Explanation
Lattice energy of a crystal is defined as the energy required to break an ionic crystal into its constituent ions in an isolated gaseous state. More ionic the compound is stronger will be its ionic bond and higher will be its lattice energy.
In $$MgN_3$$ the charge on the ionic species is greater than $$NaCl$$. So, ionic interactions in $$MgN_3$$ will be stronger.
When light strikes a photographer $$(AgBr)$$ paper, it gives energy to the electrons present in the film. These energetic electrons when strike silver ions turn them to silver atoms. So eventually, ions leave their lattice site and occupy interstitial sites. Since silver atoms are black in color so whenever light strikes a silver ion the photographic film will turn black.
Theoretical density of crystal, $$\rho = \cfrac {nM}{N_o a^3} \ g/cm^3$$
Given for bcc $$n=2; \ a=300 \times 10^{-10} \ cm, \ M=50 \ g$$
$$\therefore \rho = \cfrac {2 \times 50}{6.022 \times 10^{23} \times (300)^3 \times (10^{-10})^3}$$
$$= 6.15 \ g/cm^3$$
Hence, Option "C" is the correct answer.
Given for bcc $$n=2; \ a=400 \times 10^{-10} \ cm, \ M=100 \ g$$
$$\therefore \rho = \cfrac {2 \times 100}{6.022 \times 10^{23} \times (400)^3 \times (10^{-10})^3}$$
$$= 5.19 \ g/cm^3$$
Hence, Option "B" is the correct answer.
For NaCl, $$n=4, \ M=58.5 \ g, \ a=0.564 \times 10^{-7} \ cm$$
$$\therefore \rho = \cfrac {4 \times 58.5}{6.022 \times 10^{23} \times (0.564)^3 \times 10^{-21}}=2.16 \ g/cm^3$$
Hence, the correct option is $$\text{B}$$
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