JEE Advance - Chemistry (2015 - Paper 1 Offline - No. 5)
If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesium ions, m and n, respectively, are
1/2, 1/8
1, 1/4
1/2, 1/2
1/4, 1/8
Explanation
For ccp, Z = 4 = no. of O-atoms
No. of octahedral voids = 4
No. of tetrahedral voids = 2 $$\times$$ 4 = 8
No. of Al3+ ions = m $$\times$$ 4
No. of Mg2+ ions = n $$\times$$ 8
Thus, the formula of the mineral is Al4m Mg8nO4
4m(+3) + 8n(+2) + 4($$-$$2) = 0
12m + 16n $$-$$ 8 = 0
4(3m + 4n $$-$$ 2) = 0
3m + 4n = 2
Possible values of m and n are $${1 \over 2}$$ and $${1 \over 8}$$ respectively.
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