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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Geology (GG)
Numerical problem
the general formula of amphibole mineral is A0-1B2C5T8O22(OH)2 where A,B,C and T are cationic sites with different coordination no A=12,B=6-8,C=6,T=4.The amount of octahedral Al in an amphibole of composition Na0.6Ca2Mg3.8Al3.0Si6.2O22(OH)2 is
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Rahul kumar
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Total number of tetrahedral cations should be 8. Si will contribute to 6.2 tetrahedral cations. Remaining (8.0-6.2 = 1.8) tetrahedral cations will be contributed by Al. So 1.8 Al will be tetrahedral and remaining 1.2 (3.0-1.8) Al will be of octahedral coordination. So, the correct answer should be 1.2
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Sajan sarthak Best Answer
1.8 part will be octahedral
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3-1.8= 1.2 is octahedral aluminium
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1.8 is tetrahedral aluminium
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i was to about say sir 1.2 now.ok sir i got it
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thank u
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your preparation is going in a correct direction keep it up... definitely you can achieve
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thank u sir for your motivation
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Rahul kumar
Si will purely work as a tetrahedral cation. Al can work as tetrahedral and octahedral cation both. Part of Al (1.8) will be of tetrahedral coordination and remaining (1.2) will be of octahedral coordination in order to complete the formula of given Amphibole.