Unit cell of (BN)n, i.e. boron nitride, is similar to that of graphite except that each layer consists of alternating B and B-atoms in hexagonal rings. It crystallises in a hexagonal unit cell with a = 0.251 nm, z = 2 for (BN)n. For graphite, a = 0.2455 nm and distance between C-atoms = 0. 1415 nm. Thus, distance between B and N-atoms is
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a
0.1415 nm
b
0.1450 nm
c
0.2500 nm
d
0.1250 nm
answer is B.
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Detailed Solution
For similar structure,I(B−N)I(C−C)=a(B−N)a(C−C)I(B−N)=a(B−N)a(C−C)×I(C−C)∴I(B−N)∘=0.2510.2455×0.1415=0.145nm