
Ammonium chloride crystallizes in a body centered cubic lattice with a unit distance of 387pm. Calculate (a) the distance between the oppositely charged ions in the lattice and (b) the radius of the $N{H_4}^ + $ ion if the radius of $C{l^ - }$ ion is 181pm.
Answer
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Hint: The body centered cubic lattice is the one of the type of unit cell in which there are two atoms present in total. In this lattice the atom is present at each corner of the cube so each corner of the cube shares the 1/8 th part of the atom thereby having 1 atom. The other atom is present at the center of the cube. There are two more types of unit cells and they are the primitive unit cell and the face centered cubic unit cell.
Complete step by step solution:
(a)As we know that in the body centered cubic lattice the oppositely charged ions tend to touch each other along the cross diagonal of the cube. So we can write the following formula is:
2${r^ - }+2{r^ + }=\sqrt 3 a$
Here the${r^ - }$ is the radius of anion
${r^ + }$Is the radius of cation and
a is the unit distance = 387pm
The distance between the oppositely charged ions will be equal to ${r^ - }+{r^ + }$.
So now substituting the values in formula we get,
${r^ - }+{r^ + }={\dfrac{{\sqrt 3 }}{2}} \times 387$
${r^ - }+{r^ + }= 335.15pm$
So the distance between the oppositely charged ion in the lattice is 335.15pm.
(b) Now the distance between the ions is 335.15pm and it is given in the question that the radius of ${r^ - }$ that is chloride ion is 181pm so now we can easily calculate the ${r^ + }$that is the radius of ammonium ion. so,
${r^ - }+{r^ + }$= 335.15pm
Substituting the value of chloride radius in the above formula we fet,
$181+{r^ + }$= 335.15
So, ${r^ + }$ = 335.15-181
${r^ + }$ = 154.15pm
So the radius of ammonium ion is 154.15pm.
Note: In the primitive centered cubic unit cell the atoms are present only at the corners. Whereas in the face centered cubic unit cell the atoms are present at corners as well as at each face of the cube. So the total number of atoms present in a primitive centered cubic unit cell is 1 and in face centered cubic unit cell is 4.
Complete step by step solution:
(a)As we know that in the body centered cubic lattice the oppositely charged ions tend to touch each other along the cross diagonal of the cube. So we can write the following formula is:
2${r^ - }+2{r^ + }=\sqrt 3 a$
Here the${r^ - }$ is the radius of anion
${r^ + }$Is the radius of cation and
a is the unit distance = 387pm
The distance between the oppositely charged ions will be equal to ${r^ - }+{r^ + }$.
So now substituting the values in formula we get,
${r^ - }+{r^ + }={\dfrac{{\sqrt 3 }}{2}} \times 387$
${r^ - }+{r^ + }= 335.15pm$
So the distance between the oppositely charged ion in the lattice is 335.15pm.
(b) Now the distance between the ions is 335.15pm and it is given in the question that the radius of ${r^ - }$ that is chloride ion is 181pm so now we can easily calculate the ${r^ + }$that is the radius of ammonium ion. so,
${r^ - }+{r^ + }$= 335.15pm
Substituting the value of chloride radius in the above formula we fet,
$181+{r^ + }$= 335.15
So, ${r^ + }$ = 335.15-181
${r^ + }$ = 154.15pm
So the radius of ammonium ion is 154.15pm.
Note: In the primitive centered cubic unit cell the atoms are present only at the corners. Whereas in the face centered cubic unit cell the atoms are present at corners as well as at each face of the cube. So the total number of atoms present in a primitive centered cubic unit cell is 1 and in face centered cubic unit cell is 4.
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