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

Question

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 (A) 0.1415 nm (B) 0.1450 nm (C) 0.2500 nm (D) 0.1250 nm

Tardigrade Admin · Accepted Answer

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)=(0.251/0.2455) × 0.1415=0.145 nm

0.1415 nm

0.1450 nm

0.2500 nm

0.1250 nm

For similar structure, I(C−C)I(B−N)​=a(C−C)a(B−N)​ I(B−N)=a(C−C)a(B−N)​×I(C−C) ∴I(B−N)=0.24550.251​×0.1415=0.145nm

For similar structure,

$\frac{I ( B - N )}{I ( C - C )} = \frac{a ( B - N )}{a ( C - C )}$

$I (B - N) = \frac{a ( B - N )}{a ( C - C )} \times I (C - C)$

$∴ I (B - N) = \frac{0.251}{0.2455} \times 0.1415 = 0.145 nm$