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Grade:
        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
6 years ago

Answers : (3)

Sunil Kumar FP
askIITians Faculty
183 Points
							no of atoms ina ccp unit cell-simila to fcc=4
therefore no of B atom present=4
no of A atom=tertrahedral void and octahedral void(8+4)=12
formula of the compound=A12B4
thanks and regards
sunil kr
askIITian faculty
6 years ago
Anjali
11 Points
							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
						
3 years ago
SHIVANSH SUMAN
11 Points
							
NO. OF ATOMS IN CCP =4
THEREFORE,NO. OF ATOMS IN CLOSE PACKING =B=4
WE KNOW OCTAHEDRAL VOIDS ARE EQUAL TO THE NO. OF ATOMS IN CLOSE PACKING
THEREFORE,OCTAHEDRAL VOIDS =4
AND TETRAHEDRAL VOIDS ARE 2(NO OF ATOMS IN CLOSE PACKIN),WHICH IS EQUAL TO 8
GIVEN, 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 ATOMS
THEREFORE FORMULA OF AB=A8B4 OR A2B.
NOTE THAT 4 TETRAHEDRAL VOIDS ARE NOT OCCUPIED
 
2 years ago
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