Statement 1: If the length of the unit cell of LiCl having NaCl structure is 5.14 Ao, the ionic radius of Cl- ion is 1.82 Ao.Statement 2: Anion—anion contact is retained in LiCI structure because anions constitute the lattice.

If anion-anion contact is retained, then as shown in the adjoining figure. Interionic distance of LiCl=a/2=5.14Ao/2=2.57AoBC=AB2+AC2=(2.57)2+(2.57)2=3.63AoRadius of Cl-12BC=12×3.63=1.81AoHence, statement 1 and 2 are correct.

Statement 1: If the length of the unit cell of LiCl having NaCl structure is 5.14 Ao, the ionic radius of Cl- ion is 1.82 Ao.Statement 2: Anion—anion contact is retained in LiCI structure because anions constitute the lattice.

If anion-anion contact is retained, then as shown in the adjoining figure. Interionic distance of LiCl=a/2=5.14Ao/2=2.57AoBC=AB2+AC2=(2.57)2+(2.57)2=3.63AoRadius of Cl-12BC=12×3.63=1.81AoHence, statement 1 and 2 are correct.

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Statement 1: If the length of the unit cell of LiCl having NaCl structure is 5.14 $\overset{o}{A}$ , the ionic radius of ${Cl}^{-}$ ion is 1.82 $\overset{o}{A}$ .
Statement 2: Anion—anion contact is retained in LiCI structure because anions constitute the lattice.

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Both Statement 1 and Statement 2 are true

Both Statement 1 and Statement 2 are false.

Statement 1 is true but Statement 2 is false.

Statement 1 is false but Statement 2 is true.

answer is A.

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Detailed Solution

If anion-anion contact is retained, then as shown in the adjoining figure.

Interionic distance of $LiCl = a / 2 = 5 .14 \overset{o}{A} / 2 = 2 .57 \overset{o}{A}$

$\begin{array}{l} BC = \sqrt{{AB}^{2} + {AC}^{2}} = \sqrt{{(2 .57)}^{2} + {(2 .57)}^{2}} = 3 .63 \overset{o}{A} \\ Radius of {Cl}^{-} \frac{1}{2} BC = \frac{1}{2} \times 3 .63 = 1 .81 \overset{o}{A} \end{array}$