In a fcc lattice, A, B, C and D atoms are arranged at corner, face center, octahedral void and tetrahedral void respectively, then the body diagonal contains :
QUES. In a closed packed arrangement ,A type of atoms are at corners , B type of atoms are at face corners , C type of atoms are at tetrahedral voids and D type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed. Then what will be the formula of compound?
If A+B⇌C+D; KC=K1 and E⊖=aV; 2A+2B⇌2C+2D; KC=K2 and E⊖=bV then:
From the sum of 7a – 2b + 11 and -2a -5, subtract 7a – b + 4.