m) Mg_(8n)O_4`. As total charge on the compound is zero, hence 4 m (+3)+8 n(+2)+4 (-2)=0 or 12 m +16 n -8 =0 Substituting the given values of m and n , equation is satisfied only when `m=1/2` and `n=1/8` (as `12 xx1/2+16xx1/8-8=0`) |
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