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Question: KCl crystallizes in the same type of lattice as does \(NaCl.\) Given that \(\dfrac{{{r}_{N{{A}^{+}}}...

KCl crystallizes in the same type of lattice as does NaCl.NaCl. Given that rNA+rCl−=0.50\dfrac{{{r}_{N{{A}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.50 and rNA+rK+=0.70\dfrac{{{r}_{N{{A}^{+}}}}}{{{r}_{{{K}^{+}}}}}=0.70 , Calculate the ratio of the side of the unit cell for KClKCl to that for NaCl.NaCl.
A. 1.1431.143
B. 1.2241.224
C. 1.4141.414
D. 0.8750.875

Explanation

Solution

We first need to calculate the relation between the ionic radius and edge length of the NaClNaCl type structure. The NaClNaCl type structure is face centered structure. The length of the edges can be calculated using Pythagoras theorem.

Complete step by step answer:
In sodium chloride structure or FCC that is face centered cubic structure the atoms are present on the corners of the cube as well as the center of the each face of the cube. The atoms in case of FCC structure touches along the face diagonal. The length of the cube with edge length aa has a face diagonal as calculated from the Pythagoras theorem is 2a\sqrt{2}a. Now in case of a face centered cube the atoms touch along the face diagonal so the relation between the ionic radius and edge length will be calculated using the face diagonal length. If the cationic radius is and anionic radius is rr . Then the relation between the face diagonal and ionic radius is;
2[(r+)+(r−)]=2a2[({{r}_{+}})+({{r}_{-}})]=\sqrt{2}a
In the question we need to calculate the ratio as follow because 22 will cut in both numerator and denominator:
[(r+)+(r−)]KCl[(r+)+(r−)]NaCl=?\dfrac{{{[({{r}_{+}})+({{r}_{-}})]}_{KCl}}}{{{[({{r}_{+}})+({{r}_{-}})]}_{NaCl}}}=?
We have been given:
rNA+rCl−=0.50\dfrac{{{r}_{N{{A}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.50
We will add 11 to the both side and hence we will get:
rNA+rCl−+1=0.50+1\dfrac{{{r}_{N{{A}^{+}}}}}{{{r}_{C{{l}^{-}}}}}+1=0.50+1
Taking LCM we will get
[(rNA+)+(rCl−)(rCl−)]=1.50\left[ \dfrac{\left( {{r}_{N{{A}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)}{\left( {{r}_{C{{l}^{-}}}} \right)} \right]=1.50
We have been given
rNA+rCl−=0.50\dfrac{{{r}_{N{{A}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=0.50 and rNA+rK+=0.70\dfrac{{{r}_{N{{A}^{+}}}}}{{{r}_{{{K}^{+}}}}}=0.70
Dividing both of them we will get
rK+rCl−=0.500.70\dfrac{{{r}_{{{K}^{+}}}}}{{{r}_{C{{l}^{-}}}}}=\dfrac{0.50}{0.70}
Following the same process, adding 11 to the both side and taking LCM we will get;
[(rK+)+(rCl−)(rCl−)]=0.500.70\left[ \dfrac{\left( {{r}_{{{K}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)}{\left( {{r}_{C{{l}^{-}}}} \right)} \right]=\dfrac{0.50}{0.70}
We have two equations as:
[(rK+)+(rCl−)(rCl−)]=1.200.70\left[ \dfrac{\left( {{r}_{{{K}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)}{\left( {{r}_{C{{l}^{-}}}} \right)} \right]=\dfrac{1.20}{0.70} and [(rNA+)+(rCl−)(rCl−)]=1.50\left[ \dfrac{\left( {{r}_{N{{A}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)}{\left( {{r}_{C{{l}^{-}}}} \right)} \right]=1.50
We will divide both of them and we will get the following ratio:
[(rK+)+(rCl−)(rNA+)+(rCl−)]=1.201.5×0.70\left[ \dfrac{\left( {{r}_{{{K}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)}{\left( {{r}_{N{{A}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)} \right]=\dfrac{1.20}{1.5\times 0.70}
⇒[(rK+)+(rCl−)(rNA+)+(rCl−)]=1.143\Rightarrow \left[ \dfrac{\left( {{r}_{{{K}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)}{\left( {{r}_{N{{A}^{+}}}} \right)+\left( {{r}_{C{{l}^{-}}}} \right)} \right]=1.143

So, the correct answer is Option A.

Note: In ionic crystals the similarly charged ions never touch each other due to the electrostatic repulsion that occurs between similarly charged ions which decreases the stability of the crystal. Always ions with opposite charges touch each other so that attraction increases and crystal formed is stable.