A) \[{{A}_{3.75}}{{B}_{8}}{{C}_{3.75}}\]
B) \[{{A}_{3.75}}{{B}_{4}}{{C}_{8}}\]
C) \[{{A}_{4}}{{B}_{8}}{{C}_{3.75}}\]
D) \[{{A}_{4}}{{B}_{3.75}}{{C}_{8}}\]
Correct Answer: A
Solution :
[a] Since lattice is ccp \[{{Z}_{eff}}=4.\] \[\therefore \] Number of A ions \[=4\] \[(comer+face\text{ }centre=1+3=4)\] Number of B ions = Number of \[TVs=8\] Number of C ions = Number of \[OVs=4\] \[2TVs\]are formed at each body diagonal of cube. \[4TVs\] are formed on edge centre and body centre 2 A ions (at comer) and one C ion (at edge centre are removed) Number of A ions removed \[=2\times \frac{1}{8}\] (comer share) \[=\frac{1}{4}\] Number of C ions removed \[=1\times \frac{1}{4}\] (edge centre share) \[=\frac{1}{4}\] Number of .4 ions left \[=4-\frac{1}{4}=3.75\] Number of B ions = 8 (Since no B ion has been removed) Number of C ions left \[=4-\frac{1}{4}=3.75\] Thus, formula is: \[{{A}_{3.75}}{{B}_{8}}{{C}_{3.75}}\]
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