Question

The ratio of bridging to non-bridging oxygen atoms in the amphibole structure is:

• A. 4:11
• B. 5:6
• C. 2:7
• D. 1:2

Hint:

For chain silicates, compare "isolated" O ($4n$ for $n$ tetrahedra) with the actual O in the anionic group; the difference gives the count of bridging oxygens.

Answer 1

The Correct Option is B

Step 1: Recall the anionic unit of amphiboles.
Amphiboles are double-chain inosilicates. The fundamental silicate anion is \[ (\mathrm{Si}_4\mathrm{O}_{11})^{6-}. \]

Step 2: Count how many oxygens are shared (bridging).
If four SiO$_4$ tetrahedra were isolated, they would contain $4\times4=16$ O. In the double chain they actually have 11 O, hence the number of shared (bridging) oxygens \[ b = 16-11=5. \]

Step 3: Determine non-bridging oxygens.
Total oxygens present in the anionic unit $=11$. Therefore, non-bridging oxygens $=11-b=11-5=6$.

Step 4: Form the ratio.
Bridging : Non-bridging $= 5 : 6$.

Final Answer:\quad \[ \boxed{5:6} \]