1. Tardigrade
  2. Question
  3. Chemistry
  4. 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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Q. 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

The Solid State

Solution:

For similar structure,

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

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

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