Explanation
For bcc structure, $$4r= a\sqrt 3 \Rightarrow a = \cfrac {4r}{\sqrt 3}$$ (along body diagonal)
$$\text{Packing Fraction}= \cfrac {2 \times \cfrac43 \pi r^3}{a^3}= \cfrac {2 \times \cfrac 43 \pi r^3}{\cfrac {64 r^3}{3 \sqrt 3}}=\cfrac {\pi \sqrt 3}{8}$$
Since, the body center atom is surrounded by $$8$$ other atoms. Therefore coordination number is $$8$$.
Imperfections in the solid state compounds are known as defects. In the Frenkel defect, the cation moves from its site and occupies an interstitial site. In this type of defect, density remains constant whereas dielectric constant increases.
There are four stoichiometric defects namely, Vacancy defect, Interstitial defect, Schottky defect and frenkel defects.
Frenkel defects is also known as dislocation effect because a cation is dislocated. Silver (I) halides show frenkel defects.
So, all the statements are true.
Hence, opt. C is correct.
The lattice energy of crystal $$AB$$ is $$4X$$.
Lattice energy is the energy associated to the crystal lattice of a compound.
Mathematically,
$$E_{lattice}=k\dfrac{Q_1Q_2}{d}$$
where, $$k =$$ proportionality constant
$$Q_1$$ and $$Q_2$$ are the charges on the ions in coulombs.
$$d =$$ distance between the ion charges.
Lattice energy of $$NaCl = X$$
$$E_{lattice}(NaCl)=\dfrac{Q_1Q_2}{d}=X$$
As the ionic size remains same when $$AB$$ crystal is formed, d remains same
Lattice energy of $$AB$$,
$$E_{lattice}(AB)=\dfrac{(2Q_1)(2Q_2)}{d}$$
$$E_{lattice}(AB)=4\left(\dfrac{Q_1Q_2}{d}\right )=4X$$
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