Find the number of hexagonal faces that are present in a truncated octahedral.
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Please help Sabu decode the jail lock. Chacha Choudhary gave Sabu a formula : f_(1) = ((x)/(z)xx y), f_(2) =((f)/(v) xxu), f_(3) = ((r)/(s)xx w) Sabu can open the lock if he finds the value of 3f_(1) +f_(2) +f_(3) = key where: Number of triangular faces in a truncated tetrahedron = x Number of hexagonal faces in a truncated tetrahedron = x Number of corners in a truncated tetrahedron = z Number of square faces in a truncated octahedron = t Number of hexagonal faces in a truncated octahedron = u Number of corners in a truncated octahedron = u Number of triangular faces in a truchcated cube = w Number of octangonal faces in a truncated cube = r Number of corners in a truncated cube = s What is the KEY ?
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Find out the number of unpaired electrons in strong and weak octahedral for Cr^(3+) and Fe^(2+) ions .
Atoms of element A and B form an HCP lattice such that atoms of A are present at corners of the unit cell and atoms of B are present at the centers of both the hexagonal faces as well as inside the unit cell. All the tetrahedral voids are occupied by atoms of element C and all the octahedral voids are occupied by atoms of element D . Find out the formula of the crystalline solid.
