The question involves identifying which mineral allows the maximum substitution of aluminum (Al) for silicon (Si) in its crystal structure. This kind of substitution is common in the feldspar mineral group due to the similar ionic radii and charges of Al3+ and Si4+. Understanding the chemical composition of each option can help in determining where this substitution is most prevalent.
- Albite (NaAlSi3O8): Albite is a sodium-rich feldspar with minimal substitution of Al for Si. The chemical formula shows that it contains only one aluminum atom per molecule.
- Anorthite (CaAl2Si2O8): Anorthite is a calcium-rich feldspar with a high level of Al substitution. Its formula indicates two aluminum atoms, which allows for more Al substitution for Si.
- Oligoclase (Na0.9Ca0.1Al1.1Si2.9O8): Oligoclase is part of the plagioclase series and contains moderate aluminum substitution, more than Albite but less than Anorthite.
- Andesine (Na0.7Ca0.3Al1.3Si2.7O8): Andesine also belongs to the plagioclase series with an intermediate Al substitution level between oligoclase and anorthite.
The key to solving this question is understanding the plagioclase feldspar series, which ranges from sodium-rich albite to calcium-rich anorthite. In this series, the substitution of Al for Si increases from albite to anorthite. Therefore, the correct answer is Anorthite, as it allows for the maximum Al substitution due to its chemical composition of CaAl2Si2O8.
In summary, Anorthite allows the maximum substitution of aluminum for silicon because it has a higher aluminum-to-silicon ratio compared to the other minerals listed, which makes it the correct answer.
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