Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry

Number of octahedral voids in $$ccp,$$  is equal to effective number of atoms. In $$ccp,$$  effective number of atoms are 4 so, 4 octahedral voids. So, 1 octahedral voids per atom.

$$ZnS$$  has cubic close packed $$(ccp)$$  structure. The $${S^{2 - }}\,ions$$   are present at the corners of the cube and at the centre of each face. Zinc $$ions.$$  occupy half of the tetrahedral sites. Each zinc ion is surrounded by four sulphide $$ions$$  which are disposed towards the corners of a regular tetrahedron. Similarly, $${S^{2 - }}\,ions$$  is surrounded by four $$Z{n^{2 + }}\,ions.$$

Order of Bragg diffraction $$\left( n \right) = 2;$$
Wavelength $$\left( \lambda \right) = 1\mathop {\text{A}}\limits^{\text{o}} $$   and angle $$\left( \theta \right) = {60^ \circ }.$$
We know from the Bragg’s equation
$$\eqalign{ & n\lambda = 2d\,\sin \,\theta \cr & {\text{or}}\,\,2 \times 1 = 2d\,\sin \,{60^ \circ } \cr & \Rightarrow 2 \times 1 = 2.d.\frac{{\sqrt 3 }}{2} \Rightarrow d = \frac{2}{{\sqrt 3 }} = 1.15\mathop {\text{A}}\limits^{\text{o}} \cr} $$
( where $$d = $$  Distance between the scattering planes )

$$\eqalign{ & {\text{For }}fcc, \cr & r = \frac{{\sqrt 2 a}}{4} \cr & = \frac{a}{{2\sqrt 2 }} \cr & = 0.3535a \cr & {\text{given}}\,\,a = 361\,pm \cr & r = 0.3535 \times 361 \cr & = 128\,pm \cr} $$

Isotropical is not a crystal system type.

The appearance of colour in alkali metal halides is generally due to $$F$$-centres. $$F$$ centers in $$NaCl$$  produce a band of optical absorption towards the blue end of visible spectrum, thus the colour appears yellow under transmitted light. On the other hand, $$KCl$$  with $$F$$-centers appears magenta, and $$KBr$$  appears blue.

$$ZnS$$  has a $$fcc$$  or $$ccp$$  structure. $${S^{2 - }}\,ions$$   are present at the corners of the cube and centre of each face. $$Z{n^{2 + }}\,ions$$   occupy half of the tetrahedral sites. Each $$Z{n^{2 + }}\,ion$$   is surrounded by four $${S^{2 - }}\,ions$$   and each $${S^{2 - }}\,ion$$   is surrounded by four $$Z{n^{2 + }}\,ions.$$

$$\eqalign{ & {\text{No}}{\text{.}}\,{\text{of}}\,N{a^ + } = 4 - 1 = 3 \cr & {\text{No}}{\text{.}}\,{\text{of}}\,C{l^ - } = 4 - \left( {2 \times \frac{1}{2}} \right) = 3 \cr} $$

Superconductors are derived from compound of $$p$$ block elements.

In $$N{a_2}O$$  all tetrahedral void are occupied.