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In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solid
no of atoms ina ccp unit cell-simila to fcc=4therefore no of B atom present=4no of A atom=tertrahedral void and octahedral void(8+4)=12formula of the compound=A12B4thanks and regardssunil kraskIITian faculty
For A, no. of ov= 4, so no. of TV will also be 4 ( as equally distributed). So, A=8; B=4 (no. of atoms in fcc).... formula is A2B
NO. OF ATOMS IN CCP =4THEREFORE,NO. OF ATOMS IN CLOSE PACKING =B=4WE KNOW OCTAHEDRAL VOIDS ARE EQUAL TO THE NO. OF ATOMS IN CLOSE PACKINGTHEREFORE,OCTAHEDRAL VOIDS =4AND TETRAHEDRAL VOIDS ARE 2(NO OF ATOMS IN CLOSE PACKIN),WHICH IS EQUAL TO 8GIVEN, A ATOMS ARE EQUALLY DISTRIBUTED BETWEEN BOTH VOIDS AND A ATOMS COVER ALL OCTAHEDRAL VOIDS.THEREFORE TOTAL A ATOMS =4(OCTAHEDRAL ATOMS)+4(TETRAHEDRAL ATOMS)=8 ATOMSTHEREFORE FORMULA OF AB=A8B4 OR A2B.NOTE THAT 4 TETRAHEDRAL VOIDS ARE NOT OCCUPIED
Dear student,Please find the attached solution to the problem B = 4 (fcc lattice)Since there are 8 tetrahedral and 4 octahedral voids in fcc out of which all 4 octahedral voids are filled and an equal number of tetrahedral voids too.Hence, A = 4 + 4 = 8Hence, Formula is A8B4or, The empirical formula is A2B Hope it helps.Thanks and regards,Kushagra
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