# Solid State | Question Bank for Class 12 Chemistry

JEE Mains & AdvancedGet Solid State important questions for Boards exams. Download or View the Important Question bank for Class 12 Chemistry. These important questions will play significant role in clearing concepts of Chemistry. This question bank is designed by NCERT keeping in mind and the questions are updated with respect to upcoming Board exams. You will get here all the important questions for class 12 chemistry chapter wise CBSE. **Click Here for Detailed Chapter-wise Notes of Chemistry for Class 12th, JEE & NEET. ** You can access free study material for all three subject’s Physics, Chemistry and Mathematics. Click Here for Detailed Notes of any chapter.

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**:**There are points only at the corners of each unit. (ii) Face-centred cubic lattice

**:**There are points at the corners as well as at the centre of each of the six faces of the cube. (iii) Body-centred cubic lattice : There are points at the corners as well as in the body centre of each cube.

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**Conduction of electricity in semi-conductors**The electric conductivity of semi conductors is explained with the help of band theory. As we know, there is only a small energy gap between valence band (filled band) and conduction band (empty band) in semi-conductors. Therefore, some electrons can jump from valence band to vacant band and electric conductance is thus, possible. It can further increases with the rise in temperature since more electrons can jump to the conduction band.

**Intrinsic semi-conductors.**The extrinsic semi-conductors are formed when impurities of certain elements are added to the insulators. This process known as doping makes available electrons or holes for conductivity and thus, leads to two types of semi-conduction. These are called n-type semi-conductors and p-type semi-conductors.

**n-type semi-conductors.**n-type semiconductors are formed when impurity atoms containing more valence electrons than the atoms of the parent in insulator are introduced in it. These are called

**electron rich impurities.**For example, when traces of phosphorus (a group 15 element) is added to pure silicon (a group 14 element) on electron of each phosphorus atom (has five valence electrons) becomes free because it is not involved in any bonding with silicon atom (has four valence electrons). The unbonded electrons are free to carry the electric current. Thus, silicon containing phosphorus as the impurity is

**n-type semi-conductor.**(ii)

**p-type semi-conductors.**These are formed when impurity atoms containing lesser number of valence electrons than the atoms of the parent insulator element added to it. These are called electron deficient impurities. For example, when traces of boron. (a group 13 element) are added to pure silicon (a group 14 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of boron. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes are responsible for electrical conductivity and will result in p-type semi-conductors.

**Packing Efficiency of Simple Cubic Structure**In a simple cubic unit cell there is only one atom per unit cell. Let a be the edge length of the unit cell and r be the radius of sphere. Volume of the sphere $=\frac{4}{3} \pi r^{3}$ As the spheres at the corners are touching each other, the edge length a is equal to $2 \mathrm{r}$. Volume of the cube $=a^{3}=(2 r)^{3}=8 r^{3}$ $\%$ of the space occupied by spheres $=\frac{\text { Volume of sphere }}{\text { Volume of cube }} \times 100=\frac{\frac{4}{3} \pi r^{3}}{8 r^{3}} \times 100$ Thus, packing efficiency of simple cubic lattice is 52.4% (b)

**Packing Efficiency of Body Centred Cubic (BCC) Structure**We know that in a body centred cubic unit cell there are two spheres per unit cell. Let a be the side of the unit cell and r be the radius of sphere. volume of sphere $=\frac{4}{3} \pi r^{3}$ Volume occupied by two spheres $=2 \times \frac{4}{3} \pi r^{3}=\frac{8}{3} \pi r^{3}$

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Face diagonal $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{2 a^{2}}$ In right angled triangle ACD Body diagonal $A D=\sqrt{A C^{2}+C D^{2}}=\sqrt{2 a^{2}+a^{2}}=\sqrt{3} a$ In a body centred cubic unit cell, the spheres at the corners are not touching each other but are in contact with the sphere at the centre. As a result the body diagonal AD is equal to the four times the radius of the sphere. $\therefore A D=\sqrt{3} a=4 r$ $a=\frac{4}{\sqrt{3}} r$ Volume of unit cell $=a^{3}=\left(\frac{4}{\sqrt{3}} r\right)^{3}=\frac{64}{3 \sqrt{3}} r^{3}$ Percentage of space occupied by spheres $=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100$ $=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100=\frac{8}{3} \times \frac{22}{7} \times \frac{3 \sqrt{3}}{64} \times 100=68 \%$ Thus, packing efficiency of a bcc arrangement is $68 \% .$ (c) Face centred cubic In a cubic close packing, the unit cell is face centred cube. In a face centred cubic unit cell there are 4 spheres per unit cell. Let r be the radius of sphere and a be the edge length of the cube. Volume of sphere $=\frac{4}{3} \pi r^{3}$ Volume of four spheres $=4 \times \frac{4}{3} \pi r^{3}=\frac{16}{3} \pi r^{3}$ In a face centred cubic unit cell, the spheres at corners are in contact with sphere at the face centre but are not touching each other. Therefore, face diagonal $\mathrm{AC}$ is equal to four times the radius of sphere. $A C=4 r$ But in right angled triangle ABC (figure) $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$ $\therefore \sqrt{2} a=4 r$ $a=\frac{4 r}{\sqrt{2}}$ Volume of the unit cell $=a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3}=\frac{64}{2 \sqrt{2}} r^{3}$ Percentage of space occupied by spheres $=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100=\frac{\frac{16}{3} \pi^{3}}{\frac{64}{2 \sqrt{2}} r^{3}} \times 100$ $=\frac{16}{3} \times \frac{22}{7} \times \frac{2 \sqrt{2}}{64} \times 100=74 \%$ Thus, packing efficiency of ccp arrangement is 74%. Similarly, packing efficiency of hcp is also 74%.

**Interstitial Defects.**This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the crystal. The ions occupying the interstitial sites are called interstitials. The formation of interstitial defects in determined by the size of the interstitial ion. (iv) The "holes' occupied by electrons are called F-centres (or colour centres) and are responsible for the colour of the compound and many other interesting properties. For example, the excess sodium in $\mathrm{NaCl}$ makes the crystal appear yellow, excess potassium in $K C l$ makes it violet and excess lithium in LiClmakes it pink. Greater the number of F-centres, greater is the intensity of colour. Solids containing F-centres are paramagnetic because the electrons occupying the "holes' are unpaired.

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(b) Volume of one unit cell $=\mathrm{a}^{3}$ $=\left(3.54 \times 10^{-8} \mathrm{cm}\right)^{3}$ No. of unit cells $1.00 \mathrm{cm}^{3}$ $=\frac{1.00}{\left(3.54 \times 10^{-8}\right)^{3}}$ $=2.26 \times 10^{22}$

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