The given formula is:
\[ \text{KAl}_3\text{Si}_3\text{O}_{10}(\text{F}_{0.5}\text{OH}_x) \]
This represents a mineral where the sum of **fluorine (F−)** and **hydroxyl (OH−)** groups should balance the structural requirement.
The total content of **fluorine (F−)** and **hydroxyl (OH−)** ions must equal 2. Therefore, we have the equation:
\[ \text{F}^{0.5} + \text{OH}^x = 2 \]
Substituting **F− = 0.5**, we get:
\[ 0.5 + \text{OH}^x = 2 \]
Solving for **x**:
\[ \text{OH}^x = 2 - 0.5 = 1.5 \]
Thus, the value of **x** is: 1.5.
Group I | Group II |
P. Sillimanite | 1. First order |
Q. Quartz | 2. Second order |
R. Muscovite | 3. Greater than third order |
S. Calcite | 4. Third order variegated |