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Question: The \(KCl\) crystallises in the same type of lattice as does \(NaCl\) . Given that \(\dfrac{{{r_{N{a...

The KClKCl crystallises in the same type of lattice as does NaClNaCl . Given that rNa+rCl−=0.55\dfrac{{{r_{N{a^ + }}}}}{{{r_{C{l^ - }}}}} = 0.55 and rK+rCl−=0.74\dfrac{{{r_{{K^ + }}}}}{{{r_{C{l^ - }}}}} = 0.74 . Calculate the ratio of the side of the unit cell for KClKCl to that of NaClNaCl .
A. 1.1231.123
B. 0.8910.891
C. 1.4141.414
D. 1.7321.732

Explanation

Solution

Sodium chloride is a rock salt type substance and it forms a face centered cubic unit cell. Void formed by sodium chloride is octahedral. In order to calculate the ratio of edges we must know the relation between radius and edge length of the unit cell.

Complete step by step answer:
Sodium chloride that is NaClNaCl is a cubic structure crystal with face centered lattice type. In its crystal, lattice is formed by chloride ions and sodium ions occupy the void.
Similarly potassium chloride has the same lattice as sodium chloride. In case of potassium chloride, lattice is formed by chloride ions and potassium ions occupy the void.
The void they form is an octahedral void. And the relationship between radius and edge length in the case of octahedral void is:
2(r++r−)=a2({r_ + } + {r_ - }) = a
Where, r+{r_ + } is the ionic radius of cation and r−{r_ - } is the ionic radius of anion and aa is the edge length.
So, we have to calculate aKClaNaCl\dfrac{{{a_{KCl}}}}{{{a_{NaCl}}}} , here aKCl{a_{KCl}} is the edge length of potassium chloride lattice and aNaCl{a_{NaCl}} is the edge length of sodium chloride lattice.
So, we can write it as:
⇒aKClaNaCl=2(r++r−)KCl2(r++r−)NaCl\Rightarrow \dfrac{{{a_{KCl}}}}{{{a_{NaCl}}}} = \dfrac{{2{{({r_ + } + {r_ - })}_{KCl}}}}{{2{{({r_ + } + {r_ - })}_{NaCl}}}}
Mathematically, it can be written as:
⇒aKClaNaCl=[r+r−]KCl+1[r+r−]NaCl+1\Rightarrow \dfrac{{{a_{KCl}}}}{{{a_{NaCl}}}} = \dfrac{{{{\left[ {\dfrac{{{r_ + }}}{{{r_ - }}}} \right]}_{KCl}} + 1}}{{{{\left[ {\dfrac{{{r_ + }}}{{{r_ - }}}} \right]}_{NaCl}} + 1}}
After substituting the values:
⇒aKClaNaCl=0.55+10.74+1\Rightarrow \dfrac{{{a_{KCl}}}}{{{a_{NaCl}}}} = \dfrac{{0.55 + 1}}{{0.74 + 1}}
Now,
⇒aKClaNaCl=0.891\Rightarrow \dfrac{{{a_{KCl}}}}{{{a_{NaCl}}}} = 0.891
Hence, the ratio of the side of the unit cell for KClKCl to that of NaClNaCl is 0.8910.891 .

So, option B is correct.

Note:
Radius ratio rule is used to determine the void generally. More precisely we can say that it is used to determine the structure of ionic crystals. It has been said that anions are larger than cations in size so they occupy the corner site and form the lattice and anion has small size so it goes into the void made by cations.