Question

Which one of the following mineral pairs shows solid solubility through coupled substitution of elements?

• A. Albite - Anorthite
• B. Albite - Orthoclase
• C. Grossular - Andradite
• D. Jadeite - Aegirine

Hint:

Coupled substitution is characteristic of the plagioclase feldspar series, where two ions substitute simultaneously to preserve charge balance.

Answer 1

The Correct Option is A

Step 1: Understanding coupled substitution.
Coupled substitution occurs when two ions simultaneously substitute for two others in a crystal lattice, maintaining overall charge balance. This allows solid solution between different mineral end-members.

Step 2: Examine each pair.
- (A) Albite (NaAlSi$_3$O$_8$) and Anorthite (CaAl$_2$Si$_2$O$_8$): Solid solution exists in the plagioclase feldspar series. The coupled substitution is: \[ \text{Ca}^{2+} + \text{Al}^{3+} \leftrightarrow \text{Na}^+ + \text{Si}^{4+} \] This is a classic example of coupled substitution. - (B) Albite – Orthoclase: Both are feldspars, but substitution between Na$^+$ and K$^+$ is simple ionic substitution, not coupled substitution. - (C) Grossular – Andradite: These are garnet end-members (Ca$_3$Al$_2$Si$_3$O$_{12}$ and Ca$_3$Fe$_2$Si$_3$O$_{12}$). Their substitution is Fe$^{3+}$ ↔ Al$^{3+}$, a simple substitution. - (D) Jadeite – Aegirine: Both pyroxenes (NaAlSi$_2$O$_6$ and NaFeSi$_2$O$_6$). Substitution of Fe$^{3+}$ ↔ Al$^{3+}$ is again a simple substitution, not coupled.

Step 3: Conclusion.
Only Albite–Anorthite exhibits true coupled substitution.

Final Answer:
\[ \boxed{\text{Albite – Anorthite}} \]