Question

Find the ‘a’ radius of $N{H_3}^ + $ ion and that of $C{l^ - }$ ion (b). Distance between the oppositely charged ions in $N{H_4}Cl$ crystal lattice from the given lattice. $N{H_4}Cl$ crystallises in a body centered cubic lattice with a unit cell distance of $387{A^0}$. The radius of $C{l^ - }$ ion is $181pm$.

Answer 1

Hint:To solve this question we must know few basic information about what a unit cell is and what a body centered cubic lattice is. The smallest repeating unit of the crystal lattice is called the unit cell which is also the building block of a crystal. A unit cell can either be primitive cubic, body-centred cubic or face-centred cubic.

Complete step by step answer:
A BCC unit cell has atoms at each corner of the cube as well as an atom at the centre of the structure. The atom at the body centre fully belongs to the unit cell in which it is present.
In body centered cubic lattice oppositely charged ions remain in contact with each other along the cross-diagonal of the cube.
Since it was given in the question that $N{H_4}Cl$ crystallises in a body centered cubic lattice with a unit cell distance of $387{A^0}$.
$a = 387{A^0}$
Hence,
$\
  2{r^ + } + 2{r^ - } = \left){\vphantom{13}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{3}}a \\
  2({r^ + } + {r^ - }) = \left){\vphantom{13}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{3}}a \\
  {r^ + } + {r^ - } = \dfrac{{\left){\vphantom{13}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{3}}a}}{2} \\
   = \dfrac{{\sqrt 3 \times 387}}{2} \\
   = 335.1pm \\
\ $
Since it was given in the question that $N{H_4}Cl$ crystallises in a body centered cubic lattice with a unit cell distance of $387{A^0}$ and the radius of $C{l^ - }$ ion is $181pm$.
$a = 387{A^0}$
${r^ - } = 181pm$
In body centered cubic lattice structure,
$\
  \left){\vphantom{13}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{3}}a = 2({r^ + } + {r^ - }) \\
  \left){\vphantom{13}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{3}} \times 387 = 2({r^ + } + 181) \\
  {r^ + } = 154pm \\
\ $

Therefore the size of $N{H_3}^ + $ion is $154pm$

Note: We must note that,
In the BCC unit cell every corner has atoms.
There is one atom present at the centre of the structure.
Atom at the body centres wholly belongs to the unit cell in which it is present.