Question

The general formula of an amphibole mineral is A₀₋₁B₂C₅T₃O₂₂(OH)₂, where A, B, C, and T are cationic sites with different co-ordination numbers as stated below: A=12, B=6-8, C=6, T=4. The amount of octahedral Al in an amphibole of composition Na₀.₆Ca₂Mg_3.8Al₃Si₆O₂₂(OH)₂ is ...... 
 

Hint:

In amphiboles, the octahedral site (B-site) typically contains divalent cations such as Ca, Mg, and Al.

Answer 1

Correct Answer: 1.2

General formula: A₀₋₁B₂C₅T₈O₂₂(OH)₂

Coordination numbers:

• A = 12-coordinated
• B = 6-8-coordinated
• C = 6-coordinated
• T = 4-coordinated (tetrahedral)

Given composition: Na₀.₆Ca₂Mg₃.₈Al₃.₀Si₆.₂O₂₂(OH)₂

Identifying cation sites:

T site (tetrahedral, 4-coordinated): Si and Al

• Total T = Si + Al(tetrahedral)
• From formula: T₈ position contains Si₆.₂ + Al(in T site)

C site (6-coordinated): Mg and Al(octahedral)

• Total C = 5

B site (6-8-coordinated): Ca

• Total B = 2

A site (12-coordinated): Na

• Total A = 0-1

Determining octahedral Al:

From the given composition:

• Total Si = 6.2
• Total Al = 3.0
• Total Mg = 3.8

T site (total = 8): $$\text{Si} + \text{Al(tetrahedral)} = 8$$ $$6.2 + \text{Al(tetrahedral)} = 8$$ $$\text{Al(tetrahedral)} = 1.8$$

C site (total = 5): $$\text{Mg} + \text{Al(octahedral)} = 5$$ $$3.8 + \text{Al(octahedral)} = 5$$ $$\text{Al(octahedral)} = 1.2$$

Answer: 1.2